Op-Amp Config Question

I work out -0.767V through to -0.177V depending on the wiper position.

It's a long time since I did any homework -- do let me know what my score is when you get it marked...

This sounds like a homework problem. If that is the case, tell us what you have done so far to obtain the answer. In real life, gurus will charge at least $100/hr. To be a guru just for you. I am not willing to do that for free. For free, I might give you a small amount of guidance. Doing it for yourself is good for the soul. Clue: What is an ideal op-amp supposed to do?

Bill

-- Ferme le Bush

I can assure you this is not a homework problem. It's from a circuit that I'm trying to understand. In the past I've asked similar questions and have had a few come at me with saying I post homework problems, if it were one, it would be noted that I need assistance with homework.

I do attend school part-time at night, but I can assure everyone I'm trying to understand the math behind it. So it looks like the answers are correct (according to MultiSim), but how do I calculate Vout??

thanks

The inverting input will be at a 'virtual ground'. So with 100nA input, there must be 100nA output through the 1Mohm resistor. So if the one end of the 1Mohm resistor is at virtual ground, what must the voltage on the opposite end of the 1Mohm resistor be to get 100nA flow?

Once the voltage required on the 'other' end of the 1Mohm resistor is known, you can calculate what the op-amp output voltage must be to get that voltage on the wiper.

daestrom

Righto! The key concept is that of virtual ground. Even without a real ground, the potential difference between the two inputs is going to be zero.

Bill

-- Ferme le Bush

This is a trick question. Ideal op-amps have infinite input impedance and so can not have 100 nA flowing into the inverting input.

Bill

The input impedance is provided by the feedback resistor. See daestrom's reply, which is exactly how I worked out the figures I posted earlier (which I presume are the figures Peter says agree with MultiSim, as I don't see anyone else posted any). I didn't bother taking the 100nA into account in the potential divider, as it's 2 orders of magnitude less current than the potential divider will be grounding.

I figured out one thing I was doing wrong. In MultiSim it doesn't show the position of the pot. so I was using the wrong resistance to calculate the current because I couldn't see where the position of the pot was at.

My original thought was Vout would be equal to multiplying the current source by the 1Mohm resistor plus the value of the wherever the position of the pot was plus the 10k resistor (ignoring the 3k reistor to ground).

As I understand now, it works a little different than I thought.

So how would you calculate the overall gain of this circuit because you have a current source and not a voltage on the input. So you can't use G=Vout/Vin.

My guess would be the gain is Vout divided by the 100mV at one end of the

Not really...

True.

Not true. It merely means that there must be 100 nA also flowing away from the inverting input (through the 1Mohm feedback resistor).

Perhaps the phrase, 'flowing into the inverting input' is not quite accurate. From the input, it is flowing towards the inverting input, and being 'drained off' by the feedback resistor. Tomayto, tomahto...

daestrom

Well, you don't really have a simple amplifier, what this circuit is, is a nI to V converter. If we knew the source impedance and the final load impedance, we could figure out the power gain, but that's about all.

daestrom