Maximum DC offset voltage

Question

I want to calculate the maximum offset voltage of this circuit:

Here's what I did so far:

Ib is the bias current and Av is the op amp gain. My question is, how can I do what is asked without knowing the values of Ib+ and Ib-? I also know that the voltage source Vos can be in the reverse order.

Answer 1

Assuming that you have given us the complete problem statement, this is a simplified example where they have ignored input offset current.

You are supposed to calculate the output voltage resulting from an input offset voltage of +/- 10mV and also the output voltage resulting from an input bias current of +10uA flowing into each input. Add the two in the way that maximizes the magnitude of the output voltage (worst case).

There will be an error from the nominal input bias current even in the ideal case because the designer has failed to match the input resistances seen by each input. Sometimes that is not important (eg. CMOS-input op-amps) but in the case of such a heavy input bias current as in this example it is a good idea).

---

Since OP has long gone, might as well solve it..

The input offset is (worst case) the bias current multiplied by the resistance imbalance at the inputs (900 ohms vs. 800 ohms or 100 ohms difference). That is 1mV. So add that to the 10mV Vos and we get 11mV.

Output offset is gain times input offset or 55mV, worst-case, same answer as Trevor.

Answer 2

Your 2nd circuit shows a TL081 op-amp and this is a FET input device with a maximum offset current (that's the difference between the two input bias currents) of 2 nA. So assume all that current is flowing from one input and not from the other to calculate the offset however, this is only accurate if you have the same impedance at both inputs; you don't because you have 900 ohms at the non-inverting pin and 800 ohms at the inverting input.

So now you have to use bias currents that are different by 2 nA to work out the equivalent offset voltage. According the the data sheet, input bias currents can be as high as 20 nA so, choose individual input bias currents of 18 nA and 20 nA and do the math. Mutiply the net voltage offset (due to bias currents) by gain to get the offset at the output.

Answer 3

Your model should look something like this where OA1 is an ideal op-amp...

^{simulate this circuit – Schematic created using CircuitLab}

For D.C. and tying both inputs to ground you can simplify that to.

^{simulate this circuit}

The voltage dropped across R_4 becomes

\$V_4 = R_{3|4} / I_{b+} = 900/-0.00001 = -0.009V = -9mV\$

Using the worst case \$V_{OS} = -0.01V\$, the voltage on the negative pin of the op-amp becomes:

\$V_- = V_4 + V_{OS} = -0.009 - 0.01 = -0.019V = -19mV\$

The current through \$R_2\$ can now be calculated:

\$I_2 = V_-/R_1 + I_(b-) = -0.019/1000 + 0.00001\$

\$= -0.000019 + 0.00001 = -0.000009 = 9uA\$

The offset output voltage then becomes:

\$ V_out = V_- + R_2 * I_2\$

\$ = -0.019 + 4000 * -0.000009 = -0.019 - 0.036 = -0.055V\$

Maximum offset is therefore \$-55mV\$

NOTE: Solving for \$V_{OS} = +0.01V\$ only yields \$ V_{out} = +45mV\$.

So more accurately this circuit can be expressed as

\$V_{OUT} = 4.5*V_{i2} + 4 * V_{i1} ±(0.045,-0.050)\$