# Homework Help: OpAmp problem

1. Oct 21, 2016

### WeeChumlee

1. The problem statement, all variables and given/known data
The circuit of FIGURE 2 is known as a transimpedance circuit used
for the measurement of very small currents. Derive the relationship
between the output voltage V and the input current I; i.e. if V = kI
find k in terms of R1, R2 and Rf.

2. Relevant equations

3. The attempt at a solution

If current flows from left of Rf to right and V- is 0V then I is negative.

Then Vn (which is the voltage drop over Rf) = -IRf

R2 and Rf in parallel have the resistance

R2Rf / (R2 + Rf)

We now have a voltage divider:

Vn = {[(R2Rf / (R2 + Rf))] / [ R1 + R2Rf / (R2 + Rf)]} * Vout

Vn = R2Rf / (R1(R2+Rf) + R2Rf) * V
Vn =-IRf

I = -Vn/Rf

I = -[ R2Rf / (R1(R2+Rf) + R2Rf) * V ] / Rf

I = -R2V / (R1(R2+Rf) + R2Rf)

I * (R1(R2+Rf) + R2Rf) = -R2V

V = -[ I * (R1(R2+Rf) + R2Rf) ] / R2

V = kI

V / I =k

V / I = -(R1(R2+Rf) + R2Rf) / R2

k = -(R1(R2+Rf) + R2Rf) / R2

The next part states values:
Calculate the current I if Rf = 10 MΩ, R1 = 90 kΩ, R2 = 10 kΩ
and V –0.1 V.

I use these values and plug them into a SPICE program and see that somewhere I am out by a factor of 10.
I just can't find my mistake

If anyone can help I would be much obliged.

Thanks

2. Oct 21, 2016

### Staff: Mentor

Hi WeeChumlee, Welcome to Physics Forums.

I think you want to check the size of your current source. What was your calculated value (in Amps)?

3. Oct 21, 2016

### WeeChumlee

Calculated current was 9.998nA. (So 10nA)
This though should be 1nA . with 1nA the simulation works.
Somewhere is a factor 10 mistake I presume but I son,t see it.

4. Oct 21, 2016

### Staff: Mentor

Can you show the details of your calculation? Let's try to spot the missing 10

5. Oct 21, 2016

### LvW

Here is my recommendation (for a much shorter calculation):
If you place the resistor Rf very close to the common node of R1 and R2 you will see that these three resistors form a star network.
Hence, applying the star-triangle transformation the feedback circuit will be simplified because one of the new resistors is between opamp output and ground.
Hence, this resistor has no influence on feedback - and the remaining circuit looks very convenient (only two feedback resistors).

6. Oct 21, 2016

### Staff: Mentor

Yes that works. Another shortcut is to realize that the input current fixes the potential at the resistor junction node. Just write the node equation (KCL) and substitute the fixed potential for the node voltage. The only variables left are the current and the output voltage.

The OP did derive a correct expression for the proportionality constant, so I think the remaining issue will be related to the units used (nA versus pA for example....). I'd like to see the OP's actual calculation for k and subsequent calculation of the current for the given -0.1 V output.

7. Oct 21, 2016

### WeeChumlee

Oh man, I think I got it. Just did the same mistake over and over again somehow.
Getting a beer, leaving it alone for a while, and coming back seem to have done it.
Not got SPICE now but pretty certain I have it now.
Now have I = 9.99exp-10
That's the 10 I was looking for.
:-)

8. Oct 21, 2016

### Staff: Mentor

Yup!

9. Oct 21, 2016