Resistance and volts vs. amps question

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pricklyrobot
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Resistance and volts vs. amps question

Post by pricklyrobot » Sun Dec 28, 2008 7:55 pm

I have a homemade CV pedal, that puts out the 0 to +10V (actually I guess it's only 0 - 9V, but close enough) needed to control the Filter CV input on my Prophet-600, and it works great for that. I also have an old Digitech delay unit with a CV input to control Delay Time that I'd like to put to use, but it takes 0 to +5V. So I'm thinking I could just add a switch and a resistor to the current pedal to make it work with both.

Currently the pedal goes: 9V battery -> 100k pot -> 1/4" mono cable

For a switchable pedal, I'm thinking: 9V battery -> DPDT switch, one side goes straight to the pot, other side to resistor (of ? Ω) then -> 100k pot -> 1/4" mono cable

So firstly, does this sound like it will work; is there another component needed that I'm overlooking? If it will work what value resistor do I need to put in there?

I tried just clipping a few resistors straight on to the + terminal of a 9V battery and taking some readings with my multimeter. Problem is I'm not sure what I'm supposed to be testing (should it be the DC V or the mA?) The resistors that were cutting the current approximately in half (from about 23 mA, down to around 11.5), were only cutting the V down from 9V to about 8 or a little under. And when I found a resistor that would cut 9V down to about 4.5V, I was getting a reading of 0 when I switched over to mA again. :scratch:

My electrical and math skills are still very much in the novice range, so any suggestions, tips, explanations, etc. are much appreciated.
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Re: Resistance and volts vs. amps question

Post by nathanscribe » Sun Dec 28, 2008 8:31 pm

If you want to cut down 9V to 4.5V, you want a potential divider. Basically, that's a resistor from the positive to one end of a pot, and the other end of the pot to ground. The wiper of the pot is where you take your output from. If the value of the pot and the value of the resistor are equal, you will be able to tap off a voltage between zero and half the positive. As for which values to choose, a general purpose CV divider might use anything from 10K to 1M. Moog modulars used to use 25K pots on their attenuators I think, some folk use 50K (47K is a readily available value too) and I often use 100K. The absolute values are not as important as their relative values - that is, a 10K res and a 10K pot together will allow you to tap off the same voltage range as a 100K res and a 100K pot together. The voltage divider works proportionally.

Have a look here for easily digestible basic electronics information.

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Re: Resistance and volts vs. amps question

Post by cornutt » Sun Dec 28, 2008 9:55 pm

You're making the problem overly complicated. The pot is wired as a voltage divider, and what matters is the output voltage, not the current. Think of it this way: The resistor track inside the pot is connected to 9V from the battery at one end, and ground at the other end. The wiper touches it at some point and picks up the voltage that is present at that point. If you could extend the resistor track and make it twice as long, but stop the wiper so that it can only access from the ground end to the halfway point, you'd have a 0 to 4.5V output.

You can do this for real by adding a resistor in series to the 9V end of the pot. The resistor's value needs to be the same as the pot's value, so they will divide the voltage evenly. (If the pot is a 100K pot, you need a 100K resistor, etc.) This way, the voltage at the pot's positive end will be 4.5V. Arrange the switch so that it simply shorts out the resistor when on, so you get the 0-9V output. So the chain of connections will be:

battery--->resistor---->pot

with the switch bypassing the resistor.
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Re: Resistance and volts vs. amps question

Post by pricklyrobot » Mon Dec 29, 2008 12:40 am

nathanscribe wrote:Have a look here for easily digestible basic electronics information.
Thanks for the link. Looks like some good stuff.

I tried wiring this up to a DPDT switch, like this: 9V battery+ split straight to one side of the switch, 9V battery+ through 100k resistor to the other side of the switch. Taking readings straight out of the switch, this seems to work fine, flip the switch one way I get approx 9V, flip it the other way and I get 4.5V. The problem comes when I hook this up to the pot (also 100k). With the pot all the way open, I flip the switch one way get 9V, flip it the other way and only get about .7V (incidentally, the pot still seems to work normally in this configuration, reading about .35V at half turn, and so on). Any ideas on where I'm going wrong here?
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Re: Resistance and volts vs. amps question

Post by pricklyrobot » Mon Dec 29, 2008 1:07 am

I just tried switching it around, battery straight to pot -> two wires from wiper of the pot, one straight to one side of the switch, the other through 100k resistor to the other side of the switch. And now I'm getting proper readings. Don't know why the order made the difference. I guess this means I could use a SPDT switch instead.
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Re: Resistance and volts vs. amps question

Post by pricklyrobot » Mon Dec 29, 2008 2:58 am

Actually I don't think it's working this way either. I'm getting the right voltage reading, but still getting no detectable current on 4.5V switch setting (with the 100K resistor in line). #-o
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Re: Resistance and volts vs. amps question

Post by cornutt » Mon Dec 29, 2008 4:49 am

Two things on that current deal. First of all, maybe you know this, but... in order to read current, the meter must actually be in the circuit. You can't just probe and see amps, like you can with voltage.

Second, if I understand correctly, you have a 100K pot and a 100K resistor in series. Your current through that string is only going to be 45 uA. Most hand-held meters can't read reliably below 100 uA or so.
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Re: Resistance and volts vs. amps question

Post by pricklyrobot » Mon Dec 29, 2008 6:25 am

My problems were coming from misreading the resistor code. I had a 1 Meg resistor in there instead of 100k; guess I should've checked that with the meter. #-o

Anyhow, went back in did it the way you suggested originally: 9V battery -> 100k resistor (wired to a switch) -> 100k pot that's already in the pedal. Doing it that way, I can toggle between 9V (23 mA with the pot wide open) and 4.5V (only .1 mA with the pot wide open, but it seems to work and give me the full sweep of the Delay Time). Now I can make obnoxious boinging, pitch-shifting noises. Hooray!

I wasted several hours because of a silly mistake, but I guess it was a good learning experience in the end. Thanks both for the info and help. :thumbright:
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Re: Resistance and volts vs. amps question

Post by nathanscribe » Mon Dec 29, 2008 11:00 am

If something works perfectly first time, you've learned nothing. :wink:

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Re: Resistance and volts vs. amps question

Post by Bross » Mon Dec 29, 2008 7:37 pm

If you were really cheap you could use a SPST switch. Wire it in parallel with the resistor.

Or if you wanted to be smooth you could use a three position switch to choose 9V-with-resistor, 9V-direct, or open (no connection) so you add an off position that won't drain the battery when you're not using the pedal.

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Re: Resistance and volts vs. amps question

Post by pricklyrobot » Mon Dec 29, 2008 7:49 pm

When I originally made the pedal, I mounted the battery on the outside on the side of the pedal, so I just unhook it when I'm not using it. Not as slick as your suggestion, but it works and it was a cheap, ugly pedal to start with anyway. Maybe some day I'll make a nice one in a metal housing and wire it up all fancy-like, figure out how to make an LFO and throw that in there too.
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