Transistor amp question ---- not homework

Sep 05, 2005 5 Replies

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Peter 20 years ago

I've posted a circuit on my webpage for everyone to view since I don't think I should upload pictures here.

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I'm trying to understand how to find Vout. Last night I played around with loop equations, assumptions, and after hours of trying I had no luck.

The biggest part that is killing me is that 68ohm resistor up top because you can't just say 20 volts goes to the left part and the right part of the circuit. I know approx. what I should get because I plugged this into MultiSim, but that's cheating in my book, I should know how to calculate this circuit.

If anyone knows the best way to come up with the answer, please let me know.

Thanks in advance!

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magic 20 years ago

Could you skatch the diagram or schematics.

Tks

magic

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ehsjr 20 years ago

Here's how to "sketch" it in a post, and a hint at the bottom:

o +12 | [R1] 68 | +------------+ | | [R2] 8.25K [R5] 150 | | | e/ +-----------| Q1 2N3906 | \ | | [R3] 36.4K +---- Vout | | | / +-----------| Q2 2N3904 | e\ | | [R4] 8.25K [R6] 150 | | +------------+ | Gnd

Hint: You were concerned about the left hand side and the

68 ohm resistor. What is the lowest possible voltage, with respect to ground, at the bottom of the 68 ohm resistor? Can you analyze the circuit, if the 68 ohm resistor was replaced by a jumper wire?

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Peter 20 years ago

I believe it can be done. you pretty much ONLY have I1 on the left and I2 on the right because the emitter current of Q2 is the collector current of Q1 and the emitter current of Q1. So it's almost a parallel circuit.

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ehsjr 20 years ago

Ok, so what do you get (voltages) at the base of the transistors that way? Will B-E current flow? What will happen in terms of C-E current?

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Peter 20 years ago

ehsjr wrote in news:fDnTe.1204$d6.317@trndny05:

I think I need a few loop equations and not just pulling the 68ohm resistor out of the circuit to make assumptions. There should be a scientific approach to this. In any case, if I don't hear from someone soon, I'll pull the circuit off the webpage.

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