Push-Pull Amplifier

are u saying that i should use 2 diodes to replace the 170 ohm resistor to provide the voltage of 1.4V to reduce the crossover coupling?

I'm not saying that you should do anything. You asked what the resistive network was doing. I told you the basic method to calculate what it is doing. You did not tell us what you wanted this to do, so I cannot say if this circuit is capable of doing what you're asking this current amplifier to do. Depending on how this circuit will be used in the total system, the configuration can work fine as it is. Even the problem that I alluded to may not be noticed with how this circuit gets used.

Now, you've noticed that this circuit is operating as a class B amplifier. I presume that you wish to stop the crossover effect that happens with all class B amplifiers. Replacing the 170 ohm resistor with two diodes may bring this circuit into AB operation but it certainly won't be very far into the class A region. But I do think you will want at least one diode, very strategically placed eventually in this circuit.

Did you ever look up the current to voltage equation for a diode that I mentioned? I like the one found at www.allaboutcircuits.com. The only significant error on that page is that they don't clearly define I_s as the reverse saturation current. It's a very good site that can help with all of your designs.

by the way, i replaced the q1 and q2 with darlington transistors. any special attention need to give to the transistors?

As far as using a Darlington transistor configuration, you'll find that you will have a higher current gain so you will have to redesign the biasing network and can then get a higher input impedance. You will also now have two diode drops per transistor to bias the transistors.

I didn't mention this earlier, but you don't need the two capacitors for an input. One going to either node will work just fine. Speaking of the input impedance, this capacitor and the input impedance make an RC high pass filter that limits the lowest frequency your amplifier can drive.

Redfred is right, but I think your question means you're missing something more basic.

The resistors are used to "bias" the base of the transistor.

When the transistor Base voltage is at (or below) 0.6V it doesn't conduct between the Collector and Emitter. When the voltage goes a small bit above 0.6V current flows between Collector and Emitter. This is best seen on a graph of Base voltage versus Collector current.

So the resistor network biases the Base near this critical voltage, and the small voltage superimposed by the input signal (via the capacitor) is enough to start current flowing.

This can get very very complex, but the basic(sic) idea of biasing and adding an AC component is a good place to start (a text book will help a lot).

It is a Single supply Complimetary-Symetry PP Amp

Functions are shown on attached picture.

Note that upper Tr is PNP & the lower NPN.

Output is from Emmiters thru Cap

For detail read

CompliSymetri-push-pull-amplifier

No the upper transistor is an NPN and lower is PNP. The question is would one expect this to be in class A, B, AB, C, or D operation?

Yes Sorry for writing NPN & PNP incorrectly.

Class of amplifier will depend on biasing of transistors, Cross-over ponit selection & Pre-Amplifier class. Here it is single ended as requirements of the output stage is.

In that case a more sophisticated [in place of 3 series resistors 1K, 170R & 1K] BiasNW is needed.

Normally Class AB for Audio. NEC was the 1st to introduce it.

Thanks for correction.