Operational Amplifier Output Voltage with Equal Resistors

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Homework Help Overview

The discussion revolves around the output voltage of an operational amplifier (op-amp) circuit with equal resistors and given supply voltages of +15V and -15V. The original poster questions whether the output voltage can be -15V under these conditions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the output voltage and the resistor values, questioning the assumptions made about the circuit configuration. Some participants discuss the formulas for calculating output voltage and the implications of the op-amp's gain.

Discussion Status

The discussion is active, with participants providing insights into the behavior of the op-amp and the significance of the resistor values. There is an ongoing exploration of how to determine the voltages at the op-amp inputs and the effects of the power supply limits on the output voltage.

Contextual Notes

Participants note the importance of understanding the voltage divider formed by the resistors and the implications of cutting the wires to the op-amp inputs. There is also mention of the ideal op-amp characteristics, such as infinite input resistance and the assumption of no current flowing into the op-amp inputs.

Femme_physics
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Hi Fp! :smile:

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


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

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

Glad you're back home safely :smile: Missed it?

It's Vout = (V+ - V-) AOL.
Where the "open-loop gain" AOL is typically about 10000.

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

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).

Noted :smile:

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

+15
-15
 
Femme_physics said:
Glad you're back home safely :smile: Missed it?

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. :wink:

You seem to be back in the saddle too? :cool:


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

Uhh... :rolleyes:
I don't know. :confused:

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


Femme_physics said:
+15
-15

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


Here's the full schematic of an operational amplifier
150px-Op-amp_symbol.svg.png


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).
 
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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?
 
Femme_physics said:
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?

Where did you get the formulas?

Perhaps you can calculate V+ and V-?
 
Femme_physics said:
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?

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.
 
Where did you get the formulas?

Voltage Divider!

Perhaps you can calculate V+ and V-?

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

V+ = Vin (R2/R1+R2 )

Whereas Vin = 12V

Forget the bounds, solve the circuit, apply the bounds.

That's what am trying to do above :)
 
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.

attachment.php?attachmentid=41336&stc=1&d=1322501538.jpg


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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  • #11
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
*Necromantic bump!*

Just for verification, the answer here is "true, the output is -15 volts since V- > V+"
 
  • #13
Ooooh.
You necromancer you!

Yes. That's right.
 
  • #14
:) Thanks
 

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