# Transistor question

Discussion in 'Homework Help' started by chaosdestro0, May 1, 2011.

1. ### chaosdestro0 Thread Starter New Member

Apr 30, 2011
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I am having difficulty understanding how you would find the voltage across R2.
Let's say I had this circuit, with Vout as 4.4v and a voltage drop across the diode at 2v. From the mark scheme, it says do 9-2-4.4. I am not sure why this does this, can anyone explain?

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Do you know II Kirchhoff's law (KVL)??
For your circuit KVL must hold

Vcc = Vout + VR + Vd = 9V
VR = Vcc - Vd - Vout

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3. ### chaosdestro0 Thread Starter New Member

Apr 30, 2011
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Wait, so why does the voltage from the v out of the transistor not add to the 9v?
I have done work on it but without additional sources.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well, I don't understand you question.

5. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Because they are of opposite polarity. Look at the arrows drawn for you.

If you put two 1.5 volt batteries in series, same polarity, they add, you see 3V across both.

If you put two 1.5 volt batteries in series, but opposite polarity, they subtract, and you see about 0V across both.

6. ### jegues Well-Known Member

Sep 13, 2010
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If you are confused about the arrows, it basically indicates how one would attach a voltmeter to the circuit.

See the figure attached.

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7. ### chaosdestro0 Thread Starter New Member

Apr 30, 2011
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Now I see it, Cheers .