Simulate a dying battery mod

[nieh]

Can I add like a resister to the circuit of the fuzz pedal like on the dc jack to limit the power that gets to the pedal?

[Dr. Sherman Sticks M.D.]

i think u can do this.

though, i think what most ppl do is put a pot there so u can vary the amount of power. theres a nice writeup about dying battery sim's on beavis audio website.

i've also heard that this simulation is not exactly how an actual dying battery will sound, since a dying battery is more unpredictable....or somethin like that

[nieh]

I tried the devolt thing. I was just going to experiment with different values and put a switch in to change back to stock.

[McSpunckle]

Can I add like a resister to the circuit of the fuzz pedal like on the dc jack to limit the power that gets to the pedal?

Yes.

i've also heard that this simulation is not exactly how an actual dying battery will sound, since a dying battery is more unpredictable....or somethin like that

A dying battery is more complicated, and starve controls are more complicated than a simple voltage starve (You actually correctly used the word "power").

As a battery is dying, the voltage will drop, but the internal resistance will also go up. Internal resistance is basically just a theoretical resistor in series with the battery. It depends on the battery type, but they'll only be up in the 10s of ohms. Nowhere near the 10K of a typical starve control. But, if you just turn a starve control down a tiny bit, you can hear the difference.

Here's why it's hard to perfectly simulate: You can't get down to, say, 7 volts with a similar series resistance with the typical 10K pot used.

There are two ways to wire starve controls. Series resistance, and voltage divider. The series resistance actually doesn't affect the voltage much at all. It simply limits the amount of current available to the circuit. The voltage divider does reduce the voltage, but with a 10K pot by the time you get down to the 7 volts (assuming you have an extra resistor preventing a short circuit), you'll have about a 6K series resistance. So, yeah, the voltage is right, but you're also starving the current wayyy more than a dying battery would. Usually the pedal will gate off pretty quick if you use this method, because the pedal's hardly getting any power.

For your convenience, I didn't do any math. In my experience, the simple series resistance method is works best, and sounds most like a dying battery. But after writing this I realized I need to try about a 100 ohm pot as a voltage divider...

[eatyourguitar]

thankyou for your insight. I always knew they sounded different but there are some talented builders here who disagree. now I know the science behind it. I really dont cry for trees when I sell someone a pedal with a battery hookup. I always tell people to use zinc batteries when they show me a vintage fuzz.

[McSpunckle]

The internal resistance of Zinc-Carbon batteries is higher than Alkaline, so there's actually a chance of them sounding different. : O. But, I mean, you can put a 100 ohm resistor in series with your power supply to help prevent hum. So I don't know how much the effect -really- is. Imma test it one day...

Plus, One Spots put out almost 10 volts, so the power supply comes into play, too.

[nieh]

The internal resistance of Zinc-Carbon batteries is higher than Alkaline, so there's actually a chance of them sounding different. : O. But, I mean, you can put a 100 ohm resistor in series with your power supply to help prevent hum. So I don't know how much the effect -really- is. Imma test it one day...

Plus, One Spots put out almost 10 volts, so the power supply comes into play, too.

I messed around with some different value resistors in a row until a got the sounds I wanted. I was just going for a more gated less muffled sound. It worked out pretty well. And I put in a switch to go back to stock.

[eatyourguitar]

But cant put a 9v regulator cause it looses a 1v to 3v so a 9v regulator on 9v = 8v

[univalve]

How much Volt has a dying battery approx.? 7V?

[mathias]

But cant put a 9v regulator cause it looses a 1v to 3v so a 9v regulator on 9v = 8v

There's those adjustable 5V/12VDC regulators, where you put a resistor across two leads to raise/lower the voltage. A pot there would let you build a starve and also pump out >9V. For maximum class, I'd put a analog voltmeter on there too.

[McSpunckle]

So, I've seen a couple of pedals where the starve controls have fried, and heard about several. So I thought I'd come in and make a suggestion. It's not related to the specific problem in the thread, but I think that's been resolved.

Typical 16mm pots are 1/8 watt, like those tiny resistors. I'm not -totally- sure how much these can handle in varying positions, but when they're about all the way up, all the current in the pedal can run through like 1 or 2 ohms of the pot (it's never quite like a switch), and if you're pedal is pulling 35mA, that's 9x.035 watts, or about 1/3 of a watt going across the pot. In one pedal I've repaired, there was literally a hole blown in the carbon right at the end of the carbon element in the pot.

So, always know the current of the pedal if you're going to add a starve control, and pick the pot accordingly. Protip: isolate the starve control from the LED. The current drawn by a fuzz circuit is seriously tiny. I measured .005A (5mA) on a Gnomeratron's fuzz circuit coming off the starve control.

[multi_s]

Typical 16mm pots are 1/8 watt, like those tiny resistors. I'm not -totally- sure how much these can handle in varying positions, but when they're about all the way up, all the current in the pedal can run through like 1 or 2 ohms of the pot (it's never quite like a switch), and if you're pedal is pulling 35mA, that's 9x.035 watts, or about 1/3 of a watt going across the pot. In one pedal I've repaired, there was literally a hole blown in the carbon right at the end of the carbon element in the pot.

Do you mean with the pot in series? Check you r calculation... If the pot was turned down to 1 or 2 ohms you would not have anywhere near a 9 volt drop across it. Its true all the current runs through the pot but V = IR so say 35mA and 2 ohms as per your example.... 2ohms * .035 A = .07 Volts != 9 Volts.

(and it follows .035A^2 * 2 ohms = 2.45 mW, much less than 1/8 Watt)

[McSpunckle]

You don't lose much voltage through a series resistor. It limits current.To use ohms law, you have to factor in the rest of the resistance, not just the 2 ohms. I was just stating that to basically say "The current is still going through the pot."

So, let's math this out.

If you have 35mA of current on a 9V power supply, you can find the resistance of the circuit (between the +9V and ground) by doing R=V/I , and we end up with R=9/.035, thus R = 257.14 Ohms.

I was accounting for that 2 ohms in the pot, because it wasn't that relevant, but adding it in, we end up with, let's say, 260 ohm (added a vanity ohm).

If I = V/R, then 9V/260R=.03A, or 30mA. And if W=V*I, then 9V*30mA= .27 Watts, well over the 1/8 (.125) watt rating of the pot. It's even a bit too much for some forms of the larger pots.

Your calculation was for when the 2 ohms resistance is in parallel with the circuit, and if that was the case, the 9V power supply would have to put out 4.5A, and none of the major pedal power supplies can do that.

[multi_s]

You don't lose much voltage through a series resistor. It limits current.To use ohms law, you have to factor in the rest of the resistance, not just the 2 ohms. I was just stating that to basically say "The current is still going through the pot."

So, let's math this out.

If you have 35mA of current on a 9V power supply, you can find the resistance of the circuit (between the +9V and ground) by doing R=V/I , and we end up with R=9/.035, thus R = 257.14 Ohms.

I was accounting for that 2 ohms in the pot, because it wasn't that relevant, but adding it in, we end up with, let's say, 260 ohm (added a vanity ohm).

If I = V/R, then 9V/260R=.03A, or 30mA. And if W=V*I, then 9V*30mA= .27 Watts, well over the 1/8 (.125) watt rating of the pot. It's even a bit too much for some forms of the larger pots.

Your calculation was for when the 2 ohms resistance is in parallel with the circuit, and if that was the case, the 9V power supply would have to put out 4.5A, and none of the major pedal power supplies can do that.

You are contradicting yourself. If you agree that there is only a small voltage drop over the pot then how can you say the power dissipated by the pot is 9V*30 mA. Its impossible because if there was a 9v drop over the pot you would have 0v over the equivalent circuit resistance you mentioned (IE we expect it to be infinite, and if it was infinite you would have no current flow).

Consider
eq 1) V = I*R
eq 2) P = I*V

Substituting 1 into 2 yields

P = I*I*R

The resistance of the pot is 2 ohms, and the circuit 268 or whatever. Due to series connection the current through the pot is the current through the whole circuit. However by KVL the voltage across the pot will not be the voltage across the whole circuit. You can solve it a less direct way and get the same result using voltage divider, Vpot = Vcc * (Rpot/(Rpot+Rcirc)) then use Ppot=Ipot * Vpot.

You are right the power dissipated by the whole circuit would be .27 watts but consider how much of that is over the pot.

Also, if you check you will see my calculation was for series connection.

[Jero]

Thanks guys, I now feel dumb