question about input gain control before pulldown resistor

[ode2no1]

i'm working on a dirt pedal and decided to try an input gain control on it to allow for cleaner tones at the lowest gain setting. i used a 100k trimpot at the input and decided that i was happy with it set at halfway, so figured i'd just use a voltage divider on the input with a 50k resistor in series and a 50k to ground. my question is, if there is a 1Meg pulldown resistor after the voltage divider will that change what i am trying to accomplish? will it turn into 50k in series and 1.05Meg to ground? or will the divider be complete as i had planned because it comes before the 1Meg resistor to ground? sorry if this is a dumb question. i had it on a breadboard without the pulldown there and won't be able to mess with it for a few days so i figured i'd ask instead of going crazy thinking about it like i have for the last two days.

[bluebunny]

You don't really need the "pulldown" resistor, since the lower half of your divider is providing that function (assuming that I'm picturing your description as you see it!).  And yes, if you leave it in, it's in parallel with the other one and will affect the circuit - but not by much (1M||50K = 47.6K).

[kaycee]

I've wondered about this too? I've seen this kind of pre gain pot as both a voltage divider, and as a series resistance. I've also seen the pull down in front and after the pot. Is there any difference as far as the circuit input? I'm thinking impedance. I've found that with the pot as a voltage divider you still need a pull down or it pops on switching.

[R.G.]

bluebunny is correct. It's all just series/parallel resistors.

Imagine that a voltage divider pot is two resistors, one from the signal input and another to ground. The wiper is in the middle between them.

If you put a resistor from the wiper to ground, you are effectively making the "bottom" resistor to ground be the parallel combination of the "bottom" section of the pot resistors and the added resistor to ground. In fact, this is exactly the setup that I converted to equations to calculate the response of the tapering resistors in "The Secret Life of Pots".

I've wondered about this too? I've seen this kind of pre gain pot as both a voltage divider, and as a series resistance. I've also seen the pull down in front and after the pot. Is there any difference as far as the circuit input? I'm thinking impedance. I've found that with the pot as a voltage divider you still need a pull down or it pops on switching.

There are some simple and fast ways to calculate the effects of this kind of resistor combinations. In my first circuits class, one entire lecture was spent on Thevenin/Norton equivalent circuits and how to combine them as needed. I'm just going to state a few consequences of this way of thinking on pots and added resistors, but there's a fair amount of math and explanation hiding behind it.

If you have two resistors set up as a voltage divider, R1 on top and R2 to ground, the voltage at the junction **in the absence of significant loading on this junction** with a voltage Vin applied to the top is

Vout = Vin * R2/(R1+R2)

This is the familiar voltage divider equation. But how does it act if there is a load? The answer is that it acts just like a voltage of Vin*R2/(R1+R2) with a series resistance of R1 parallel to R2.

This is the Thevenin version of the circuit. A pot is nothing except an arrangement that sets up R1 and R2 so that R1+R2 is a constant, that being the nominal resistance of the pot. R1 decreases the same amount R2 increases, and vice versa.

So yes, you are right, a voltage divider pot acts like a voltage reducer, plus a resistance in series with it. The resistance in series is not fixed though. It's 0 (for the ideal pots you can't really buy!) at both maximum and minimum, and in the electrical middle it's 1/4 of the total value of the pot.

There is a hidden component here, and I think it may account for your "needs a pulldown or it pops" comment. The contact of the wiper on the resistor element is not perfect. There is some resistance internally in series with the wiper contact, and this is variable and may have erratic contact, especially as the pot wears. I'm surprised that it would be much bigger than 1M, which is generally an OK value for a pulldown, but maybe. In general, pots set up as input or output level dividers don't need pulldowns. Maybe there was something else going on in the circuit.

[ode2no1]

Thanks for the responses guys. By the way, am I correct in using two 50k resistors to emulate a 100k pot set at noon? I feel like I am, but I'm second guessing myself and thinking maybe I should be using two 100k resistors.

[PRR]

> am I correct in using two 50k resistors to emulate a 100k pot set at noon?

Yes.

If it is a Linear pot. ("Trim pots" are almost always linear.)

If you find 47K instead of 50K, nobody will know the difference.

No added 1Meg pull-down is needed -- the bottom 50K does that excellently at no added cost or load.

[PRR]

FWIW: two 68K resistors is a VERY common thing-- found on the input of most Fender and Fender-like amplifiers as a cut-down for hot pickups. 68K is not very different from 50K. Try it, it should be fine.

[ode2no1]

I'm actually in front of my breadboard right now. Let's see if I can explain a problem I'm having without any real way to upload a picture....I want to use a spdt to switch between straight to the input cap and a lower gain setting with the voltage divider before the input cap. Problem is then I have a 50k resistor to ground hanging there when I'm in the normal input mode. I know I can do this with a dpdt and get that resistor out of the circuit, but is there any way to do this with an spdt?

[ashcat_lt]

I don't believe so.  You could do it so that it has the full 100K of the two resistors, which might be a little better.

Edit - Actually (!!!) if you can live with the other 50K in series with the input, you could just lift the ground end and defeat the voltage divider.  Then, of course, you'd have a 50K antenna connected to your input, but...

Edit again - err...I think that I can see how you could lift the ground end and short that bottom resistor.  Bottom of  the  resistor to the common, top of the resistor one outside lug, and ground to the other.  Then you don't have it acting like an antenna when defeat the divider, but still have the 50K in series with the signal.  How much that matters depends on just about everything else in the circuit.

[peterg]

If you have room for a switch why not install at pot? Same space - more flexibility

[ode2no1]

After testing some more it doesn't really matter too much that the signal gets loaded down by that resistor. I made a tweak somewhere else that left me without dull clean signal going thru the pedal. Sounds pretty much the same on or off now with the gain all the way down, which is a relief. The reason I don't want to use a pot is that it just isn't necessary. I just wanted to be able to get into that cleaner range and then back to stock. Thanks so much for the help guys. I'm excited to get this pedal into a box.

[PRR]

> why not install a pot? Same space - more flexibility
> it doesn't really matter too much that the signal gets loaded down by that resistor.

Good thoughts/observations.

> I don't believe so. ... Edit - Actually (!!!) if ....  Edit again - err...I think that I can see how .....

Right. I'm not sure I follow your text explanation, but on the chalkboard I see this:

[ashcat_lt]

Right. I'm not sure I follow your text explanation, but on the chalkboard I see this:

That's not what I what I was talking about, but it's even better!