The lab I am working on is about ideal and practical op amps. We started off wiring up an ideal op amp for just one step and measured the output voltage before setting up the practical op amp. In the lab report I am suppose to discuss what I observed when doing this and explain why the ideal op amp integrator is not a practical circuit. It then says to note the amount of feedback that exists in the ideal integrator. Then it asks for the importance of adding a feedback resistor in parallel with the capacitor creating a practical ideal integrator.

I know the difference between the ideal and practical, but I guess I just don't understand what I am looking at. The voltage output for the ideal was 11.5V and with adding the feedback resistor which is double the input resistor I saw 13.1mV. Can someone help me realize what went on. Keep in mind this is my second circuits class and I haven't taken a circuits and differential equation class for about a year so things might need to be broke down into baby steps.
Thanks.

 

Where did you find an ideal op amp to wire up in a lab course?

 

I guess we do not have one we just use the lm741 and wire it up as seen in figure 1 and call it ideal? Cause there is no real life ideal op amp right?

 

It is likely that the ideal/non-ideal label is not referring to the op-amp itself but rather to the circuit design which assumes what the op-amp that is used will be. So using a 741 in a circuit that was meant for an ideal op-amp can lead to some really bad performance results.

 

I think that is what my professor is doing, which makes sense in showing that solving it with the assumptions of an ideal op amp on paper doesn't work in real life.

 

Under what conditions did you measure the output? It's an integrator, so you expect it's output to be a function of the history of its input.

 

I measured the output with an oscilliscope connected to ground and the output. The op amp is biased at +/-12V, C=1nF, R1=100Kohms, R2=200kohms. Or are you asking for the equation?:
Vout(t)=-(1/(RC)) integral[Vin(t)dt - Vc(0)] from 0 to t (this is for the ideal op amp)

 

The voltage output for the ideal was 11.5V and with adding the feedback resistor which is double the input resistor I saw 13.1mV. Can someone help me realize what went on

Adding a feedback resistor the circuit turns into a inverting amplifier, so the output falls down to GND (0 V).

Note that the second circuit is also a low-pass active filter.

 

The lab I am working on is about ideal and practical op amps. We started off wiring up an ideal op amp for just one step and measured the output voltage before setting up the practical op amp. In the lab report I am suppose to discuss what I observed when doing this and explain why the ideal op amp integrator is not a practical circuit. It then says to note the amount of feedback that exists in the ideal integrator. Then it asks for the importance of adding a feedback resistor in parallel with the capacitor creating a practical ideal integrator.

I know the difference between the ideal and practical, but I guess I just don't understand what I am looking at. The voltage output for the ideal was 11.5V and with adding the feedback resistor which is double the input resistor I saw 13.1mV. Can someone help me realize what went on. Keep in mind this is my second circuits class and I haven't taken a circuits and differential equation class for about a year so things might need to be broke down into baby steps.
Thanks.
View attachment 99553 View attachment 99554

By Ideal versus Practical I take it you refer to one with just the cap as feedback versus one with a resistor?
Were the voltages you refer to DC or AC, If AC what frequency?

 

It is likely that the ideal/non-ideal label is not referring to the op-amp itself but rather to the circuit design which assumes what the op-amp that is used will be. So using a 741 in a circuit that was meant for an ideal op-amp can lead to some really bad performance results.

< >Dang. Will we ever get away from that LM741 in schools? </ > It is still a good example of a bad example in op amps. The "Acme product" everybody wants to compare their product to for "lower offset ... less noise ... etc.
.

 

By Ideal versus Practical I take it you refer to one with just the cap as feedback versus one with a resistor?
Were the voltages you refer to DC or AC, If AC what frequency?

In this part of the lab we measured them in DC.

 

In this part of the lab we measured them in DC.

So what was the DC input voltage? 0V? 10 V? What?

 

Can someone help me realize what went o

The ideal integrator is likely oscillating. The integration capacitor shorts the opamp to ground and most opamps, especially a 741, would not be able in that case.

The ideal of paralleling the capacitor with a large resistor is to better define the upper end of the gain. However, in this case your choice of that resistor has too low of a value. Use a large value resistor instead.

 

BTW, when the circuit is working, its output will change linearly with time. Either increase or decrease depending on the flow of current.

 

So what was the DC input voltage? 0V? 10 V? What?

+/-12V

 

I would venture a guess that the input voltage is quite low, in the 13mv range.

 

In this part of the lab we measured them in DC.

The capacitor is an open circuit at DC. Maximum gain in the op amp and max voltage out. This the difference with the resistor across the cap.