Establishing Vref and bias currents of opamps.

[demonstar]

I've been trying to understand best I can about how to bias opamps using a potential divider. I've read the article at Geofex 5 times each day for the last week and pourred over datasheets and pages and pages about opamps and now I still don't fully understand it. How exactly are the values calculated? For example I understand that the bigger the divider resistors the smaller the current flow through the divider but higher risk of noise. So I understand those choice of components. However why in the dist+ is there a bias resistor of 1M but the DOD 250 clone a 470K? Do they provide different bias currents? Also how is the is the decoupling cap of the divider calculated. I understand that it's got to be a high value but I'm keen to know why?

The article at geo talks about the current flow if the differential input is clearly grounded but I fear in reality it's not. How does this affect it?

I've been fiddling around all day on the bread board and still haven't clearly worked this out. There seem to be so many conflicting articles on this forum and all over the internet. It's seems to be a gray area. I'm sorry if I'm missing the obvious but if anyone could help me out or point me in the correct direction it would be very much appreciated. The areas I've been looking at in particular are non inverting ac coupled opamp applications such as the dist+ and DOD etc.

Thankyou!

[R.G.]

I've been trying to understand best I can about how to bias opamps using a potential divider. I've read the article at Geofex 5 times each day for the last week and pourred over datasheets and pages and pages about opamps and now I still don't fully understand it. How exactly are the values calculated? For example I understand that the bigger the divider resistors the smaller the current flow through the divider but higher risk of noise. So I understand those choice of components. However why in the dist+ is there a bias resistor of 1M but the DOD 250 clone a 470K? Do they provide different bias currents? Also how is the is the decoupling cap of the divider calculated. I understand that it's got to be a high value but I'm keen to know why?

I think what you're missing is that the opamp inputs only "eat" as much current as they want.

The hypothetical ideal opamp has inputs that are infinite impedance - they sense voltage without letting any current flow into or out of them. That is of course not possible in the real world. But real opamps do come more or less close to it. A CMOS opamp comes closest - their inputs may be pulled from positive to negative rail or vice versa by currents too small to accurately measure, pico-amperes or femto-amperes. JFET opamps may need a few nano-amperes  of current to pull them around. Bipolar opamps are the laggards, needing maybe 100's of nano-amps to pull them to wherever they go.

But whichever style you use, the opamp only takes in as much current as it needs to bias itself. This is always far, far less than the rest of the circuit uses. And that's where the difference in biasing resistors comes in - the rest of the circuit. In the D+, the 1M resistor is pretty much the total input loading that the input signal sees. The same is true for the DOD250. In neither case does the opamp itself put the requirement there for that resistor value. It's the other stuff, the input signal, capacitors, frequency response, etc. that sets the resistor value.

For the biasing divider, the biggest issue is to make sure that at least ten times as much current flows through the biasing resistors as flows into or out of their junction. This is so the biasing currents in or out of the junction can be ignored when computing the resistors, and so that as the rest of the circuit pulls more or less current, the bias voltage set by the resistors will not change.

For the decoupling cap: the bias string resistors appear in parallel to any signal which leaks through at their junction. You want the capacitor to look like a short circuit compared to the resistors at the lowest signal frequency which could possibly come into the bias voltage junction. The easy way to do that is to note that the resistors are less than 1/10 the impedance of the resistors coming to them for biasing, so if the cap decouples the biasing resistors at the lowest frequency, then it more than adequately decouples the signals coming in from the bias resistors to the various inputs it biases.

So the decoupling cap is C = 1/(2*pi*F*R) where R is the parallel value of the biasing string resistors, and F is the lowest frequency you will have, usually 82 Hz (guitar) or 42Hz (bass) or 20 Hz (full audio). For a 10K/10K bias string and decoupling for bass at 42Hz, the cap is at least C = 1/2*pi*5K*42 = 0.758uF; 1uF would be fine. 10uF is about the same price, so unless you have a reason for going with the minimum cap, use a 10uF. Or a 22. Or a 4.7. You just have to have enough.

The article at geo talks about the current flow if the differential input is clearly grounded but I fear in reality it's not. How does this affect it?

Sorry - I couldn't translate that. Which statement in which article?

I've been fiddling around all day on the bread board and still haven't clearly worked this out. There seem to be so many conflicting articles on this forum and all over the internet. It's seems to be a gray area. I'm sorry if I'm missing the obvious but if anyone could help me out or point me in the correct direction it would be very much appreciated. The areas I've been looking at in particular are non inverting ac coupled opamp applications such as the dist+ and DOD etc.

It's gray because it's not a single-point answer, it's a "how much is enough" answer.

For non-inverting AC coupled inputs with opamps, the right way to look at it is that you pick the input bias resistor to be as high an input impedance as you need, 470K, 1M, whatever. That being chosen, you pick the signal input cap to the + input to let the lowest frequency you want to let in. That's C = 1/2*pi*Rin*F.

Now you need the bias string resistors from + to ground to let at least ten times the bias current through that 1M biasing resistor flow through them. The biasing resistor only lets a DC bias  current of ... let's guess at 100na for the particular opamp you use... flow. So the resistors have to let at least 1uA flow from +9V to ground so we can ignore the DC bias current through the 1M or 470K bias resistor. (notice that the opamp will eat a constant biasing current; it doesn't care if you used 1M or 470K, it'll still eat 100na)

So the two resistor bias divider only needs to be 9V/1uA = 9M or 4.5M each to make this barely work. We want it to more than barely work and we don't want the noise from those two 4.5M resistors so we use two 10Ks - or 47Ks - or 100Ks or whatever we like and happen to have in the box at the time. Let's say we use 47K. That makes the AC impedance of the two bias dividers be 23.5K, and we need a decoupling cap of C = 1/2*pi*23.5K*82Hz = 0.082uF. Anything bigger will be fine. 0.1uF would be OK-ish, but we can use a 1uF electro and be far, far more decoupled. We're not trying to just barely make it, we're trying to be sure there's plenty enough.

[demonstar]

R.G. thats super!!! thankyou soooooo much! You've totally cleared it up for me. I think the thing I didn't realise and I'm not sure everyone else does too is the fact that the opamp only pulls the current it needs and that the bias resistor is just there for other reasons. Also that the reason for having 10% of the current flowing through the bias resistor is just so that Vb doesn't move around.

Now I can go and finish designing my pedal and continue my research and learn more about opamps and I'm going to start learning about setting impedance of pedals.

Thanks once again!

[ubersam]

Wow, that is very good information. Thanks. Just a question...

...For non-inverting AC coupled inputs with opamps, the right way to look at it is that you pick the input bias resistor to be as high an input impedance as you need, 470K, 1M, whatever. That being chosen, you pick the signal input cap to the + input to let the lowest frequency you want to let in. That's C = 1/2*pi*Rin*F.

The "Rin" in the formula above, would that be the input bias resistor (between biasing string junction and opamp input)?

[R.G.]

Yes.

[ubersam]

Thanks again!