# Op Amp Circuit problem

1. Feb 13, 2013

### gl0ck

2. Relevant equations
I think I figured out the 1st question. The result I got is Vout=2*Vin+2V.
Which means twice as big wave as the Vin and shifted with 2V.
Is this enough or I have to write equations like 2nd -> Vout = 2*Vin+2V?
or
2) 2kV/s is the same as 2V/ms, or 20V in 10ms

3) Vout “clips” at ±8VIn fact it would be rather worse than this, as the output would not quite make it to the ±8V supplies.

4) Short-circuit across R1 means that negative feedback has gone and the Op-Amp is simply comparing its input to 0V, multiplying the difference by the raw Op-Amp gain (very large) and thus saturating Vout at the (positive) power supply of +15V

5) Short-circuit across R2 makes the circuit into a unity-gain buffer with R1 as its load resistor, so Vout = Vin

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Last edited: Feb 13, 2013
2. Feb 13, 2013

### rude man

see above in red

3. Feb 13, 2013

### gl0ck

shows a non-inverting Op-Amp circuit. *Its input waveform, Vin is a square wave with minimum and maximum values of 0V and +5V respectively. *R1 and R2 are both 1kΩ resistors. *For all of this assignment, apart from (3) below, assume that the Op-Amp’s power supplies are ± 15V.
The resistors values are 1kOhm
and the power supply is +-15V

4. Feb 14, 2013

### rude man

I think I figured out the 1st question. The result I got is Vout=2*Vin+2V.
Which means twice as big wave as the Vin and shifted with 2V.
Is this enough or I have to write equations like 2nd -> Vout = 2*Vin+2V?
or
That depends on whether the bias current flows into the + or - input. If into the + there is no effect. If into the - input your output expression is correct.

2) 2kV/s is the same as 2V/ms, or 20V in 10ms
So draw the input and output waveforms.

3) Vout “clips” at ±8VIn fact it would be rather worse than this, as the output would not quite make it to the ±8V supplies.
Correct.

4) Short-circuit across R1 means that negative feedback has gone and the Op-Amp is simply comparing its input to 0V, multiplying the difference by the raw Op-Amp gain (very large) and thus saturating Vout at the (positive) power supply of +15V

With an ideal op amp and the input going from exactly 0 to +5V, the output will obviously be hard-over +15V when the input is at +5V. But what about when the input is at 0V?

5) Short-circuit across R2 makes the circuit into a unity-gain buffer with R1 as its load resistor, so Vout = Vin
That is correct.