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35. Class B and Class AB Amplifiers
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- Опубликовано: 14 авг 2024
- The problem with amplifiers is that as efficiency improves, faithfulness to the original signal diminishes. Overcoming this lack of fidelity requires the use of multiple transistors which employ dual power supplies or a complementary symmetry configuration. This video covers complementary configuration circuits using a single power supply to achieve Class AB and Class B operation. Efficiency if the circuit, biasing, and current mirrors are discussed.
The Best half an hour I have spent watching a RUclips video.
Unlike some videos dealing with amplifier circuits, you dont leave out the basics and you lay out the algebra in a way that makes it easy to understand.
Electronics should be approachable, understandable. I try my best to get both in my videos. Thanks for watching.
I'm glad you liked it and thank you for watching.
Really like your videos. I'm a development engineer in the uk. Really needed to finally understand BJT amps. Now, much more confident. Thanks for the uploads..
Thank you for taking the time to watch and I am glad that the videos were informative.
For this title means "power amplifiers " i even studied fundementals of microelectronic in which we`ve got remarkable digestivable concepts but boldy none of them can reach to your great and increadible video lectures ...Thanks man .
Thanks for the time you spend making these video's!!! I have learned more from you and others on youtube than I did in college. I find that making the output stage to be straight forward, it is the VAS stage of the amp that I have problems making, I like using a split power supply so I can use a op amp for the pre amp stage because finding matched transistors for a reasonable price is impossible.
Thank you so much, my teachers redesign the course and desing a wrong circuit schematic and all the people in the lab burn their arduino, so now we needed to design ourselves the amplifier but no one show the math behind it.
Thanks for this video and all your video's man, you explain everything really well and showing it on the scope makes it so easy to understand. Thanks a bunch.
Thank you. The goal is to make the material accessible. Nice to know that I succeed sometimes :)
This video is excellent and easy to understand. Kudos Sir!
This is a great effort to teach novices like me . Thank you very much
You are welcome!, learn anyway you can. Thanks for watching
Another great video. Thank you.
At the negative cycle, the npn transistor should be in its off state, therefore disconnecting the 10V source from the pnp transistor, so where does the current comes from that allow negative voltage to appear at the output?
That 4 resistor voltage divider is getting me confused... Shouldnt the voltage at first base be 10V-VR1? That in this case is 5.32
Thanks for this great video, now I can explain class ab to a baby
Love all your vids
Love it! Thank you! Next you should do a video on bootstrapping techniques for a class AB amplifier. Not that I need help with that. :-)
the best !!! ⭐⭐⭐⭐⭐
You should remake this video with better sound or just a better mic :)
14:35 I realize that this circuit is mainly to exemplify a sort of class B (partly AB) output stage, but.... having two separate input caps raises some issues regarding the output transistors getting out of sync. One problem will occur at startup when C2 charging will lag C1, resulting in both Q1 and Q2 on at the same time, shorting the supply for some period of time.
I got a basic understanding now. Thanks,
29:20 "efficiency [of AB] goes up a little bit [compared to B]". Power consumption difference of the bias network is insignificant. The main difference in efficiency is due to the fact that in class AB the transistors are on for a larger proportion of the time, and conduct some current directly from one rail to the other, at least for part of the cycle. Hence AB is less efficient than B. (When comparing B and AB "push pull" amplifiers as shown in this video.)
Great video!
How would you calculate the AC Gain (hFE), rE', Rin and Rout?
RMS-power is a nonsensical name, what you mean is not the rms-value of p(t) (p(t) = the instantaneous value of the power) but the _average_ power of p(t). Also note you wrote on the board a formula for Pout with units V^2 which is incorrect.
Cool video and tanks. I am doing my on ab amplifier but the redidtir krpp warm upp and búin. What am í doing rhong?
When no input signal do transistor have current?
Very Very welll explanation.
Thanks for the video
I couldn't understand why to use the 2 capacitors that are connected to the source
Why do we need class AB?
to get rid of the crossover section (the part where Vo=0v in the B class) in other words so that the output signal matches the input
@@assilmahfoudi2824
♥️🙏
What is the minimum value of the DC biasing current in class A output stage? Explain you answer.
Hi, The circuits are common collectors (emitter followers) so at best the gain is unity. Rout would be RE||RL AC resistance in to the base is r'e*beta. Overall input resistance would be r'e*beta||RB1||RB2. Hope that helps.
Thanks
Excellent tutorial.
great video and this is probably a stupid question but is the sine wave from a signal gen? if so at what frequency?
What an explanation! But where is the amplification done by transistors?
Great work sir
I have noticed that most people don't include an emitter resistor in their Class AB Amplifier designs.
I have found that it is really needed to eliminate thermal runaway. How do you determine the value
of these emitter resistors?
You would want to select extremely small values of resistance. Since the circuit already has no voltage gain, adding too much resistance decreases gain more. Resistances as small as 1 or 2 ohms would suffice to give some negative feedback and prevent thermal runaway. The first step would be to add diodes to the base biasing network of course. Since the Av of a CC is RL/(RL+r'e), adding more resistors to the emitter changes the equation to RL/(RL+r'e+RE). You would have to decide how much was to much for voltage loss. Anyway, a couple of ohms will do the trick so the hit to Av won't be too bad. Hope that helps.
Thanks so much for your help.
John
In your circuit diagram, you have a Vcc which forward biases the pn diode, how come that it doesn't conduct for the full cycle? Did you assumed that the Vcc is not present there?
awesome content ! thank u
U helped me so much I like it thank u
Hello:
I am back into electronics after a hiatus. Would you consider doing a video on a Class AB amplifier
with a common-emitter driver. I have been searching for a good document or video that explains
how to design and analyze such a circuit. I have a schematic in mind but I couldn't attach it to this
comment.
Although I listened to that part many times, I still don't understand how both nodes end up being equal to 318mV at 8:47. The lower node should be (10000/21360)*10=4.682V and the upper node should be (11360/21360)*10= 5.318V as far as I can see.
What is the most linear class of amplifier
what is the least energy efficient class of amplifier
Class A without a doubt - 25% theoretical maximum efficiency; practical efficiency is even worse.
Class A with a microwave oven transformer?
Hello:
I have 2 questions with regard AB power amp circuits.
(1) If you look at the Fairchild datasheet for a TIP29 power transistor, how do you determine the DC gain? It says that if Vce = 4V, Ic = 0.2A then the DC gain has a min value of 40. However, if Vce = 4V and Ic = 1A, the the DC gain has a min value of 15 and a max value of 75.
(2) For power amplifiers how do you arrive at a value for r'e? Can you use 25mV / Icq?
Thanks very much.
You can't know the DC gain with any certainty. To minimize the variation in DC gain (and fix it at a specific value) you would use voltage divider biasing and a resistor on the emitter. For part 2, Yes!
Hey thanks for the video.
Just a quick question about the design of the of the Class AB amplifier. How do i find the values of the two resistors and the capacitors. I have found the KVL around the Resistor and the diodes, which is; R = (Vcc - 1.4V)/(2*I) . But i dont know what to put for the current in the KVL. Also, my load resistor is 8 Ohms. Thanks you
Hello, Treat the each transistor and base resistor set as if it were it's own circuit. So, to find voltage on the base of Q1,
Vbase q1=Vcc*RB2/(RB1+RB2).
You can rearrange the equation to get the values needed for 0.7 volts on the base.
You want the capacitors to be electrically "invisible" to the AC signal so a low Xc is needed. As a rule, the Xc value should be a maximum of 1/10 the resistance you are trying to bypass or couple to. The capacitor for your 8 ohm load resistor would be calculated using the lowest frequency going through it. You Xc should be 0.8 ohm (preferably less) at that frequency.
Thanks for the reply. I tried using the values in your circuit but I did not get the 10 Kohms value for the resistors. I dont know what im doing
Love these videos. They are so well done!
I am stil not sure how to use a transistor datasheet to arrive at a DC gain estimate. As there is such a great range.
If you look at the Fairchild datasheet for a TIP29 power transistor, how do you determine the DC gain? It says that if Vce = 4V, Ic = 0.2A then the DC gain has a min value of 40. However, if Vce = 4V and Ic = 1A, the the DC gain has a min value of 15 and a max value of 75.
Thanks
Hi again, DC amplification in a transistor is used principally in switching, the circuit is either fully on or completely off. The data sheet gives you broad parameters for the expected outputs given a range of inputs. Other than setting the DC q-point, that's it for transistor DC amplification.
Very nice video.. would u like to know what is the model number of pnp and non bjt ?
Hello. The transistors are 2N3904 and 2N3906. And thank you for viewing my videos.
I designed a class AB amplifier with dual voltage driving PNP and NPN transistor. The final was the amplifier works perfect but when the power is switched off the positive voltage gets drained but the negative voltage stays what's the solution. Pls
That's a question I can't really answer without schematics. Measure for voltage drops - there shouldn't be any with the circuit off. Good luck and thanks for watching.
@@TheOffsetVolt thanks for your reply i increased the grounding resistance of the two pair transistor in the input stage. Now its ok . I have to learn more by practical work. Thanks for the reply.
Sir please make a video to explain the circuitry of class b amplifier where you teach us the need of each and every component required... Please I would be exceptionally thankful to u.
Wow!
What could cause 0v output in the circuit for the class AB
When the PNP turns on do the electrons flow to themselves?
3:40 no negative output? it do have , system is on Q point
just watch it in few minutes and this make me realize what the hell is going on. others cannot explain it like what a model you set at initial. this make me know how a push pull amp work.
plz explain curren tmirror using mosfet bulk driven
Thanks for watching my videos. I will add this to my list - which never seems to get shorter :)
ASMR starts at 17:56
24:55 "any current that goes through D2 is going to go into the transistor, be multiplied beta times...". That part of the explanation is wrong. The current through D2 does not go through Q2, nor vice versa. (The D2 current and the Q2 base current sum to make the current through RB4.) Instead, the current through D2 produces a particular voltage drop that is similar to the VBE required for a similar current in Q2. These amounts are not guaranteed to match exactly, so RB1 and RB4 would need to be adjusted to get a voltage across D1 and D2 that sets the desired ICE in the transistors. And at least the diode drop and required VBE will track with temperature, but only if the diodes and transistors are thermally connected (so not in the circuit as built, contrary to 27:06). It is indeed sort of a current mirror, though it's missing emitter resistors that would normally be used to set the current stablely.
thankyou sir
alot easier to tell them Vrms squared divided by R load.
How can Power out can be .7Vout*.7Vce
Hi, I am not sure what you are asking.
I have the same question, you wrote Pout= 0.707*Vout*0.707*Vceq => Pout=0.25Icsat*Vcc
According to this Voutpk=Vceq/Rl
And Icsat=Voutpk
hello thanks for video. I have homework about ab amplifiers and i watched that video but i didnt find answer to my questions i need Rb1 and rb2 and my teacher gave me that formule rb1=(vcc-vd1)/(idmin + (iq+ILmax/1+b)) that formula gives me rb1=rb2 and i just dont know ILmax in that formula if i found it i can get rb1 thanks for helping me and if you give your mail adress i can show you my problems.
thanks for reply my load is a resistor , actually my circle is individual same with yours i have simetric dc source +12 -12 and ac source but ac source is connecting middle of two diodes i have to find Rb1 and Rb2 like i said, the formule is there , i tried sometings but i m not sure it is true , by the way sorry my english
I do n't understand the significance of the Viking.
I understand the Viking more than the PNP function.