Electronic – Input Bias Compensation Resistor Calculation for an Integrator

The transdiode circuit has a formula that is: -

Vout = -26mV\$\times ln(\dfrac{V_{IN}}{I_S\times R_{IN}})\$

\$I_S\$ might be ~10nA and so for an input of 1mV and 10k resistor for Rin, Vout will be about -60mV.

If you consider that the input bias current of the OP27 might be 40nA (worst case) and this were not balanced by the resistor in the non-inverting lead, it would generate an offset voltage error at the input of 10k x 40nA = 0.4mV.

So now you have an efective input of 1.4mV or 0.6mV (polarity is unknown of the bias current).

Assuming the effective input becomes 0.6mV, the effective output is about -46mV.

That's a big difference (if you want it to be) but, of course, if the error went the other way the output would be about -69mV and a smaller difference.

If your baseline input didn't need to go below 10mV then an effective input of 10.4 mV wouldn't be so bad. 10mV on the input produces about -120mV and 10.4mV in produces an output of about -121mV.

Your third image is not a (source) follower, rather it is a common-source amplifier and generally is configured to provide gain > 1.

A source follower looks like this:

Note that there is no drain resistor at the top and that the output is taken from the source terminal rather than the drain.

\$V_{out}\$ will follow \$V_{in}\$ with a drop of roughly the gate threshold voltage of the MOSFET, \$V_{th}\$. This may be somewhere around 5V depending on the device, so one might reasonable choose a BJT emitter follower instead:

You can see the circuit is essentially identical excepting the transistor type. \$V_{out}\$ will follow \$V_{in}\$ with a drop of \$V_{BE}\$, typically only 0.7 V.

The input impedance of an emitter follower is relatively high, and it's output impedance is relatively low. So placed between a resistive voltage divider and the rest of the circuit (such as an amplifier input) has the effect of stabilizing the bias voltage developed across the divider against variation due to changes in current drawn from the divider.