To those watching confused about the 26mV, it is the thermal voltage of a bipolar junction transistor. It depends on temperature and is assumed to be 26mV at room temperature.
Finally a video where you're not simply asked to find operating point and are asked a more real world problem of designing something as oppose to just finding currents and voltages
I have a question, assuming this same circuit, to know the maximum amplitude i can have on input before i reach Collector saturation current and start clipping the output ... do i just add the amplitude of the input to my base voltage and calculate a new base current and multiply by the gain to figure out if i will reach Ic Sat?
A good exposition, but there are a couple of snags. The value for re is both temperature dependant and non-linear, so the gain of the circuit (Rc/re) will depend on temperature and be non-linear. Also the bypass capacitor needs to have a reactance much smaller than re=6.5R for the frequencies we are interested in. That will be huge if we want to amplify down to 20Hz. With a quiescent current of 4mA and a collector resistor of 650R, the quiescent collector bias point is just (4mA x 650R) = 2.6V below the positive rail, i.e. at 21.4V. That leaves very little headroom for output swings, a mere 5.2Vpk-pk compared with the 24V available from the supply. Both of these issues can be addressed by splitting the emitter resistor and only by-passing part of it. That gives a larger effective emitter resistance and allows a larger collector resistance. The cost is a higher output impedance, but that is to be expected for a class-A amplification stage. If you're willing to lose the 10V of headroom, then you can continue to use the same Rb1, Rb2. That leaves about 14V for the output signal, so you can increase Rc to (7V / 4ma) = 1.75K. To maintain an ac gain of 100, the effective Re now needs to be 17.5R, of which 6.5R can be assumed to be the base-emitter resistance, as before. So you split the 2.325K Re into 11R and 2.3K with the latter being bypassed by the emitter capacitor as before. Note that the bypass capacitor makes a high-pass RC-filter with the 17.5R effective emitter resistance (not the 2K3 emitter resistor), so it needs to be around 900μF to get a -3db cut-off frequent at 10Hz. Personally, I'd not be so ambitious with the gain and work with a much lower base bias point. A gain of 20 using a collector current of 0.8mA could use a 12K collector resistor (setting the collector to 24.0 - 9.6V = 14.4V) and an unbypassed emitter resistor of 600R. That would swamp out the non-linear base-emitter resistance of about 30R, giving much more linearity because of the negative feedback. I'd suggest setting the emitter at around 4V meaning a total emitter resistor of (4V / 0.8ma) = 5K, made up of 600R unbypassed and 4.4K bypassed. The bypass capacitor would be 27μF for a cut-off of 10Hz. Working at the lower current would allow larger Rb1 and Rb2, so planning for HFE_min of 100, we would have Rb2
Hi Rex. Thanks so much for your lengthy and detailed comment. These are good things to address once a new learner has a general idea of how the heck a transistor works at all.
@@ElectronXLab I quite agree, and I'm sorry I only just found your video. I read a lot of the comments and it seems you have a very wide range of abilities among your viewers. Hopefully I won't have confused the beginners too much, and perhaps I'll have given your more able viewers something more to think about. Cheers.
This 0.026 Volts comes by manipulating the diode current equation which is exponential in form. You remove exponents to get a DE and afterwards solve the DE with initial values of t = 0 and I = 0 to reach value of 0.026. Please read Jeremy Everard's book Fundamentals of RF Circuit Design with Low Noise Oscillators published by JW&S. Hope this helps!
I think there is an error in the calculation for Zin. hie which is approximately 656Ω should be in parallel with 28K || 20K. This would make Zin closer to 621Ω not 4.17K as your calculation.
I know I'm late in the day but... calculating the impedance (around 8 min 20 sec into video) from 28K ‖ 20K ‖ 101x6.5 (which is 656) comes to 621 Ohms? I calculated this using the standard method ie the reciprocal of the sum of reciprocals for each of the 3 resistors. Where am I going wrong in order to get the 4.17 K you calculated? (I see others have asked the same question but no reply, so I remain confused). Having taught for many years (pharmacology, not electronics ... and that probably shows) I'd just say that it's worth joining the dots for individual steps. I see from the comments the 26mV caused confusion in this respect. Otherwise I commend you in a very practical approach. So many videos on circuit analysis and yours is one of the few that actually starts with the design.
I think I may have seen the error, it seems to be an arithmetic mistake. If you multiply 6.5 by 1001, and calculate the values all in parallel, you get around 4.17k
It is simply an arithmetic error. In fact, if the entire emitter resistor is bypassed, then the ac input impedance of the transistor is just 6.5R x HFE. Not only is that a minimum of 650R, but it depends on temperature (because re does) and the HFE of the transistor. It's far better to have some emitter resistance unbypassed to improve linearity and reduce the sensitivity of the circuit to the transistor's parameters.
Hold on, how did you get your input impedance of 4.17k? If Rb1, Rb2, and 101*re are in parallel to each other, your equivalent resistance should be 622 ohms. It seems you multiplied re by 1001 and posted that number.
Also, another question about the input impedance -- isn't the emitter capacitor a complex impedance, and so shouldn't the input impedance be 36.8+654.4j? If not, what would be the reason for considering it a resistance instead of a reactance? Also, is the choice of the correct value for that capacitor target frequency dependent?
Usually your input signal is at a frequency high enough so that the capacitors reactance is negligible since it’s proportional to 1/f so it’s so small u can just neglect it
Howdy again. I was wrong. You and JohnAudioTech are correct. Rc is the output impedance. Then the output signal is ½ the unloaded and most power is transferred. I remembered JAT's reply wrongly. Regards.
I'd like to know too -- I get 622R. Is there some other parallel resistor equation we should be using? I am curious, because I'm really struggling with a lot of resources teaching this stuff, and each one seems to leave out some detail. In this case, I am flumoxed by this step.
To find the impedence, you short dc supplies to ground. So when you see from the input side, you get Rb1 || Rb2. Since you are tracing the ac path, The signal goes through Ce and to ground and hence Re is bypassed. Also you have a ac resistance through base emitter ( diode in forward bias ) i.e rb . Now when you go from base to emmiter, the current changes by a factor of (1+ beta) .. so does the resistance. So finally all the resistances will be in parallel.
@@jephthai It must be a simple calculation error on his part. The resulting resistance across three (or any number of) resistors in parallel must be less than the lowest value. I get the same as you, 622 ohms.
Way cool got into 26mv/ma for Re (aka Re', bulk emitter resistance, etc.) as a part of calculating voltage gain. I learned it in the early 70's from Tektronix app notes. In my later years a routine question to ask in an interview I drew the circuit and asked the candidate to tell me what the voltage gain was. As soon as they uttered something about beta I knew they weren't there.
The BJT's Base-emitter diode is literally a diode, so we will handle it as a diode. The diode's current is a function of the voltage across the diode: Id = f(Vd) The function is: Id = Is * (exp(Vd / (k*T/q)) - 1), where `Is` is the reverse-voltage leakage current of the diode (typically really small, around uA-s). The curve is exponential, the Vd is the voltage across the diode, and the k*T/q is the Thermal voltage of the diode, where the k is the Boltzmann constant, the T is the temperature in Kelvins, (26.85°C is 300 K ), q is the charge of the electron. If you calculate, you will get 25.875 ~ 26 mV-s. But how does this correlates with the operating point of the transistor? At the operating point, DC current flows through the diode, and this causes some voltage to drop. If we take this voltage and divide it by the current, we got a 'dynamic' resistance, which can replace our diode in the operating point. If you do the math, and derivate the diode's equation by the voltage (d (Id) / d(Vd)), and you take the reciprocal of it, you will get r = Vth / Id, where Vth is the thermal voltage (26 mV) we discussed before, and Id is the diode's current in the operation point.
Howdy. Nice. One comment though. According to JohnAudioTech the Zout is 2 x Rc. I believe JAT is correct. One could test this checking the output level unloaded and then loaded. The load that causes a -3dB drop is the Zout. Most energy is transferred. JAT suggest the Irb2 is about 20 x the base current. Then Irb1 is almost the same as Irb2. And for practical purposes the Zin is Rb1 II Rb2. High Regards.
Howdy again. The statement that most power is transferred when Zload = Zout is true when the collector resistor is fixed. However. If we design the circuit with a Rc = say 1/10 x Zload the Pload will be higher. What is this ? Well. The Zload = Zout is the overall most energy efficient design of the whole circuit. With Rc = 1/10 x Zload the Zout = 1/5 x Zload. The load power will be higher but the total energy consumption of the whole circuit will be way higher. But yes. The increased energy consumption usually has little meaning. Unless, of course, one aims at low power designs. Usually it is recommended that Zout is 1/5 - 1/10 times Zload. The load will basically be voltage driven for best linearity. If the circuit is designed for Zout = Zload the load is more current driven. Frequency dependent variations in Zload will result in variations squrared in Pload. Regards again.
What is the little re on the diagram? Your talking I am looking at the diagram and suddenly your talking about little re. Where did the 26mv come from?
I know this was published a while ago, but I was looking for how you come up with design requirements and where in the transistor data sheet do you find the needed data to ensure the transistor selected meets the design requirements
Think there is an error with Zin you wrote 20k || 28k || 656.5 = 4.17K When I learnt about parallel resistors I was always taught that the effective resistance must be smaller than the smallest resister. How can you get 4.17K when you effectively have a 656 resistor in the mix?
There is a difference between ICmax= 8mA and ICq=4mA.. ICmax is the current that put the transisitor in the saturation region. so we use this current to calculate the RC when Vce=0.. Why Vce=0 , This is because in staturation Vce=0. And since we need to let the Q point in the middle of the load line so we divide (ICmax /2) to get ICq which is 4mA.. This is an assumption value you can put Icq= 5 mA ,1 mA or 2mA ... Then IC(sat) will be twice the value of ICq... It is all about your designe . Finally, Active region is also called a liner region because any effect on Ic will change Vce and IB..And If you look at the load line you will find that in order to center the Q point he put Vce =VCC/2 and also ICq will be ICmax/2 ... so there is a linearty here
The choice of collector current directly sets the output impedance and then indirectly sets the input impedance. The output impedance is equal to the collector resistor and the voltage across that is normally going to be around half the supply voltage, so the collector current is chosen to be about half the supply voltage divided by the desired output impedance. The input impedance is generally dominated by the ac emitter resistance multiplied by the transistor β and the ac emitter resistance is the collector resistance divided by the voltage gain, so the input impedance is going to be roughly the output impedance divided by the gain and multiplied by β.
The dc input impedance would indeed be (Beta+1) x (re + RE). That helps to keep the operating points stable. However, the ac input impedance assumes that RE has been bypassed by a large capacitor which has a small impedance compared with re. That means the ac input impedance is effectively (Beta+1) x (re), which is approximately 650R.
I'm not stupid, but I do resent how some people assume others watching this "KNOW" how you obtain your calculations. "re" is a prime example, RC being 650 ohm is yet another. Perhaps walking your audience through each step might make this video more informative and less intimidating. I once learned and majored in Electronic Instrumentation 30 years ago, but this is a waste of time.
@@selassiay1 re = (k * T / q)/Ic, where k = Boltzman constant, T = absolute temperature (room temperature in Kelvin), q = electron charge (both k and q are constants you can google). When you multiply the numerator you get 26mV. NB: As for where re comes from, one way is to derive it through differentiation of the Ebers-Moll equation with respect to collector current. This equation describes one of the many ways to model the behaviour of BJTs.
If Vce = 12V and VRE = 9.3V, then Vc (for DC) sits at 21.3V, there is literally only 2.7V positive headroom left for the AC to swing. I think you should've started in a way/had a design requirement so Vc would be VCC/2.
@@minhdang108 Thermal voltage = kT/Q where k = Boltzman constant, T = absolute temperature and Q is the charge of an electron (also a constant). We usually take 25 degrees celcius as temperature. Filling this all into the equation leads to 26mV. I'm not sure if he explained it anywhere, so here it is.
You actually want it to be half-way between Vcc and the lowest voltage on the collector that keeps it out of saturation (say approximately the voltage you've set on the base). Of course, it doesn't matter if you're only dealing with a few hundred millivolt signal swings, but in general, it's as easy to design for a decent amount of headroom as it is to end up with very little. I'd normally pick a collector current that gave the best noise figures as long as that allowed large enough base bias resistors to get a usable input impedance. Leaving aside a few volts (say 3 to 4) for the base/emitter bias points, that gives me the value of the collector resistor for maximum available swing. For a 24V supply, I'd want 4V at the base and 14V at the collector with quiescent current. The other values follow from that.
Can u point out how to determine the values of Coupling capacitors and the bypass capacitor to a preferred high pass frequency . And another request is can u do a tutorial for multi-stage bjt amplifier. Thank you.
For capacitor values, I recommend the following rule-of-thumb formula. (All results in uF.) Ce= 160K / (20 Hz x (RE / 10))= 160K / (20 x 233)= 34: Use 39 uF, standard value. Cin= 160K / (20 Hz x (Zin / 10))= 160K / (20 x 622)= 12.8: Use 15 uF, standard value. Cout= 160K / (20 Hz x (Zout / 10))= 160K / (20 x 65)=123 uF: Use 150 uF, standard value. You may adjust these values as necessary. Larger capacitors increase lower frequencies. Smaller capacitors decrease lower frequencies. (For more precise formulas, consult an engineering textbook.) Capacitors narrow the bandpass of this amplifier. The emitter capacitor causes the gain of this amplifier to vary according to frequency. High frequencies receive the most amplification. In this circuit, the other capacitors enhance this high-pass effect. The 160K figure above is roughly ((1/2π) x 10^6): Close enough to determine typical capacitor values. By "typical capacitor values," I mean those that one can actually buy.
@@w9xk The emitter bypass capacitor needs to have a reactance smaller than re, which is 6.5R across the frequencies we are interested in. The circuit shown will need a capacitor of Ce = 160K / (20Hz x 6.5R) = 1,200μF to have a 3db loss at 20Hz. With 39μF, the emitter impedance would be (6.5 + 204j) ohms at 20Hz giving a gain of just over 3 with an almost 90° phase shift. The ac input impedance of the circuit can be as little as 650R, not the 4.17K suggested.
It would be helpful if you refer to the spec sheet for transistors. how do you know what I(sat) is? spec sheet for a 2n2222 doesn't show I(sat). Otherwise it is a very good tutorial. makes me remember what I for got.
Hello, nice tutorial, but I have few questions. Where 26 mV in first calculation came from? And how parallel of 20k || 28k || 656.5 can be 4.17k ... I find BIG negative on this tutorial - direct write of numbers and results without formulas, because I suppose, most of people will design the amplifier with different values. This is big problem of this video in view of education, where the key and main focus should be the understanding, which bunch of numbers definitely does not meet. But take it please as a constructive critique from a teacher. I am mechanical eng. teacher who studied electrotechnical high school more than 20 yeas ago and wanted to blow the dust. Thank you.
Thanks for the comments. You're right about the parallel combination of resistors. I'm not sure how I came out with 4.17k - I think I used 6565 ohms instead of 656.5 ohms...but even then, the answer is not quite right. Correct answer is 622 Ohms
Very nice tutorial, much better at AC analysis/design than other common emitter design explanations that I've found. However, I'm a bit confused at the input impedance calculation of 4.17K. That calculation is 20K || 28K || 6.5 * 101 (or 656.5). That would be less than 656.5 and not nearly 4.17K. I did back out your result: I think that you used 6500 instead of 656.5 in the calculation. Am I right or did I misunderstand something?
Robert Dunn I was just scratching my head too over that Zin because it wasn't adding up for me , so he probably messed up somewhere in that calculation , aside from that great tutorial.
Thanks for the tut. What if we don't have Avoc? Can I just arbitrarily allocate voltage drops in the resistors, but keeping in mind that Vce is in the middle of the load line?
You can pick any value for (Avoc) I think it's decided by the value of your collector Resistor divided by the value of the internal resistance of the transistor which you get by using this formula : 25mV/IE. Your dividing The Collector Resistor by the Internal Resistance of the transistor because of the bypass capacitor which shorts out RE. Avoc = -100 Avoc = -RC/re Avoc = -RC/650 Ohms RC = Avoc x re RC = 100 x 650
how do you power 2 for stereo using one 9volt battery? i tried it and it goes mono. needs to have a power supply to separate the power going to the collectors. emitter is common.
I don't understand the value given for Re. The equation using saturation current is 8 = (24 - 0.2)/(650 + Re) . When I solve for Re, I get -647.025 Thanks, Rich
hello, sorry to bother you, I hope you will notice my question. How does the common emitter amplify current if the emitter isn't connected to the output, but to the ground and the rest of the circuit?
Thank you for the excellent vid. Got a noob question. The input at the base of the transitor will include both AC(signal) voltage and DC(24v) ? So what is the actual voltage at the base?
The drawing at 1 Second has what is called a blocking capacitor; This only allows alternating voltage to pass; The DC does not go past that as he has it; This is not an unusual concept even though most think of a return path;
The final voltage at the base of the transistor will be the sum of both the AC (signal) and DC (bias) voltages. That is to say, if the DC bias voltage required is 10V (as in the video) and the input AC signal is 10mV peak-to-peak, then the highest voltage possible at the base would be 10.01V and the lowest would be 9.99V. This changing base voltage causes a small change in base current which in turn causes a huge change in collector current (how huge depends on hFE or ß) which is how the BJT amplifier basically amplifies.
Where would i go about starting my design if my inital conditions are Vin= 0.1 V and Vout=5 V (Peak to peak Values). I will be using a 9 V battery as VCC and a 170 beta transistor. Help would be much appreciated
You would want to specify the desired input and output impedances, and be realistic - you can't have Rin/Rout bigger than the beta in any usable design. Also a gain of 50 is tough with a single stage and an output swing more than half the supply voltage. I would suggest aiming to set the voltage on the base around 3V and on the collector 6V. If we wanted high input impedance and low noise, I'd suggest using something like a BC184 with a collector current around 0.2mA giving a beta of 100 minimum at that collector current and decent noise figures. That gives a collector resistor of (9 - 6)V / 0.2mA = 15K. The emitter resistor would then be (3V - 0.6V) / 0.2mA = 12K, but that would give a gain of 15/12, so we need to bypass most of the emitter resistor with a capacitor to get an ac gain of 50. So the ac emitter resistance has to be 15K/50 = 300R. But the intrinsic emitter resistance is 26mV/0.2mA = 130R, so we have to use a resistance (Re1) of 170R to make a total of 300R. Therefore we have a remaining resistance (Re2) about (12k - 300R) = 11.7K of the dc emitter resistance. That has to be bypassed to ground by a capacitor whose reactance is less than 300R at the frequencies used. C = 1/(2πfR) which gives 27μF to have a cut-off below 20Hz. The entire emitter impedance would be a 180R resistor (Re1) in series with the combination of a 12K resistor (Re2) bypassed by 27μF (Rc). The transistor's ac input impedance will be (beta x Re) = 100 x 300R = 30K, so there's no point in using base resistors much bigger than that. Using Rb1 = 100K and Rb2 = 51K would give 3V at the base, as designed. The current through Rb2 would be about 60μA and that makes the base current insignificant (Ic / beta = 0.2mA/100 = 2μA). That should give up to 6V pk-pk available at the output (swinging from 3V to 9V). That can be trimmed by changing Rb1 (100K) slightly to make Vc = 6V. The gain should be around 50 for audio frequencies. That can be trimmed by altering the 180R emitter resistor slightly. The input impedance is around 15K at ac. Use an input capacitor of at least 1μF. The output impedance is 15K, so it needs a coupling capacitor of at least 1μF as well. I'm uncomfortable with the size of re (130R) compared with the size of Re1 (180R) as it would lead to non-linearities and some distortion, but it's hard to get a better ratio of Re1:re without increasing the supply voltage or reducing the gain. You could lower the base voltage to around 2V by increasing Rb1, giving more voltage swing and allowing higher Rc and Re1, with better linearity, but the emitter voltage is then just 1.4V and the collector bias point would be more variable with choice of transistor. Of course, if you're just making a single circuit, you can change the resistors slightly to get the voltages you want, but that doesn't scale very well if you're making several of them.
Hello. The amplifier only amplifies the one part of the signal. I have used Electronics Workbench and my simulator gets a half of wave in osciloscope. Do you have more videos?? Please i want to mount my first circuit and this is my chance. Thanks. Can you give hints about how it works? Please, theorical material written. Thanks for your video.
Your Beta was the same as your A voc. I think your A voc is your desired gain. Is Beta what the actual chip gain is or H fe? Also in this example, you said you wanted a gain of 100 and yet in the end you only got a gain of around 60. Shouldnt you have got the 100 as calculated? What did I miss here?
Is there a way to calculate this when you know what your load impedance is? (or how to alter the design to account for the input/output impedance that changed the desired gain in this example)
You really need a multi-stage amp for that kind of control. One stage for controlling input impedance, one for gain and one for controlling output impedance.
Greetings, Great video!! I would like to ask 2 questions: 1. In 2:50 you calculate re = 26mV/4mA. How is the 26mV produced? 2. Why don't we use the DC/AC loadlines to calculate the Rc/Re values?
1) The base emitter junction is a PN or diode junction. The AC resistance of a diode can be determined from taking the derivative of Shockley's diode equation with respect to voltage: d(Id)/dVd where Id is the current through and Vd is the voltage across the diode. The inverse of this derivative is the AC resistance of the diode and that emitter resistance is a diode resistance: re = 26mV/IE 2) In a hand-wavy way I did use the loadlines by trying to bias as near to the center of the loadline as I can.
@@ElectronXLab I validated your design and found that Vce = 12.1V. Which means that the Q-point is on the middle of the load line between Icmac=8ma and VCE=Vcc=24V But some desings lectures say that Av= -Rc/RE .. How it is come? And Some designs say That RC should be great than RE. And Also some designs divide RE to two REs in order not to depend on re on their calculations I'm little bit confused from the different ways of designing an amplifiers
@@hero96559 I am afraid the choice of Vce = 12V in the video might not be a good Q point. Vc=12V (to the ground, NOT Vce) is a better choice. If we selected Vce@12V, it means that we can only have around 2.5V for your upper margin of your ac output since Vc(to the ground) is already up to 21.5V. Besides, a general selection of Re shouldn't be too high, as most of the hfe gain of the transistor is above 100, it make the Vb voltage way too high as well. Anyway, the calculation here seems no problem, but the design parameters provided is not a practical one!
Can someone explain the derivation of 26mv in the calculation of re? I did look through the comments but still do not understand where that value came from.
Looks like this value is to compensate for Intrinsic emitter resistance. It attempts to account for properties that can cause distortion (such as operating temperature).
Yes but you want the base current (which also flows through Rb1) to be negligible compared to the current flowing through Rb2. That ensures that the base current has little effect on the base bias voltage. Good design will aim to minimise the effect of transistor parameters on the circuit performance and reproducibility.
I have a few questions. Is the choice of -100 completely arbitrary...could i get a different gain if say i choice -50 instead? another thing is where does the 8mA come from in equation number 3 and why is equation 4 divided by 10.
. Design an amplifier circuit with the following specifications: • The overall voltage gain should not be less than 350. • There must be phase difference of 180° among input and output voltage. • The input impedance should be in the range of 50 𝑘Ω and 60 𝑘Ω. • The output impedance should be as small as possible, preferably lower than 5 𝑘Ω. • You cannot use Operational Amplifier. You can only use BJT or FET or both for your design.
@@AmirKhan-fp7xo Why would anybody try to solve it? A closed-loop gain of -350 is ludicrously large for a single stage and you haven't specified the frequency response or supply voltage, nor the required output swing. Using a common dual opamp like the NE5552 would allow a two-stage solution with a flat response across the audio range and a minimum of fuss in the circuit layout.
Sir is it possible for you to make a video on how to design RF transmitter and receiver module circuit diagram with explaining it's working how the signal is transferred from transmitter to receiver. Topics to be studied for designing your own RF transmitter and receiver module circuit diagram eg. ( 433 MHz RF transmitter and receiver module)
@@zizogo3417 i think he use by computer system .but i don't know that computer system . .......... do you know me i am Ethiopian man may be you know ...we are the the beginning of the world ...but i don't know for where are you from..???????????????????????????????????????????????????????????????????????????????????????????????????????????????????
Wow... the ease with which you just rattle all this off makes me seriously question whether EE is right for me... I hardly followed your logic at all, and though I understood what you meant at several points along the way, overall I'm just mystified. Thanks for putting this out there.
good video! one question: why the circuit is set to Ic = 4mA in the first place? I believe setting to a lower current value like 2mA or even 1mA could produce the same voltage gain. And consume less power when running...Thanks
That keeps the output impedance low (but unfortunately lowers the input impedance as well). As you suggest, we'd normally design for currents significantly less than 4mA, allowing higher emitter resistances, which increase the input resistance, at the cost of a higher collector resistance for the same gain, and hence higher output resistance.
@@RexxSchneider So he's using an Ic=4ma to keep the output impedance low? He mentioned something in the video about using that value so that the transistor is biased correctly but as long as it is reverse biased correctly there are a range of collector currents that he could use. Still not sure why's he using that Ic unless he's requiring that for whatever load he connects to the amplifier.
@@ricomajestic If the gain of the circuit is a given requirement, then that determines the ratio between the collector resistance and the emitter resistance. You want the quiescent voltage on the collector to be about half-way between the supply voltage and the voltage at the emitter of the transistor as it clearly cannot swing beyond those limits. So the voltage across the collector resistor is fixed by that. That means if you design for a particular collector current, it determines the value of the collector resistor, and therefore the output impedance. For example, with a 9V supply and allowing about 1V at the emitter, you have about 8V of signal peak-to-peak before clipping. So you want half of that, i.e. 4V, across the collector resistor. With a collector current of 4mA, you have a 1K collector resistor, so the output impedance is 1K. If you dropped the collector current to 0.4mA, you would have a 10K collector resistor and so the output impedance would be 10K. Generally speaking, once you have a set of resistor values calculated, you can scale them all up or down by the same amount (within reason) and still have a circuit with approximately the same bias points and gain. But both the input impedance and output impedance will scale up or down by the same factor.
Why does everyone in the comments want all the simple calculations spoon fed. He literally writes down the equations. Simply solve them and you get the value he writes.
thank you but is your transistor is 2N3904 and does it word with DC input ? and the value of capacitors isn't exist in the video at least write it in description
The 10 is somewhat arbitrary; the purpose is to make sure that the resistance of RB2 is much less (i.e., 10x less) than the resistance seen looking in to the base
Can someone please help me understand part 4!? Where does the selection of RB2 must be less than (beta)(R_E)/10 come from?? In my textbooks, I see only RB2 should be (V_eq)/(9*i_B). Please and thank yoU!!
Strictly speaking RB2 does not have to be less than (beta)(R_E)/10. If you are using the shortcut of assuming I_B is so small as to be ignored, than RB2 needs to be at least 10 times bigger than the resistance looking in to the base.
You're trying to ensure that the current through Rb2 is at least 10 times the base current drawn by the transistor. If it isn't, then the HFE of the transistor starts to have an increasingly large effect on the base bias voltage, and therefore on the emitter-collector current, so different transistors of the same make will give different collector voltages, potentially causing clipping and a lack of reproducibility of the circuit.
That's the thermal voltage and is a property of the PN junction. It is temperature dependent, so the value of 26mV is only valid when the temperature is about 25C. VT = kT/q. Where k is Boltzmann's constant, T is the temperature in Kelvin and q is the charge on an electron.
A couple of design restrictions that I set are IC=4mA and to bias in the middle of the load line. For IC=4mA to be in the middle of the load line requires that the saturation current (IC(sat)) be two times that -> 8mA. IC(sat) = (VCC-VCE(sat))/(RC+RE). I've already determined RC, so I just have to solve that equation for RE. It's important to note that re and RE are different resistors. RE is the external biasing resistor connected at the emitter and re is the internal resistance in the emitter of the BJT.
Thanks for the reply, but when I do the math I get a completely different number. This is what I have 8=23.8/(650+RE) when I do this I get 0.2929, sorry maybe I'm dense, I just dont see how your getting 2325??
That should be 8mA or 0.008. Then, I think you made a mistake in your arithmetic somewhere. 0.008=23.8/(650+RE) -> 0.008(650+RE) = 23.8 -> 650+RE = 2975 -> RE = 2325 Ohms
Why did you select BETA as 100 in step 4 while calculating parameters? For the sake of simplicity? or did I miss sth?By the way Thanks a lot for these useful videos !!!
Beta is a characteristic of a BJT. The value can vary significantly from transistor to transistor even ones with the same partner. For this example, I just picked the value of beta for the sake of simplicity (just like you guessed).
Thank you for sharing your knowledge on RUclips. I have a question. In determining the value of small re, why wouldn't you use Ic(sat) current instead of the quiescent? Also, I plugged all of the formulas into a spread sheet and entered various Vcc's and Icq's and beta's. Everything worked for the sample values you gave but when I changed values I began to get negative RE values for the emitter resistor....for example, use 9 Vcc and 200 beta.
You use the quiescent current because that is the current that is actually going through the transistor and re is based on the state of the transistor. You were getting negative values probably because the values you were putting in are values in which the the transistor can't actually operate.
@@ElectronXLab Ok, that makes sense. I notice some designers always make RE smaller than RC. For example they start out with Vce being half of VCC, then VRC being 3/4 of what's left and VRE being the other 1/4. They choose a Icq and go from there. I was wondering if one method is better than the other or is there more than one way to skin a cat......
For people way behind or not up to date on the math and references and variables should explain also explain the type of material and the reason of why and what happens when those components aren't correct or not processing or working properly would be helpful i can solder and put together parts need to know why and the purpose plus the results of incorrect input
That would only be true if VCE(sat) is zero (it's not) and if there was no RE resistor. Ic(sat) = (voltage across RC) / RC. Voltage across RC = (VCC - voltage at collector) and voltage at collector, Vc = IE(RE) + VCE(sat), so Ic(sat) = (VCC - IE(RE) - VCE(sat))/RC and if you approximate IE = IC, and rearrange that equation, you get Ic(sat) = (VCC - VCE)/(RC+RE)
@@ElectronXLab And since Vce_sat is around 200mV for most small signal transistors, you can neglect it compared to Vcc. That gives you the useful equation: Ic_sat = Vcc / (Rc + Re)
This is the thermal voltage at the junction that is potential caused by the thermal motion of electrons. It's temperature dependent, but about 26mV at room temperature.
To those watching confused about the 26mV, it is the thermal voltage of a bipolar junction transistor. It depends on temperature and is assumed to be 26mV at room temperature.
Finally a video where you're not simply asked to find operating point and are asked a more real world problem of designing something as oppose to just finding currents and voltages
To be fair, I do have a video on how to find the operating point too. But I agree, design problems are more real world.
I have a question, assuming this same circuit, to know the maximum amplitude i can have on input before i reach Collector saturation current and start clipping the output ... do i just add the amplitude of the input to my base voltage and calculate a new base current and multiply by the gain to figure out if i will reach Ic Sat?
A good exposition, but there are a couple of snags.
The value for re is both temperature dependant and non-linear, so the gain of the circuit (Rc/re) will depend on temperature and be non-linear. Also the bypass capacitor needs to have a reactance much smaller than re=6.5R for the frequencies we are interested in. That will be huge if we want to amplify down to 20Hz.
With a quiescent current of 4mA and a collector resistor of 650R, the quiescent collector bias point is just (4mA x 650R) = 2.6V below the positive rail, i.e. at 21.4V. That leaves very little headroom for output swings, a mere 5.2Vpk-pk compared with the 24V available from the supply.
Both of these issues can be addressed by splitting the emitter resistor and only by-passing part of it. That gives a larger effective emitter resistance and allows a larger collector resistance. The cost is a higher output impedance, but that is to be expected for a class-A amplification stage.
If you're willing to lose the 10V of headroom, then you can continue to use the same Rb1, Rb2. That leaves about 14V for the output signal, so you can increase Rc to (7V / 4ma) = 1.75K. To maintain an ac gain of 100, the effective Re now needs to be 17.5R, of which 6.5R can be assumed to be the base-emitter resistance, as before. So you split the 2.325K Re into 11R and 2.3K with the latter being bypassed by the emitter capacitor as before. Note that the bypass capacitor makes a high-pass RC-filter with the 17.5R effective emitter resistance (not the 2K3 emitter resistor), so it needs to be around 900μF to get a -3db cut-off frequent at 10Hz.
Personally, I'd not be so ambitious with the gain and work with a much lower base bias point. A gain of 20 using a collector current of 0.8mA could use a 12K collector resistor (setting the collector to 24.0 - 9.6V = 14.4V) and an unbypassed emitter resistor of 600R. That would swamp out the non-linear base-emitter resistance of about 30R, giving much more linearity because of the negative feedback. I'd suggest setting the emitter at around 4V meaning a total emitter resistor of (4V / 0.8ma) = 5K, made up of 600R unbypassed and 4.4K bypassed. The bypass capacitor would be 27μF for a cut-off of 10Hz. Working at the lower current would allow larger Rb1 and Rb2, so planning for HFE_min of 100, we would have Rb2
Hi Rex. Thanks so much for your lengthy and detailed comment. These are good things to address once a new learner has a general idea of how the heck a transistor works at all.
@@ElectronXLab I quite agree, and I'm sorry I only just found your video.
I read a lot of the comments and it seems you have a very wide range of abilities among your viewers. Hopefully I won't have confused the beginners too much, and perhaps I'll have given your more able viewers something more to think about. Cheers.
How did you get 26mV to find the value of R(e) at first?
Check out 'EBERS - MOLL' Physics of diodes.
I think it's a generally accepted value but not a named constant
This 0.026 Volts comes by manipulating the diode current equation which is exponential in form. You remove exponents to get a DE and afterwards solve the DE with initial values of t = 0 and I = 0 to reach value of 0.026. Please read Jeremy Everard's book Fundamentals of RF Circuit Design with Low Noise Oscillators published by JW&S. Hope this helps!
At 2:52 where did you get 26mV?
THE 26mV COMES FROM APPLYING DIFERENTIAL CALCULUS AND,ALGEBRA TO THE SHOCKLEY EQUATION FOR THE DIODE.
I think there is an error in the calculation for Zin. hie which is approximately 656Ω should be in parallel with 28K || 20K. This would make Zin closer to 621Ω not 4.17K as your calculation.
Awesome explanation...
This is what I have been searching for quite long.
Can you please explain how to rule out the value 26mV?
It's the thermal voltage of bjt
I know I'm late in the day but... calculating the impedance (around 8 min 20 sec into video) from 28K ‖ 20K ‖ 101x6.5 (which is 656) comes to 621 Ohms? I calculated this using the standard method ie the reciprocal of the sum of reciprocals for each of the 3 resistors. Where am I going wrong in order to get the 4.17 K you calculated? (I see others have asked the same question but no reply, so I remain confused).
Having taught for many years (pharmacology, not electronics ... and that probably shows) I'd just say that it's worth joining the dots for individual steps. I see from the comments the 26mV caused confusion in this respect.
Otherwise I commend you in a very practical approach. So many videos on circuit analysis and yours is one of the few that actually starts with the design.
am waiting for Mr David responce for that too with respect
I think I may have seen the error, it seems to be an arithmetic mistake. If you multiply 6.5 by 1001, and calculate the values all in parallel, you get around 4.17k
It is simply an arithmetic error. In fact, if the entire emitter resistor is bypassed, then the ac input impedance of the transistor is just 6.5R x HFE. Not only is that a minimum of 650R, but it depends on temperature (because re does) and the HFE of the transistor. It's far better to have some emitter resistance unbypassed to improve linearity and reduce the sensitivity of the circuit to the transistor's parameters.
Hold on, how did you get your input impedance of 4.17k? If Rb1, Rb2, and 101*re are in parallel to each other, your equivalent resistance should be 622 ohms. It seems you multiplied re by 1001 and posted that number.
I got the same thing as you.
W2aew has some good biasing videos.
Also, another question about the input impedance -- isn't the emitter capacitor a complex impedance, and so shouldn't the input impedance be 36.8+654.4j? If not, what would be the reason for considering it a resistance instead of a reactance? Also, is the choice of the correct value for that capacitor target frequency dependent?
Usually your input signal is at a frequency high enough so that the capacitors reactance is negligible since it’s proportional to 1/f so it’s so small u can just neglect it
Howdy again.
I was wrong. You and JohnAudioTech are correct. Rc is the output impedance. Then the output signal is ½ the unloaded and most power is transferred. I remembered JAT's reply wrongly.
Regards.
Wait a minute, how on earth did you get input impedance of 4.17K?
I'd like to know too -- I get 622R. Is there some other parallel resistor equation we should be using? I am curious, because I'm really struggling with a lot of resources teaching this stuff, and each one seems to leave out some detail. In this case, I am flumoxed by this step.
To find the impedence, you short dc supplies to ground. So when you see from the input side, you get Rb1 || Rb2. Since you are tracing the ac path, The signal goes through Ce and to ground and hence Re is bypassed. Also you have a ac resistance through base emitter ( diode in forward bias ) i.e rb . Now when you go from base to emmiter, the current changes by a factor of (1+ beta) .. so does the resistance. So finally all the resistances will be in parallel.
@@jephthai It must be a simple calculation error on his part. The resulting resistance across three (or any number of) resistors in parallel must be less than the lowest value. I get the same as you, 622 ohms.
@@ahmadshamim3289 totally agree
he multiplied 6.5 by 1001 instead of 101
Way cool got into 26mv/ma for Re (aka Re', bulk emitter resistance, etc.) as a part of calculating voltage gain. I learned it in the early 70's from Tektronix app notes. In my later years a routine question to ask in an interview I drew the circuit and asked the candidate to tell me what the voltage gain was. As soon as they uttered something about beta I knew they weren't there.
How did you get 26 M V over 4 M A?
26mV is "thermal voltage" of a BJT.
r"e=.026/Ie
I hate when people forgets details like that, specially at teaching.
@@omarcusihuaman4261 Happens because some people are actually self promoting not teaching, er, like this guy.
The BJT's Base-emitter diode is literally a diode, so we will handle it as a diode. The diode's current is a function of the voltage across the diode: Id = f(Vd)
The function is: Id = Is * (exp(Vd / (k*T/q)) - 1), where `Is` is the reverse-voltage leakage current of the diode (typically really small, around uA-s). The curve is exponential, the Vd is the voltage across the diode, and the k*T/q is the Thermal voltage of the diode, where the k is the Boltzmann constant, the T is the temperature in Kelvins, (26.85°C is 300 K ), q is the charge of the electron. If you calculate, you will get 25.875 ~ 26 mV-s.
But how does this correlates with the operating point of the transistor? At the operating point, DC current flows through the diode, and this causes some voltage to drop. If we take this voltage and divide it by the current, we got a 'dynamic' resistance, which can replace our diode in the operating point.
If you do the math, and derivate the diode's equation by the voltage (d (Id) / d(Vd)), and you take the reciprocal of it, you will get r = Vth / Id, where Vth is the thermal voltage (26 mV) we discussed before, and Id is the diode's current in the operation point.
Howdy. Nice.
One comment though. According to JohnAudioTech the Zout is 2 x Rc. I believe JAT is correct. One could test this checking the output level unloaded and then loaded. The load that causes a -3dB drop is the Zout. Most energy is transferred.
JAT suggest the Irb2 is about 20 x the base current. Then Irb1 is almost the same as Irb2. And for practical purposes the Zin is Rb1 II Rb2.
High Regards.
Howdy again.
The statement that most power is transferred when Zload = Zout is true when the collector resistor is fixed.
However. If we design the circuit with a Rc = say 1/10 x Zload the Pload will be higher. What is this ?
Well. The Zload = Zout is the overall most energy efficient design of the whole circuit. With Rc = 1/10 x Zload the Zout = 1/5 x Zload. The load power will be higher but the total energy consumption of the whole circuit will be way higher.
But yes. The increased energy consumption usually has little meaning. Unless, of course, one aims at low power designs.
Usually it is recommended that Zout is 1/5 - 1/10 times Zload. The load will basically be voltage driven for best linearity. If the circuit is designed for Zout = Zload the load is more current driven. Frequency dependent variations in Zload will result in variations squrared in Pload.
Regards again.
What is the little re on the diagram? Your talking I am looking at the diagram and suddenly your talking about little re. Where did the 26mv come from?
The input impedance should be 621 ohm and not 4.17k. That change would cause the final output voltage gain with the load to reduce.
I know this was published a while ago, but I was looking for how you come up with design requirements and where in the transistor data sheet do you find the needed data to ensure the transistor selected meets the design requirements
Carl Davis Yeah, data sheets are required to find the standard values.
Think there is an error with Zin you wrote 20k || 28k || 656.5 = 4.17K When I learnt about parallel resistors I was always taught that the effective resistance must be smaller than the smallest resister. How can you get 4.17K when you effectively have a 656 resistor in the mix?
This is certainly a calculation mistake. Zin should be 621 ohm
hi there, why does your collector current (Ic) need to be 4ma, and thus your Ic(sat) be 8ma?
I was wondering the same thing!
There is a difference between ICmax= 8mA and ICq=4mA.. ICmax is the current that put the transisitor in the saturation region. so we use this current to calculate the RC when Vce=0.. Why Vce=0 , This is because in staturation Vce=0.
And since we need to let the Q point in the middle of the load line so we divide (ICmax /2) to get ICq which is 4mA..
This is an assumption value you can put Icq= 5 mA ,1 mA or 2mA ... Then IC(sat) will be twice the value of ICq... It is all about your designe .
Finally, Active region is also called a liner region because any effect on Ic will change Vce and IB..And If you look at the load line you will find that in order to center the Q point he put Vce =VCC/2 and also ICq will be ICmax/2 ... so there is a linearty here
The choice of collector current directly sets the output impedance and then indirectly sets the input impedance. The output impedance is equal to the collector resistor and the voltage across that is normally going to be around half the supply voltage, so the collector current is chosen to be about half the supply voltage divided by the desired output impedance.
The input impedance is generally dominated by the ac emitter resistance multiplied by the transistor β and the ac emitter resistance is the collector resistance divided by the voltage gain, so the input impedance is going to be roughly the output impedance divided by the gain and multiplied by β.
The input impedance should be (Beta+1) x (re + RE)
But isn't Re bypassed by the capacitor Ce in the AC model?
@@SopanKotbagi 3:30
The dc input impedance would indeed be (Beta+1) x (re + RE). That helps to keep the operating points stable. However, the ac input impedance assumes that RE has been bypassed by a large capacitor which has a small impedance compared with re. That means the ac input impedance is effectively (Beta+1) x (re), which is approximately 650R.
I'm not stupid, but I do resent how some people assume others watching this "KNOW" how you obtain your calculations. "re" is a prime example, RC being 650 ohm is yet another. Perhaps walking your audience through each step might make this video more informative and less intimidating. I once learned and majored in Electronic Instrumentation 30 years ago, but this is a waste of time.
That is exactly what I felt. I am glad that I am not the only one who is confused.
What about the 26mv, was it from the data sheet of the transistor?
@@selassiay1 26mv is common bjt threshold voltage
@@selassiay1 re = (k * T / q)/Ic, where k = Boltzman constant, T = absolute temperature (room temperature in Kelvin), q = electron charge (both k and q are constants you can google). When you multiply the numerator you get 26mV. NB: As for where re comes from, one way is to derive it through differentiation of the Ebers-Moll equation with respect to collector current. This equation describes one of the many ways to model the behaviour of BJTs.
nununoodle
BJTs don’t have a threshold voltage. 26mV is the thermal voltage.
If Vce = 12V and VRE = 9.3V, then Vc (for DC) sits at 21.3V, there is literally only 2.7V positive headroom left for the AC to swing. I think you should've started in a way/had a design requirement so Vc would be VCC/2.
I don't understand 26mV , from Where ?
@@minhdang108 Thermal voltage = kT/Q where k = Boltzman constant, T = absolute temperature and Q is the charge of an electron (also a constant). We usually take 25 degrees celcius as temperature.
Filling this all into the equation leads to 26mV.
I'm not sure if he explained it anywhere, so here it is.
@@trevortjes Thank you.
You actually want it to be half-way between Vcc and the lowest voltage on the collector that keeps it out of saturation (say approximately the voltage you've set on the base). Of course, it doesn't matter if you're only dealing with a few hundred millivolt signal swings, but in general, it's as easy to design for a decent amount of headroom as it is to end up with very little. I'd normally pick a collector current that gave the best noise figures as long as that allowed large enough base bias resistors to get a usable input impedance. Leaving aside a few volts (say 3 to 4) for the base/emitter bias points, that gives me the value of the collector resistor for maximum available swing. For a 24V supply, I'd want 4V at the base and 14V at the collector with quiescent current. The other values follow from that.
Can u point out how to determine the values of Coupling capacitors and the bypass capacitor to a preferred high pass frequency . And another request is can u do a tutorial for multi-stage bjt amplifier. Thank you.
For capacitor values, I recommend the following rule-of-thumb formula. (All results in uF.) Ce= 160K / (20 Hz x (RE / 10))= 160K / (20 x 233)= 34: Use 39 uF, standard value. Cin= 160K / (20 Hz x (Zin / 10))= 160K / (20 x 622)= 12.8: Use 15 uF, standard value. Cout= 160K / (20 Hz x (Zout / 10))= 160K / (20 x 65)=123 uF: Use 150 uF, standard value. You may adjust these values as necessary. Larger capacitors increase lower frequencies. Smaller capacitors decrease lower frequencies. (For more precise formulas, consult an engineering textbook.) Capacitors narrow the bandpass of this amplifier. The emitter capacitor causes the gain of this amplifier to vary according to frequency. High frequencies receive the most amplification. In this circuit, the other capacitors enhance this high-pass effect. The 160K figure above is roughly ((1/2π) x 10^6): Close enough to determine typical capacitor values. By "typical capacitor values," I mean those that one can actually buy.
@@w9xk The emitter bypass capacitor needs to have a reactance smaller than re, which is 6.5R across the frequencies we are interested in. The circuit shown will need a capacitor of Ce = 160K / (20Hz x 6.5R) = 1,200μF to have a 3db loss at 20Hz. With 39μF, the emitter impedance would be (6.5 + 204j) ohms at 20Hz giving a gain of just over 3 with an almost 90° phase shift.
The ac input impedance of the circuit can be as little as 650R, not the 4.17K suggested.
what voltage is the 26mV?
It would be helpful if you refer to the spec sheet for transistors. how do you know what I(sat) is?
spec sheet for a 2n2222 doesn't show I(sat). Otherwise it is a very good tutorial. makes me remember what I for got.
Hello, nice tutorial, but I have few questions. Where 26 mV in first calculation came from? And how parallel of 20k || 28k || 656.5 can be 4.17k ... I find BIG negative on this tutorial - direct write of numbers and results without formulas, because I suppose, most of people will design the amplifier with different values. This is big problem of this video in view of education, where the key and main focus should be the understanding, which bunch of numbers definitely does not meet. But take it please as a constructive critique from a teacher. I am mechanical eng. teacher who studied electrotechnical high school more than 20 yeas ago and wanted to blow the dust. Thank you.
yeah i have the same question.... 20||28||656.5 = 4.17....
Thanks for the comments. You're right about the parallel combination of resistors. I'm not sure how I came out with 4.17k - I think I used 6565 ohms instead of 656.5 ohms...but even then, the answer is not quite right. Correct answer is 622 Ohms
@@ElectronXLab I have just noticed that because it is obvious, that parallel combination has to be always less then smallest resistance.
Very nice tutorial, much better at AC analysis/design than other common emitter design explanations that I've found. However, I'm a bit confused at the input impedance calculation of 4.17K. That calculation is 20K || 28K || 6.5 * 101 (or 656.5). That would be less than 656.5 and not nearly 4.17K. I did back out your result: I think that you used 6500 instead of 656.5 in the calculation. Am I right or did I misunderstand something?
Robert Dunn I was just scratching my head too over that Zin because it wasn't adding up for me , so he probably messed up somewhere in that calculation , aside from that great tutorial.
No you are right, since RB1 and RB2 are much larger than 650 you could just call the input impedance 650 ohms
yeah i was completely stuck on that trying to figure it out, he did 6.5*1001 instead of 6.5*101
The outro with the hexbug, very nostalgic
LOL. Thanks for watching all the way to the end!!
I’m sorry please excuse my question but what exactly does a emitter do ?
Thanks for the tut. What if we don't have Avoc? Can I just arbitrarily allocate voltage drops in the resistors, but keeping in mind that Vce is in the middle of the load line?
You can pick any value for (Avoc) I think it's decided by the value of your collector Resistor divided by the value of the internal resistance of the transistor which you get by using this formula : 25mV/IE. Your dividing The Collector Resistor by the Internal Resistance of the transistor because of the bypass capacitor which shorts out RE.
Avoc = -100
Avoc = -RC/re
Avoc = -RC/650 Ohms
RC = Avoc x re
RC = 100 x 650
Really cool explanation. The best !!!
U can say is the thermal voltage, so, it does not matter volt in base?
how would this change if you knew the load resistance?
why you used 26mv in your calculation of re?
Sorry, probably I miss something but in order to calculate Zin you wrote 20k || 28k || 656.5 = 4.17K ??!?!?!?!? seems to give 621.5 ohms, Am I wrong ?
I agree. The result cannot be larger than the smallest value.
Mostly helpful thank you, I'm taking away that the magic 26mV is an accepted value for the re resistance, the AC resistance of a diode junction.
when calculating re you used 26 mV. Where did you get this value from?
26mV where did you get that ?
Just awesome .Thanks sir. Alot informative.
From Pakistan .
how do you power 2 for stereo using one 9volt battery? i tried it and it goes mono. needs to have a power supply to separate the power going to the collectors. emitter is common.
I don't understand the value given for Re. The equation using saturation current is 8 = (24 - 0.2)/(650 + Re) . When I solve for Re, I get -647.025
Thanks,
Rich
what should be the conditions if we want to use a speaker
hello, sorry to bother you, I hope you will notice my question. How does the common emitter amplify current if the emitter isn't connected to the output, but to the ground and the rest of the circuit?
How did you calculate RE you just skipped the steps?
are you dumb? just rearrange the equation, so damn simple. If you cant do that yourself, good luck 🤣
Thank you for the excellent vid. Got a noob question. The input at the base of the transitor will include both AC(signal) voltage and DC(24v) ? So what is the actual voltage at the base?
At the base there would be ur original signal with an offset voltage
The drawing at 1 Second has what is called a blocking capacitor; This only allows alternating voltage to pass; The DC does not go past that as he has it; This is not an unusual concept even though most think of a return path;
The final voltage at the base of the transistor will be the sum of both the AC (signal) and DC (bias) voltages. That is to say, if the DC bias voltage required is 10V (as in the video) and the input AC signal is 10mV peak-to-peak, then the highest voltage possible at the base would be 10.01V and the lowest would be 9.99V. This changing base voltage causes a small change in base current which in turn causes a huge change in collector current (how huge depends on hFE or ß) which is how the BJT amplifier basically amplifies.
good video---but I am not sure where you got the 26ma in step one????
Hello Brother. The videos were very useful. The presentation is also good. However, I am delighted to solve some more examples of BJT's AC analysis.
Where would i go about starting my design if my inital conditions are Vin= 0.1 V and Vout=5 V (Peak to peak Values). I will be using a 9 V battery as VCC and a 170 beta transistor. Help would be much appreciated
You would want to specify the desired input and output impedances, and be realistic - you can't have Rin/Rout bigger than the beta in any usable design. Also a gain of 50 is tough with a single stage and an output swing more than half the supply voltage.
I would suggest aiming to set the voltage on the base around 3V and on the collector 6V. If we wanted high input impedance and low noise, I'd suggest using something like a BC184 with a collector current around 0.2mA giving a beta of 100 minimum at that collector current and decent noise figures.
That gives a collector resistor of (9 - 6)V / 0.2mA = 15K.
The emitter resistor would then be (3V - 0.6V) / 0.2mA = 12K, but that would give a gain of 15/12, so we need to bypass most of the emitter resistor with a capacitor to get an ac gain of 50.
So the ac emitter resistance has to be 15K/50 = 300R. But the intrinsic emitter resistance is 26mV/0.2mA = 130R, so we have to use a resistance (Re1) of 170R to make a total of 300R.
Therefore we have a remaining resistance (Re2) about (12k - 300R) = 11.7K of the dc emitter resistance. That has to be bypassed to ground by a capacitor whose reactance is less than 300R at the frequencies used. C = 1/(2πfR) which gives 27μF to have a cut-off below 20Hz. The entire emitter impedance would be a 180R resistor (Re1) in series with the combination of a 12K resistor (Re2) bypassed by 27μF (Rc).
The transistor's ac input impedance will be (beta x Re) = 100 x 300R = 30K, so there's no point in using base resistors much bigger than that. Using Rb1 = 100K and Rb2 = 51K would give 3V at the base, as designed. The current through Rb2 would be about 60μA and that makes the base current insignificant (Ic / beta = 0.2mA/100 = 2μA).
That should give up to 6V pk-pk available at the output (swinging from 3V to 9V). That can be trimmed by changing Rb1 (100K) slightly to make Vc = 6V. The gain should be around 50 for audio frequencies. That can be trimmed by altering the 180R emitter resistor slightly. The input impedance is around 15K at ac. Use an input capacitor of at least 1μF. The output impedance is 15K, so it needs a coupling capacitor of at least 1μF as well.
I'm uncomfortable with the size of re (130R) compared with the size of Re1 (180R) as it would lead to non-linearities and some distortion, but it's hard to get a better ratio of Re1:re without increasing the supply voltage or reducing the gain. You could lower the base voltage to around 2V by increasing Rb1, giving more voltage swing and allowing higher Rc and Re1, with better linearity, but the emitter voltage is then just 1.4V and the collector bias point would be more variable with choice of transistor. Of course, if you're just making a single circuit, you can change the resistors slightly to get the voltages you want, but that doesn't scale very well if you're making several of them.
Your input impedance calculation is a mystery to me....
Hello. The amplifier only amplifies the one part of the signal. I have used Electronics Workbench and my simulator gets a half of wave in osciloscope. Do you have more videos?? Please i want to mount my first circuit and this is my chance. Thanks. Can you give hints about how it works? Please, theorical material written. Thanks for your video.
Your assuming there is no base current because there is no signal input am I right?
As in you bias the circuit under operating conditions?
*Non operating conditions
Why do we have two vOmin values in calss A amplifiers?
Your Beta was the same as your A voc. I think your A voc is your desired gain. Is Beta what the actual chip gain is or H fe? Also in this example, you said you wanted a gain of 100 and yet in the end you only got a gain of around 60. Shouldnt you have got the 100 as calculated? What did I miss here?
Is there a way to calculate this when you know what your load impedance is? (or how to alter the design to account for the input/output impedance that changed the desired gain in this example)
You really need a multi-stage amp for that kind of control. One stage for controlling input impedance, one for gain and one for controlling output impedance.
@@ElectronXLab That makes sense, thank you!
Greetings,
Great video!!
I would like to ask 2 questions:
1. In 2:50 you calculate re = 26mV/4mA. How is the 26mV produced?
2. Why don't we use the DC/AC loadlines to calculate the Rc/Re values?
1) The base emitter junction is a PN or diode junction. The AC resistance of a diode can be determined from taking the derivative of Shockley's diode equation with respect to voltage: d(Id)/dVd where Id is the current through and Vd is the voltage across the diode. The inverse of this derivative is the AC resistance of the diode and that emitter resistance is a diode resistance: re = 26mV/IE
2) In a hand-wavy way I did use the loadlines by trying to bias as near to the center of the loadline as I can.
Thanks a lot for your response!!! Highly appreciated! :)
@@ElectronXLab I validated your design and found that Vce = 12.1V. Which means that the Q-point is on the middle of the load line between Icmac=8ma and VCE=Vcc=24V
But some desings lectures say that Av= -Rc/RE .. How it is come?
And
Some designs say That RC should be great than RE.
And
Also some designs divide RE to two REs in order not to depend on re on their calculations
I'm little bit confused from the different ways of designing an amplifiers
@@hero96559 I am afraid the choice of Vce = 12V in the video might not be a good Q point. Vc=12V (to the ground, NOT Vce) is a better choice. If we selected Vce@12V, it means that we can only have around 2.5V for your upper margin of your ac output since Vc(to the ground) is already up to 21.5V. Besides, a general selection of Re shouldn't be too high, as most of the hfe gain of the transistor is above 100, it make the Vb voltage way too high as well. Anyway, the calculation here seems no problem, but the design parameters provided is not a practical one!
Can someone explain the derivation of 26mv in the calculation of re?
I did look through the comments but still do not understand where that value came from.
Looks like this value is to compensate for Intrinsic emitter resistance. It attempts to account for properties that can cause distortion (such as operating temperature).
@@richardeberhardt1301English man
Wait, don’t you want current to flow into the base since that’s the direction the current flows anyways?
Yes but you want the base current (which also flows through Rb1) to be negligible compared to the current flowing through Rb2. That ensures that the base current has little effect on the base bias voltage. Good design will aim to minimise the effect of transistor parameters on the circuit performance and reproducibility.
I found the video useful but the 26mV did throw me.
I have a few questions. Is the choice of -100 completely arbitrary...could i get a different gain if say i choice -50 instead? another thing is where does the 8mA come from in equation number 3 and why is equation 4 divided by 10.
. Design an amplifier circuit with the following specifications:
• The overall voltage gain should not be less than 350.
• There must be phase difference of 180° among input and output voltage.
• The input impedance should be in the range of 50 𝑘Ω and 60 𝑘Ω.
• The output impedance should be as small as possible, preferably lower than 5 𝑘Ω.
• You cannot use Operational Amplifier. You can only use BJT or FET or both for your
design.
Can any one solve this plzz
@@AmirKhan-fp7xo Why would anybody try to solve it? A closed-loop gain of -350 is ludicrously large for a single stage and you haven't specified the frequency response or supply voltage, nor the required output swing. Using a common dual opamp like the NE5552 would allow a two-stage solution with a flat response across the audio range and a minimum of fuss in the circuit layout.
Which book did you refer?
What if ICq= 15mA..this will derive that re=1.67 ohm... And then RC= 167 ohm
I didn't see any RC values and class AMPLIFIER
Very helpful. Thanks
Sir is it possible for you to make a video on how to design RF transmitter and receiver module circuit diagram with explaining it's working how the signal is transferred from transmitter to receiver.
Topics to be studied for designing your own RF transmitter and receiver module circuit diagram eg. ( 433 MHz RF transmitter and receiver module)
Many details are not explained well.
hello,
thanks for the tutorial.
can you tell me, what program that you use to draw circuit schematic like the one in the video (CE-amp.)
I just use OneNote and a wacom tablet
I don't know how you manage to draw such a circuit with only OneNote, but they look awesome.
anyway thanks for replying.
@@zizogo3417 i think he use by computer system .but i don't know that computer system . ..........
do you know me i am Ethiopian man may be you know ...we are the the beginning of the world ...but i don't know for where are you from..???????????????????????????????????????????????????????????????????????????????????????????????????????????????????
@@ElectronXLab can RE higer value than RC ? please answer as soon as possible
How saturation current ic = 8ma
Wow... the ease with which you just rattle all this off makes me seriously question whether EE is right for me... I hardly followed your logic at all, and though I understood what you meant at several points along the way, overall I'm just mystified. Thanks for putting this out there.
good video! one question: why the circuit is set to Ic = 4mA in the first place? I believe setting to a lower current value like 2mA or even 1mA could produce the same voltage gain. And consume less power when running...Thanks
good question
That keeps the output impedance low (but unfortunately lowers the input impedance as well). As you suggest, we'd normally design for currents significantly less than 4mA, allowing higher emitter resistances, which increase the input resistance, at the cost of a higher collector resistance for the same gain, and hence higher output resistance.
@@RexxSchneider So he's using an Ic=4ma to keep the output impedance low? He mentioned something in the video about using that value so that the transistor is biased correctly but as long as it is reverse biased correctly there are a range of collector currents that he could use. Still not sure why's he using that Ic unless he's requiring that for whatever load he connects to the amplifier.
@@ricomajestic If the gain of the circuit is a given requirement, then that determines the ratio between the collector resistance and the emitter resistance. You want the quiescent voltage on the collector to be about half-way between the supply voltage and the voltage at the emitter of the transistor as it clearly cannot swing beyond those limits. So the voltage across the collector resistor is fixed by that. That means if you design for a particular collector current, it determines the value of the collector resistor, and therefore the output impedance.
For example, with a 9V supply and allowing about 1V at the emitter, you have about 8V of signal peak-to-peak before clipping. So you want half of that, i.e. 4V, across the collector resistor. With a collector current of 4mA, you have a 1K collector resistor, so the output impedance is 1K. If you dropped the collector current to 0.4mA, you would have a 10K collector resistor and so the output impedance would be 10K.
Generally speaking, once you have a set of resistor values calculated, you can scale them all up or down by the same amount (within reason) and still have a circuit with approximately the same bias points and gain. But both the input impedance and output impedance will scale up or down by the same factor.
What is "little r-e'? There is no component labeled "little r-e in the schematic.
It's a part of the model of the transistor, so you won't see it in the schematic
Thanks so much@@ElectronXLab
Why does everyone in the comments want all the simple calculations spoon fed. He literally writes down the equations. Simply solve them and you get the value he writes.
Does this work for low input impedance low output impedance? I need an input of 150ohms and output of 620 ohms
I don't follow how beta was determined
thank you but is your transistor is 2N3904 and does it word with DC input ?
and the value of capacitors isn't exist in the video at least write it in description
26mV?
From were you get 26 v
26mV is the thermal voltage of the BJT and it's a constant (well, a temperature dependent constant) for the transistor
@@ElectronXLab 🙏 thanks for the reply sir. Any video on how to design a RF transmitter and receiver module for remote control ie rc car etc.
for Rb2 calculation RB2
The 10 is somewhat arbitrary; the purpose is to make sure that the resistance of RB2 is much less (i.e., 10x less) than the resistance seen looking in to the base
Can someone please help me understand part 4!? Where does the selection of RB2 must be less than (beta)(R_E)/10 come from?? In my textbooks, I see only RB2 should be (V_eq)/(9*i_B). Please and thank yoU!!
Strictly speaking RB2 does not have to be less than (beta)(R_E)/10. If you are using the shortcut of assuming I_B is so small as to be ignored, than RB2 needs to be at least 10 times bigger than the resistance looking in to the base.
You're trying to ensure that the current through Rb2 is at least 10 times the base current drawn by the transistor. If it isn't, then the HFE of the transistor starts to have an increasingly large effect on the base bias voltage, and therefore on the emitter-collector current, so different transistors of the same make will give different collector voltages, potentially causing clipping and a lack of reproducibility of the circuit.
How do I rearrange step 3 without VCE(sat)? I suck at algebra, help appreciated. I just need this dang emitter resistor!
You could assume that it is an ideal transistor where VCE(sat) is 0
Good work
It somewhat helps, although Im trying to figure out how to design an amp that generates a 20volt sine wave.
If you are in the comments to check where he got the 26 mV.... then I am sorry, I also don't know
re: little Re, where did 26mv come from?
That's the thermal voltage and is a property of the PN junction. It is temperature dependent, so the value of 26mV is only valid when the temperature is about 25C. VT = kT/q. Where k is Boltzmann's constant, T is the temperature in Kelvin and q is the charge on an electron.
I dont understand how you got 2325 ohms on part 3?
Please explain.
A couple of design restrictions that I set are IC=4mA and to bias in the middle of the load line. For IC=4mA to be in the middle of the load line requires that the saturation current (IC(sat)) be two times that -> 8mA. IC(sat) = (VCC-VCE(sat))/(RC+RE). I've already determined RC, so I just have to solve that equation for RE.
It's important to note that re and RE are different resistors. RE is the external biasing resistor connected at the emitter and re is the internal resistance in the emitter of the BJT.
Thanks for the reply, but when I do the math I get a completely different number.
This is what I have 8=23.8/(650+RE) when I do this I get 0.2929, sorry maybe I'm dense, I just dont see how your getting 2325??
That should be 8mA or 0.008. Then, I think you made a mistake in your arithmetic somewhere. 0.008=23.8/(650+RE) -> 0.008(650+RE) = 23.8 -> 650+RE = 2975 -> RE = 2325 Ohms
OK thanks, yes I wasn't converting ma to a so I was dividing/multiplying by 8 not .008 :(
@@ElectronXLab im still not getting it
is I_csat always 2*I_c ???????
yes, if you want to center the Q point across the load line
Why did you select BETA as 100 in step 4 while calculating parameters? For the sake of simplicity? or did I miss sth?By the way Thanks a lot for these useful videos !!!
Beta is a characteristic of a BJT. The value can vary significantly from transistor to transistor even ones with the same partner.
For this example, I just picked the value of beta for the sake of simplicity (just like you guessed).
thank you sir, waiting for more videos to come :)
Thank you for sharing your knowledge on RUclips. I have a question. In determining the value of small re, why wouldn't you use Ic(sat) current instead of the quiescent? Also, I plugged all of the formulas into a spread sheet and entered various Vcc's and Icq's and beta's. Everything worked for the sample values you gave but when I changed values I began to get negative RE values for the emitter resistor....for example, use 9 Vcc and 200 beta.
You use the quiescent current because that is the current that is actually going through the transistor and re is based on the state of the transistor.
You were getting negative values probably because the values you were putting in are values in which the the transistor can't actually operate.
@@ElectronXLab Ok, that makes sense. I notice some designers always make RE smaller than RC. For example they start out with Vce being half of VCC, then VRC being 3/4 of what's left and VRE being the other 1/4. They choose a Icq and go from there. I was wondering if one method is better than the other or is there more than one way to skin a cat......
@@billwilde8725 can you send me the video that telling that
@@hero96559 Sorry, that was 2 years ago, it's lost in the shuffle now....
@@billwilde8725 So did you get an answer for your quetion ?
For people way behind or not up to date on the math and references and variables should explain also explain the type of material and the reason of why and what happens when those components aren't correct or not processing or working properly would be helpful i can solder and put together parts need to know why and the purpose plus the results of incorrect input
isnt the Ic(sat) = Vcc / Rc ?
if not how did you find Ic(sat) ?
That would only be true if VCE(sat) is zero (it's not) and if there was no RE resistor. Ic(sat) = (voltage across RC) / RC. Voltage across RC = (VCC - voltage at collector) and voltage at collector, Vc = IE(RE) + VCE(sat), so Ic(sat) = (VCC - IE(RE) - VCE(sat))/RC and if you approximate IE = IC, and rearrange that equation, you get Ic(sat) = (VCC - VCE)/(RC+RE)
@@ElectronXLab oh i see. Thank you.
Would you take the max value for the saturation voltage from a data sheet ?
@@ElectronXLab And since Vce_sat is around 200mV for most small signal transistors, you can neglect it compared to Vcc. That gives you the useful equation:
Ic_sat = Vcc / (Rc + Re)
Vce value?
Where did 26mv come from ?
Chet Lund come from the equation re=26mv/ I_E Its define as the voltage in the emisor inside the bjt
What is re how find of npn transistor
its the resistance to ground looking from base
26 mV...????
This is the thermal voltage at the junction that is potential caused by the thermal motion of electrons. It's temperature dependent, but about 26mV at room temperature.
Good job, thank you.
hey can i know if RE is higher than RC it this acceptable
Nice job
Why do you use Ic(sat) =8mA BTW what Is Ic(sat)
Ic sat is when the transistor is fully on inother words saturated and it cannot amplify any more.
Would the saturation voltage line up with the saturation current say if you were using a graph from a BJT data sheet to find your quiescent point ?
can RE> RC?
Either do something right or don't it at all. You come up arbitrary with numbers without explaining.
26mV just appeared from nowhere!