Voltage Divider Question

[Electron Tornado]

I am building a distortion pedal that is based on the Distortion + and the DOD OD250, both of which use the 741 op amp. My question is about the voltage divider used to help bias the op amp. The Dist+ uses two 1M ohm resistors with a 1uf cap, while the OD250 uses two 20k ohm resistors and a 10uf cap. Are there any pros or cons in using either of these two voltage divider circuits? From what I've read, I gather that depending on the resistors used and the resistance of the load, there can be issues with the current required by the load. However, I'm thinking that this may not be critical in this application since both circuits work despite using different resistors.

Also, how were the capacitor values and voltage ratings selected?

Here are links to the schematics:

http://www.diystompboxes.com/pedals/MXRDIST2.GIF
http://www.montagar.com/~patj/dodoverd.gif



I know I've read something about this on one of these forums, but I can't seem to find the thread.   

[JKowalski]

It's not really important. Using smaller value resistors gives you better stability, since the Vref is then lower impedance and less susceptible to pick up noise. The capacitor doesn't really make much difference either, it just provides some extra DC stability to the Vref, so that signals that accidently get coupled to Vref are somewhat filtered out. When you use an op amp input, you don't really need it to be that low impedance because the op amp will draw miniscule current into it's input and will not load the Vref basically at all.

Generally less resistance + more capacitance would be better, but is often not really necessary.

[isildur100]

This is a point i still don't completely understand. Sometimes biasing is performed using resistors in the Mega-ohm range, and other times using only 10k resistors. Some say using low value resistors bring loading problems and using high value resistors brings more noise... Low values provides more current, high values less. For my own experiments I mostly use building blocks from others and sometimes tweaking the bias in some ways gives better results and has an impact on the sound.

From what I have seen mostly is that the biasing of the first stage is often done using high value resistors for impedance reasons, and for the following stages it varies

[Electron Tornado]

It's not really important. Using smaller value resistors gives you better stability, since the Vref is then lower impedance and less susceptible to pick up noise. The capacitor doesn't really make much difference either, it just provides some extra DC stability to the Vref, so that signals that accidently get coupled to Vref are somewhat filtered out. When you use an op amp input, you don't really need it to be that low impedance because the op amp will draw miniscule current into it's input and will not load the Vref basically at all.

Generally less resistance + more capacitance would be better, but is often not really necessary.

Thanks for the info.

I recall reading in a forum thread that noise could be an issue as well. Can you point me to any sources that explain the relationship between the higher resistance and increased noise?

[.Mike]

R.G. made up a page that explains how to pick the resistors for voltage dividers when biasing opamps:

http://www.geofex.com/circuits/biasnet.htm

It doesn't have any information about noise, but it is very informative.

Mike

[R.G.]

It doesn't have any information about noise, but it is very informative.

Darn, I left that out! 

The trick is that the capacitor(s) decoupling the divider network to ground shunt the noise to ground. The cap is a low impedance for all audio frequencies, and the two resistors making the divider are a larger impedance. The thermal noise of the divider resistors is shunted to ground and is not a huge effect for any normal choices of resistors and caps.

[PRR]

> how were the capacitor values and voltage ratings selected?

Voltage rating is something over what you expect to get.

For 9V battery divided to half, your cap must stand 4.5V. Or, to be safe for battery eliminator supply, 5V-6V in case the wart puts out more than 9V.

If you want it to last for many thousands of hours, you should add another 10%-50% safety factor (they don't last forever AT rated volts). 7V-10V rating.

In almost any "pedal" design, the smallest practical (leaded and widely sold) caps will be 16V rated so you don't actually have to think about it.

The value..... this is actually a tricky thing.

If you are just biasing ONE input, in a low-gain system, the main thing is to clean the noise off the power supply. If power is a battery, hum/buzz is absent and hiss is quite small. While you could use less, figure the cap which has impedance less than divider impedance at 20Hz. For 10K resistors, about 1uFd. For 1Meg resistors, about 0.01uFd. Then round up-up-up. I'd use at least 10uFd with 10K. However the leakage of 10uFd aluminum electrolytic may sag a 1Meg divider (also it would take 10 seconds to charge-up), so I'd use 0.1uFd (or use 100K resistors).

With wall-power, you may also have to clean hum/buzz. It should already be clean, but it must be much cleaner at the input-bias than at the chip power pin. This means the cap must bypass the ressitors for much lower than 120Hz. Again, bypass for 20Hz and then round-up generously.

When several inputs share a divider, you have to think about sneak paths. To get NO sneakage you would need an infinite cap. To get "negligible" sneakage you must think a LOT about how the system really works and what is "negligible". However, for most pedal work, 100uFd is "nearly infinite enough" and not expensive enough to hurt.

> "Too lazy to look it up?"

That's a problem. However, BJT input chips (741, 4558, 5532, etc) in audio can be OK with 100K bias paths. Sometimes much more. You want the actual input resistor to be large enough that your source is not too loaded. Then keep your divider somewhat smaller so as to not add much more error. This does lead to 10K-50K dividers.

FET input (TL072) have negligible leakage. Bias paths well over 1Meg make no trouble.

Yes, the MXR Distorion + has 741 with 1.5Meg equivalent bias path in one side. The worst-case 741 bias current is 1.5uA. This computes to a 2.25V error. However the room-temp spec is under 0.1uA, a 0.15V error. And really we do not care if the input sits at 4.500V or at 4.3V. Either way, the cap in the gain-set forces DC gain to unity, the cap after the 741 wipes the DC so the diodes feel zero DC. Also the 1meg in the gain-set is a (imperfect) cancellation of the error in the 1.5Meg at the other input. So this does work until it is so cold your fingers break off. And since 741 is kinda the standard that all other chips are better than, 1meg may be generally workable for one input.

[Electron Tornado]

Thank you to everyone for their helpful replies.

The trick is that the capacitor(s) decoupling the divider network to ground shunt the noise to ground. The cap is a low impedance for all audio frequencies, and the two resistors making the divider are a larger impedance. The thermal noise of the divider resistors is shunted to ground and is not a huge effect for any normal choices of resistors and caps.

The value..... this is actually a tricky thing.

If you are just biasing ONE input, in a low-gain system, the main thing is to clean the noise off the power supply. If power is a battery, hum/buzz is absent and hiss is quite small. While you could use less, figure the cap which has impedance less than divider impedance at 20Hz. For 10K resistors, about 1uFd. For 1Meg resistors, about 0.01uFd. Then round up-up-up. I'd use at least 10uFd with 10K. However the leakage of 10uFd aluminum electrolytic may sag a 1Meg divider (also it would take 10 seconds to charge-up), so I'd use 0.1uFd (or use 100K resistors).

R.G.:  I have a filter in place that filters the incoming power. Is the capacitor in the voltage divider there as an added power filter or strictly to get rid of thermal noise from the resistors in the divider?


PRR:  I'm not sure how you are getting your initial values for the caps and then "rounding up" by a factor of 10.

[PRR]

> I'm not sure how you are getting your initial values for the caps

Do you know how to "get a value" for a coupling cap?

[Electron Tornado]

> I'm not sure how you are getting your initial values for the caps

Do you know how to "get a value" for a coupling cap?

I know the basics of how to find cap and resistor values for use in a filter, and this seems like it's that type of application. My question was really more about the need to then "round up up up" by a factor of 10.

[R.G.]

R.G.:  I have a filter in place that filters the incoming power. Is the capacitor in the voltage divider there as an added power filter or strictly to get rid of thermal noise from the resistors in the divider?

The power supply filter is quite different from the filter cap on the bias voltage divider. It's a matter of intent.

Both the power supply and the bias voltage are in theory supposed to be *voltage sources*, in the pure sense, such as being able to supply a literally infinite amount of current in either the outward or inward directions to keep the voltage at their outputs constant. No "give", sag, or springiness at all. We know that is impossible - it would require a source of zero impedance, and Mother Nature has decreed that no real world sources have truly zero impedance. Something about the Second Law, I think. 

But the lower the impedance the source, the better. And the bias voltage is supposed to be a *different* voltage source than the power supply voltage. They *both* need to be low-source-impedance taps of *different* DC voltages. Long ago, people (EE people, that is) noticed that you don't have to have low impedance all all frequencies. Your audio equipment doesn't care about the impedance of its power supply at 100MHz. You only have to have the impedance be low for freuqencies that matter to your application. In the most general sense of audio, that's 20Hz to 20kHz. For guitar, that's often as restricted as 80Hz to 7kHz. So with no need to get to low impedance at DC, you can fake a good DC bias voltage very cheaply by making it a high impedance at DC (the resistor divider does this) and make it much lower impedance at the frequencies of interest with a shunt capacitor to ground. The capacitor looks like an every decreasing impedance with increasing frequency. And if the capacitor's impedance is much smaller than the loads on the bias tap, this all works out.

By that line of reasoning, you can get a perfectly usable *different* DC bias voltage with a low impedance by using only two resistors and a cap if you're willing to use a big enough cap. That is an economic win that has forced all bias voltages to be Resistor/Resistor/Capacitor unless there is some other highly specialized factor requiring it to be different. The capacitor on the bias voltage is not filtering in the sense the power supply filtering is - it's impedance lowering. And it happens almost by accident that the cap value which makes the bias voltage work well is many times bigger than needed for the thermal noise of the bias resistors.

Finally there's that "how big is big enough". Like the answer to "how much is enough money?", it's " just a little bit more." You can't go wrong with a bigger cap (in most cases). You CAN go wrong with a minimally selected cap, so it pays to go bigger when caps are cheap, and they are. That's the origin of PRR's advice.

The smallest cap you can use successfully is the one which has an impedance  [ Xc = 1/(2*pi*F*C) ] which is less than 1/10 of the sum of all the parallel loads connected to the bias voltage point. If that's one 1M resistor, you can get by with a cap that's less than 100K at the lowest frequency you're interested in , presumably something between 40 and 82Hz. If it's a bunch of 10K resistors, your cap may need to be less than 100 ohms at 40 Hz: that's about 40uF, and really it should be more like 100 to 470 to get much, much lower, as Prr mentioned. You want your bias voltage *quiet*.

I'm simplifying heavily here. This issue touches on about nine semester hours of EE that may be needed if you want a complete understanding. Or you can accept the handwaving, and get accurate enough answers.

[Electron Tornado]

Thanks, R.G., that helps clarify things a bit and gives me something to chew on.

Thanks again to you and PRR for the helpful replies.

[PRR]

> I know the basics of how to find cap and resistor values for use in a filter

Good. I wasn't sure where to start; I've learned not to assume builders know theory.

> the need to then "round up up up" by a factor of 10.

In most signal filtering, we want "small" loss at some frequency. A 0.22ufd cap into a 10K resistance will give about 70% of response at guitar's lowest note as it does mid-band.

In supply or reference filtering, we want "large" loss. 70% is hardly worth doing. By picking a cap 10 time larger, we'll knock the offending signal down to about 10%, a real change.

We could use a sharp pencil to get the "exact" amount of filtering needed to "meet specs". But this is DIY, we don't do specs. Anyway in 9V systems the cost of "rounding-up generously" is pennies at most. What is the cost difference 16V at 2uFd versus 10uFd or 20uFd?

[Electron Tornado]

Thanks, PRR, now I understand what you were saying.

[Electron Tornado]

In most signal filtering, we want "small" loss at some frequency. A 0.22ufd cap into a 10K resistance will give about 70% of response at guitar's lowest note as it does mid-band.

In supply or reference filtering, we want "large" loss. 70% is hardly worth doing. By picking a cap 10 time larger, we'll knock the offending signal down to about 10%, a real change.

We could use a sharp pencil to get the "exact" amount of filtering needed to "meet specs". But this is DIY, we don't do specs. Anyway in 9V systems the cost of "rounding-up generously" is pennies at most. What is the cost difference 16V at 2uFd versus 10uFd or 20uFd?

Thinking about this brings up another question - how does changing component values affect the degree of filtering? Changing a value will move the dB vs freq curve left or right until the desired amount of filtering for a certain frequency is achieved. Changing the actual slope of the db vs freq curve would require a higher order filter. Am I correct?

[JKowalski]

Thinking about this brings up another question - how does changing component values affect the degree of filtering? Changing a value will move the dB vs freq curve left or right until the desired amount of filtering for a certain frequency is achieved. Changing the actual slope of the db vs freq curve would require a higher order filter. Am I correct?

Yup