I'm having difficulty finding input resistance because for some reason there is no virtual ground across the inputs, so the current flowing through each is not the same. What am I doing wrong?

 

I am afraid you have forgotten to give the power supplies a value.

 

Wouldn't it be more difficult to isolate Vin if there is output voltage? I mean, it's still drawing current as there is a 1V source... Regardless, there is still no virtual ground because input currents are still different.

 

In general the input currents WILL be different. The difference is called the input offset current.

Take a look at this:

http://www.analog.com/static/imported-files/tutorials/MT-038.pdf

 

Hi geft,

it would be wise to reduce the gain or the input voltage - or to increase the power supply voltages.
Why?
Because with 1 Volt input signal and a gain of (1+14)=15 you operate the device outside of its linear amplification range.

 

Hi geft,

it would be wise to reduce the gain or the input voltage - or to increase the power supply voltages.
Why?
Because with 1 Volt input signal and a gain of (1+14)=15 you operate the device outside of its linear amplification range.

VCC and VEE were 15V and -15V originally, as can be seen by my previous post above. But I don't really see how the output is any way related to input resistance.

I'm trying to work out offset current to see if it's what's causing the reading to be off...

 

Can you tell me how you define this input resistance ?

 

I'd assume it to be the resistance across the input nodes of the op-amp. It should be simple to measure, but I must be doing something wrong since I'm not getting the correct number (about 2MΩ for the uA741).

 

Your approach is very wrong as other have said. In any real world situation this would not have been the way to go. Your circuit is an inverting standard opamp stage. As you have wired it up. The opamp will drive the negative input to a virtual ground. If you then connect a DC source to the other end of R1. What will the current in R1 be then In opamp circuits. The input impedance for the system, do not have to be the same as the opamp input impedance.

 

Try to use AC sweep analysis. And Rin = Vin/Iin = 1Meg.

 

The opamp will drive the negative input to a virtual ground. If you then connect a DC source to the other end of R1. What will the current in R1 be then

Does that mean I should just short the feedback loop and remove R1? Or short both inputs together?

In opamp circuits. The input impedance for the system, do not have to be the same as the opamp input impedance.

What do you mean? Which one should I be looking for?

Try to use AC sweep analysis. And Rin = Vin/Iin = 1Meg.

Thanks, but why does it work and mine doesn't? Shouldn't an op-amp work with DC as well?

 

Geft, in post#8 you have no feedback - in contrast to your circuit in post#1.
Of course, you can do both - find the input resistance with or without feedback.
But you should know what you really want!
In case of feedback it is important - as I have mentioned before - that the output is not in saturation.

 

Geft, in post#8 you have no feedback - in contrast to your circuit in post#1.

Sorry, I thought virtual ground only exists in op-amp circuits with feedback, so I assumed it's a given. The picture is something I took from Google to illustrate input resistance...

In case of feedback it is important - as I have mentioned before - that the output is not in saturation.

Ah, now I see what you meant. I didn't know this before. Does that mean 16V voltage rails will be sufficient?