Transistor PA
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- Опубліковано 8 лип 2024
- This is a full tutorial explaining how to design a push-pull, class AB transistor power amplifier. This assumes that the viewer understands concepts such as basic transistor and diode operation, ohms law and a bit about RC filters. This design process is supposed to be somewhat flexible. It can accommodate larger or smaller power designs.
this is pure, pure gold...
EXCELLENT PRESENTATION,LEARNED SOME THING NEW ON POWER AMPLIFIERS THAT I DID NOT WHEN STUDYING THEM IN COLLEGE.
I HAD TROUBLE FOLLOWING ALONG WHEN YOU FLIP BACK AND,FORTH WITH THE SCHEMATIC DRAWING.
SOME TIMES THE GROUNDING OF THE CIRCUITS CAN CAUSE UNWANTED POSITIVE FEEDBACK.
Really good explanation, thanks!
Glad you enjoyed it!
The Tube Roaster always wondered why some of my amplifiers with negative feedback started to pull 5 amps straight through the output resistors even when the two bases of the push pull were connected together! Could that be caused by oscillation? Gonna try the 22pf cap soon
With output resistors i meant transistors**
It's possibly an oscillation. Make sure that your voltage amplifier is compensated with enough capacitance between collector and base. You could also try to drop the amount of feedback or disconnect it to see if that stops it. If you have an oscilloscope, try probing around looking for oscillation in the 100s of kHz region. If it's not oscillation, it might be some form of shoot-through where both output devices are being fully turned on at once.
The Tube Roaster thank you very much! Unfortunetely i do not have an oscilloscope. It is actually a differential amplifier with a current mirror followed by a common emitter so the open loop gain is quite high. I will add some capacitance there and see if its fixed. Thanks
hello thank you for your effort sir , Its really valuable
I want to know the book that you're referring to please
The book is called: Transistor Circuit Techniques by G. J. Ritchie
Hello Mr!, In the voltage amplifier why Ic6 and Ic5 are 3mA?
It's mainly to set the quiescent collector voltage (from the voltage drop across R3 and R4) the to slightly higher than 1/2 Vcc to achieve maximum signal swing before clipping. The reason it has to be slightly higher than 1/2 Vcc is that the lower end of the amplified diode will drop some voltage (at least 1.2V - 1.4V) before it feeds the lower half of the output stage. The idea is to get the output to clip as symmetrically as possible for maximum signal amplitude. When I demo the circuit, you can actually see the bottom always clips first. The symmetry is close enough but not exact because it depends on what voltage the amplified diode is set at.
@@thetuberoaster8321 OK, I thought it was because of the 0.9mA needed on TR3 and TR4 basis. Thanks for the explanation.
@@thetuberoaster8321
Could you make a video using topology C (with differential amplifier)?
@@adrianolopes4717 I'm basically working on a video now with a differential input amp.
@@thetuberoaster8321
I'll be waiting, thank you very much.
Hello Mr!, In 15:42,
why VR5 = 14-1.24? wouldn't VR5 = 12.6-1.24 because of the negative feedback?
14V is the optimal voltage we want the quiescent output voltage at for max swing. 12.6-7V is the voltage across R5 because we want R6 to have 1.24V dropped across it. (Diode drop plus VRe6 to set collector current in TR6) The only reason we need to know this voltage is to make sure the voltage divider has enough juice to feed the transistor base and stay at the same voltage.
why The output power P = V² /8xRL ! from where the 8 came?
and Ic peak = VCC / 2xRl ! why 2 xRl ?
TBH I don't actually know where that comes from. It has also been a long time since I've looked at this circuit. The book shows the expected Vcc²/2xRl but then later it also shows Vcc²/8xRl for the same output stage calling it maximum average output power.
@@thetuberoaster8321 i can see in the book that we assume that Vout = Vcc /2
and Vout rms = Vout/sq2 so Pout avg= Vout²/R = Vcc²/(2²xsq2²xR)