Operational Amplifier Output Voltage with Equal Resistors

[Femme_physics]

Homework Statement

http://img51.imageshack.us/img51/1912/vout.jpg
Given the supply voltages of the operational amplifier are (plus and minus)15V, as well it is given that R1 = R2 = R3. We get that Vout is -15V

True or false?

Explain.

The Attempt at a Solution

False because it depends on the value of R.
http://img535.imageshack.us/img535/9167/falsei.jpg

 

[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 V_out makes no sense to me.
An operational amplifier has a different formula for V_out.

It's V_out = (V₊ - V_-) A_OL.
Where the "open-loop gain" A_OL is typically about 10000.
Furthermore V_out 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?

 

[Femme_physics]

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

Glad you're back home safely Missed it?

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

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

Furthermore V_out 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

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

+15
-15

 

[I like Serena]

Glad you're back home safely 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.

You seem to be back in the saddle too?

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

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

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

+15
-15

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


Here's the full schematic of an operational amplifier

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

You have yet to determine V₊ and V_-, before you can calculate V_out.
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).

 

[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?

 

[I like Serena]

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-?

 

[Quinzio]

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.

 

[Femme_physics]

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 :)

 

[gneill]

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.

 

[Femme_physics]

Oh, I think I understand. Since an (ideal) Op-Amp Rin -> Infinity , the current will go directly to the ground point and we're not bothering with the transistor in this case? Therefor the answer is false because the transistor is out of the picture.

Such as when current doesn't flow in R2:

http://img266.imageshack.us/img266/6051/such1.jpg

So this will be the runway in our case ->http://img193.imageshack.us/img193/430/such2.jpg

 

[gneill]

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.

 

[Femme_physics]

*Necromantic bump!*

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

 

[I like Serena]

Ooooh.
You necromancer you!

Yes. That's right.

 

[Femme_physics]

:) Thanks