Ideal Op-Amp Voltage sign error

[CoolDude420]

Homework Statement

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

The correct answer is -2V.

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??

Homework Equations

The Attempt at a Solution

 

[Charles Link]

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.

 

[CoolDude420]

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.

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.

 

[Charles Link]

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.)

 

[NascentOxygen]

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.

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 V_L – V_R 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 V_out / V_in will turn out to be the same as what you obtain above. Try it!

 

[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?

 

[Charles Link]

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?

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 $\pm 12 \, volts$.

 

[rude man]

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.

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.

 

[CWatters]

I believe your problem is that you wrote..

I_B = V_o/R_f

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 I_B and V_o it should be..

I_B = (V_- - V_o)/R_f

Then
V_- ≈ 0
so
I_B = -V_o/R_f

 

[CWatters]

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.

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.

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?

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 V_rf + V_o = 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.

 

[CWatters]

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.

This is what I'm getting at..

 

[Charles Link]

This is what I'm getting at..

View attachment 108510

What @CWatters is saying is basically what I mentioned in post #4, that the op-amp is a 3 port device and the output voltage (pin 6) is referenced to the non-inverting input (pin 3). Pin 3 is essentially shared between input and output.