Engineering Solve Op Amp Circuit Problem Homework Equations

[gl0ck]

Homework 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

 

[rude man]

Homework 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
That depends on whether the bias current flows into the + or - input. If into the + there is no effect. If into the - there will be an output offset. Can't verify your answer since you didn't give us the values of R1 and R2.

2) 2kV/s is the same as 2V/ms, or 20V in 10ms
Impossible to answer (2) and (3) unless R1 and R2 are given values.

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.
If the input is

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

and what about negative outputs? Actually, with an ideal op amp and the input going from exactly 0 to +5V, that question cannot be answered! The output is equally likely to stay at +15V or switch between + and - 15V. (Why?). But you have the right idea here.


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.

see above in red

 

[gl0ck]

see above in red

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

 

[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.