# VCCS Analysis

Discussion in 'Homework Help' started by tAllann, Oct 27, 2013.

1. ### tAllann Thread Starter New Member

Oct 26, 2013
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0

I solved for Vo in terms of Vi and I get the negative of the correct answer according to the book. Someone told me that the TA set the node to the left of Vo = to -Vo, and this explains why my answer is the negative, but I don't understand why the node adjacent to Vo is -Vo instead of just Vo.

In this problem, gm = 2m / 1 ohm, and I get that Vo = -2 Vi, but the book gets that V0 = 2 Vi.

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2. ### WBahn Moderator

Mar 31, 2012
23,847
7,378
gm = 2m/Ω = 2mA/V

Vo = -(gm*Vx)*2kΩ

Vx = Vi/2

Vo = -((2m/Ω)*(Vi/2))*2kΩ

Vo = -2Vi

You're right, they're wrong.

I have NO idea why they are putting Vo on the top-right node of the left hand circuit (that's what you are talking about, right?).

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3. ### LvW Well-Known Member

Jun 13, 2013
779
105
tAllan,
the circuit shown equals the classical equivalent small-signal ac circuit diagram for a common emitter transistor amplifier. As you will know, this amplifier has inverting gain characteristics.
Thus, for a positive input voltage, the output is negative, indeed.

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4. ### tAllann Thread Starter New Member

Oct 26, 2013
11
0
But can't you solve the circuit without knowing this with simple analysis? Is my analysis wrong?

5. ### WBahn Moderator

Mar 31, 2012
23,847
7,378
Of course you can solve it without relying on this. You did. I did.

He is just pointing out that this is a circuit that happens to match a common circuit that you will run into and, if you DID happen to know that, that it confirms that the output voltage is inverted relative to the input voltage.