Operational amplifier

1. Sep 20, 2011

Femme_physics

Last edited by a moderator: May 5, 2017
2. Sep 20, 2011

I like Serena

Hi Fp!

You'll be happy to know that I'm back in my regular office, yep!

The resistors can be treated as being in series.
You have that right.

However, your formula for Vout makes no sense to me.
An operational amplifier has a different formula for Vout.

It's Vout = (V+ - V-) AOL.
Where the "open-loop gain" AOL is typically about 10000.
Furthermore Vout can not be higher or lower than the voltage applied to the operational amplifier (in your problem that is given as +15 V respectively -15 V, but this is not drawn in your diagram).

Can you tell me what the voltages at the + and at the - of the operational amplifier are?

Last edited: Sep 20, 2011
3. Sep 20, 2011

Femme_physics

Glad you're back home safely Missed it?

You say "typically", but how do we find its actual value?

Noted

+15
-15

4. Sep 20, 2011

I like Serena

Not really, but it is good to be moving again.
I do feel a lot better, having had a lot of rest, exercise and sun.
And of course, having finished a few loose ends.

You seem to be back in the saddle too?

Uhh... :uhh:
I don't know.

But wait! We don't need it for this problem.

Noooooooo.
Did you really think it would be that simple?

Here's the full schematic of an operational amplifier

In your case Vs+ is +15 V and Vs- is -15 V.
These form the power supply of the operational amplifier.

You have yet to determine V+ and V-, before you can calculate Vout.
You should do that for instance with KVL through the resistors...
You should assume that no current is drawn by the operational amplifier (other than from its power supply).

Last edited: Sep 20, 2011
5. Nov 27, 2011

Femme_physics

So, I think I got it now that I know the formulas.

So,

Vout = Vin (R2/R1+R2 )

Vout = 12 (R2/R1+R2)

Hmmmm........now how can I solve it without being give the value of R?

6. Nov 27, 2011

I like Serena

Where did you get the formulas?

Perhaps you can calculate V+ and V-?

7. Nov 27, 2011

Quinzio

These formulas may be for some basic circuits with Opamps inside.
Forget them for a while.
An OpAmp is a mathematical machine which accomplishes this : $Vout = (v^+ - v^-) G$ where G is the gain and it tends to infinity for an ideal OPAmp.
All you have to know is that.
Translate all the circuits into formulas and then play just with numbers.

In your case you have the annoyance of the Power supply bounds. It translates in
$Vout = (v^+ - v^-) G$ like before
then $Vout'= min(15,max(-15,Vout))$ that's no mind twister, it just says that Vout cannot pass the bounds.
Forget the bounds, solve the circuit, apply the bounds.

8. Nov 28, 2011

Femme_physics

Voltage Divider!

Ah, I guess I see what was wrong with my formulas now. It really should be:

V+ = Vin (R2/R1+R2 )

Whereas Vin = 12V

That's what am trying to do above :)

9. Nov 28, 2011

Staff: Mentor

Suppose for a moment that you cut the wires to the op-amp inputs. You're left with a voltage divider consisting of three equal-values resistors connected to 12V at one end and ground at the other. Call the nodes where the resistors connect Va and Vb. You should be able to tell by inspection whether Va will be greater than, less than, or equal to Vb.

You can even say that the amount by which Va and Vb differ doesn't matter! ANY difference in voltage between the op-amp inputs will drive the op-amp output to either the positive supply level or the negative supply level, depending upon whether Va > Vb or Va < Vb.

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10. Nov 28, 2011

Femme_physics

Last edited by a moderator: May 5, 2017
11. Nov 28, 2011

Staff: Mentor

Yes, no current ever flows into (or out of) an ideal op-amp's input leads. So the only path for current to flow is through the external resistors.

12. Jul 16, 2012

Femme_physics

*Necromantic bump!*

Just for verification, the answer here is "true, the output is -15 volts since V- > V+"

13. Jul 16, 2012

I like Serena

Ooooh.
You necromancer you!

Yes. That's right.

14. Jul 16, 2012

:) Thanks