hello

I have an assignment that requires testing an inverting op amp and finding the bandwidth, then changing the resistor value and again testing. the second setup should show that the output produced is being clipped as a result of being out of bandwidth.

I have to simulate an amplifier and determine the voltage gain
measure and record the bandwidth of the amp
measure and record the spectral content of the output signal
increase the input voltage amplitude and repeat the above two questions
comment on the results
change resistor value to give a gain of -50 and repeat above

then, I have to construct the circuit on breadboard
measure and record the gain and bandwidth, draw frequency response curve
change the value of the feedback resistor to give a gain of -100
measure and record the gain and bandwidth of the amp, draw the frequency response curve.
comment on the results
compare the results from the simulation to the real world circuit.

I draw the response curve by taking voltage values at the output over a range of increasing frequencies. then put those values against frequency into excel and plot a graph. it should look like a bode plot for a filter circuit.

im stuck on the working of the circuit. I know what gain is and how to find the resistors to get the gain. I know that with no resistor the gain will be really high and so the bandwidth will be very narrow and the amp will be unstable. so bringing the gain down would widen the bandwidth.

my lecturer talks a little too fast for me and I feel like im an idiot for not getting everything he says. so I just agree and try to find the answer myself. I cant seem to find the answer on internet though.

I have a power supply connected to the op amp. +15 and -15, so the maximum output would be 30vpk-pk. if the input is 10 volts and the output is 30volts then there is a gain of 3. I know the output would be inverted because a positive value is going into the negative input.
if the gain is -100, then the input of 10volts times -100 is -1000. I don't know what this will do to the output as such, does it only give out 30vpk-pk.

I have just read some stuff about gain/bandwidth product. the 3dB drop off point is the bandwidth, this multiplied by the gain. I have calculated that the frequency for the 3dB drop is 1.6kHz. so this 1.6*-100 should give -160,000? but this is frequency or voltage. I don't think it is voltage. so 160kHz is the gain/bandwidth product. 1.6kHz is the 3dB drop off point. so at this point there should be no voltage coming out the op amp. the 10 volts input should remain the same throughout. the output should be whatever the gain times the input is, so a gain of 10 should give 100 volts output? but 30 volts is maximum? because of the +15 and -15. though there is no clipping of the waveform until the gain goes to -100?

 

Perhaps I am missing something. But, in my limited experience, horizontal clipping occurs when you hit the maximum output voltage. Meaning that you amplified the input as far as amplifier setup allows. You can even use this effect to change sine wave into square wave.

Bandwidth deals with frequency and I am not sure how the sort of clipping I described above has relationship with bandwidth.

 

I have a power supply connected to the op amp. +15 and -15, so the maximum output would be 30vpk-pk. if the input is 10 volts and the output is 30volts then there is a gain of 3. I know the output would be inverted because a positive value is going into the negative input.
if the gain is -100, then the input of 10volts times -100 is -1000. I don't know what this will do to the output as such, does it only give out 30vpk-pk.

You are correct. Your output will be clipped at the -15 volt level. Like the picture here:

To be even more precise on the subject, you will never reach -15 volt, which means you will never have 30 volt peak to peak. You are using +15 and -15 volt to power the op amp, so +15 and -15 volt goes in. However, some of this voltage is consumed by the op amp itself. Meaning it is not available to amplify the signal. If you have a poorly designed/constructed op amp, you may waste as much as 4-5 volts of those 15 volts. Which means the highest output you can get is +11/-11 volts and your spread is 22 volts peak-to-peak. This is just an example. Since I don't know the model of op amp you will use, I can not look the exact information from it's datasheet.