- #1

andthenwecan

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## Homework Statement

http://img846.imageshack.us/img846/955/29278644.jpg [Broken]

## The Attempt at a Solution

Basically I want to know whether my working method was correct as I have doubts with some of it.

I first did voltage division at the two base resistances, and I get 3V at the base. Since we know there's a base-emitter drop of 0.7V, the emitter voltage of Q1 is 2.3V which is also the base voltage of Q2, therefore the emitter voltage of Q2 is 1.6V.

This means we can calculate the emitter current of Q2 easily. After obtaining the emitter current of Q2, using the Ie = Ib + Ic relation and substituting for Ib = Ic/B, we can easily getting the collector current

*ic2*.

http://img542.imageshack.us/img542/8632/80235300.jpg [Broken]

Now here is where I'm a bit confused as to how to proceed and get

*ic1*

I thought I can get the current from Vcc to Vb, let's call it

*ix*, going back to the Vcc node, can I use that as a KCL point? So basically if I have

*ix*leaving that node, and let's call the other current

*il*which is going through Rc, those two must add up to give 0.

Once I have

*il*, then il = ic1 + ic2 and therefore I can solve for ic1 and hence also find the voltage at Vc which is the DC voltage they asked for.

http://img6.imageshack.us/img6/664/12595537.jpg [Broken]

Thanks.

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