Hello
What's the output voltage Vout in terms of the input voltage Vg in this attached circuit? My answer I'm not sure about is Vout = 2*Vg - 5
View attachment 86419
Thanks.
Nope.
Here's what I got:
(Vg-2.5)/5K=(2.5-Vx)/10K ; Vx= op-amp output.
Rearranging: Vx= 7.5-2*Vg
Vout=Vx-2.5=5-2*Vg
One way to check this, is to think about what the output would look like if then non-inverting input were grounded-then you'd have a simple inverting amp, so the output would have to be negative with respect to the input (not what your equation shows)
How would you assign the voltages to input terminals? Both have to be equal as I understand from the ideal OpAmp rules, v+ = v-
But which one equals which voltage? So both are 20 or 10 V?
View attachment 86454
The supply voltage is 5 V. But how would this affect the analysis?
That circuit is a degenerate case. You can't analyze that using an ideal opamp model.
I tried it with PSpice and it did not complain. The voltages on V+ V- and Vout are: 20V, 10V and -15V respectively.
How can I tell if it's a degenerate case?
Sounds fantastic. -15 volts output with a single +5 V supply.
-analoger-,
if you really expect a serious answer you should present a complete schematic with all supplies and signal sources (indicate dc or ac).
By the way - it looks surprising to see a voltage source connected to the opamp output. Your own invention?
I tried it with PSpice and it did not complain. The voltages on V+ V- and Vout are: 20V, 10V and -15V respectively.
How can I tell if it's a degenerate case?
~analoger~
Some further hints:
* The dc voltage of 2.5 volts at the noninv. input wiil be amplified by (1+10/5)=3 and, thus, appears at the output as +15 volts;
* If you like to suppress any dc voltage at he output simply use a capacitor.
Oooo, new math: 3*2.5=15!!!
Barry, thanks - but you should take into account that here in Germany we are short before sleeping time. Nevertheless, blame my pocket calculator.
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