Hello.

I've just been reading in my book about ideal op amps and think I understand that. Now it's moved onto other more realistic models of the op amp - the offsets model and the finite gain model.

Offsets:

Finite gain:

Because the book doesn't clarify, I was assuming that bias current and input current are the same thing.

The book bills the offsets model as "accounting for non-zero bias current", which I can see that it does, as the input current is certainly equal to the bias current in that case.

However, it says the finite gain model does not account for non-zero bias current. But the input currents are non-zero.

So what is the bias current doing, and what distinguishes it from the portion of the input current which just goes through the input resistor?

Sorry if this is a stupid question!

Thanks, Jon.

 

The bias current for the input transistors in an opamp is very low.

Some opamps have NPN input transistors and some have PNP input transistors so the bias current could be sinking or sourcing.

Some opamps have Jfet inputs that have no bias current, just extremely small leakage current that increases when the temperature increases.

The input bias current of a transistor turns it on. Usually negative feedback keeps it from saturating so it operates as a linear amplifier.

If the input has a coupling capacitor then the bias current flows in the input resistor.

 

A basic practical op-amp with a simple input structure such as a BJT or FET long tailed pair will require some bias current to correctly set the differential input stage bias conditions.

The simplified model for a non-ideal amplifier models the input side as a resistance which accounts only for the input current drawn from the source in response to the input signal amplitude variations. The bias current will probably also be drawn from the source (if it's not AC coupled) or via other means such as additional resistances tied from input terminals to ground.

 

Okay thanks for your replies guys