# Trouble understanding how class a driver can feed class AB push pull pair.

Discussion in 'The Projects Forum' started by damien83, Feb 24, 2016.

1. ### damien83 Thread Starter New Member

Sep 17, 2015
14
1
Hi people just need some help to understand this circuit please.

I believe that this is a class A pre-amp driving a class AB push pull power amp.

I have two questions regarding this circuit -

I was under the impression that a push pull pair requires + and - voltages but the schematic only has + 0 to 9 V.

Also - even if this did have a split power supply (And this is my main question as it's the one that I most need the answer to )

How would a class A driver (Which I think is only capable of amplifying a signal which goes from 0 upwards (( no negative voltages to be seen))
drive a class AB push pull pair? Or does this work as the class A driver collector emitter junction be set at half supply voltage which would essentially be 0 Volts as seen by the push pull pair. Ie - If you had a +/1 30V power supply then +30v would be zero as it's in the middle of 0 - 60?

Any input appreciated.

2. ### AnalogKid Distinguished Member

Aug 1, 2013
4,709
1,301
Your second guesses are correct. While absolute voltage values are critical to getting a circuit to function, the relative values are what lead to understanding. There is nothing about a class A stage that limits it to amplifying in one direction. For example, if you replaced the class A NPN transistor with a PNP and reversed the voltage to -9V and 0V, the stage would amplify exactly the same. In fact, since PNP transistors were the first ones invented, many transistor circuits in the 50's and 60's were all PNP. The typical small 6-transistor radio had a 9 V battery, with the + terminal as ground. It's all a matter of orientation.

What makes a stage class A is not the "direction" of its output, but whether or not the output transistor ever is cut off. In class AB, B, and C stages there are times when one of the output transistors is not conducting any current. In a class A stage, whether it is biased by a resistor or an active current source, the output transistor is conducting at all points in the output waveform, just more or less depending on the output voltage and the load. Nothing about that last sentence is limited by voltage polarity.

A push pull stage is called that because it can push (source) electrons into the load, and pull (sink) electrons from the load. For this to work, the load must be somewhere between the two supply rails so the relative voltage (and hence the current direction) between the amp output and the load can swing both ways. Again, voltages are relative. With a single power supply design, the zero-signal output voltage should be approx. 1/2 the supply voltage. The output capacitor charges up to this voltage so the DC or average value across the speaker coil is zero V. In your circuit, the speaker is returned to V+ rather than ground, but from an AC standpoint these are identical.

ak

3. ### damien83 Thread Starter New Member

Sep 17, 2015
14
1
Thanks analog,

So I can run a class A driver from a +/- power supply? The same power supply that I feed my push pull stage with?

If i have a +/- 30V supply, (0 - 60V) all I have to do is drop half the voltage over the collector resistor for the class A driver and it will be set at 0V in relation to the push pull stage?

I am beginning to grasp the concept of relativity. Ie - "Ground" is a made up reference voltage in some cases.

4. ### crutschow Expert

Mar 14, 2008
13,523
3,390
Yes, ground is just an arbitrary (0V) reference point.

Running from ±30V supplies will require some other changes to the circuit since 0V (ground) is now -30V (for example the volume control should be connected to the new ground, not -30V).

Note that C1 has the incorrect polarity shown.

5. ### damien83 Thread Starter New Member

Sep 17, 2015
14
1

What I'm really after is more volutme at the speaker.

If my op amp is only pushing 15ma of current, this is going to effect the volutme/power of the amplifier right?

even though I have a push pull stage that is hooked up in darlington pair configuration (voltage gain of 200A max) If I am only deliverying a measly 0.015 A from my op amp, this is going to stop high volume?

Should I just opt for a high current op amp or should I just be done with it and create a valve pre-amp? Even the high current op amps will only accept a power supply voltage of 60V and my transfofrmer is 0-60 V

I have only just realized that the V X A = W applies to this so even if I have a Vcc of 60 V then I'm only goint to be getting about 10W out.

I have a 15V op amp capable of 30ma. So 15V X 0.015 only a few watts.
My power amp stage is capable of 100W easy.
]

Is this right? i feel I have been focusing on voltage rather than current required to drive a load.

6. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
1,463
372
Last edited: Feb 28, 2016
7. ### crutschow Expert

Mar 14, 2008
13,523
3,390
The op amp just has to deliver voltage to the volume control.
As long as the volume control has a high enough resistance so maximum current is less than 15mA, then you op amp should be able to drive the circuit to full volume.

And that is not the best circuit if you are looking for high power.
How much do you want?