What does the cap in a voltage divider/voltage reference do?

[John Lyons]

I believe the cap is a resevoir for keeping the voltage more constant.
I ran across a circuit that did not have a cap there.
Does the absence of this cap make the voltage fluctuate more at high demands?
What would difference would we hear in this case?

This is the circuit but the '77 250 version had no 10uf cap.



Thanks

John

[petemoore]

  If you're using a battery, basically there's your BFC, adding acap to filter ripple where there is none is to no avail.
  If you're using a PS, that cap will help to reduce any ripple, and provide the circuit with 'less wavy/more ripple free' 9v 'dc'.

[Mark Hammer]

My guess is that the cap makes only very minimal difference in the illustrated instance.  Where the cap has much greater impact on the stability of the Vref would be in those instances where you have multiple parts of a more complex circuit, each having discrepant current needs at any given point in time.  The poster child for this would be any modulation pedal that has an audio path and an LFO, especially if there is a BBD or perhaps some FET-based phase shift stages.  In that instance, the current demands of the LFO might throw off the Vref going to the other circuit portions, hence the greater need to stabilize/smooth deviations in the Vref by means of a cap, using the method shown.

My understanding of the details of circuits is limited, so there may well be a need or benefit to the added cap in the DOD250 circuit, but like I say, since that humble 741 isn't "fighting" any other semiconductors in the circuit, the need for the 10uf cap is considerably reduced, and is probably mostly a matter of addressing poorly regulated wallwarts.

[gaussmarkov]

basicaudio,

[mark hammer posted before i was finished--but i will post my whole response anyway. ]

you've seen this thread, right:  Decoupling caps ...?

as you say, the cap holds the voltage closer to constant.  AFAIK, there are two sources of variation in the voltage.  one is the source that petemoore is talking about, ripple in the power supply.  i don't think batteries have ripple as a general rule.  the ripple comes from converting AC to DC imperfectly.  so, as petemoore says, the issue doesn't arise with batteries.

the second source of variation in voltage is the demands of the circuit itself combined with the internal resistance of the power supply.  this is usually more of a problem with old batteries than with power supplies that have much higher current ratings than the circuit they are supplying.  in multi-transistor circuits for example, each transistor makes its its own demands.  if there is appreciable internal resistance then each demand is reflected in power supply voltage swings for all of the transistors on that supply rail.

in the circuit that you reference, there are no transistors.  instead, there is an ic.  i have seen people say that ic's are generally manage this problem better than transistors, but i don't know why that is.  my guess is that they have some kind of on-board/internal supply capacitor.  in addition, the ic is competing with anything else.

in many applications, i bet the capacitor is there just for insurance--not because someone did a calculation or an experiment and realized it was necessary.  in the old fuzz faces, where you don't see such decoupling caps, it seems to matter.  as stm@runoffgroove pointed out to me, this absence sets up the "dying battery" effect for which fuzz faces are known.  R.G. explains various places that a dying battery has high internal resistance that accounts, at least in part, for this effect.  If one put in a large decoupling cap, this would reduce the influence of that internal resistance appreciably.

that's the best i know.  someone with a better understanding may clarify, extend, or correct what i have said.  and that would be great!

gm

[John Lyons]

Ok, mark and gaussmarkov, thanks,  that makes sence and was what I was thinking along the lines of.
I have read that decoupling thread, nice one!

I wasn't seeing the 10uf as much of a decoupling cap but it's only 22k away from the +9v so it sure is...
For some reason I was thinking it was reacting more with how hard the pedal was driven and taking current away from the 9v source.

So when the bias resistors are 1M/1M and 1 meg to the input pin of the 741, how does this change things?

John

[heyniceguy]

if you use 1M, then you severely limit the current delivery of your Vref network. Lets say right now, you are using 22k/22k. Lets also say that your load for Vref is worst case 0 ohms. Then your current delivery to your Vref network is a maximum of 9V/22k = 0.4mA. Now lets say you use 1M/1M with the same load network. 9V/1M = 0.009 mA, much less current delivery. So you could say it "starves" your Vref of current the larger resistor network you use.

i had a situation when i started out where i used 1M/1M to set Vref. and it resulted in a most horribly splatted wet fart.

[Mark Hammer]

So when the bias resistors are 1M/1M and 1 meg to the input pin of the 741, how does this change things?

John

I've been pondering that very thought for a while, since it is probably THE most robust difference between the Distortion+ and a bunch of other look/workalikes, such as the 250.  Certainly the difference between a 470k bias resistor from the noninverting input to Vref (as in the 250) and a 1M resistor in the same spot (as in the Dist+) is unlikely to yield any substantial difference.  But there is a whack of difference between the bias current available when the Vref is produced from a 1M pair, compared to a 22k pair.

At least one of the peculiarities of the Dist+ is its inability to just about EVER deliver up a clean tone.  Seriously.  Take the diodes out, and turn the gain way down, and you *still* get a dirty sound.  Not a "distorted" sound in the classic OD/Fuzz sense, but rather something you'd rather not run your CD player through if you were attempting to convert your CDs to 8-tracks.  In part, "the Distortion+ tone" is a product of boosting THAT sound and *then* clipping it, not just the clipping diodes alone.

I, for one, would be curious about what the more technical folks here might be able to chime in with as regards the impact of restricting the bias current in a 741.

[Harry Palms]

It provides a low impedance path to ground for stray signal voltages which might otherwise shift the operating point of your bias supply.

[John Lyons]

Now we're getting somewhere! Thanks heyniceguy

In the MXR Distortion + a 1M/1M and 1M is used as the voltage reference to the chip. So this is a major compromise and is starving the Vref correct?
The Dist+ is a pretty ugly and brittle sounding circuit compared to the OD 250 and especially the '77 version (which uses 22k/22K and 470K as the vref, with NO cap). Have the cap there can only help depending on power supply issues but it seems this Vref values may be a cause of some harsh sound.
So the larger voltage reference resistances ( 1M vs. 22K)  will limit current and cause some "junk" sound . Yeah?

EDIT:
Ok, I'm glad that I kept on this and did a new thread here. I was after the differences between the 250 and dist+ etc etc.
Mark had brought up the Vref differences a couple of times in another thread and I thought there had to be more difference in the tone other than a couple slight  cap value changes.
And the cap at Vref is a voltage "shock absorber" of sorts for the bias.

Thanks folks I think I'm seeing the picture now!

John

[JonFrum]

It provides a low impedance path to ground for stray signal voltages which might otherwise shift the operating point of your bias supply.

That was my "pulled out of my a$$" guess. Then again, sometimes a component is just copied from another circuit, and it really doesn't belong there from first principles. I dunno....

[R.G.]

It's there for some of all of the above.

A resistor divider network to establish a reference voltage is always a compromise. Regular readers of GEO will recognize much of the material here from my article on calculating resistors for bias networks. I recommend reading that first.

A bias network sets a particular bias voltage, but at the same time sets both a DC and an AC impedance of that voltage. You may want the impedance of your bias source to be high, or low, or high for DC or AC and low for AC or DC.

First the math - what are we dealing with here?

A two resistor divider from a given voltage V0 makes a second voltage Vr = Y0* R2/(R1+R2) if we name the top resistor R1 and the bottom one R2. If you load that voltage Vr, it looks to any load like it has an internal impedance equal to R1 and R2 in parallel, or Zout = (R1*R2)/(R1+R2). At least at DC. This is one of the results of Thevenin's theorm.

If you're biasing the input of an amplifier and you want a high input impedance, you may want R1 and R2 to be big so that the 9nput impedance is also big, the parallel combination of two resistors. In an opamp input, you would want R1 and R2 to be on the order of 1M - 4.7M to give the input a nice high impedance. You would also use as large a resistor as you can get away with on the base input of a transistor amplifier or MOSFET stage, for the same reason. In this case, you WANT the voltage to be presented to the input, base, or gate with high impedance so it can be easily shoved around by the signal voltage.

If on the other hand you're trying to make the resulting voltage be a quiet place, you want the impedance to be low. So you use low value resistors for R1 and R2, and also may use a capacitor to tie the resultingt voltage to signal ground for frequencies where the capacitor looks like a short circuit. In this "quiet place" scenario, you want any signal to find it very difficult to shove the reference voltage around. Low resistors and big (i.e. low impedance) caps to ground are what you want here. Any time you tie more than one stage to the same reference, you will automatically want the reference to be low impedance to avoid the stages interacting through the reference.

There is one more consideration - noise.

If you may any reference voltage from resistors, the thermal noise of the two resistors is combined in an RMS fashion at the voltage. If you shunt this point to ground with a lower impedance capacitor, the noise is divided down by the ratio of the parallel value of the resistors to the impedance of the cap. So a resistor-resistor-capacitor will always be a lower noise voltage than one without a capacitor. The low noise way to bias with high impedance is to make a low impedance reference voltage with low impedance resistors and caps, then connect one high value resistor from there to the thing being biased. This minimizes any excess voltage or current noise on the resistor, and gives you potentially high impedance with as low a noise as physics... err, Mother Nature will allow.

So the answer is - the cap lowers the impedance of the reference voltage, and lowers the noise on the reference voltage. This ls almost always what you want to do, even if you have to use a third resistor to bias an opamp input or a transistor base.

The only time you would not want a low impedance, quiet reference voltage is... um... I can't think of one.

[Paul Perry (Frostwave)]

An embarassing experience:
I once used 1Meg resistors for a 1/2 Vcc bias..... but, the op amp I was using had enough input leakage to pull the bias right off....... and of course, the leakage varied with temperature, day of the week, and between individual op amps. Made for interesting debugging (bearing in mind, using a DMM or a scope probe with a meg or so impedance is going to push the voltage to a whole other place as well).

It's worth remembering, because this is the insiduous kind of fault, where if you are UNlucky, it 'works' perfectly the first time..... explains a lot of "difficult" circuits one finds around the place

[MartyMart]

An embarassing experience:
I once used 1Meg resistors for a 1/2 Vcc bias..... but, the op amp I was using had enough input leakage to pull the bias right off....... and of course, the leakage varied with temperature, day of the week, and between individual op amps. Made for interesting debugging (bearing in mind, using a DMM or a scope probe with a meg or so impedance is going to push the voltage to a whole other place as well).

It's worth remembering, because this is the insiduous kind of fault, where if you are UNlucky, it 'works' perfectly the first time..... explains a lot of "difficult" circuits one finds around the place

+1 on this one, happened to me twice and is a real PITA to degug !!
Now I use metal film resistor "pairs" of between 10 and 47k and usually a 22uf cap for Vref
This seems to be stable/quiet and provides enough current for Vref ( unless you WANT it starved ! )

MM.

[John Lyons]

Thanks everyone, very informative!
I stumbled opon RG's infor on the subject last night. Funny how I had looked at it a long while back and it didn't make sense to me at that point in my circuit making. ( it made sence, just that I didn't know how to apply it to what I was doing )

John