Another opamp question

It would have no effect on an ideal opamp, except to add some Johnson noise.

In real life, it will increase output DC offset and noise and reduce gain and bandwidth, but not dramatically with your numbers.

There is some signal at the inverting input; if there weren't, there would be no output. So some current does flow in your added resistor.

Huh? why not bread board it? That's what I'd do. Try 10 ohms to ground on the inverting input. Try hitting it with fast edges/pulses. (then stick a 1 nF cap to ground.)

George H.

Doesn't do a thing. The inverting input is already at "virtual" ground.

I once cured a production line OpAmp noise-pickup problem by bypassing the summing node with a capacitor (inverting mode). ...Jim Thompson

Only if there's no offset voltage.

...Jim Thompson

We did that once to solve a 2.4GHz TDM signal getting into a bipolar opamp.

Except where it's not a virtual ground, of course. So, at high frequencies, where the op-amp's gain is low. A resistor reduces HF response, probably only slightly.

Repeat the same experiment with a capacitor (or R+C), and you'll see why capacitance on the inverting input node is generally a no-no -- but why it can be useful, too (namely, compensating for op-amp gain roll-off). Peaking quickly turns into oscillation, so use it carefully. :)

Tim

Any more pedants?

Not at all pedantic, just giving additional information.

Tim

uencies,

bably

It'll cut the closed-loop BW in half by forcing the amp to work at a noise gain of 20. Must be restful, working on stuff where you can ignore that. ;)

To the OP: the 10k input resistor and the 10k resistor to ground form a 2:1 voltage divider with a Thevenin resistance of 5k, which is what goes into the op amp gain formula. So you actually have a 20x amp with a 2:1 voltage divider on the front. The DC gain is still 10, but the bandwidth is halved and the offset voltage is doubled. (The op amp's voltage noise contribution is doubled as well, though at 5k you might not have to care.)

Cheers

Phil Hobbs

Hmm, OK. I can imagine it might help in some instance. In general I'm (now) conditioned to keep C there to a minimum. I'm guessing it wasn't 1 nF. :^)

George H.

Thanks very much. Yeah, I sensed that I was wrong about it not affecting the gain.

Well, I needed to knock down an input audio signal so when it goes low it doesn't exceed the absolute maximum rating for the voltage on any pin below the negative rail. The resistor divider seemed to be the easiest way to do that. I don't really care about bandwidth or noise because I'm basically turning it into a square wave, rectifying, etc.

Would it be the same explanation for what the signal source sees as the input impedance of the amp? At DC it's still

Dumbass. Of course it can't be less than 10K. So maybe there's no change.

At low frequency it's still 10k because the loop keeps the SJ at ground, so no current flows through the grounded resistor.

Neglecting the effect of strays and the op amp's Cin, and looking well outs ide the loop BW, R_in would be about 19k. This is because the 100k R_f appe ars in parallel with the bottom leg of the voltage divider.

In the neighbourhood of f_3dB there will be some whoopdedoos in Zin.

Cheers

Phil Hobbs

IIRC, it was _2nF_ >:-}

I'll post an analysis. ...Jim Thompson

Huh, OK. No big deal if you don't have time to post anything. Years ago I spent some time C-loading TIA's and compensating them, sorta checking my math with the circuit.

George H.

you need to read up on the concepts of virtual ground and summing junction.

Adding the 10k as you described has no effect on the closed loop gain.

Really?

So you don't think it affects the bandwidth? Closed-loop gain is a function of frequency after all.

Cheers

Phil Hobbs

A gumdrop 1 MHz opamp only has a gain of 50 at 20 KHz. The OP was running at a closed-loop gain of -10. So he's already out of the ideal-opamp zone at audio frequencies. There will be a substantial gain error from the ideal, and adding the 10K will make it worse.

Bootstrapping a photodiode off the summing point of an opamp looks strange at first, but it works.

Friends don't teach friends that opamp inputs are always zero.