Why don't BJT IC opamps use resistive degeneration in the input stage?

[godfrey]

Originally Posted by DF96

Resistors use a lot of silicon?

Possibly, but it's hard to imagine that chip real estate is such a big issue. Perhaps there's some difficulty in creating decent quality or close tolerance resistors. I notice that specifications that rely on resistors tend to be quite badly defined (e.g. quiescent current or max output current).

Better to use three or four matched BJTs than a resistor.

That does seem to be the approach sometimes taken, but I'm still wondering why.

[SY]

Close tolerance is hard but close matching is easy. Of course, the resistors also degrade noise specs since their Johnson noise is uncorrelated.

edit: I see that one of the smart guys has already mentioned the noise issue.

[godfrey]

Originally Posted by SY

Close tolerance is hard but close matching is easy.

Thanks, SY - That's good to know.

So for an opamp circuit like the one below, we don't know exactly what the idling currents of the two input stages are, but they will be closely matched (when the common mode input voltage is close to zero).

The main feature of the circuit is that the common mode input voltage range extends about 0.5V beyond the supply voltage in both directions.

The benefit of biasing the input stages with resistors is that as the common mode input voltage moves from rail to rail, control is handed over gradually from one input stage to the other, with the total transconductance remaining constant.

I'd been wondering how guilty to feel about the resistors, but it should be fine.