Opamp bias currents

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Ideal opamps have no current flowing into their inputs. This is not true for real world opamps. The input impedance for these is in the order of Mohm to Tohm (tera-ohms). The current flowing into the opamp will offset the voltages on it's inputs. Usually these currents are in the nA range but that depends on the temperature.

I didn't find a good derivation of the formulas for biasing resistors, so here they are.


Inverting amplifier

Inverting opamp with bias currents drawn.

The above schematic shows the inverting opamp. In and Ip are current sources which symbolize the opamp bias currents. In this case any current flowing into the negative input (In) will offset the voltage on it. We will assume that the current In is the same as Ip (we model this opamp's input impedance as an impedance Rd between the input terminals). We have to find the value of R3 so that the current In = Ip will offset the positive input by the same value. This way the voltage difference between the terminals is net zero and the error caused by the bias currents is minimized.

Uwe == Uin is the input voltage, Uwy == Uout is the output voltage.

First we have to find the relation between the currents. From KCL:

I1 + I2 = In  ->  I1 = In - I2

Now for the opamp output voltage:

Uout = Ud * A  -> Ud = Uout/A

Then we find Ud, the voltage between the opamp inputs. For that we use the superposition principle. It's simple - for each voltage we treat all other sources as zero (Current sources give no currents, voltage sources are clamped to ground). We work out the formula for the value that interests us. Then we sum all the formulas. As an example let's calculate Ud.

First we zero all voltages except Uout, and find Ud:

Ud = Uout * R2/(R1+R2)

Then, zero all sources except Uin and find Ud:

Ud = Uin * R1/(R1+R2)

Then, zero all sources except In and find Ud:

Ud = In * R1R2/(R1+R2)

Similarly for Ip

Ud = - IpR3

This formula has a negative sign because the higher Ip, the smaller Ud.

Now we have to sum all these terms above:

Ud = Uout * R2/(R1+R2) + Uin * R1/(R1+R2) + In * R1R2/(R1+R2) - IpR3

Substitute Ud and reorganize the equation:

-Uout/A - Uout R2/(R1+R2) = Uin R1/(R1+R2) + In R1R2/(R1+R2) - IpR3
-Uout(1/A + R2/(R1+R2)) = Uin R1/(R1+R2) + In R1R2/(R1+R2) - IpR3

We want our opamp to output 0V when the input is 0V. Therefore we assume these values for Uin and Uout and see how the math works out:

Uin = 0, Uout = 0  -> 0 = In R1R2/(R1+R2) - IpR3

At the beginning we assumed that In is practically the same as Ip

In ~= Ip 
R3 = R1R2/(R1+R2)

This is the value of R3 we are looking for.