I thought this was only in effect when the inputs were grounded and you need 5mV between them for the output to be zero. So what you are saying is this offset is always multiplied by the gain?
Yes, for a gain stage, input offset voltage causes an offset at the output which is equal to the input offset voltage multiplied by the stage gain.
Here's an op-amp differential voltage amplifier:
It accepts two voltages on its two inputs, nodes A and C, and generates an output voltage on node E, relative to the 0V rail, that is equal to the differential input voltage (i.e. V
A-V
C) multiplied by the stage gain, which is R3/R1 (and R4/R2), which is 100.
The simplified description of this circuit neglects four sources of error that are present in the real world:
- It assumes that the resistors are perfect and have zero resistance error;
- It assumes that the op-amp is ideal and therefore has zero input offset voltage;
- It assumes that the op-amp is ideal and therefore has infinite open-loop gain;
- It assumes that the op-amp is ideal and therefore has infinite input impedance.
First I'll run through an example scenario using all of those assumptions, then I'll repeat it with the second assumption removed - I'll use an input offset voltage of 5 mV.
My input conditions, for both tests, will be a differential input voltage of 0.02V (20 mV) on top of a common-mode voltage of 10.01V:
V
A = 10.02V
V
C = 10.00V
V
AC (V
A - V
C) = 20 mV
V
OUT should be 20 mV × 100 = 2.0V.
Assuming zero input voltage offset
Because there is negative feedback (via R3), the op-amp's behaviour is to adjust its output voltage until its differential input voltage (V
BD) is exactly zero. This means that after settling delays, V
D will be equal to V
B.
V
B is determined only by V
A and R2 and R4. We can calculate it using the voltage divider formula, which gives:
V
B = V
A / ((R2 / R4) + 1)
= 10.02 / ((1 / 100) + 1)
= 10.02 / 1.01
= 9.920792V.
The next step is to calculate I
R1, which is determined by the voltage across R1, i.e. V
CD.
Since the op-amp has zero input voltage offset, V
D = V
B, so V
CD = 10.00 - 9.920792 = 79.208 mV. Use Ohm's Law to calculate I
R1:
I
R1 = V
CD / R1 = 79.208 mV / 1000 ohms = 79.208 µA.
Since R1 and R3 are in series (no current flows into the op-amp input), the same current flows through R3, producing a voltage across R3 of V = I R = 79.208 µA * 100,000 ohms = 7.9208V. Subtract that from V
D to get V
E: 9.920792V - 7.9208V = 2.0V (ignoring rounding errors).
So V
OUT is 2.0V. That's how it's supposed to work.
Assuming 5 mV input voltage offset
Most of the calculations are the same as before, including V
B = 9.920792V.
The op-amp will adjust its output voltage to bring the inverting ("-") input voltage equal to the non-inverting ("+") input voltage, but because of the 5 mV input offset, it thinks the inputs are equal when there's actually 5 mV difference between them.
That means that V
D is actually either V
B + 5 mV, or V
B - 5 mV, depending on the direction of the offset. I'll assume the first one, so:
V
D = V
B + 5 mV
= 9.920792 + 0.005
= 9.925792V.
Now calculate V
CD: V
CD = 10.0 - 9.925792 = 74.208 mV. This is 5 mV lower than before, because V
D is 5 mV higher than it should be.
Now calculate I
R1, from V
CD and R1's resistance:
I
R1 = V
CD / R1
= 74.208 mV / 1000 ohms
= 74.208 µA.
Now calculate the voltage across R3 (i.e. the difference between V
D and V
E) with 74.208 µA flowing through R3:
V
DE = I
R3 × R3
= 74.208 µA × 100,000 ohms
= 7.4208V
Finally, calculate V
E by subtracting that voltage from V
D:
V
E = V
D - V
DE
= 9.925792 - 7.4208
= 2.505V. (Should be 2.0V)
So the 5 mV input offset voltage has been multiplied by the stage gain (100) and is causing a 500 mV offset at the output.
