Positive (?) feedback in an opamp

[npelov]

Hi,

I was looking at LM399 Datasheet, typical applications section on page 6 (portable calibrator). The LT1001 opamp has a resistor from output to (+) input. What is this resistor used for. It doesn't change the voltage on the (+) pin. If you put 8.8k and the 19k+3k in parallel it's about 6k. Is the 5k for canceling the input bias current? If yes, isn't it supposed to go in series with the (+) input? And why it's not 6k?

[Andy Watson]

The LT1001 opamp has a resistor from output to (+) input. What is this resistor used for.

Ignore the 200k. For a stable voltage from the reference zener it would be advantageous to drive it with a stable, constant current source. Imagine if you had a stable voltage, say 3.05V higher than the zener voltage - 3.05 V / 5 k = 610uA. It appears that you have just such a voltage at the output of the op-amp! (and it should be very stable )

The 200k is there to ensure that the circuit starts-up when you first power it.

[npelov]

Ow, so the 200k is not really driving the zener . Stupid me ... It doesn't provide enough current, right? So that's why when I put second multimeter the voltage changes . So the 5k is actually driving the Zener relying on the far better output of the opamp, rather than the power supply. Well, that means that this circuit is no good for low voltage outputs. I thought I can put a divider in the input and have 1 volt instead of 10. Well I can do that, but on the output and I'll need a buffer.
Thanks!