This was a challenge problem that my instructor gave. Here is what a class AB amplifier looks like:

Class AB Amplifier

I've looked everywhere to find out about internal impedance but I got nothing. This is the problem I'm trying to solve:

Determine the maximum peak to peak signal that can be applied to a amplifier and the efficiency rating or nstage for the amp. Assume the amp is driven by a sine source and has an internal impedance of 600 ohms.


  • \$\text{hfe} = \text{hFE} = 100\$
  • \$V_{be} = 0.7\$
  • \$E_s = 32.093~V_{p-p}\$ at 1 kHz
  • \$R_L = 8~\Omega\$
  • \$R_1 = R_2 = 1~k\Omega\$

I've tried all that I've been taught so far. I took the internal impedance as \$Z_{input}\$. Then, I used \$Z_{input}\$ and \$E_s\$ to find \$V_{in}\$, since \$V_{in} = V_{out}\$ is the voltage at the load. Finally, I used that answer to find the efficiency as \$P_L / P_{dc} \cdot 100 \$.

I calculated an efficiency over 200%, which must be wrong. What am I doing wrong?

  • \$\begingroup\$ You will notice that if you put some more care in writing the question it will receive much better feedback. I've edited it as an example. \$\endgroup\$
    – clabacchio
    Commented Oct 6, 2014 at 12:56
  • \$\begingroup\$ Sorry I was in a rush for class an this is my first post so forgive me for my negligence but your advice is well noted. \$\endgroup\$
    – Deathkamp
    Commented Oct 6, 2014 at 13:00
  • \$\begingroup\$ I've only done a few push pulls, but those caps on the input bother me. Will this work? \$\endgroup\$ Commented Oct 6, 2014 at 13:35
  • \$\begingroup\$ Sorry I took a look at my circuit and u r right there is no caps just what you see above since I changed it. Inputs signal connects directly to circuit with no Rx resistor but I still don't understand how an internal impedence affects the vload or how much voltage the amp can be put through \$\endgroup\$
    – Deathkamp
    Commented Oct 6, 2014 at 14:58
  • \$\begingroup\$ What is Es, and why is it specified to 5 significant digits? It seems it might be the input voltage, but that doesn't make sense since that is what you are supposed to solve for. \$\endgroup\$ Commented Oct 6, 2014 at 15:05

1 Answer 1


Finding the peak input voltage that still results in linear output drive is not too hard.

Think of how either Rl is projected back to the base of the transistor, or how Rb is projected to the emitter. You should be able to do that knowing the transistor gain. After that, it's a voltage divider to find what either the drop on Rl or Rb are. Remember that the B-E junction takes away a little from the overal available voltage. Then accounting for the 700 mV B-E drop, you get one from the other.

Once you know the maximum base voltage, you note that everything above applies to the other side, and the P-P voltage is twice the maximum.

  • \$\begingroup\$ I dont understand what you mean by projected back, and then you say transistor gain? what gain? power, voltage or current?Please be more specific, on what i exactly have to do because I am all out of options? \$\endgroup\$
    – Deathkamp
    Commented Oct 6, 2014 at 21:22
  • \$\begingroup\$ @Death: No, I'm not going to do your homework for you. It should be obvious what "Gain" of a bipolar transistor means. If there is 1 kOhm on the emitter of a transistor with suitable collector voltage, what is the apparent impedance when driving the base? That's what I mean by a resistance projected from emitter to base. This would have been gone over in your class. If not, you should be able to derive it yourself. It's really very straight forward. \$\endgroup\$ Commented Oct 6, 2014 at 22:52
  • \$\begingroup\$ We did not go over this it's a challenge problem that's why I am here not hw. I don't understand how impedences affect the peak to peak of a transistor when the vp to peak is given? So I am to find what is already given?thats why I am confused because I don't know what effect a internal impedance has on the vp peak to peak input. Can u walk me through the steps I will calculate just tell me what to do and why. \$\endgroup\$
    – Deathkamp
    Commented Oct 7, 2014 at 3:07
  • \$\begingroup\$ Ohh I think I know how to do it based on ur previous post. If I find the base voltage using the Ib formula vee-vbe/rb+hfe*re I could double that to get vb mAx an use av or voltage gain to estimate how much can actually be loaded? I think. \$\endgroup\$
    – Deathkamp
    Commented Oct 7, 2014 at 3:17

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