Simulate a dying battery mod

[multi_s]

i am not trying to make anyone feel smart/dumb. There is a power dissipation consideration but its just not at the pots min. heres a plot of a pots dissipation over its range assuming 300 ohm circuit res and 1k pot. Blue is Current through the pot (and circuit), red is power dissipated by the pot. Theres a bad assumption that the circuit will behave like resistor, which, if there are any semiconductors involved, it probably wont but for illustrative purposes its ok.



You see the peak of the power (red) plot is not at the pot min but somewhere in the middle. Here it only goes up to ~68 mW but im sure under some combinations you might unexpectedly go beyond rating. (and that its minimum is actually at the pots minimum R)

[McSpunckle]

Ok, did it in a circuit simulator and figured out what you were talking about. The voltage difference between the two sides of the resistance determine the wattage... which was kinda obvious looking back since the resistance is relative to the components position in the circuit...

You win. My bad. : D

Which leaves me perplexed as to what the hell happened to these starve controls. SHRUG.

OK, UPDATE. You can use whatever pot you want. and fuck what I said before.

-edit- Wow, I was totally not accounting for anything in my posts in this thread. O.o Turns out the simple series resistance method has a voltage divider effect on the circuit... so you actually get a voltage drop.

God damnit... now I feel silly. Partly because I wasn't accounting for things I already knew...

NOW I KNOW THAT. So there's good news!

Still, though. Damn.

[MEC]

Every time I have had a pot burn up on a starver it was due to a ground issue. I'm not 100% sure what caused it and I'm not even going to pretend to know math. What I do know is that once everything was grounded proper I got no more smoky pots.

[multi_s]

Ok, did it in a circuit simulator and figured out what you were talking about. The voltage difference between the two sides of the resistance determine the wattage... which was kinda obvious looking back since the resistance is relative to the components position in the circuit...

You win. My bad. : D


Still, though. Damn.

ya im not trying to 'win' or anything. just commenting on what i thought was some misinformation. There could be something else im over looking but i just dont think the pot is burning when its at min, but obviously if you have burnt pots they are burning somehow...

as a conclusion you could say the maximum power will always occur when Rpot = Rcirc, regardless of Vcc. This is probably a more useful fact when trying to figure out if the pot is rated high enough. Since if you know ~Rcirc (by the method you mentioned earlier) you can roughly assume the current when you have 2*Rcirc and then find the power Rpot will dissipate accordingly. I wont do the math unless you want me to show it all the way but basically if you write

P =I*V

With I = Vcc/(Rpot+Rcirc) and V = Vcc*Rpot/(Rpot+Rcirc) then we have

P = [Vcc/(Rpot+Rcirc)] * [Vcc*Rpot/(Rpot+Rcirc)]

as a general plan you can find maximums and minimums of a function by taking its derivative and equating the result to zero. In this case we were interested in the maximum value of P that can occur right? So find the change in P with respect to change in Rpot (aka dP/dRpot)... so if you solve dP/dRpot = 0 you will see that this occurs when Rpot = Rcirc. I think also its actually just proof of the maximum power transfer theorem or something like that. Anyways in the plot thats when it happens, when Rpot = 300ohms = Rcirc. Try it in the simulator you were using with some other values to see when max power occurs. it should hold that it occurs when Rcirc = Rpot regardless of their value or Vcc.

If you think about it though, a pedal with a higher current draw has lower Rcirc right? So Rpot at max power dissipation will be low as well. If you were using say a 1k or 10k pot, the angle of rotation that yields such a low value would be closer to Rmin then in the above plot. So one might conclude once the pot burns that it was because it was at Rmin even though its probably not true, it was in fact at Rcirc (which may be close to Rmin in particular pedals with high current draw bum bum bum everybody 'wins')

[modernage]

Are you guys talking about fantasy league baseball?

[eatyourguitar]

so not having a ground on the amp leads to DC coming up the guitar cable into the pedal and up the starve control?

[jfrey]

Side question: Does starve have any decent effect on Devi's?

[eatyourguitar]

[youtube]http://www.youtube.com/watch?v=PMBs3Qlg4Y0[/youtube]
and there it is

[jfrey]

Question. The way it shows here:
http://www.beavisaudio.com/Projects/DBS/

In the last image (for adding starve to a pedal) will it also starve when you're running from battery? Or just from an adapter?

Also, if there isn't much effect with a certain value pot, would increasing the pot value help?

[McSpunckle]

Question. The way it shows here:
http://www.beavisaudio.com/Projects/DBS/

In the last image (for adding starve to a pedal) will it also starve when you're running from battery? Or just from an adapter?

Also, if there isn't much effect with a certain value pot, would increasing the pot value help?

1) It'll starve from a battery or power supply. Either way, yer good.
2) Yup! There are pedals where starving won't work, but for most analog circuits-- especially fuzz/distortion-- you're good.

[MEC]

Question. The way it shows here:
http://www.beavisaudio.com/Projects/DBS/

In the last image (for adding starve to a pedal) will it also starve when you're running from battery? Or just from an adapter?

Also, if there isn't much effect with a certain value pot, would increasing the pot value help?

With a Voltage Divider type starver, you will always stay between whatever your minimum voltage is (as set by the fixed resistor between ground and lug1) and the amount of voltage the adapter supplies. Lowering the fixed resistor value will allow you to reach lower voltages. Increasing the pot value will just make for finer tuning. Depending on where you set your minimum a 10-100k pot should be suffcient.

I find that most of the usable* tones for guitar playing fall between 6-9 volts and for noise, between 3-9 volts.

* term is highly subjective.

[jfrey]

With a Voltage Divider type starver, you will always stay between whatever your minimum voltage is (as set by the fixed resistor between ground and lug1) and the amount of voltage the adapter supplies.

I don't understand. In that picture Lug 1 is going to DC+ on the board, not ground.

[multi_s]

[
With a Voltage Divider type starver, you will always stay between whatever your minimum voltage is (as set by the fixed resistor between ground and lug1) and the amount of voltage the adapter supplies. Lowering the fixed resistor value will allow you to reach lower voltages. Increasing the pot value will just make for finer tuning. Depending on where you set your minimum a 10-100k pot should be suffcient.

Increasing the pot value will allow you to get lower voltages given a fixed resistor from the pot to ground. If anything it will make the tuning less fine as well since you now have a larger voltage range over the same range of angular rotation. Also dont forget the circuits resistance is in parallel with the bottom half of the divider so you have to factor that in when calculating the result.


go here for some simulation...

http://www.falstad.com/circuit/#%24+1+4 ... 05+0+-1%0A

on the table on beavis i believe he is just giving the value if you use the 10k pot, i dont think he is implying that the output is actually completely determined by the fixed resistor. The range will change if you use a different value pot with those R values and also he measures it unloaded. When you plug a circuit it you will almost definitely not have those output values (you should get something less).

[multi_s]

also try the mothers day mega mod.

http://www.falstad.com/circuit/#%24+1+5 ... 25+0+-1%0A

[MEC]

With a Voltage Divider type starver, you will always stay between whatever your minimum voltage is (as set by the fixed resistor between ground and lug1) and the amount of voltage the adapter supplies.

I don't understand. In that picture Lug 1 is going to DC+ on the board, not ground.

I apologize, the link you gave just went to the main page. I didn't read your previous post and assumed you were talking about building a stand alone starver and not building one into a pedal. I would recommend the stand alone starver because then you can use it on multiple pedals and you don't risk messing anything up or devaluing your pedal.

Here is the image I was referencing:

Increasing the pot value will allow you to get lower voltages given a fixed resistor from the pot to ground. If anything it will make the tuning less fine as well since you now have a larger voltage range over the same range of angular rotation. Also dont forget the circuits resistance is in parallel with the bottom half of the divider so you have to factor that in when calculating the result.

on the table on beavis i believe he is just giving the value if you use the 10k pot, i dont think he is implying that the output is actually completely determined by the fixed resistor. The range will change if you use a different value pot with those R values and also he measures it unloaded. When you plug a circuit it you will almost definitely not have those output values (you should get something less).

Here is the table:
Resistor Value Minimum Voltage
2.2k 1.8 volts
4.7k 3.1 volts
10k 4.9 volts
15k 5.8 volts

When using the diagram I posted above in response to jfrey and according to the above table the fixed resistor sets the minimum amount of voltage and the adapter sets the maximum amount of voltage.

If you are using an adapter that puts out 9v with a 10k pot and A 4.7K fixed resistor your maximum voltage will be 9v (pot all the way off) and your minimum voltage will be 3.1 volts (pot all the way on).

If you are using an adapter that puts out 9v with a 100k pot and A 47K fixed resistor your maximum voltage will be 9v (pot all the way off) and your minimum voltage will be 3.1 volts (pot all the way on). In this case the 100k pot should offer finer tuning.

Keep in mind this is for a stand alone starver not connected to a circuit. If you were to build this into a pedal you would get wide variations depending on the circuit.
There are a lot of variables, as you and others have listed in this thread, and it's obvious you posses far more electronics knowledge than I ever will, but if you're just looking for a cheap and easy way to reduce the voltage supplied to a pedal the Beavis method works fine in my opinion.