# power for the load in push-pull amplifier

1. what is the power needed for the load resistor?
2. Is there any problem? and what might be the solution for it?

I really appreciate you help, thank you!

• Books have been written about this subject, maybe you should read one of those ? Google: "audio ampifier design book" and there are plenty of links. – Bimpelrekkie Oct 17 '16 at 21:00

2) The circuit in your other question is the improved version of the one in this question.

Let's take it as reference:

If you look at the circuit carefully, voltage gain is 1. But since the output impedance seen from the common-emitter point is about a few miliOhms, this circuit can act as a power amplifier.

1) The power needed for the load depends on the input, of course. But be careful, VCC and VEE are limiting values for the output signal before clipping. Applying maximum 8Vpp = 4Vp = 2.83Vrms as input can be a good choice.

Suppose we apply 8Vpp = 2.83Vrms as input. So the output signal at the common-emitter point will be Vin = 8Vpp = 2.83Vrms. Since R9 and RL form a voltage divider, the voltage across the load will be 2.83 x 100 / 112 = ~2.5Vrms. So the power dissipated by the load will be 2.5²/100 = ~63mW.

• so we have 8vpp at the output because the amplification of this circuit is 1 right? – Hilton Khadka Oct 18 '16 at 13:26
• @HiltonKhadka Yes, voltage gain is 1 and if you apply 8Vpp input then you'll have 8Vpp at the output. But note that the limiting input values are determined by VCC and VEE. Of course you can apply VCC - VEE = 5 - (-5) = 10Vpp, but beyond this value, the higher the input, the more clipping at the output. – Rohat Kılıç Oct 18 '16 at 14:15

If you take a 9 volt battery with 2 * 1K(or more of equal value) resistors connected in series and have the center tap between the two resistors as your ground with the +5v connected to the +9v and the -5v connected to the -9v. Enjoy your music!