Class AB Push-Pull Amplifier

Thread Starter

elec_eng_55

Joined May 13, 2018
214
Hi:

I have bread-boarded several different Class AB Push-Pull amplifiers and
found that even though the maximum peak output voltage is supposed
to equal Vceq, it doesn't. It seems that for example, if Vceq = 4 volts, then
the maximum attainable output is 2.83V peak.

It seems that if you need to have a 4 volt peak output, you have to increase
Vcc.

I must be missing something here.

Thanks,

David
 

MrChips

Joined Oct 2, 2009
21,663
This is generally true since there is a voltage drop across the transistors.
Can you show us a circuit schematic?
 

MrAl

Joined Jun 17, 2014
7,757
Hi:

I have bread-boarded several different Class AB Push-Pull amplifiers and
found that even though the maximum peak output voltage is supposed
to equal Vceq, it doesn't. It seems that for example, if Vceq = 4 volts, then
the maximum attainable output is 2.83V peak.

It seems that if you need to have a 4 volt peak output, you have to increase
Vcc.

I must be missing something here.

Thanks,

David

Hi there,

One of the problems is that the upper transistor bias usually comes from the same power supply source that powers the amplifier itself and thus that powers the upper transistor collector too. When you have that situation then the max output is really one diode drop down from +Vcc because the highest voltage in the system is +Vcc and the base emitter drop constitutes one diode drop. This means the transistor can not enter saturation and that keeps the voltage drop at around one diode drop. Depending on load, this can easily be 0.7v or even higher.

Now if you were willing to use a higher than +Vcc voltage source like 1 or 2v higher than the main positive power rail, you could get the upper to saturate, and that could get you down to a drop of around 0.2v for some transistors and even lower for some really good transistors. Without that increase in bias however, there's no way to do it without going to a PNP upper transistor stage and that gets more complicated.
The good thing is that this increased voltage rail does not have to power the whole output load, just the input bias of the output stage which would be much less of a power requirement. So a 10 watt amp might only need a 1 watt power source for the bias even though it requires at least a 10 watt power source for the collector power rail.
This is also a problem when trying to use N channel mosfets for an upper. In that case the bias might have to be as much as 10v higher than the main power rail, but the power requirement of the bias supply is very low.
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
Hi there,

One of the problems is that the upper transistor bias usually comes from the same power supply source that powers the amplifier itself and thus that powers the upper transistor collector too. When you have that situation then the max output is really one diode drop down from +Vcc because the highest voltage in the system is +Vcc and the base emitter drop constitutes one diode drop. This means the transistor can not enter saturation and that keeps the voltage drop at around one diode drop. Depending on load, this can easily be 0.7v or even higher.

Now if you were willing to use a higher than +Vcc voltage source like 1 or 2v higher than the main positive power rail, you could get the upper to saturate, and that could get you down to a drop of around 0.2v for some transistors and even lower for some really good transistors. Without that increase in bias however, there's no way to do it without going to a PNP upper transistor stage and that gets more complicated.
The good thing is that this increased voltage rail does not have to power the whole output load, just the input bias of the output stage which would be much less of a power requirement. So a 10 watt amp might only need a 1 watt power source for the bias even though it requires at least a 10 watt power source for the collector power rail.
This is also a problem when trying to use N channel mosfets for an upper. In that case the bias might have to be as much as 10v higher than the main power rail, but the power requirement of the bias supply is very low.
This is generally true since there is a voltage drop across the transistors.
Can you show us a circuit schematic?
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
Thanks for your response.

Man I wish textbooks would elaborate when they make statements like Voutmax = Vcc.
It makes you want to give up sometimes.
 

MrAl

Joined Jun 17, 2014
7,757
Thanks for your response.

Man I wish textbooks would elaborate when they make statements like Voutmax = Vcc.
It makes you want to give up sometimes.
Hello,

You apparently quoted two posts in #6 but did not insert any reply :)

Books very often use different levels of detail. Some go more in depth than others. Dont let it get you down, just try to figure out what else can be happening.
If you have a question like this it could very well mean that they did not go into enough detail (yet). This happens a lot because books only have so many pages in which to convey the information the author thinks is important for the subject material at the time. It is very rarely enough information really, just enough to make a point of some kind that may very well be just a rough approximation.

You should still be happy however, because you have the web ... something we didnt have when i was growing up.
 
Last edited:

Jony130

Joined Feb 17, 2009
5,180
First you need to understand how this circuit work. And notice the fact the when the input voltage increases the current through Q1 decreases.

https://forum.allaboutcircuits.com/threads/class-b-push-pull-amplifier.85631/#post-615183
https://forum.allaboutcircuits.com/threads/compensating-diodes-used-on-push-pull-amplifier.101749/#post-766381

And the situation for the max voltage at the output is :

212.png

VL_max = Vcc - VR1 - Vbe - Vc = VL_max = Ie *RL = (Vcc - Vbe - Vc)/(RL + R1/(β+1)) ≈
≈ (8V - 0.6V - 4V)/(8Ω + 165Ω/(50 +1))*8Ω ≈ 2.42V



This is why in real power amplifiers we do do not drive the output stage from a voltage source directly. Instead , we are using CE amplifer and "current drever".
https://forum.allaboutcircuits.com/threads/audio-amplifier-design-basic-questions.104719/page-6 (Q3 ) or TR1 in MrChips schematisc.
 

AnalogKid

Joined Aug 1, 2013
8,536
Why do you have Q1 = TIP31C and Q2 = TIP32C?
You can replace those with two diodes.
True, but mounting "normal" diodes on a heatsink is a pain. Given what these transistors cost, having one that is easy to mount and whose Vbe characteristic perfectly matches that of an output transistor is a good thing. The temperature and Vbe tracking are much better than with either signal diodes or rectifiers in the bias string, and allow adjusting the output stack idle current to a lower than usual value. Been there, did that, amp still runs.

ak
 

crutschow

Joined Mar 14, 2008
25,262
You can increase the positive voltage output of the top NPN by adding a capacitor bootstrap to the base-bias circuit as shown below (from Nuts and Volts Magazine).
That will allow the positive peak output to approach the V+ supply voltage.

upload_2018-6-5_13-4-2.png
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
Hello,

You apparently quoted two posts in #6 but did not insert any reply :)

Books very often use different levels of detail. Some go more in depth than others. Dont let it get you down, just try to figure out what else can be happening.
If you have a question like this it could very well mean that they did not go into enough detail (yet). This happens a lot because books only have so many pages in which to convey the information the author thinks is important for the subject material at the time. It is very rarely enough information really, just enough to make a point of some kind that may very well be just a rough approximation.

You should still be happy however, because you have the web ... something we didnt have when i was growing up.
True, but mounting "normal" diodes on a heatsink is a pain. Given what these transistors cost, having one that is easy to mount and whose Vbe characteristic perfectly matches that of an output transistor is a good thing. The temperature and Vbe tracking are much better than with either signal diodes or rectifiers in the bias string, and allow adjusting the output stack idle current to a lower than usual value. Been there, did that, amp still runs.

ak
You can increase the positive voltage output of the top NPN by adding a capacitor bootstrap to the base-bias circuit as shown below (from Nuts and Volts Magazine).
That will allow the positive peak output to approach the V+ supply voltage.

View attachment 153716
 

Audioguru

Joined Dec 20, 2007
11,249
Most old school books wrongly show no driver transistor and the input signal feeding the junction of the diodes. Then the input signal turns off the output transistors and their bias resistors struggle to turn them on.

Bootstrapping increases the maximum output level of an amplifier and since it makes the base pullup for the NPN output transistor a constant current source it increases the load impedance of the driver transistor which increases the open loop gain which increases the negative feedback which reduces distortion.
 

MrAl

Joined Jun 17, 2014
7,757
Hi,

Bootstrapping works for audio circuits but not for logic circuits. If the output has to stay high for some length of time, the bs cap discharges. That's the only drawback i think.
Some years ago we actually did an audio mosfet design using bootstrapping, and we talked about it on the Nuts & Volts site. That was a challenge because the boost in voltage had to be greater. It should still be there, and someone actually built it and tested it.
 
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