Finding Rf in op-amp circuit with Is=10mA and Vout=2V

[Metamorphose]

What is Rf --> Op_Amp

1. Consider the amplifier circuit shown. What value of Rf will yield vout = 2V when Is = 10 mA and Ry = 2Rx = 500Ω



2. The way I did this was by employing KCL: (Is that applicable?)

The Attempt at a Solution

Rx(Is) + Rf(Is) = Vout

I'm actually not quite sure how to do this one. It's left me in a bit of a pickle.

 

[gneill]

1. Consider the amplifier circuit shown. What value of Rf will yield vout = 2V when Is = 10 mA and Ry = 2Rx = 500Ω

I don't see how it's possible. The potential at the op-amp + input will be negative, and feedback should drive the - input to the same value. Thus Vout will be less than that (so long as only positive resistance values are permitted).

 

[NascentOxygen]

Perhaps someone has introduced a typo—and that output of 2V should actually be shown as –12V?

 

[CWatters]

As gneil said...

Should that be Vout = -2V or Is = -10mA ?

 

[rude man]

R_y is obviously a negative dc resistance!

 

[rude man]

As gneil said...

Should that be Vout = -2V or Is = -10mA ?

No combination of + or - 10mA and + or - 2V is possible.

 

[CWatters]

Humm, you're right.

If Is were -10mA then V+=+5v. For the amp to be in a linear mode V- ≈ V+. However V- can't be +5V if Vo is only 2V.

If Is were -10mA and Vo was -2V you have same problem. V- can't be +5V if Vo=-2V.

 

[rude man]

Humm, you're right.

If Is were -10mA then V+=+5v. For the amp to be in a linear mode V- ≈ V+. However V- can't be +5V if Vo is only 2V.

If Is were -10mA and Vo was -2V you have same problem. V- can't be +5V if Vo=-2V.

Yup. Point is, in the noninverting mode, the gain can't be less than 1.

Unless you have active feedback (a second op amp configuration feeding the output of the first op amp back to its input.)