A little question about inverting opamp

[Agung Kurniawan]

Hi guys
What is the purpose of +in to Vb resistor on inverting opamp?
And what is the difference if I connect the +in straight to Vb?

[vigilante397]

When using an op-amp in an inverting configuration the positive input is used to bias the amplifier.

This is an excellent read if you want to learn way more than you will ever use about op-amps:

https://ocw.mit.edu/courses/media-arts-and-sciences/mas-836-sensor-technologies-for-interactive-environments-spring-2011/readings/MITMAS_836S11_read02_bias.pdf

The bit about biasing is a couple pages in. Hope this helps

[DrAlx]

To minimise errors in output bias voltage you try and make the DC resistance presented to both inputs the same.
If the left input resistor couples signal in via a capacitor then you can ignore it and the DC resistance presented to the inverting input is just the feedback resistor. Otherwise the DC resistance is the parallel combination of the feedback resistor and all other resistance hanging on the inverting input.
Same applies to the non-inverting input.
Rb is mainly something you would need to consider if the opamp has a bipolar input stage with large input bias currents. Opamps with FET inputs have much smaller input bias currents and you can then not bother with Rb.

[Agung Kurniawan]

Thanks guys

[Agung Kurniawan]

Thanks guys

[merlinb]

Rb is mainly something you would need to consider if the opamp has a bipolar input stage with large input bias currents.

Even then you're usually talking about millivolt levels of output offset, so the extra resistor is almost never required in a guitar pedal. Millivolts of offset don't make a hill o' beans difference to the 4.5V output.

[antonis]

There is a rumor hanging around that some perfectionists try to minimize off-sets by making input bias resistor value same as the parallel combination of R_F//R_G..

Some more headcase guys take into account Vb source equivalent resistance as well..

But, as Merlin said, ordinary pedal signal doesn't mind if it rides on 4500mV or on 4527mV..

[GibsonGM]

In application notes, I've often seen the value of Rb calculated as:   Rb = Rf//Rg, and have followed that with no problems. 

Now I see that it may be overkill  But that's how I do it, so later on if anyone sees my work they think I did math and all. 

[antonis]

It actually does no harm, Sir... 
(except in case of non-inverting amp with relative low value uncoupled feedback resistor because of significantly lowering input impedance..)

IMHO, we shouldn't bother a lot about off-sets if we don't mess with high pressision circuits..

[amptramp]

There is a rumor hanging around that some perfectionists try to minimize off-sets by making input bias resistor value same as the parallel combination of R_F//R_G..

Some more headcase guys take into account Vb source equivalent resistance as well..

But, as Merlin said, ordinary pedal signal doesn't mind if it rides on 4500mV or on 4527mV..

The input bias resistor should only be the parallel combination of R_f||R_g when there is DC coupling to the previous stage where the output resistance of the previous stage low (like another op amp stage).  If it is capacitively coupled, just use R_f.  Taking the source equivalent resistance into account doesn't make someone a headcase.  It just makes you thorough.  And OCD.  OK, I'll go along with headcase.

[antonis]

Taking the source equivalent resistance into account doesn't make someone a headcase.  It just makes you thorough.  And OCD.  OK, I'll go along with headcase.

Plz tell me that you didn't take it seriously, Ron... 

[R.G.]

Even then you're usually talking about millivolt levels of output offset, so the extra resistor is almost never required in a guitar pedal. Millivolts of offset don't make a hill o' beans difference to the 4.5V output.

Depends.

In one-opamp setups, or setups with near unity gain, nope, no issue at all.

Things can get really mysterious in ... um ... eclectic    one-opamp designs where you're biasing the (+) input from a low-impedance bias and trying to run that single stage as a high gain clipper (i.e. with a large feedback resistor) and decide to wash off some bass  coming into the (-) terminal through the inpu resistor by blocking it with a cap. Now ibias times rbias for the (-) input is generated across the feedback resistor only. And the feedback resistor is both large and variable.

Ask me how I know this so well.    

But if there is a string of opamps where that Ibias times Rbias can get passed on and muliplied, you can get to worrisome levels of offsets, especially since as we know that all offsets **will** add up in the worst possible direction instead of cancelling, according to Edsel Murphy.  Note that parallel setups where the outputs of many opamps are summed, the offsets will be summed too. It doesn't take a lot of gain x offsets to make an offset as big as an input signal, so you can truly hose some section that needs to be DC accurate, like an envelope detector.

This same issue can make for a hard-to-find [klik] if you switch anything inside the pedal, as in changing gains or routing.

For strings of opamps you can completely stop the build up of offsets by either worrying a lot about the DC source impedance of the bias currents to each opamp, or putting a cap after the opamp. Many commercial circuits will have caps after opamps in places where it's not clear that they'd ever be blocking any DC. That's often what this is for. And you only need to introduce this before places where a many-mV offset would cause issues. Often this is once per pedal, even in complicated pedals.

But again, in one-opamp pedals, there is usually no issue with letting DC bias source impedance run amok.

[anotherjim]

A while ago I was really pleased to be told not to worry about the offset effects, and to leave out the +input resistors (call it Rin+?) completely. Actually, fitting a DC blocking capacitor between stages is a handy way of inserting a fixed high pass filter in a more rational place than the input or output.

I found so many op-amp designs that showed an Rin+ resistor, but did not explain how they were calculated. Reverse engineering the drawn values to discover the rational was thwarted - I found many inconsistencies.
Some had Rin+ = Rin-, regardless of what Rf was.
Some had what seemed an arbitrary value, like 10k, in every op-amp, regardless of any Rin- or Rf relationship.

Here's a fun one. You will find what is essentially the same circuit with those same values in many other IC datasheets...


R4 and R6 appear to have been calculated according to some rule.
R4 might seem that it should be 10k, but it's 15k.

I don't think I have enough virtual sausages to reward whoever can sort it out

[R.G.]

I was a NS groupie during the time that those circuits were being published. I don't know that he wrote it but that circuit was during Robert Widlar's tenure at National. He was encyclopedic - and sneaky.

I can opine that the different-from-theoretical value for R4 might come from one of two directions, one more plausible than the other.

Less-plausible, but possible; the early semiconductor processing was far from perfect, semiconductors were $OLID GOLD, so the app engineers liked to show you how very, very good their products were, so some really clever guy might, just might, have figured out that the LM101 worked better somehow (bandwiidth, speed, accuracy, etc.) with a hint more input bias on the (-) side than the (+) side. That is, better press by sneakily knowning more about your silicon.

More pliusible: offset bias poirs a tiny, tiny, offset through the diodes that actually do the rectification and make them turn on/off better in the app.

[PRR]

> R4 might seem that it should be 10k

Or =D2. Which could be 60r to infinity.

And at dead-zero, neither diode conducts so it is just R1 20K.

Jung's OpAmp CookBook shows many rectifiers, the first few without citation and some showing the exact same thing as the NatSemi circuit, without derivation of the bias-balance resistor.

I want to read the Burr-Brown book, later.

Idiot-assistant is taking a number like 7.82k, but I suspect I am not asking right.

[DrAlx]

If diodes don't conduct then can we just treat them as infinite resistance and so ignore them.
Doing that

R1|| (R2+ R3+ R7) is roughly R4  (=14.45k)

EDIT: Probably should be   R1 || (R2 + R3 + (R6 || R7)) = 13.39k

By the same reasoning, R5 should be R7 || R6 || (R1+R2+R3) = 8.69k

[antonis]

Or maybe C4 reactance at high frequencies is taken into account..

(I know that shouldn't be involved in DC offset, but...)

[DrAlx]

Or maybe C4 reactance at high frequencies is taken into account..

Not sure what you mean as the offset error is a DC effect.

Thinking about what R.G. said about the offset errors usually adding up in a bad way...

Using R4 = 15k  (i.e. 1.61k larger than above calculated value of 13.39k) and R5=6k2 (i.e. 2.49k smaller than above calculated value) then does that make the overall offset cancel out better or make it worse?

[antonis]

Not sure what you mean as the offset error is a DC effect.

Another politically correct way to imply a possible bad design...

(Nobody's perfect, even on company's official data sheets suggestions..)


Anyway, my previous agreement with Merlin's view about ingoring small off-set variations is restricted το DIY pedals..
(otherwise we should implicate T network bias trimming or other complicated ways - but it shouldn't be Rock n Roll anymore..) 

[anotherjim]

Almost the same thing (no integrator), but with bias resistors probably according to the rule, IF I can work the paths right.

Both rectifiers are incomplete circuits - we don't know what's driving!

If low impedance DC coupled source, then R1 is alone to IC1 and R1||R2 makes R6 ideally 5k.  IC2 has R3||R4||R5 so R7 ideal is 2.5k.
Bloody hell! This one works out!

But if input is AC coupled, then you see the path differently, and R1 can find supply back via R4+(R3||R5)?

So the first circuit looks according to what Dr Alx worked out IF AC coupled?

As for the diodes, I'm thinking bias current would tend raise the -in pin, output responds driving low to get it back down via Rf and thus keep D1 just in conduction and D2 will be off during idle. Is that anything like right?