Thank You! I've wanted to understand electronics since I was a kid and thanks to internet and people like you it is now possible. The transistor amplifier design has been a real struggle for me but the pieces are starting to fall in their places. Among some written material, your videos have been an invaluable resource and by far the best videos I've come across altogether.
Another excellent lesson Alan. Appreciate your efforts preparing and delivering very well organized and insightful lesson plans, always looking forward to them. Much obliged, sincere thanks.
I thought this video was great! It was easy-to-follow, and you covered the most important topics in a clear and concise way. You're a great teacher! source: someone getting his M.A. in Integrated Circuits at UCLA
Wow- amazing clarity not often found in electronics videos. You’re a great teacher. Thanks so much for all the time and effort that goes into your videos. If you’re looking for a new topic for one of your fundamental videos, you might consider an active load on a common emitter amplifier. I’ve been trying to wrap my mind around this topic for some time now. I can’t find a tutorial that explains it clearly, the way you do. Thanks again.
@@w2aew Awesome, can you give me the video numbers? I searched your videos for the keywords but couldn’t find anything. I did watch one video you made on current sources, and although you mentioned active loads briefly, there were no details about how they work. In that video you mention that if a current source has a high output impedance, you can use it to achieve high gain in an amplifier. I’d love to get details about how to achieve that. Thanks so much for your reply.
@@jrausa1 Yeah, I guess I haven't gotten into the details of an active load specifically. I'll have to put that on my list. Here are a few of the videos I was thinking of that touch briefly on the topic: ruclips.net/video/xR0RfmmRhDw/видео.html ruclips.net/video/2IbWvq7Abv4/видео.html ruclips.net/video/mejPNuPAHBY/видео.html
A PROfound example of understanding a subject down to it's core, thereby "eliminating all guesswork"! It will be a banner day when I apply the formula's to obtain the answer, and then, confirm via scope! Thank You!
Excellent demonstration of the corner frequencies by scoping the frequencies at the -3dB. Many thanks for this, it kind of everything made sense right now!
A very informative video, thanks. For everyone that wants to use this circuit on higher frequencies: it is advised to remove the capacitor from collector to ground.
Thanks, very clear explanation. Seeing this video I remembered I learned this when I was studying electronics about 20 years ago. I thing I'll do some calculations and measurements myself, just for fun.
Thank you for your prompt response. I understood. Power is proportional to the square of voltage, or current. Thank you very much for all your excellent videos. They are priceless.
For Rosa Rojas (since I can't reply to the comment...) It's all there in the formulas - simple math. Gain is -gm*Rc. So, either make Rc 5k vs 10k, or decrease gm by a factor of two (by increasing the 2.2k to something closer to 4.4k).
+w2aew I assume that this applies in case we are capactively coupling this amplifier to the next stage, what we are not doing that? That means the miller capacitance shall limit the upper 3dB point? Also, you said that when the reactance of the capacitor at output is equal to Rc at 5:46 minutes that is when we get the upper 3dB point. That means the amplitude is halved at output? or some other reason?
+Yueeiuyoo It is when the output power is halved - not the output voltage. In many amplifiers, the miller capacitance will be the bandwidth limiting factor.
+w2aew when you say that the impedance of the capacitor has to be equal to the resistance of the collector resistors for the upper -3dB point, how is this conclusion arrived at?
I have been going through older videos to learn some and I played along. Pretty cool to be able to do hands on with you. I learned something. 73, ad0am
I Would Say This Is your BEST VIDEO So Far!!! Info on Basic transistor amplifiers Is Very Hard To Find,,,,These Basic Concepts on Transistor Amplifiers were Long Lost In My Head Until Your Video Came Along!! I'm going to Breadboard this Lab To Verify The Calculations Thanks Once Again w2aew!!! PS**** How about a Video on Fixed Bias,Self Bias and Beta Independent Transistor Amplifiers,,,A Super Job Once Again Alan!!! Keep Those Videos Coming,,I'm Starving for some more Electronics Theory
The cutoff frequency is the frequency either above or below which the power output of a circuit has fallen to a given proportion of the power in the passband. Most frequently this proportion is one-half the passband power, also referred to as the 3 dB point since a fall of 3 dB corresponds to approximately to half power. As voltage ratio, this is a fall to approximately 0.707.
Thanks for the great explanation. I misunderstood the low frequency cutoff. And you've corrected that with this video. However, a more accurate explanation of Fl would be that there are multiple low pass filters and the one involving Ce and re' dominates. There is actually four low pass filters. The input circuit with Cb and 50ohms || R1 || R2 || b*re', the ouput circuit with Cc and Rc, and 2 at the emitter, Ce and Re and Ce and re'. The Ce and re' filter should be designed to dominate to make calculations simple. Since the re' filter cuts off at 39 Hz it dominates the Re filter which cuts off at 1.5 Hz. Obviously the higher low pass filter will dominate unless they are close together. Then the filter calculations becomes complicated. Since 39Hz >> 1.5Hz we can ignore the influence of the Re low pass filter.
There are two capacitors at play and hence two low-pass filters to consider. One is formed from the 47μF emitter bypass capacitor and the parallel resistance of 86R (the intrinsic emitter resistance) and 2k2, which is 83R. That has a cutoff frequency of 41Hz, very close to the 39Hz obtained by ignoring the 2K2 resistance. The other is the 4.7μF input capacitor and the parallel combination of 56K || 20K || β.Re where Re is the effective total emitter resistance. Note that the 50R load on the signal generator plays no part in that calculation. That is complicated because β is unknown, but could be as low as about 40 for a 2N2222 operating at 300μA, and because Re is the complex impedance of the emitter resistance and the 47μF capacitor. Normally, we should design so that β.Re is significantly greater than the parallel combination of 56K || 20K (i.e. 15K) at the frequencies of interest, but 40 x 83R = 3K3 and that will dominate the input impedance unless you select your transistor to have a β significantly more than 180 at 300μA. With the transistor used, the cutoff frequency associated with the input capacitor could be 10Hz if β=40. Fortunately, the ~40Hz cutoff from the emitter bypass will dominate in this case.
If you look at the bias point of the transistor - mainly the collector and emitter voltages, you can get a good estimate how far the collect voltage can swing before it either hits the positive supply rail or saturates the transistor. Once you determine this swing, then divide by the gain of the amplifier to determine the input signal swing.
@@w2aew I wanted to knoe the anwser for this same question. But why cant I just increase the output of the function generator until the output of the circuit starts clipping and then read off the value of the signal from the function generator to find the max input to the circuit. Is this approach incorrect?
quick question, i have a function generator in which you can set the impedance , would that be equivalent to that load termination that you have or is that in addition to the generators impedance.. ?
It does not change the generators output impedance. This setting tells the generator what the expected load impedance is so that it can generate the correct voltage. I did a video on this: ruclips.net/video/tClE8s6RZdg/видео.html
Brilliants!!!! Every kid should have a teacher like this in every subject. Should you go for a gain of 117 or do it in two stages at 10 each? When do you need multiple stages.
If you go for a gain of 100+ in one stage, you will inevitably run into the problem of distortion with a large signal. I don't mean clipping, but when the gain depends on Rc/Re and Re is the non-linear intrinsic emitter resistance, you are bound to get distortion. To design that out, you have to have an unbypassed emitter resistor significantly greater than the intrinsic emitter resistance of the transistor at whatever emitter current you are using. That means that because the intrinsic emitter resistance = 25mV/Ic, the emitter resistor must be at least 10 times that resistance, which equals 250mV/Ic. That means you need at least 250mV across the emitter resistor. Since the collector resistor will need to be 100 times the emitter resistor for a gain of 100 and the collector current is almost the same as the emitter current, you will need 100 times the voltage drop across the collector resistor that you have across the emitter resistor. That means you need to drop 100 x 250mV = 25V across the collector resistor. Since the quiescent collector bias point needs to be about half the supply voltage, you will need a 50V supply rail (and a transistor that can handle a 50V supply rail). That is the inescapable logic of designing a common emitter stage: to increase the gain while maintaining low distortion, you have to increase the supply voltage.
While you have the most splendid and clear tutorials on these things where others go into details that just make the core harder to understand, it will be quite powerful if you made a video explaining construction of a simple amplifier e.g the 741. This shall involve current mirrors, long tailed pair and much more. Making measurements at each point and why design was done a certain way will be very helpful. The only problem is that such a video may require more work to be produced.
+Yueeiuyoo I do spend some time talking about the internal circuitry of the LM358 op amp in my video describing cross-over distortion: ruclips.net/video/VgodYtiD_F0/видео.html
your videos are very helpful, i've been able to confirm a lot about typical transistor operation in general and become more familiar with the dynamics of a transistor given the various configurations... that elevator buzzer (boop or beep or ding) has had me looking around the room a couple times though... heheh
Vacuum tubes have a MU spec and GM transconductance spec. I'm not sure what internally inside the vacuum tube is determining the MU amount and the transconductance amount but I think is how much electrons are on the cathode coating ? Because the cathode is what is releasing the electrons which is the current which is the MU and transconductance?
I would love to see someone competent (thus commenting on this channel ;) do a transistor tutorial. Example: Given a gain requirement and input voltage/current/frequency, design an amplifier using an NPN transistor. I've watched dozens of videos entitled "how to bias a transistor" but each and every one of them analyzes an existing circuit. What I need to know (and I believe many others) is the step by step process to designing a transistor amplifier circuit. Would also be nice to cover the different classes of amplification as well (ie: how one class differs from the next by way of the biasing or however?) Also watched tons of videos on Transistor theory - ie: materials, doping, etc... Like to see it kept to a practical (how to) view. Some people start with a ratio of Rc to Re (ie: gain), some start with biasing the Base and Re, then simply state that Ie is the same as Ic (why? how?)... All very confusing. A logical process would be welcome. There must be one - this is electronics/math not voodoo.
+Digger D A couple of comments: - IC is almost the same as IE because IE is equal to IC + IB, and IB is small with respect to IC (IB = IC / Beta). - I already did a video on biasing and amplifier class of operation: ruclips.net/video/c6cmkm3UPUI/видео.html - As for a tutorial on designing a transistor amplifier, you can't get much better than the one that Shahriar did: ruclips.net/video/Y2ELwLrZrEM/видео.html
***** Thanks, it's starting to come along. Next to determine how to decide on values for the voltage divider on the bias (ie: many possibilities but what logic does one use to decide what values of resistance to use, I suspect it depends on the current required at the base?). IB * Beta determines IC and ratio of Rc over Re determines voltage gain? But I would have thought there would be a process to determine each. Most of the vids I've watched they seem to pull a value out of the air. Thanks again for the reply and links.
***** I watched Shahriar's video but he too pulled values out of thin air (Ic = 2 mA??, why did he just pick a value?) and gm was not defined etc... I did however just discover that Rbe is a constant of 25 or 26 ohms, I'll go with 26 as I just saw the math for that result. Now this makes more sense as I can now calculate Ib and thus choose my bias resistors to set the current into the base and knowing the Beta value (generally as published) I can then calculate what my Ic will be. Ic = beta X Ib. It now becomes a resistor network problem. As for the voltages over Rc, CE and Re that is pretty easy to determine. It's like trying to learn the secrets of a mystical lodge who's members are sworn to rumors only. ;) Even finding a book that explains this in a step by step logical manner is difficult. I've spent a week watching many videos, reading several articles and asking questions but it's starting to come together. Transistors have always been a stopping point for me in analogue electronics as I just could not find the information needed to design a circuit. So for a common Emitter amplifier I think I'm ok, now for all the other configurations... Thanks again for your help and I did watch your bias video earlier but it too didn't quite answer all my questions but it did create some new and better questions. Every component and resulting value in a circuit has some logic behind it and sometimes I find people skip over something that has become common knowledge to them but for us folk that are trying to absorb the info, it leaves a huge gap. Once again thanks for your help.
+Digger D Sometimes the initial choices for things like IC are arbitrary, other times the initial choices are governed by requirements of the application - such as: input impedance must be X, or load resistor or output impedance must be Y, etc. Or, there may be a limit on power dissipation for the circuit, or the designer might choose a bias level based on some optimized transistor characteristics. The initial constraints will often help decide the starting point (whether it is bias point, a resistor value or two, transistor selection, etc.) However, the initial choices are governed, the art of the remaining part of the circuit design is determining the circuit structure and component selection to achieve the desired goals while keeping within the constraints given in the initial performance desires or requirements. It's not an exact science, every application is different.
Excellent video. I second 321reh comment: this is one of your best and the beginning is just a little fast. How did you get the thermovoltage and the transconductance? A little more info on that would be amazing. Also, how did you came up with all the values for the resistors and capacitors. It's pretty complicated to match all the constraints. Can you recommend a method?
Fascinating set of videos. Thank you for the time you have spent putting all of these together. Could you please complete an analysis, using the measured voltages to ascertain the beta of the transistor used? IC = 310uA, IE=311uA meaning IB=10uA, assuming the circuit is working in the linear region. Does this mean the beta of this particular transistor works out at something like 30? Thanks again.
Since we’re likely talking about beta on the order of 100-200 or more, we would have to be very careful with our measurements. For example, if you’re measuring the voltages across RC and RE to determine those currents, you would *have to* carefully measure those resistor values first because their tolerance (how much they can be off of their ideal values) will dramatically affect the calculated beta.
Yes i understand that the DC offset won't be on the secondary, but won't the AC waveform have an AC offset not starting at the zero crossing point. The AC waveform might not start exactly at the zero crossing point might have some offset on the secondary?
on the topic of uhf and emf tourture from a distance through direct sound or "organized stalking". what is the best way to protect my self from these "psychotronic attacks"?
Sir, the formula to calculate dB is dB = 10 log (base 10) of V2/V1 log (base 10) of 0.707 is equal to - 0.15. Please, could you explain how this is equivalent to - 3 db?
Why is it necessary to use the 4.7 uF coupling capacitor on the input? I understand it would block any dc components of the signal, but if we're applying a simple sine wave from a signal generator, then there would be no dc component, correct? Or am I missing something? Best practice just to be sure the signal is pure?
The 4.7uf cap is used at the input to *preserve* the DC bias at the base that is established by the 56k/20k resistor divider. If the capacitor wasn't there, the 50 ohm terminator for the signal generator would be in parallel with the 20k resistor and the bias voltage on the base would then be dragged very close to ground.
Very nice! Is the only purpose of the 220pF cap to ground to set the high frequency point? would you include this in the final design of an audio anp? Without this cap, the gain does not drop with frequency, why would this be bad, maybe becuase high frequencies would also be amplified? Thanks fot the help and the great video. (Also, the point of the 50ohm source resistance? Doesnt the generator already have a 50ohm output impedance?)
Yes, the 220pF sets the high frequency rolloff. It was used as an example here. Without it, the frequency response would be wider, but the gain would eventually drop off due to other device capacitances. The 50ohm resistor is properly terminating the output of the signal generator. Again, not important at audio frequencies, but it ensures that the voltage provided by the generator matches what it is set to.
Thank you very much for the response. So in an actual circuit, with a speaker as the load, will you keep the capacitor in to set the high frequency cutoff? Or do you rely on the capacitance of the speaker to do this job? Thanks again for the info. As for the 50 ohm resistor, why would you want to terminate the input signal? Don't you usually do that when you have an open circuit somewhere and you don't want any reflections? Thanks again!
It all depends on the amplifier design, whether or not it is necessary to add some high-frequency rolloff. Many/most applications wouldn't care. At audio frequencies, reflections are not an issue. The main reason I added the termination resistor is so that the displayed amplitude on the generator accurately reflects the voltage applied to the circuit input. The generator allows you to specify the load impedance, and it defaults to 50 ohms. Since it has a 50 ohm output impedance, the voltage generated by the generator is 2x the value that is set in order to account for the voltage divider effect that comes from the 50 ohm load. I could have optionally specified the actual input impedance of the circuit and omitted the 50 ohm termination, but it was easier to simply use the termination and 50 ohm load setting.
@@pierreretief The output impedance of the circuit shown is set by the collector resistor at 10K. You're not going to be driving many speakers with that sort of output impedance. With the exception of Quad Electrostatic speakers, loudspeakers present an inductive, not a capacitive load to the amplifier, so you can't count on rolling off high frequencies that way. Normally, a capacitor is used to feed back from the output to the input of an inverting amplifier, which sets the upper roll-off frequency and ensures overall stability of the amp.
How do I combine these two waves? Like I want a 360 wave where two sine waves are 180degree phase shift from each other at the same time for the output, is there a circuit for that?
How can the bias V_B voltage be 1.28V in the first place? I couldn't read properly the value of the lower resistor of the base divider (it's either 26k or 20k). In either case (and if we neglect the base current) it's 5*(26/82)=1.59V or 5*(20/76)=1.316V. Moreover if it's indeed 26k according to my calculation the BJT should be in saturation: V_B=1.59V -> V_E=1.59-0.6=~1V -> I_E = 1/2.2 =~ 0.45mA -> V_CE=5 - 0.45*(10+2.2) < 0 !
The lower resistor value is 20k, so your 1.316V calculation is correct. It's only off by 30mV or so (about 2% off), which is due to resistor tolerances and base current contribution through the 56k resistor.
The amplifier doesn't "stop" working for small signals, so there really isn't a limit on how small the input can be. However, factors such as the signal frequency, the required bandwidth, the noise in the signal and amplifier, etc. will all affect how well the amplifier works.
The Bypass Capacitor across the emitter resistor/cathode resistor on transistors or tubes will also Filter out the AC Ripple? because in tube amplifiers if you don't use the Bypass capacitors the tubes AC heater and power supply AC ripple will get amplified because the bypass capacitor is not filtering out the fluctuations?
The bypass capacitor on the emitter or cathode is to increase the gain by allowing the intended signal to bypass the resistor. It is not for supply filtering.
Hi, I was looking at video 273 and thought it would be good idea to visit the other videos you suggested. So it hit me, I saw something on your scope that I have always had showing up on every simulation and test of the collector output that I have seen. The peaks had wider span then the troughs. I understand it was the exponential behavior of the BJT as the linear range is not entirely linear but rather somewhat almost linear. Is there another reason you can think of? From a BJT class A amp, i probably think this should always be the case, right?
Yes, it is more prevalent the larger the signal and is primarily the result of the lower gain owing to lower incremental transconductance as the collector current decreases (which is what's happening as the output signal rises). The reduction in gain results in compressing the signal peaks. The opposite occurs at the troughs (at least until you get close to saturation).
How do you Hi Pot test an amplifier circuit? If an amplifiers input has DC offset voltage can it amplify the DC offset voltage on the amplifiers output? What is amplifier input and output AC offset voltage mean (not DC offset )?
Hey Alan I really enjoyed this video and the ones related to it. These lessons have helped me with guitar pedal builds and understanding the math more behind the circuits. would you be interested in breaking down a fuzz face circuit? I’m mainly using npn transistors for the amplifier instead of pnp. Thanks again!
@@w2aew Ahhh well fuzzcentral.ssguitar has a Fuzz Face NPN Silicon schematic that I would love to get your take on. I’m using most of the same values and have modded the bias point of both transistors as well as collector voltage
@@wingate8 OK, I think this is the page you're talking about: fuzzcentral.ssguitar.com/fuzzface.php Looks like a fairly simple two-stage amplifier with a little bit of negative feedback. The pot adjusts the gain of the second stage which influences how much (if at all) the 2nd stage is driven into clipping for a given input signal level.
How do i calculating AV in a multi stage amp for example a 2 stage CE amp's AV(tot) ? I know this is hard to put in words with no resistor values etc so if you can up load a video about this it would be much appreciated
great video. just wondering if you have any other videos explaining the capacitors types to use at what location as I not sure do I use film capacitors on the input and electrolytic on the output etc.
Even if an amplifier hits the 3db drop off thus "showing the top of bandwidth", isn't it true it still has usable gain though until it hits zero db entirely?
@@w2aew Thanks! I did a bit more googling and found that they call the range where it is above 3db drop, the "power bandwidth" vs the "unity gain bandwidth", which is the "entire" bandwidth overall
Excellent videos! Thank you. I'm curious about what starting frequency you used to determine the 3db down points. Did you sweep to the max amplitude frequency for a starting point? EG. if you started at 100 Hz or 60 kHz and adjusted the frequency to find 3db down, you would come up with different corner frequencies. Does this make sense? Can you clarify for me? Thanks
I started with a frequency that would be fairly far away from the corners based on the component values. Or, doing a quick sweep or set of measurements at various frequencies, it's easy to find something in the passband, and slide up and down to find the edges.
Nice tutorial, can you also upload some RF (100MHz to 500MHz) amplifier tutorial with few mW input to few watt output , using bjt and how to calculate the power gain, frequency response of the amplifier and also the input and output impedance. thanks in advance
First of all yo really know your formulae. What would you reccomend in a hi-fi application a single transistor or an op-amp for premium sound quality and low distortion, which is best?
I believe I finally understand how those two filters work. They both work the same way. They both would be considered RC low pass filters. It's just that the filter at the emitter increases the output signal above it's cutoff frequency while the filter at the output attenuates the output signal above it's cutoff frequency. The equation for the cutoff frequency of both is fc = 1/2πRC. Together they limit the range of frequencies that will be amplified. I'm wondering. Would this type of set up be used for a radio transmitter? Longwave for example. Would the two filters be set to filter out all frequencies except those between 153 KHz and 279 KHz? And if so what would be the benefit? Thanks.
Yes, you're definitely gaining a good understanding. For RF circuits, often a tuned load is used to create a bandpass filter around the signal of interest. The main goal is to minimize the transmission of harmonic and inter-modulation distortion components.
Just want to ask you regarding the high and low pass frequencies; so does that mean the range of frequencies between which the gain can be found is 38Hz and 72Khz?
I would be nice if you could explain how can we make this circuit have half of the gain already measured by removing a capacitor or by making some other changes. Is it possible to make some arrangements and have a gain of approximately -55?
Just add an extra 100R emitter resistor between the emitter and the 2K2 resistor and bypass just the 2K2. That will get you in the right ballpark for gain, as well as helping to linearise the output, (thus reducing the distortion), improving the frequency response down to about 20Hz, and raising the collector bias point a little so that you can get more headroom from your 5V supply.
Thank you for your explanation, I would like to ask how did you solve for Vt? or is it always 26mV? I am trying to solve an assignment for the output voltage and frequency but I was not given the value for the capacitors how can I find the frequency? Also, does the voltage gain formula change? In class it was given as the output voltage/input voltage (Vout/Vin) not -Rc/re
The Vt value (thermal voltage) is calculated via kT/q (en.wikipedia.org/wiki/Boltzmann_constant#The_thermal_voltage) At room temperature, it is about 26mV. If they didn't give you values for the capacitors, maybe they're assuming that they're large enough that they can be ignored (effective low Z at the operating frequency). The gain formula doesn't really change, but it could be represented a different way. Since gm=1/re, you often see the formula written as AV=-gm*RC. If feedback is used, such as adding an emitter resistor, then the formula does change.
@@w2aew I appreciate your response. So if the capacitors are large enough that they are ignored, how can the frequency be obtained? Also given the emitter resistor what will the gain formula be?
@@w2aew it’s a common emitter circuit with an NPN transistor like the one in your video with an input of 100mV@50Hz and a load of 1k ohm. Base resistors of 10kohm and 2.2kohm, collector resistor of 3.6kohm, emitter resistor of 180ohm and 820ohm, the 820ohm is short with a capacitor. The output resistor is 51kohm. Vcc is +10v and beta is 200
@@sarahmajin5358 Sounds like you have what you need to calculate the operation point, collector current, etc. in order to figure out the gain. Not sure what you mean when you say the load resistor is 1k and the output resistor is 51k. Do you mean that the collector has 3.6k to 10V, then there is a 51k resistor in series between the collector and the 1k load resistor? Another question - when you say you need to solve for frequency, I"m not sure what you're looking for. You said that the input signal is at 50Hz, so that is the frequency. The amplifier won't change this. Or, are they asking for the frequency response? If they don't give you capacitor values, maybe they're asking for you to describe the equation that determines the frequency response (i.e. the emitter RC determines the high-frequency corner, etc.).
Any chance you doing a video for UHF frequency range, explaining more how to include parasitics. I always watch advertisements, before you videos, till the end so you get full ca-chugn ;-)
because @-3dB point, output gain is reduced to 70.71% of its maximum value then it start to decay with 20dB so you can say upto 3dB your output will be significant or meaningful which will not be the case beyond 3dB point.
There's something else I'm wondering about. I know that an amplifier like this has to be connected to the signal source and the load through a coupling capacitor. My understanding is that the purpose of these capacitors is to filter out the dc component of the signal so that only ac enters and leaves the amplifier. With this is mind it seems to me that the higher the capacitance the better in order to keep the reactance as low as possible so that the capacitors behave like a short circuit. Is this correct or is it more complicated than that? Thanks.
In general, yes, you want to use values that make the reactance low compared to the circuit impedance at the lowest frequency of interest. Making the caps unnecessarily large may result in delays or stability at startup when the cap has to charge up to the DC difference between the two circuits. Also, very large value caps might not work as well at the highest frequency of interest due to parasitic inductance. Also the very large values might mean the use of less desirable capacitor types like electrolytics.
Is it ok for an amplifier IC chip output a DC offset voltage or is this bad because the DC offset voltage can damage the speaker? If the output transformer primary has a DC offset voltage from the amplifiers IC chips output, will the output transformer secondary have a DC offset voltage also since the primary has a DC offset voltage?
I have had a hard time finding a concrete answer on this, but can a small signal transistor like a 2N3904 effectively amplify a larger signal(say, around 1.3Vpp) and generally, at what point in signal amplitude does it become necessary to step up to a power transistor?
It certainly can. Remember, voltage alone does not equal power. Power is voltage*current. Typically a lot depends on the load that the amplifier needs to drive. Transistor power is primarily dictated by the collector-emitter voltage times the collector current.
@w2aew My aim is to have an MCU driven Bluetooth receiver feeding audio into an old tube radio I restored from 1938. The raw audio output from the MCU has some slight audio distortion(which generally seems localized to the lower frequencies) I messed about a lot with different means of matching the impedance. It seems the best results I've gotten so far was shunting a 560 ohm resistor from the input to the chassis on the radio, but the low frequency problem remains. So I was looking into adding a common emitter stage in between to see if that helped out
@w2aew I was also wondering what technique you use to measure small signal base current when you have a voltage divider on the base? I have yet to purchase an AC current probe for my scope(I do have some SCTs laying around that I could likely make work) but that would only be a viable solution on a breadboard. I suppose I could subtract the RB2 current from RB1 and end up with base current
@@ColossusEternum Very difficult to comment on the cause of your distortion without knowing what your circuit looks like, or what the input of the audio-input of your old radio.
@w2aew I understand that completely. I'm just going to have to keep researching and experimenting with it. Right now I'm not terribly concerned about that specific circuit(I was trying to run before I could walk lol) it's just a little daunting when you build an amplifier because there are all these ways to adjust the circuit, but they all have a range of consequences. I've been meaning to draw a table in which I list all the factors that each component has an influence over.(like base resistors influencing Vb, Ib, reducing hFe dependence, input impedance etc.
Is there any advantage in replacing the base bias resistors (56k / 20k) with,say, a100k potentiometer / preset to optimize the operating point of the transistor? And how did you decide to use 56k / 20k anyway? Other values could have been used to get the 1.28v
The advantage would be, as you said, being able to easily dial-in the optimum operating point. As for how I picked the bias resistors on the base... ...I usually use values between 10k and 100k since they provide a good tradeoff between a nice reasonably high input impedance and relatively low base-current effects on bias. Of course, other design constraints for the overall circuit and application can sometimes result in me using much larger or smaller values. If there are no such circumstances, then I typically go with the 10-100k values. After that, it's usually whatever I can lay my hands on. I happen to have a pile of 20k resistors on my bench, so I started with that, then figured the 56k to set the bias I wanted at the base and emitter. Of course many other combinations would work well too.
You want the current flowing through the bias chain to be significantly greater than the quiescent base current. So if you have 300μA collector current and you pick a transistor with a β of 100, you get 3μA base current. The current flowing through the 56K + 20K from a 5V supply is 66μA, which is a teeny bit low, given that the 2N2222 could have a β as low as 40, thus taking a base current of 7.5μA. I'd personally want slightly lower base bias resistors or a higher gain transistor to make sure the output operating point couldn't vary too much depending on the transistor β.
Just at the end of the video I saw a Bird 43 and realized you are an OM right ? Congratulations for this nice training and video very clear and crisp ! One question : I would like o use this amplifier between a super-regenerative detector and a LM386 amplifier. To avoid saturating the LM 386 input, I think the best would be to limitate the gain of this amplifier between 2 to 10 max. How is that possible ? Thank you ! Best 73 F1CUJ
Yes, it is very easy to limit the gain to lower values as you need. It is usually done by using emitter degeneration. This is explained clearly in my latest video: ruclips.net/video/VWY2WQcKJgk/видео.html
Is it possible to use a single BJT to obtain an 8 Vp-p Symmetrical Output Swing and voltage gain between 800V/V and 1000 V/V at 100 Hz by capturing an oscilloscope screen shot of input and output voltage? (Its a class project)
It's possible, but you'll need to use a VCC supply of 30V or more. Shortcut way to think about this... Av = gm*Rc. gm = Ic / Vt, Vt is 26mV at room temp. Combining these equations, you can show that Av = V(Rc)/Vt. Thus, in order to achieve voltage gain Av of 1000, you much drop at least 26V across the Rc collector resistor at the quiescent point. In order to swing 8Vpp around that and not saturate the transistor, you'll need another 4+ volts of headroom, thus at least 30V VCC available.
hello sir, so the frequency response of a common emitter amplifier does only depend on bypassed capacitor and output Rc coupled capacitor?? please reply as soon as.
In this specific case, yes. But not always. There can be many different factors that affect the frequency response. A very common limit is often imposed by the base-collector capacitance which tends to roll off the input signal. There is no single universal answer.
Hi, I do not understand why the voltage on the collector is 2 volts. If we have Ve = 0.685v and we have Re = 2200 ohms, then it is Ie = 310 micro amps. 310 micro amps through Rc = 10000 ohms, we get the voltage at 3.1. 5v - 3.1 V = 1.9 V! So Vc is 1.9 v! Is it because of parasitics, or because this 47 uF cap, or Ie is not equal to Ic? Excellent video, thanks!
It's more likely due to the tolerances of the emitter and collector resistors. I can't really make it out at 6:57, but they might well be 5% resistors. If Re were 5% high and Rc were 5% low, we would get: Re = 2.31K, so Ie = Ic = 297μA. Rc = 9.5K, then Vc = 5V - (297μA x 9.6K) = 5V - 2.8V = 2.2V Does that look most likely?
@@MrFKD The traditional way to define circuit bandwidth (high pass, low pass, bandpass, or notch) is to look for which frequency results in a response that is reduced to 1/2 the power level of the in-band response. Since Power is a defined by voltage-squared divided by resistance, the half-power point corresponds to the point where the voltage is reduced (divided by) the square root of 2. V/sqrt(2) is the same thing as V*0.707, or approximately 70% of the amplitude in-band. This is also known as the 3dB point, because a 3dB reduction in power is another way of saying that the power is divided by 2. See my video on decibels for more information. ruclips.net/video/1mulRI-EZ80/видео.html
Hi Alan, I have recalculated all the values and agree with what you have measured. However, I am upset with the calculation of beta: Ib = Vcc/R1+R2 = 0,06 mA Beta= Ic/Ib = 0,3/0,06 = 4,35 it is very low ! I have checked the 2N 2222 data sheet and beta with IC=0,3 A is about 35 Did I made a mistake in my calculations ? As far as I understand we can't force the beta of the transistor ? Thanks in adavance
Your Ib value is the BIAS current through R1 and R2, that is correct. The Ic current is approx 0.3mA, that is correct too. You *can't* force the beta value, it is what it is. So, the BASE current will be determined by Ic/Beta, or approximately 8.5uA. The general guideline is that you want the BIAS current through R1 & R2 to be >> base current. In this case, it is about 8x larger, so that's not bad.
@@Bremontval The β is at least 35 for a 2N2222 at 300μA emitter current, so the base current can be no more than 300μA/35 = 8.5μA. The current through the base divider chain is 5V/(56K + 20K) = 5V/76K = 66μA. I think we can agree that 66μA is about 8 times larger than 8.5μA.
Thank You! I've wanted to understand electronics since I was a kid and thanks to internet and people like you it is now possible. The transistor amplifier design has been a real struggle for me but the pieces are starting to fall in their places. Among some written material, your videos have been an invaluable resource and by far the best videos I've come across altogether.
Same in my case.
Me too
Another superb video, even if it is from almost 10 years ago! Next I'm watching #113, 114, and 185 in preparation for watching #273. Thanks!
Cool - brushing up on your BJT circuits, I see!
@@w2aew Yes. 50 years after learning basic transistor circuits, I'm trying to tutor my 12 year old grandson. Grandpa is rusty.
Another excellent lesson Alan. Appreciate your efforts preparing and delivering very well organized and insightful lesson plans, always looking forward to them. Much obliged, sincere thanks.
I thought this video was great! It was easy-to-follow, and you covered the most important topics in a clear and concise way. You're a great teacher!
source: someone getting his M.A. in Integrated Circuits at UCLA
Wow- amazing clarity not often found in electronics videos. You’re a great teacher. Thanks so much for all the time and effort that goes into your videos. If you’re looking for a new topic for one of your fundamental videos, you might consider an active load on a common emitter amplifier. I’ve been trying to wrap my mind around this topic for some time now. I can’t find a tutorial that explains it clearly, the way you do. Thanks again.
I do have some videos on transistor current sources which talk a bit about using them as an active load.
@@w2aew Awesome, can you give me the video numbers? I searched your videos for the keywords but couldn’t find anything. I did watch one video you made on current sources, and although you mentioned active loads briefly, there were no details about how they work. In that video you mention that if a current source has a high output impedance, you can use it to achieve high gain in an amplifier. I’d love to get details about how to achieve that. Thanks so much for your reply.
@@jrausa1 Yeah, I guess I haven't gotten into the details of an active load specifically. I'll have to put that on my list. Here are a few of the videos I was thinking of that touch briefly on the topic:
ruclips.net/video/xR0RfmmRhDw/видео.html
ruclips.net/video/2IbWvq7Abv4/видео.html
ruclips.net/video/mejPNuPAHBY/видео.html
@@w2aew Thanks again, Alan.
A PROfound example of understanding a subject down to it's core, thereby "eliminating all guesswork"!
It will be a banner day when I apply the formula's to obtain the answer, and then, confirm via scope! Thank You!
Alan makes this look so easy!
Excellent demonstration of the corner frequencies by scoping the frequencies at the -3dB. Many thanks for this, it kind of everything made sense right now!
A very informative video, thanks. For everyone that wants to use this circuit on higher frequencies: it is advised to remove the capacitor from collector to ground.
Thanks, very clear explanation. Seeing this video I remembered I learned this when I was studying electronics about 20 years ago. I thing I'll do some calculations and measurements myself, just for fun.
Thank you for your prompt response. I understood. Power is proportional to the square of voltage, or current. Thank you very much for all your excellent videos. They are priceless.
As always, excellently explained. You have a very nice way of explaining things. Keep it up!
Important stuff. I'll have to watch this many times to get these formulas down. Thanks. I'm gonna build this and check it out some more.
For Rosa Rojas (since I can't reply to the comment...) It's all there in the formulas - simple math. Gain is -gm*Rc. So, either make Rc 5k vs 10k, or decrease gm by a factor of two (by increasing the 2.2k to something closer to 4.4k).
+w2aew Why have capacitor at the collector ... to ground?
+Yueeiuyoo A low pass filter - to remove high frequency components from the output.
+w2aew I assume that this applies in case we are capactively coupling this amplifier to the next stage, what we are not doing that? That means the miller capacitance shall limit the upper 3dB point?
Also, you said that when the reactance of the capacitor at output is equal to Rc at 5:46 minutes that is when we get the upper 3dB point. That means the amplitude is halved at output? or some other reason?
+Yueeiuyoo It is when the output power is halved - not the output voltage. In many amplifiers, the miller capacitance will be the bandwidth limiting factor.
+w2aew when you say that the impedance of the capacitor has to be equal to the resistance of the collector resistors for the upper -3dB point, how is this conclusion arrived at?
Your videos and the knowledge your share is addictive. Thanks for sharing.
Finally I have found the frequency equations for BJT CE Amp. Thanks a lot
This is the best video ever. I've done so much work to understand amplifier circuit and this helped a lot.
I have been going through older videos to learn some and I played along. Pretty cool to be able to do hands on with you. I learned something. 73, ad0am
Glad to hear that you're finding my videos useful.
Thanks once again Alan. I always learn something from your videos.
I Would Say This Is your BEST VIDEO So Far!!! Info on Basic transistor amplifiers Is Very Hard To Find,,,,These Basic Concepts on Transistor Amplifiers were Long Lost In My Head Until Your Video Came Along!! I'm going to Breadboard this Lab To Verify The Calculations Thanks Once Again w2aew!!! PS**** How about a Video on Fixed Bias,Self Bias and Beta Independent Transistor Amplifiers,,,A Super Job Once Again Alan!!! Keep Those Videos Coming,,I'm Starving for some more Electronics Theory
Excellent exposition. Good effort expended in you videos . Well done. Thank you.
The cutoff frequency is the frequency either above or below which the power output of a circuit has fallen to a given proportion of the power in the passband.
Most frequently this proportion is one-half the passband power, also referred to as the 3 dB point since a fall of 3 dB corresponds to approximately to half power. As voltage ratio, this is a fall to approximately 0.707.
Thanks for the great explanation. I misunderstood the low frequency cutoff. And you've corrected that with this video. However, a more accurate explanation of Fl would be that there are multiple low pass filters and the one involving Ce and re' dominates. There is actually four low pass filters. The input circuit with Cb and 50ohms || R1 || R2 || b*re', the ouput circuit with Cc and Rc, and 2 at the emitter, Ce and Re and Ce and re'. The Ce and re' filter should be designed to dominate to make calculations simple. Since the re' filter cuts off at 39 Hz it dominates the Re filter which cuts off at 1.5 Hz. Obviously the higher low pass filter will dominate unless they are close together. Then the filter calculations becomes complicated. Since 39Hz >> 1.5Hz we can ignore the influence of the Re low pass filter.
There are two capacitors at play and hence two low-pass filters to consider.
One is formed from the 47μF emitter bypass capacitor and the parallel resistance of 86R (the intrinsic emitter resistance) and 2k2, which is 83R. That has a cutoff frequency of 41Hz, very close to the 39Hz obtained by ignoring the 2K2 resistance.
The other is the 4.7μF input capacitor and the parallel combination of 56K || 20K || β.Re where Re is the effective total emitter resistance. Note that the 50R load on the signal generator plays no part in that calculation. That is complicated because β is unknown, but could be as low as about 40 for a 2N2222 operating at 300μA, and because Re is the complex impedance of the emitter resistance and the 47μF capacitor. Normally, we should design so that β.Re is significantly greater than the parallel combination of 56K || 20K (i.e. 15K) at the frequencies of interest, but 40 x 83R = 3K3 and that will dominate the input impedance unless you select your transistor to have a β significantly more than 180 at 300μA. With the transistor used, the cutoff frequency associated with the input capacitor could be 10Hz if β=40. Fortunately, the ~40Hz cutoff from the emitter bypass will dominate in this case.
Another SUPERB video! Thanks!
Another excellent tutorial, thanks Alan!
@w2aew How do you know the limits of the amplitude you can have at the input before you start clipping your output?
If you look at the bias point of the transistor - mainly the collector and emitter voltages, you can get a good estimate how far the collect voltage can swing before it either hits the positive supply rail or saturates the transistor. Once you determine this swing, then divide by the gain of the amplifier to determine the input signal swing.
@@w2aew I wanted to knoe the anwser for this same question. But why cant I just increase the output of the function generator until the output of the circuit starts clipping and then read off the value of the signal from the function generator to find the max input to the circuit. Is this approach incorrect?
@@GLUEALLSTAR You certainly can do that. I was assuming that you wanted to know how to determine the clipping level without testing it.
quick question, i have a function generator in which you can set the impedance , would that be equivalent to that load termination that you have or is that in addition to the generators impedance.. ?
It does not change the generators output impedance. This setting tells the generator what the expected load impedance is so that it can generate the correct voltage. I did a video on this:
ruclips.net/video/tClE8s6RZdg/видео.html
Ahhh thanks!!!!
Brilliants!!!! Every kid should have a teacher like this in every subject. Should you go for a gain of 117 or do it in two stages at 10 each? When do you need multiple stages.
If you go for a gain of 100+ in one stage, you will inevitably run into the problem of distortion with a large signal. I don't mean clipping, but when the gain depends on Rc/Re and Re is the non-linear intrinsic emitter resistance, you are bound to get distortion.
To design that out, you have to have an unbypassed emitter resistor significantly greater than the intrinsic emitter resistance of the transistor at whatever emitter current you are using.
That means that because the intrinsic emitter resistance = 25mV/Ic, the emitter resistor must be at least 10 times that resistance, which equals 250mV/Ic.
That means you need at least 250mV across the emitter resistor.
Since the collector resistor will need to be 100 times the emitter resistor for a gain of 100 and the collector current is almost the same as the emitter current, you will need 100 times the voltage drop across the collector resistor that you have across the emitter resistor.
That means you need to drop 100 x 250mV = 25V across the collector resistor.
Since the quiescent collector bias point needs to be about half the supply voltage, you will need a 50V supply rail (and a transistor that can handle a 50V supply rail).
That is the inescapable logic of designing a common emitter stage: to increase the gain while maintaining low distortion, you have to increase the supply voltage.
While you have the most splendid and clear tutorials on these things where others go into details that just make the core harder to understand, it will be quite powerful if you made a video explaining construction of a simple amplifier e.g the 741. This shall involve current mirrors, long tailed pair and much more. Making measurements at each point and why design was done a certain way will be very helpful.
The only problem is that such a video may require more work to be produced.
+Yueeiuyoo I do spend some time talking about the internal circuitry of the LM358 op amp in my video describing cross-over distortion:
ruclips.net/video/VgodYtiD_F0/видео.html
your videos are very helpful, i've been able to confirm a lot about typical transistor operation in general and become more familiar with the dynamics of a transistor given the various configurations... that elevator buzzer (boop or beep or ding) has had me looking around the room a couple times though... heheh
This was a easy to follow video. I find the wires a little distracting. I just wish my free simulator was working as well as your reality.
Thanks for your videos so didactic. I should have had a teatcher like you when I was a student ... thanks again
Vacuum tubes have a MU spec and GM transconductance spec. I'm not sure what internally inside the vacuum tube is determining the MU amount and the transconductance amount but I think is how much electrons are on the cathode coating ? Because the cathode is what is releasing the electrons which is the current which is the MU and transconductance?
I would love to see someone competent (thus commenting on this channel ;) do a transistor tutorial. Example: Given a gain requirement and input voltage/current/frequency, design an amplifier using an NPN transistor. I've watched dozens of videos entitled "how to bias a transistor" but each and every one of them analyzes an existing circuit. What I need to know (and I believe many others) is the step by step process to designing a transistor amplifier circuit. Would also be nice to cover the different classes of amplification as well (ie: how one class differs from the next by way of the biasing or however?) Also watched tons of videos on Transistor theory - ie: materials, doping, etc... Like to see it kept to a practical (how to) view. Some people start with a ratio of Rc to Re (ie: gain), some start with biasing the Base and Re, then simply state that Ie is the same as Ic (why? how?)... All very confusing. A logical process would be welcome. There must be one - this is electronics/math not voodoo.
+Digger D A couple of comments:
- IC is almost the same as IE because IE is equal to IC + IB, and IB is small with respect to IC (IB = IC / Beta).
- I already did a video on biasing and amplifier class of operation:
ruclips.net/video/c6cmkm3UPUI/видео.html
- As for a tutorial on designing a transistor amplifier, you can't get much better than the one that Shahriar did:
ruclips.net/video/Y2ELwLrZrEM/видео.html
***** Thanks, it's starting to come along. Next to determine how to decide on values for the voltage divider on the bias (ie: many possibilities but what logic does one use to decide what values of resistance to use, I suspect it depends on the current required at the base?). IB * Beta determines IC and ratio of Rc over Re determines voltage gain? But I would have thought there would be a process to determine each. Most of the vids I've watched they seem to pull a value out of the air. Thanks again for the reply and links.
***** I watched Shahriar's video but he too pulled values out of thin air (Ic = 2 mA??, why did he just pick a value?) and gm was not defined etc... I did however just discover that Rbe is a constant of 25 or 26 ohms, I'll go with 26 as I just saw the math for that result. Now this makes more sense as I can now calculate Ib and thus choose my bias resistors to set the current into the base and knowing the Beta value (generally as published) I can then calculate what my Ic will be. Ic = beta X Ib. It now becomes a resistor network problem. As for the voltages over Rc, CE and Re that is pretty easy to determine. It's like trying to learn the secrets of a mystical lodge who's members are sworn to rumors only. ;) Even finding a book that explains this in a step by step logical manner is difficult. I've spent a week watching many videos, reading several articles and asking questions but it's starting to come together. Transistors have always been a stopping point for me in analogue electronics as I just could not find the information needed to design a circuit. So for a common Emitter amplifier I think I'm ok, now for all the other configurations... Thanks again for your help and I did watch your bias video earlier but it too didn't quite answer all my questions but it did create some new and better questions. Every component and resulting value in a circuit has some logic behind it and sometimes I find people skip over something that has become common knowledge to them but for us folk that are trying to absorb the info, it leaves a huge gap. Once again thanks for your help.
+Digger D Sometimes the initial choices for things like IC are arbitrary, other times the initial choices are governed by requirements of the application - such as: input impedance must be X, or load resistor or output impedance must be Y, etc. Or, there may be a limit on power dissipation for the circuit, or the designer might choose a bias level based on some optimized transistor characteristics. The initial constraints will often help decide the starting point (whether it is bias point, a resistor value or two, transistor selection, etc.) However, the initial choices are governed, the art of the remaining part of the circuit design is determining the circuit structure and component selection to achieve the desired goals while keeping within the constraints given in the initial performance desires or requirements. It's not an exact science, every application is different.
Amazing video! Honestly a God send!
Great video. Many thanks.
Great video, man!
Excellent video. I second 321reh comment: this is one of your best and the beginning is just a little fast. How did you get the thermovoltage and the transconductance? A little more info on that would be amazing.
Also, how did you came up with all the values for the resistors and capacitors. It's pretty complicated to match all the constraints. Can you recommend a method?
so basically, the critical or cutoff frequency is +- 3dB, .707 and 1.414 are turning out to be a magical number in the wold of alternating current
please make video on tuned amplifier
Excellent video and NICE O'Scope! I'd love to have one of those babies!
Fascinating set of videos. Thank you for the time you have spent putting all of these together. Could you please complete an analysis, using the measured voltages to ascertain the beta of the transistor used? IC = 310uA, IE=311uA meaning IB=10uA, assuming the circuit is working in the linear region. Does this mean the beta of this particular transistor works out at something like 30? Thanks again.
Since we’re likely talking about beta on the order of 100-200 or more, we would have to be very careful with our measurements. For example, if you’re measuring the voltages across RC and RE to determine those currents, you would *have to* carefully measure those resistor values first because their tolerance (how much they can be off of their ideal values) will dramatically affect the calculated beta.
@@w2aew Many thanks.
Yes i understand that the DC offset won't be on the secondary, but won't the AC waveform have an AC offset not starting at the zero crossing point. The AC waveform might not start exactly at the zero crossing point might have some offset on the secondary?
Only the AC portion of the waveform will be present on the secondary - no offset unless there is one applied on the secondary.
You helped me a lot. Thanks mate.
Awesome explanation!
on the topic of uhf and emf tourture from a distance through direct sound or "organized stalking". what is the best way to protect my self from these "psychotronic attacks"?
Another great video!. Thx for sharing that.
Sir, the formula to calculate dB is
dB = 10 log (base 10) of V2/V1
log (base 10) of 0.707 is equal to - 0.15.
Please, could you explain how this is equivalent to - 3 db?
When taking a voltage ratio, the formula is 20 * LOG10 (V2/V1)
also whats the best way to detect these attacks? thank you for your time.
Why is it necessary to use the 4.7 uF coupling capacitor on the input? I understand it would block any dc components of the signal, but if we're applying a simple sine wave from a signal generator, then there would be no dc component, correct? Or am I missing something? Best practice just to be sure the signal is pure?
The 4.7uf cap is used at the input to *preserve* the DC bias at the base that is established by the 56k/20k resistor divider. If the capacitor wasn't there, the 50 ohm terminator for the signal generator would be in parallel with the 20k resistor and the bias voltage on the base would then be dragged very close to ground.
Can't the transformers core or secondary windings cause an offset on the secondary or what can cause an offset on the secondary besides a DC voltage?
thnku so much 4 d upload.....how r capacitor values designed??
Very nice! Is the only purpose of the 220pF cap to ground to set the high frequency point? would you include this in the final design of an audio anp? Without this cap, the gain does not drop with frequency, why would this be bad, maybe becuase high frequencies would also be amplified? Thanks fot the help and the great video. (Also, the point of the 50ohm source resistance? Doesnt the generator already have a 50ohm output impedance?)
Yes, the 220pF sets the high frequency rolloff. It was used as an example here. Without it, the frequency response would be wider, but the gain would eventually drop off due to other device capacitances. The 50ohm resistor is properly terminating the output of the signal generator. Again, not important at audio frequencies, but it ensures that the voltage provided by the generator matches what it is set to.
Thank you very much for the response. So in an actual circuit, with a speaker as the load, will you keep the capacitor in to set the high frequency cutoff? Or do you rely on the capacitance of the speaker to do this job? Thanks again for the info. As for the 50 ohm resistor, why would you want to terminate the input signal? Don't you usually do that when you have an open circuit somewhere and you don't want any reflections? Thanks again!
Does your generator have an output impedance of 50 ohm (set internally) ?
It all depends on the amplifier design, whether or not it is necessary to add some high-frequency rolloff. Many/most applications wouldn't care.
At audio frequencies, reflections are not an issue. The main reason I added the termination resistor is so that the displayed amplitude on the generator accurately reflects the voltage applied to the circuit input. The generator allows you to specify the load impedance, and it defaults to 50 ohms. Since it has a 50 ohm output impedance, the voltage generated by the generator is 2x the value that is set in order to account for the voltage divider effect that comes from the 50 ohm load. I could have optionally specified the actual input impedance of the circuit and omitted the 50 ohm termination, but it was easier to simply use the termination and 50 ohm load setting.
@@pierreretief The output impedance of the circuit shown is set by the collector resistor at 10K. You're not going to be driving many speakers with that sort of output impedance. With the exception of Quad Electrostatic speakers, loudspeakers present an inductive, not a capacitive load to the amplifier, so you can't count on rolling off high frequencies that way. Normally, a capacitor is used to feed back from the output to the input of an inverting amplifier, which sets the upper roll-off frequency and ensures overall stability of the amp.
How do I combine these two waves? Like I want a 360 wave where two sine waves are 180degree phase shift from each other at the same time for the output, is there a circuit for that?
How can the bias V_B voltage be 1.28V in the first place? I couldn't read properly the value of the lower resistor of the base divider (it's either 26k or 20k). In either case (and if we neglect the base current) it's 5*(26/82)=1.59V or 5*(20/76)=1.316V. Moreover if it's indeed 26k according to my calculation the BJT should be in saturation: V_B=1.59V -> V_E=1.59-0.6=~1V -> I_E = 1/2.2 =~ 0.45mA -> V_CE=5 - 0.45*(10+2.2) < 0 !
The lower resistor value is 20k, so your 1.316V calculation is correct. It's only off by 30mV or so (about 2% off), which is due to resistor tolerances and base current contribution through the 56k resistor.
Great video!!!
is it compulsory to use non-polar capacitor or we can use polar capacitor.what will be the effect on output with both types of capacitor.
It depends on whether the circuit has sufficient DC offset between the two sides of the capacitor to prevent reverse biasing a polarized cap.
Can u do anything on scalar waves? Or have both waves of the opposite phase shift in one output?
what is the minimum signal that can be amplified with this circuit. how to check this property or how we can calculate it?
The amplifier doesn't "stop" working for small signals, so there really isn't a limit on how small the input can be. However, factors such as the signal frequency, the required bandwidth, the noise in the signal and amplifier, etc. will all affect how well the amplifier works.
The Bypass Capacitor across the emitter resistor/cathode resistor on transistors or tubes will also Filter out the AC Ripple? because in tube amplifiers if you don't use the Bypass capacitors the tubes AC heater and power supply AC ripple will get amplified because the bypass capacitor is not filtering out the fluctuations?
The bypass capacitor on the emitter or cathode is to increase the gain by allowing the intended signal to bypass the resistor. It is not for supply filtering.
Hi, I was looking at video 273 and thought it would be good idea to visit the other videos you suggested. So it hit me, I saw something on your scope that I have always had showing up on every simulation and test of the collector output that I have seen. The peaks had wider span then the troughs. I understand it was the exponential behavior of the BJT as the linear range is not entirely linear but rather somewhat almost linear. Is there another reason you can think of? From a BJT class A amp, i probably think this should always be the case, right?
Yes, it is more prevalent the larger the signal and is primarily the result of the lower gain owing to lower incremental transconductance as the collector current decreases (which is what's happening as the output signal rises). The reduction in gain results in compressing the signal peaks. The opposite occurs at the troughs (at least until you get close to saturation).
How do you Hi Pot test an amplifier circuit? If an amplifiers input has DC offset voltage can it amplify the DC offset voltage on the amplifiers output? What is amplifier input and output AC offset voltage mean (not DC offset )?
A DC offset on an amplifier input doesn't necessarily mean that it is DC coupled all the way through. I've never heard the term AC-offset.
Hey Alan
I really enjoyed this video and the ones related to it. These lessons have helped me with guitar pedal builds and understanding the math more behind the circuits. would you be interested in breaking down a fuzz face circuit? I’m mainly using npn transistors for the amplifier instead of pnp. Thanks again!
I don't know what a "fuzz face" circuit is.
I’m not sure in my reply’s are going through
@@wingate8 RUclips doesn't allow links in comments/replies apparently.
@@w2aew
Ahhh well fuzzcentral.ssguitar has a Fuzz Face NPN Silicon schematic that I would love to get your take on. I’m using most of the same values and have modded the bias point of both transistors as well as collector voltage
@@wingate8 OK, I think this is the page you're talking about:
fuzzcentral.ssguitar.com/fuzzface.php
Looks like a fairly simple two-stage amplifier with a little bit of negative feedback. The pot adjusts the gain of the second stage which influences how much (if at all) the 2nd stage is driven into clipping for a given input signal level.
How do i calculating AV in a multi stage amp for example a 2 stage CE amp's AV(tot) ? I know this is hard to put in words with no resistor values etc so if you can up load a video about this it would be much appreciated
+Denis JJ You simply multiply the gain of each stage, making sure to account for the loading affects of the stages on each other.
Can this circuit be modified to work at 300-1,000MHz or is only good for Audio frequencies?
Great analysis.. Thanks
great video. just wondering if you have any other videos explaining the capacitors types to use at what location as I not sure do I use film capacitors on the input and electrolytic on the output etc.
I don't have one on this topic, but it's a good idea. I'll add it to my list.
Thanks. I'll look out for it.
Even if an amplifier hits the 3db drop off thus "showing the top of bandwidth", isn't it true it still has usable gain though until it hits zero db entirely?
Sure, it all depends on whether the usable gain is suitable for the application.
@@w2aew Thanks! I did a bit more googling and found that they call the range where it is above 3db drop, the "power bandwidth" vs the "unity gain bandwidth", which is the "entire" bandwidth overall
Please, what's the value of the frequency that you choose in the function generator in the beginning?
Looks to be about a 500us period, or 2kHz.
@@w2aew thanks a bunch man... you're the best
Excellent videos! Thank you. I'm curious about what starting frequency you used to determine the 3db down points. Did you sweep to the max amplitude frequency for a starting point? EG. if you started at 100 Hz or 60 kHz and adjusted the frequency to find 3db down, you would come up with different corner frequencies. Does this make sense? Can you clarify for me?
Thanks
I started with a frequency that would be fairly far away from the corners based on the component values. Or, doing a quick sweep or set of measurements at various frequencies, it's easy to find something in the passband, and slide up and down to find the edges.
Nice tutorial, can you also upload some RF (100MHz to 500MHz) amplifier tutorial with few mW input to few watt output , using bjt and how to calculate the power gain, frequency response of the amplifier and also the input and output impedance. thanks in advance
+Freddie Homecillo If I ever do a project like that, I'll be sure to document it in video.
First of all yo really know your formulae. What would you reccomend in a hi-fi application a single transistor or an op-amp for premium sound quality and low distortion, which is best?
Great demo
thankyou
I believe I finally understand how those two filters work. They both work the same way. They both would be considered RC low pass filters. It's just that the filter at the emitter increases the output signal above it's cutoff frequency while the filter at the output attenuates the output signal above it's cutoff frequency. The equation for the cutoff frequency of both is fc = 1/2πRC. Together they limit the range of frequencies that will be amplified. I'm wondering. Would this type of set up be used for a radio transmitter? Longwave for example. Would the two filters be set to filter out all frequencies except those between 153 KHz and 279 KHz? And if so what would be the benefit? Thanks.
Yes, you're definitely gaining a good understanding. For RF circuits, often a tuned load is used to create a bandpass filter around the signal of interest. The main goal is to minimize the transmission of harmonic and inter-modulation distortion components.
Just want to ask you regarding the high and low pass frequencies; so does that mean the range of frequencies between which the gain can be found is 38Hz and 72Khz?
Yes, the gain is relatively flat (constant) between those frequencies, and is reduced as you go outside this frequency range.
Brilliant job! Thanks :)
I would be nice if you could explain how can we make this circuit have half of the gain already measured by removing a capacitor or by making some other changes. Is it possible to make some arrangements and have a gain of approximately -55?
Just add an extra 100R emitter resistor between the emitter and the 2K2 resistor and bypass just the 2K2. That will get you in the right ballpark for gain, as well as helping to linearise the output, (thus reducing the distortion), improving the frequency response down to about 20Hz, and raising the collector bias point a little so that you can get more headroom from your 5V supply.
Thank you for your explanation, I would like to ask how did you solve for Vt? or is it always 26mV?
I am trying to solve an assignment for the output voltage and frequency but I was not given the value for the capacitors how can I find the frequency?
Also, does the voltage gain formula change? In class it was given as the output voltage/input voltage (Vout/Vin) not -Rc/re
The Vt value (thermal voltage) is calculated via kT/q (en.wikipedia.org/wiki/Boltzmann_constant#The_thermal_voltage)
At room temperature, it is about 26mV.
If they didn't give you values for the capacitors, maybe they're assuming that they're large enough that they can be ignored (effective low Z at the operating frequency).
The gain formula doesn't really change, but it could be represented a different way. Since gm=1/re, you often see the formula written as AV=-gm*RC. If feedback is used, such as adding an emitter resistor, then the formula does change.
@@w2aew I appreciate your response. So if the capacitors are large enough that they are ignored, how can the frequency be obtained?
Also given the emitter resistor what will the gain formula be?
@@sarahmajin5358 Without knowing what circuit you're analyzing, it is impossible to say.
@@w2aew it’s a common emitter circuit with an NPN transistor like the one in your video with an input of 100mV@50Hz and a load of 1k ohm.
Base resistors of 10kohm and 2.2kohm,
collector resistor of 3.6kohm,
emitter resistor of 180ohm and 820ohm, the 820ohm is short with a capacitor.
The output resistor is 51kohm.
Vcc is +10v and beta is 200
@@sarahmajin5358 Sounds like you have what you need to calculate the operation point, collector current, etc. in order to figure out the gain. Not sure what you mean when you say the load resistor is 1k and the output resistor is 51k. Do you mean that the collector has 3.6k to 10V, then there is a 51k resistor in series between the collector and the 1k load resistor? Another question - when you say you need to solve for frequency, I"m not sure what you're looking for. You said that the input signal is at 50Hz, so that is the frequency. The amplifier won't change this. Or, are they asking for the frequency response? If they don't give you capacitor values, maybe they're asking for you to describe the equation that determines the frequency response (i.e. the emitter RC determines the high-frequency corner, etc.).
Great video !!!!
Thanks a lot ...
Any chance you doing a video for UHF frequency range, explaining more how to include parasitics. I always watch advertisements, before you videos, till the end so you get full ca-chugn ;-)
in what range can we hve bandwith?? what r d limitations there???
interesting video, i like that scope!!! i dont think i have any of those functions on mine.
First Great video. I want to ask why during frequency gain, we calculate for -3DB ( approx -30%) point ? Why not bit more than -3DB or less ?
because @-3dB point, output gain is reduced to 70.71% of its maximum value then it start to decay with 20dB so you can say upto 3dB your output will be significant or meaningful which will not be the case beyond 3dB point.
There's something else I'm wondering about. I know that an amplifier like this has to be connected to the signal source and the load through a coupling capacitor. My understanding is that the purpose of these capacitors is to filter out the dc component of the signal so that only ac enters and leaves the amplifier. With this is mind it seems to me that the higher the capacitance the better in order to keep the reactance as low as possible so that the capacitors behave like a short circuit. Is this correct or is it more complicated than that? Thanks.
In general, yes, you want to use values that make the reactance low compared to the circuit impedance at the lowest frequency of interest. Making the caps unnecessarily large may result in delays or stability at startup when the cap has to charge up to the DC difference between the two circuits. Also, very large value caps might not work as well at the highest frequency of interest due to parasitic inductance. Also the very large values might mean the use of less desirable capacitor types like electrolytics.
Is it ok for an amplifier IC chip output a DC offset voltage or is this bad because the DC offset voltage can damage the speaker? If the output transformer primary has a DC offset voltage from the amplifiers IC chips output, will the output transformer secondary have a DC offset voltage also since the primary has a DC offset voltage?
DC will not pass though the transformer. So, if there is DC on the primary, it will not be on the secondary.
Thank you sir..
Thank your the best!!!
can u make a video of common gate amplifier...it will be very helpful.
THANKS IN ADVANCE
I have had a hard time finding a concrete answer on this, but can a small signal transistor like a 2N3904 effectively amplify a larger signal(say, around 1.3Vpp) and generally, at what point in signal amplitude does it become necessary to step up to a power transistor?
It certainly can. Remember, voltage alone does not equal power. Power is voltage*current. Typically a lot depends on the load that the amplifier needs to drive. Transistor power is primarily dictated by the collector-emitter voltage times the collector current.
@w2aew My aim is to have an MCU driven Bluetooth receiver feeding audio into an old tube radio I restored from 1938. The raw audio output from the MCU has some slight audio distortion(which generally seems localized to the lower frequencies) I messed about a lot with different means of matching the impedance. It seems the best results I've gotten so far was shunting a 560 ohm resistor from the input to the chassis on the radio, but the low frequency problem remains. So I was looking into adding a common emitter stage in between to see if that helped out
@w2aew I was also wondering what technique you use to measure small signal base current when you have a voltage divider on the base? I have yet to purchase an AC current probe for my scope(I do have some SCTs laying around that I could likely make work) but that would only be a viable solution on a breadboard.
I suppose I could subtract the RB2 current from RB1 and end up with base current
@@ColossusEternum Very difficult to comment on the cause of your distortion without knowing what your circuit looks like, or what the input of the audio-input of your old radio.
@w2aew I understand that completely. I'm just going to have to keep researching and experimenting with it. Right now I'm not terribly concerned about that specific circuit(I was trying to run before I could walk lol) it's just a little daunting when you build an amplifier because there are all these ways to adjust the circuit, but they all have a range of consequences. I've been meaning to draw a table in which I list all the factors that each component has an influence over.(like base resistors influencing Vb, Ib, reducing hFe dependence, input impedance etc.
These videos are very helpful. I have an additional non video related questions. Any way to send you a private message?
Is there any advantage in replacing the base bias resistors (56k / 20k) with,say, a100k potentiometer / preset to optimize the operating point of the transistor? And how did you decide to use 56k / 20k anyway? Other values could have been used to get the 1.28v
The advantage would be, as you said, being able to easily dial-in the optimum operating point. As for how I picked the bias resistors on the base... ...I usually use values between 10k and 100k since they provide a good tradeoff between a nice reasonably high input impedance and relatively low base-current effects on bias. Of course, other design constraints for the overall circuit and application can sometimes result in me using much larger or smaller values. If there are no such circumstances, then I typically go with the 10-100k values. After that, it's usually whatever I can lay my hands on. I happen to have a pile of 20k resistors on my bench, so I started with that, then figured the 56k to set the bias I wanted at the base and emitter. Of course many other combinations would work well too.
You want the current flowing through the bias chain to be significantly greater than the quiescent base current. So if you have 300μA collector current and you pick a transistor with a β of 100, you get 3μA base current. The current flowing through the 56K + 20K from a 5V supply is 66μA, which is a teeny bit low, given that the 2N2222 could have a β as low as 40, thus taking a base current of 7.5μA. I'd personally want slightly lower base bias resistors or a higher gain transistor to make sure the output operating point couldn't vary too much depending on the transistor β.
That was a nice video.
Just at the end of the video I saw a Bird 43 and realized you are an OM right ?
Congratulations for this nice training and video very clear and crisp !
One question : I would like o use this amplifier between a super-regenerative detector and a LM386 amplifier. To avoid saturating the LM 386 input, I think the best would be to limitate the gain of this amplifier between 2 to 10 max.
How is that possible ?
Thank you !
Best 73
F1CUJ
Yes, it is very easy to limit the gain to lower values as you need. It is usually done by using emitter degeneration. This is explained clearly in my latest video:
ruclips.net/video/VWY2WQcKJgk/видео.html
OK Alan. I have watched the video. It's very clear ! Thank you very much
Is it possible to use a single BJT to obtain an 8 Vp-p Symmetrical Output Swing and voltage gain between 800V/V and 1000 V/V at 100 Hz by capturing an oscilloscope screen shot of input and output voltage? (Its a class project)
It's possible, but you'll need to use a VCC supply of 30V or more. Shortcut way to think about this... Av = gm*Rc. gm = Ic / Vt, Vt is 26mV at room temp. Combining these equations, you can show that Av = V(Rc)/Vt. Thus, in order to achieve voltage gain Av of 1000, you much drop at least 26V across the Rc collector resistor at the quiescent point. In order to swing 8Vpp around that and not saturate the transistor, you'll need another 4+ volts of headroom, thus at least 30V VCC available.
*****
Thank You for the help
hello sir, so the frequency response of a common emitter amplifier does only depend on bypassed capacitor and output Rc coupled capacitor?? please reply as soon as.
In this specific case, yes. But not always. There can be many different factors that affect the frequency response. A very common limit is often imposed by the base-collector capacitance which tends to roll off the input signal. There is no single universal answer.
Love this.
Hi, I do not understand why the voltage on the collector is 2 volts. If we have Ve = 0.685v and we have Re = 2200 ohms, then it is Ie = 310 micro amps. 310 micro amps through Rc = 10000 ohms, we get the voltage at 3.1. 5v - 3.1 V = 1.9 V! So Vc is 1.9 v! Is it because of parasitics, or because this 47 uF cap, or Ie is not equal to Ic? Excellent video, thanks!
It's more likely due to the tolerances of the emitter and collector resistors. I can't really make it out at 6:57, but they might well be 5% resistors.
If Re were 5% high and Rc were 5% low, we would get:
Re = 2.31K, so Ie = Ic = 297μA.
Rc = 9.5K, then Vc = 5V - (297μA x 9.6K) = 5V - 2.8V = 2.2V Does that look most likely?
At 8:21 why did you exactly choose 70% of the output amplitude(1.41 volts)?..
Because that is the 1/2-power point - the traditional way to specify bandwidth.
@@w2aew i don't really understand can you explain more, please ?
and i don't know what's the relation between 3db and 70%; i mean why 3db means 70%
@@MrFKD The traditional way to define circuit bandwidth (high pass, low pass, bandpass, or notch) is to look for which frequency results in a response that is reduced to 1/2 the power level of the in-band response. Since Power is a defined by voltage-squared divided by resistance, the half-power point corresponds to the point where the voltage is reduced (divided by) the square root of 2. V/sqrt(2) is the same thing as V*0.707, or approximately 70% of the amplitude in-band. This is also known as the 3dB point, because a 3dB reduction in power is another way of saying that the power is divided by 2. See my video on decibels for more information.
ruclips.net/video/1mulRI-EZ80/видео.html
@@w2aew i get it now...thank you so much
hi W2aew what can i do if you in put in higher and the out put is lower ?? my circuit was a common emitter
+Jorge Gouveia I'm sorry, but your question is not clear.
its fine man no problem
Hi Alan,
I have recalculated all the values and agree with what you have measured.
However, I am upset with the calculation of beta:
Ib = Vcc/R1+R2 = 0,06 mA
Beta= Ic/Ib = 0,3/0,06 = 4,35 it is very low !
I have checked the 2N 2222 data sheet and beta with IC=0,3 A is about 35
Did I made a mistake in my calculations ?
As far as I understand we can't force the beta of the transistor ?
Thanks in adavance
Your Ib value is the BIAS current through R1 and R2, that is correct. The Ic current is approx 0.3mA, that is correct too. You *can't* force the beta value, it is what it is. So, the BASE current will be determined by Ic/Beta, or approximately 8.5uA. The general guideline is that you want the BIAS current through R1 & R2 to be >> base current. In this case, it is about 8x larger, so that's not bad.
Thank you Alan for the explanations. But I don't see where you gave the Beta value in the video ?
How did you calculate it is 8 times larger ?
@@Bremontval The β is at least 35 for a 2N2222 at 300μA emitter current, so the base current can be no more than 300μA/35 = 8.5μA.
The current through the base divider chain is 5V/(56K + 20K) = 5V/76K = 66μA.
I think we can agree that 66μA is about 8 times larger than 8.5μA.