# Op-amp true or false

1. ### Femme_physics

2,555
True or false:

In the following circuit no current flows through R since Vout equals Vin.

I say TRUE because it's the same as Voltage follower

My classmate disagrees with me, says it's false, saying "if vout passes through a resistor, then the voltage on that input is lower than vout"

Who's right?

### Staff: Mentor

It's an underdefined problem as stated.

Are you assuming an "ideal" opamp? Or one that has input bias currents? And you should probably label the + and - inputs of the opamp, although that wouldn't change the answer to the problem.

3. ### I like Serena

6,194
I believe you are right. :)

Btw, Vout does not "pass" through a resistor.
A "current" passes through a resistor, and only if the voltages on both sides of the resistor are different.
In this case those voltages are the same.

However, I am assuming the op-amp has its + and - inputs configured like a voltage follower.
If they are reversed, the circuit is unstable and Vout will either go up to Vcc+ or go down to Vcc-.
Either way, there will no (significant) current flowing through the resistor since the op-amp has a resistance that is near infinity.

EDIT:
What do you mean by "underdefined"?

4. ### Femme_physics

2,555
I didn't compose the questions, I didn't draw the image but took it directly from the source. I was confused about the lack of + and - inputs myself. And they didn't say if it's ideal or not, HOWEVER in our course IIRC we were told to assume all Op-Amps are ideal.

As I suspected! Thank you

So, if it's an inverter though, the it's no longer a Voltage follower, so I can't use the same Vout = Vin law. That part I understand, though I wouldn't be sure what to make of it if the inputs were indeed reversed.

5. ### I like Serena

6,194
In this case the circuit is not configured like the way an inverting amplifier or a voltage follower should be configured, so you should not use their behavior.

To figure out what the circuit does, you need to apply the basic op-amp law that says:
$$V_{out} = (V_+ - V_-) A_{OL}$$
where ##A_{OL}## is the so called open-loop-gain which is typically about 1000000 and where ##V_{out}## is limited to the range Vcc- to Vcc+.
From there you can try and "solve" the circuit.

After solving the circuit, you should consider what would happen if ##V_{in}## would rise just the tiniest fraction, say ##10^{-10} V##.
Either the circuit "pushes back" or it "diverges" to another solution where ##V_{out}## is either Vcc+ or Vcc-.

Last edited: Feb 29, 2012
6. ### Femme_physics

2,555
I'll make sure I have this printed for reference. Thanks a lot Klaas! :)

7. ### Femme_physics

2,555
I have another question for true and false.

For an ideal operative amplifier there's infinite resistance at the entrance and an infinite amplification in an open circuit

I answered True, because I know the first sentence is true, but what is it about infinite amplification in an open circuit? I thought amplification is limited to the formulas ( Vout = ...) and the parameters of the op-amp. I didn't think it can ever get "infinite".

8. ### I like Serena

6,194
Yes, an ideal op-amp does have infinite amplification.

The formula ##V_{out}=(V_+ - V_-) A_{OL}## gives the amplification.
In a real-life op-amp the amplification ##A_{OL}## is about 1000000.
In an ideal op-amp it is infinite.

9. ### rude man

5,664
An ideal op amp has zero offset voltage, zero offset current, zero bias currents, infinite gain, zero output impedance. So your answer is correct. Why? Well ...

Zero bias current → zero current thru R, therefore zero voltage across R, therefore output = - input.

And since + input = - input, output voltage = Vin. So it's a voltage follower.

Last edited: Mar 1, 2012

### Staff: Mentor

And zero cost!

11. ### I like Serena

6,194
Doesn't that depend on your point of view?
Since I'm producing ideal op-amps (virtually) I'm charging an infinite cost. ;)

12. ### rude man

5,664
Thass' right! I once had a prof who made up a table of ideal parameters and that was his entry too!