Compensating diodes used on push pull amplifier

Thread Starter

Saviour Muscat

Joined Sep 19, 2014
151
Hello
Hope you are all good,
Can someone explain how does the compensating diodes work during the positive and negative half cycles input signals of push pull amplifier?
Thanks and regards
SM
 

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Thread Starter

Saviour Muscat

Joined Sep 19, 2014
151
I am confused , it is a little bit hard to understand, the circuit that I am studying don't have a dual supply but a single supply as the image attached. I need how the circuit works when given an ac signal(during negative and positive half cycles). I don't need diagrams but refer to the image attached. I hope for your understanding.
thanks
 

KL7AJ

Joined Nov 4, 2008
2,229
I am confused , it is a little bit hard to understand, the circuit that I am studying don't have a dual supply but a single supply as the image attached. I need how the circuit works when given an ac signal(during negative and positive half cycles). I don't need diagrams but refer to the image attached. I hope for your understanding.
thanks
Those aren't compensating diodes; they're forward bias diodes.
 

Jony130

Joined Feb 17, 2009
5,213
Well, but the circuit work almost exactly the same. Without any input signal both transistor conduct because of this diodes in the circuit.
For positive half cycles upper transistor deliver current the the load by charging capacitor. And for negative half cycles lower transistor discharge capacitor. And this is why we have a negative voltage at load.

181.jpg

Those aren't compensating diodes; they're forward bias diodes.
But what if we mount these diode on the same heat sink as a output transistor?
 

Thread Starter

Saviour Muscat

Joined Sep 19, 2014
151
I didn't expected a good reply like this ,I understood completely many thanks. I hope this is the last question in this thread.
How the bias diodes (not compensating diodes) bias the transistor more 0.6V to eliminate crossover distortion? Especially during positive and negative half cycles.
Many thanks to everyone for your support?
 

#12

Joined Nov 30, 2010
18,217

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Jony130

Joined Feb 17, 2009
5,213
How the bias diodes bias the transistor more 0.6V to eliminate crossover distortion? Especially during positive and negative half cycles.
Pleas notice that BJT's need at least 0.5V to 0.7V of forward base-emitter bias voltage before they will go into conduction. In our push pull amplifier both of the transistors will be nonconducting of OFF, when input signal is in range +/- 0.5V. And this form's a "deadzone" in the input and produces crossover distortion in the output.

dd.PNG

So, to avoid this crossover distortion, it is necessary to add a small amount of forward bias to take the BJT's in the verge of conduction or slightly beyond. And this is why we add this two diode's into the circuit. The job for the diodes is to provide a small amount of bias voltage, to take the BJT's in the verge of conduction or slightly beyond. And this is how we eliminate this "deadzone".
dd2.PNG
Now even with no input signal both transistor conducting the small amount of DC current.
And this help as eliminate the crossover distortion.

Also try read this
http://forum.allaboutcircuits.com/threads/sine-wave-distortion.90332/ (post 16 )
 

alfacliff

Joined Dec 13, 2013
2,458
the temprature charistics of the diodes compensates for heat in the transistors, when the transistors get hot, the diodes do too, reducing bias, and reducing current.
if you measure the foreward drop across a diode, it changes with heat.
 

MrChips

Joined Oct 2, 2009
22,912
Jony130 explains it very well.

In a class B amplifier only one of the two transistors is active. This results in cross-over distortion.
In a class AB amplifer you want the two transistors to start sharing the load current at cross-over. Hence the transistors are biased into conduction with the aid of the two bias diodes.
 

#12

Joined Nov 30, 2010
18,217
To avoid this crossover distortion, it is necessary to add a small amount of forward bias to take the BJT's in the verge of conduction or slightly beyond. And this is why we add this two diode's into the circuit. The job for the diodes is to provide a small amount of bias voltage, to take the BJT's in the verge of conduction or slightly beyond. And this is how we eliminate this "dead zone".
My post #7 details how to adjust the current through the diodes to minimize the dead zone. Attaching the diodes to the heat sink of the power transistors helps the voltage across the diodes stay equal to the voltage you want from base to base. I would also like to say I've never seen this circuit without at least a few tenths of an ohm in series with each emitter, but that's getting into fine details about temperature tracking.
 

Thread Starter

Saviour Muscat

Joined Sep 19, 2014
151
Thank you very much for your support, yesterday I was at work I can't reply the same day sorry
I tried to understand the operating principle with your help and I went to conclusion aided with the picture attached.
I marked in red in the picture with the voltages without the input signal applied in the circuit, As I concluded is that the two transistors are forward bias by 0.6V as I calculated in red.
Finally I marked in green with the input signal applied of 2V peak to peak to the circuit. I understood that during the positive half cycle, the signal can only pass through the upper diode only and this produces a voltage drop(or superimposed) of 11.6V(10+0.6+1) with respect to ground and this voltage is fed to the upper transistor then if you subtract the input voltage of 11.6V by 10V(the voltage between the two transistors) the result is that the upper transistor has an input bias 0f 1.6V .
During the negative half cycle the signal passes only through the lower diode and produces a voltage drop (or superimposed) of 8.4V(10-0.6-1) and this voltage is fed to the lower transistor. then if you subtract the input voltage of 10V by 8.4V the result is that the lower transistor is biased by 1.6V.
I am not 100% sure what I'm concluded, please kindly can you confirm if I am correct? or otherwise make a correction what I said wrong.
Your support is much appreciated!
Thanks and good day
Saviour Muscat
 

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MrChips

Joined Oct 2, 2009
22,912
You have the right idea. However you cannot analyze the circuit by considering voltages alone. You have to also consider currents.

Each transistor is configured as an emitter follower (common collector) amplifier. This produces current gain and unity voltage gain.

The output current flows through the load. Hence the impendance of the load must also be considered. To analyze the circuit, one calculates the base current. The emitter current is the base current x (1 + beta). Hence this produces current gain. Because this output stage is the power stage, one is interested in the output power. Hence one calculates the power delivered to the load.
 

Jony130

Joined Feb 17, 2009
5,213
Hi, I encourage you to read this very carefully.
http://forum.allaboutcircuits.com/threads/voltage-divider-bias.71104/#post-494950
http://forum.allaboutcircuits.com/threads/class-b-push-pull-amplifier.85631/#post-615183

As for your circuit. Without the input signal voltage at T1 base is equal to 10.6V and 9.4V for T2.
Output capacitor is charge to 10V and input cap is charged to 10V also.
Id1 = Id2 = (20V - 2Vd)/(R1 + R2) = 18.8V/7.8K = 2.41mA (I ignore the base current).
Now let as apply the input signal (2Vpp= peak to peak) to the circuit.
The input signal "modulates" the DC voltage and currents in the circuit in the "rhythm" of AC input voltage.
During the positive half cycle input DC voltage rise from 10V to 12V and back to 10V in the "rhythm" of AC input voltage.
T1 base become more positive, D1 and R1 current decreases but D2 and R2 current increasing.
T1 base current comes from 20V battery and R1 resistor. Input source current flow in the path:
plus Vin--->Cin--->D2--->R1--->GND--->minus Vin.
And the voltage across load resistor is equal to
Vload = Ve1 - V_Cout = 11.95 - 10V = +1.95V
For negative half cycle we have very similarity situation but this time T2 base becomes more negative, D2 and R2 currents decreases but R1 and D1 current increase. And Vin source current is flow in the path:
plus Vin ---> minus_20V supply--through 20V supply--plus 20V supply--->R1--->D1--->minus Vin.
And the output voltage is equal to Vload = Ve - V_Cout = 8.05V - 10V = -1.95V

push pull.png
 

Thread Starter

Saviour Muscat

Joined Sep 19, 2014
151
Thank you for the detailed explanation it was what I expected
My last questions

Hi, I encourage you to read this very carefully.
plus Vin--->Cin--->D2--->R1--->GND--->minus Vin.
I think it should be R2 not R1

For negative half cycle And Vin source current is flow in the path:
plus Vin ---> minus_20V supply--through 20V supply--plus 20V supply--->R1--->D1--->minus Vin.
I am confused regarding this current path.



Last questions
I think you assumed Vbe of T1 to be 0.65V during positive half cycle if not please can you tell me how to calculate this voltage (T1Vbe+T2Vbe)= 0.65+0.55=1.2V vice versa
(T1Vbe+T2Vbe)= 0.55+0.65=1.2V


Million thanks
SM
 

Jony130

Joined Feb 17, 2009
5,213
I think it should be R2 not R1
Yes, of course
Vin--->Cin--->D2--->R2--->GND--->minus Vin

For negative half cycle And Vin source current is flow in the path:
plus Vin ---> minus_20V supply--through 20V supply--plus 20V supply--->R1--->D1--->minus Vin. I am confused regarding this current path.
Why? And I forgot about Cin.
How about this one ?
plus 20V --->R1--->D1--->Cin--->Vin--->gnd (minus 20V)

Last questions
I think you assumed Vbe of T1 to be 0.65V during positive half cycle if not please can you tell me how to calculate this voltage (T1Vbe+T2Vbe)= 0.65+0.55=1.2V vice versa
(T1Vbe+T2Vbe)= 0.55+0.65=1.2V
Yes, you are right, I simply assumed this value. If you want to know the exact value you need to measure it in real circuit.
 
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