# An OPAMP with input offset voltage

Discussion in 'Electronics Homework Help' started by fgda, May 26, 2013.

1. ### fgda

2
0
May 26, 2013
1. The problem statement, all variables and given/known data

2. Relevant equations
Superposition
Non-inverting OPAMP: V_out = (1 + R/R)V_in = 2*V_in2
Inverting OPAMP: V_out = -V_in
V_os = 2.5mV

3. The attempt at a solution

1. V_in enabled, V_os disabled

V_out1 = -V_in

2. V_os enabled, V_in disabled

To get V_out2, use the non inv. equation where V_in2 is equal to V_1 + V_os
where V_1 is the node between R and R at the middle bottom of the circuit.
V_out2 = 2*(V_1 + V_os)

3. V_out = V_out1 + V_out2

Is this the correct approach?

2. ### fgda

2
0
May 26, 2013
I'm pretty sure that step 2 is correct. The professor did an example in class that was similar enough to that case. But in step 1, I have input voltage at both OPAMP terminals so I do not know what to do. I'd rather take the OPAMP as a black box with a characteristic equation if possible.

3. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
OK I'll have a go. I have to assume that the op-amp has infinite input resistance.

The formulas for the voltages at the two op-amp inputs are:
1. . V_inverting_input = (Vin + Vout) / 2
2. . V_non_inverting_input = Vout / 6

This circuit has one stable condition because there is more negative feedback than positive feedback.

Vin = +2.2V and the input offset voltage is 0.0025V. The question doesn't state which way the inputs are offset so I'll state the offset as +/- 0.0025V.

3. . Vout / 6 = ((Vout + 2.2) / 2) +/- 0.0025

After some simple rearrangements this formula yields
4. . Vout = - (3.3 +/- 0.0075)

Assuming no op-amp input offset, Vout = -3.3V. Plugging 2.2V and -3.3V into formulas 1 and 2 gives -0.55V for both op-amp inputs.

With the input offset voltage included, Vout = -3.2925 or Vout = -3.3075 depending on the polarity of the offset.

BTW shouldn't this be in the homework forum?

Disclaimer: I think I have a complete understanding of the problem, and I think these explanations and formulas are comprehensively correct, accurate, and perfectly reasoned, but I could be completely wrong!