Try this:

Without the diodes, you get class B operation. Only one transistor would be turned on at a time. There would also be crossover distortion because the input signal has to exceed +0.7 or -0.7 (the Vbe drops) before the corresponding transistor conducts.

The diodes make it class AB if your lucky. Part tolerances with no signal will make one of the output transistors to conduct more.
Those diodes should be thermally mounted to the transistors. As Vbe changes with temperature, the diode drop changes too.

Hi,

Really better would be to use two more transistors as diodes so they track better.
Course, there are better solutions yet.

 

I know I'll be a little of subject but it's still on topic. I've read all the explanations on what biasing is suppose to be but even I don't quite know what it actually means. Is there a simple way to put it that doesn't require wording that goes over my head.
Brzrkr

 

Biasing means applying the appropriate amount of voltage or current to the device to bring it into the operational mode that the application requires.

For example, we have a detector made of silicon that requires a bias of +1800V to bring it into a mode that results in optimal signal-to-noise ratio.

 

For example, we have a detector made of silicon that requires a bias of +1800V to bring it into a mode that results in optimal signal-to-noise ratio.

So whatever the bias voltage is makes it work at optimal performance ? Anything over or under effects the out come of it's performance.
Brzrkr

 

So whatever the bias voltage is makes it work at optimal performance ? Anything over or under effects the out come of it's performance.
Brzrkr

Hi,

The diodes are there to make the base bias voltages one diode drop above the emitter.
One diode drop is about 0.7v, so if the emitter is at 2v then the base must be driven at 2+0.7 where this 0.7 comes from the base emitter diode.

So we actually have four diodes to consider:
1. The base emitter diode of the top transistor.
2. The diode drop of the top diode
3. The base emitter diode of the bottom transistor
4. The diode drop of the bottom diode

Now when 1 and 2 are equal and 3 and 4 are equal we have the transistor bases biased correctly.
When the input is at 5v the emitter is at 5v thanks to the top diode (and bottom diode too).
Then, consider what happens when we short out the top diode. Now when the input is 5v the top emitter at 5v minus the base emitter drop and so we end up with 4.3v at the emitter instead of 5v. So the top transistor is not biased correctly now.

So the diode voltage drops match the base emitter drops when biased correctly.
That way the emitters follow the input closely rather than be off by 0.7v.

You should probably look up "bipolar transistor voltage follower" to see how the base emitter drops the input by 0.7v without the diode.

 

In a nutshell, the voltage on the base of the transistor has to be about 0.7V higher than the emitter (for NPN transistor) for the transistor to start conducting, i.e. we are moving the transistor's operating mode from Class-B mode to Class-A mode (call it Class-AB).

In Class-B mode, the two pushpull NPN/PNP output transistor pair are idling when there is no input. This results in severe cross-over distortion.

By biasing the NPN/PNP pair into the conducting mode (Class-A), we reduce the amount of cross-over distortion.