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.

 

[rezeew]

Great job solving the first question! It is always good to show your work and equations to fully explain your solution. For the second question, yes, it would be helpful to also write out the equation to show how you arrived at your answer. For the remaining questions, your explanations are correct and show a good understanding of the concepts involved in op amp circuits. Keep up the good work!