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My professor mentioned it briefly in class, and I've been googling it and all I find is the CC output stage. Can anyone show me a picture of this configuration and maybe describe how it works? Can you provide a source?

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  • \$\begingroup\$ Do you know what a common-emitter circuit is? Do you know what a push-pull circuit looks like? I suggest that you Google both terms and look not at the web sites but the images tab. You will see many circuits. The circuit you're after is a combination of two CE stages into a push-pull stage. Also next time, as the professor for an example circuit. It takes only a minute to draw one. \$\endgroup\$ – Bimpelrekkie Jan 18 '18 at 11:11
  • \$\begingroup\$ I searched images but only got either cc ones or complicated ones. I tried to draw it. But what I have drawn doesn't seem to work. \$\endgroup\$ – Myla Izaman Jan 18 '18 at 11:17
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    \$\begingroup\$ Edit your question, click the circuit diagram editor button, and draw it there. Then we can see where you're at. \$\endgroup\$ – Dampmaskin Jan 18 '18 at 11:20
  • \$\begingroup\$ @MylaIzaman The common-emitter topology for high power requires thermal sensing of the output transistors, of COURSE you only find complicated ones. \$\endgroup\$ – Whit3rd Jan 19 '18 at 9:09
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A BJT (NPN in this example) connected in common emitter configuration looks like this:

Putting two of these with opposite polarity together is a push-pull stage:

The name comes from the fact that the output is actively pushed and pulled in both polarities.

There are a great many details about how to drive the two transistors so that they each turn on when needed, but not both turn on to fight each other. Just getting the DC quiescent operating point right isn't trivial. This is sometimes called biasing. That would be good to look up.

Also look up class A amplifier, class B amplifier, and class AB amplifier. These classes have to do with the overall strategy of when each of the transistors are intended to conduct. The classes are a tradeoff between smooth transitions and wasted power.

Those details are beyond your simple question, but this should give you things to go look into.

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  • \$\begingroup\$ I'm pretty sure you have the transistors mixed up in the second picture, it'd be pretty tricky to bias that to get them to work in their respective linear regions \$\endgroup\$ – cactus1 Sep 7 '18 at 20:43
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    \$\begingroup\$ @cact: No, that's a PNP/NPN push/pull pair. Yes, it is tricky to bias, as I mentioned in the answer. \$\endgroup\$ – Olin Lathrop Sep 8 '18 at 17:03

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