# Ideal Op-Amp Voltage sign error

1. Oct 30, 2016

### CoolDude420

1. The problem statement, all variables and given/known data

Above is my attempt of the answer and the problem q.

In the solutions given to us, it has this line:

I understand that V- is 0V due to the characterisitcs of an op-amp. But I don't understand where is the - V0 coming from. Any ideas??

2. Relevant equations

3. The attempt at a solution

2. Oct 30, 2016

The minus terminal of this op-amp configuration is often referred to as a virtual ground. Essentially $V_+=V_- = 0 \, Volts$. Now do you see why $V_o$ is negative? $I_B$ is positive and there is a voltage drop through the resistor.

3. Oct 30, 2016

### CoolDude420

Yeah. V0 voltage drop is from output to the 0V point at the - input of op-amp. So current is actually flowing in the opposite direction to what I have.

4. Oct 30, 2016

One thing that you may find a puzzle in these op-amps is once the current makes it to the output terminal, where does it go from there? The voltage at the output is determined by the current flow in the feedback resistor, but you can put in any "load" resistor you want, and the output voltage doesn't change. (Here your load resistor is infinite, but you could make it $R_L=100 \, \Omega$ and you get the same output voltage.) For these op-amps, the current into the circuit doesn't look like it equates with what comes out of the circuit. And the answer is that any differences are taken up by what can be unequal flows of current on the $\pm \, 12 \, volt$ bias lines. $\\$ (Incidentally, your ground connection is incorrect and belongs at the "+" input. The standard op-amp has 3 ports: two input pins (pins 2 and 3) and one output (pin 6). The output voltage gets referenced to the + input pin. The standard op-amp does not contain two output ports.)

Last edited: Oct 30, 2016
5. Oct 30, 2016

### Staff: Mentor

No, you have it correct.

You marked current as flowing from L to R, so the voltage difference that gives rise to this current must
be VL – VR and then you use this with Ohm's Law.

Though it wouldn't matter if you had drawn the current arrow to point from R to L, the resultant
equation for Vout / Vin will turn out to be the same as what you obtain above. Try it!

6. Oct 30, 2016

### CoolDude420

One more thing.

Say we had a 5V source between the non inverting input of the op-amp and ground. Then by the ideal characteristics of the op-amp the voltage at the inverting output will be 5V as well. Now, my question is since voltages can't just be there, they obviously have to be across something. What is that 5V across? I mean I know its measured in relation to GND but where is the ground connection?

7. Oct 30, 2016

It could never occur without blowing out the op-amp or at least pinning the output to $\pm 12 \, volts$.. The 5V source would have a resistance R associated with it, and the current would be I=5V/R. $V_-$ would continue to be a virtual ground. The voltage drop would occur in the feedback resistor (just as in your problem above), and the maximum output voltage is limited to (approximately) the bias voltage of $\pm 12 \, volts$.

8. Nov 3, 2016

### rude man

You meant inverting INPUT, right? In this case the output voltage is 5V - 1K x 2 mA = 3V, referred to the ground of the noninverting 5V source.

9. Nov 5, 2016

### CWatters

I believe your problem is that you wrote..

IB = Vo/Rf

That is not consistent with the "definitions of +ve" current and voltage on your drawing. To be consistent with your definition (arrows and signs) for IB and Vo it should be..

IB = (V- - Vo)/Rf

Then
V- ≈ 0
so
IB = -Vo/Rf

10. Nov 5, 2016

### CWatters

That's only true if the op-amp is operating as a linear amplifier. I agree with Rudeman and think Charles is wrong.

V+ = 5V
V- = 5V
Vo = 5 - (2mA * 1k) = 3V

Appears to be valid and stable.

The 5V on the V+ terminal is across the 5V voltage source you added.
The 5V on the V- terminal is across the 2mA current source. It's also across the series combination of Rf and the op-amp output (eg Vrf + Vo = 5V)

Remember that an op-amp amplifies the difference between the V+ and V- input. So as long as both are referenced to the same ground it won't normally make a difference to the output voltage if that ground changes. Perhaps look up common mode rejection. This circuit is slightly unusual in that it has a current source input and that means the output does change.

Last edited: Nov 5, 2016
11. Nov 5, 2016

### CWatters

This is what I'm getting at..

12. Nov 5, 2016