quick check

Discussion in 'Homework Help' started by ihaveaquestion, May 11, 2009.

1. ihaveaquestion Thread Starter Active Member

May 1, 2009
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figred out

Last edited: May 12, 2009
2. t_n_k AAC Fanatic!

Mar 6, 2009
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I would take a different approach - provided you have discussed op amp theory in class.

Top LH amplifier is a x2 gain inverting stage
Bottom LH amplifier is a x3 gain non-inverting amplifer
The RH amplifier is a summing / inverting stage with gain of x2.

Anyway whatever your method, I have Vo=-2V

3. t_n_k AAC Fanatic!

Mar 6, 2009
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Apologies - Vo=+2V

4. ihaveaquestion Thread Starter Active Member

May 1, 2009
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I don't get 2..

5. t_n_k AAC Fanatic!

Mar 6, 2009
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This is where you went wrong.

Vout2 = 1V+(1/20k)*40k

6. ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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nope got it

Last edited: May 13, 2009
7. ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
got it cool

Last edited: May 13, 2009
8. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
For the amp (lower left hand) in question....

The voltage at the negative terminal is 1V which we are agreed on.

The current in the 20k from neg. terminal to ground is 50uA towards ground. Given this is an ideal op-amp, the same current must flow in the 40k feedback path - from the output terminal to the negative terminal. So there is a voltage drop of 2V across this resistor with the polarity being additive to the voltage at the neg. terminal. So for this amp, Vo = 1V + 2V = 3V.