Sag pot sparking and smoking

[Bobandy]

I am having an issue with my build. I wanted to add an adjustable voltage pot and have burned through 3 already. I have had a wire running from the power jack to the outside lug of a pot and have the other outside lug to ground.  The wiper runs to the power of my circuit but every time I turn the knob towards ground it starts sparking and lighting  up. I have burned through a 5k,20k, and 50k. Any suggestions on what may be the cause?

[nocentelli]

Just put the sag pot in series from the +9v to the power supply point on the circuit using lug2 and 3. What circuit are you using this with?

[PRR]

Turning a potentiometer connection to GROUND should not spark/smoke.

Agree we need the whole story.

[R.G.]

Through careful processing, you have discovered a way to make a LEP - a Light Emitting Potentiomenter.

Sadly, the expected rated operational hours on these are not great. After a short half-life, they transmute into Darkness Emitting Diodes, or DEDs.

Poteniometers are limited in how much current they can pass. If you wire them up as a variable resistor (and "sag resistors"  do this) you must limit the current through them to no more than the current that would make the entire potentiometer dissipate its rated power.

For instance, if you have a 10K, 100mW pot, you would find that since P = I²R, the current it can pass without burning up is I  = SQRT(P/R) = 3.5ma. That is the safe current for **any part of the conductive element**.

If you have this hooked up as a variable resistor and turn the resistance down until the remaining section of the pot from one end to the wiper is carrying more than 3.5ma, it overheats that small section of the pot, and can - and does, as you found out - burn it out.

[vigilante397]

I always hook it up like this:

Lugs 2&3: +9v
Lug 1: board power

In my understanding (correct me if I'm wrong) hooking it up

Lug 3: +9v
Lug 2: board power
Lug 1: ground

then turning the pot down means connecting a resistor (your pot) straight between voltage and ground, which means (depending on the value of the pot) a possibly uncomfortable bit of current across your pot.

[Bobandy]

OK that makes more sense. I don't remember where I got the schematic for the sag control but it had the last lug going to ground. I'll see if I can scrounge up another pot and get it working. How much difference will the pot make?  I saw somebody recommend a 5k but I might only have a 50k left.

[GibsonGM]

Well, Vigilante...what happens when your variable resistor's resistance approaches low levels??   Do the math...if you are at 100 ohms, you have:

9/100=90 mA

.09A*9 = .81W!   

Of course, this will get worse as you turn the pot down and decrease the resistance.

Doesn't seem very good for a .5W rated pot, LOL....the limiting resistor sure sounds like a good idea!    We often 'get away' with doing some of this stuff, but at some point the blue smoke comes out...

Edit:  Using more than about 160R, 1/2W, for a series resistor with your 1/2W pot will limit the current to safe levels so you can play with the pot.    <<   This assumes there is no limiting resistance AFTER the pot...of course, the maximum current that can flow in the circuit is just that...if it is less than the amount required to exceed the pot's rating, you're good to go!! 

[PRR]

> if you have a 10K, 100mW pot, you would find that since P = I2R, the current it can pass without burning up is I  = SQRT(P/R) = 3.5ma. That is the safe current for **any part of the conductive element**.

Assuming 0.125 Watt rated pot, and 9.0 Volts, the maximum current for several values computes to:

1K - 11mA
5K - 5mA
10K - 3.5mA
50K - 1.6mA
100K - 1.1mA

A small 1-transistor boost may be under 1mA.

Some elaborate circuits pull 50mA (and up and up).

LEDs are commonly run at 2mA to 10mA (and why would you want to "sag" an LED?).

So some notion of what you are powering is needed.

You can get 2-Watt pots. You can get 100-Watt pots. (Bring money.) You can buffer a small pot with a transistor (transistors still burn out, but usually don't smoke when they do). But shoveling bigger parts at a problem can get costly, some thinking could be a better path.

[guidoilieff]

from experience I can say there is a large amount of voltage going trough (ground connected to the 9v when the pot reaches the end). Maybe you made a mistake with some resistor?

[vigilante397]

from experience I can say there is a large amount of voltage current going...

[vigilante397]

Well, Vigilante...what happens when your variable resistor's resistance approaches low levels??   Do the math...if you are at 100 ohms, you have:

9/100=90 mA

.09A*9 = .81W!   

I agree, that's why I don't think one of the pot's lugs should go to ground. By grounding one lug you remove the rest of the circuit and the power supply only "sees" the resistance of the pot. If you wire the pot up as a variable resistance then it just becomes an additional current limiting resistor "seen" in addition to the rest of the circuit.

As a disclaimer I should mention the only pedal I really utilize a sag on is the Fuzz Factory ("stability" control), which uses a 5k pot wired as I mentioned above and I've never had any problems personally

[Jdansti]

Glad y'all figured it out. When I first read the title of this thread, I thought it was about drugs. 

[amz-fx]

Though it may not be the way that you intend, only 2 connections to the pot are required for a sag control. Only simple transistor circuits respond well to a sag resistance so there is no good reason to use it with an opamp and certainly not a digital circuit such as an echo/reverb. Some cmos circuits do odd stuff with resistance in the voltage supply, but they will be low current draw, and should not present a problem.

Also, as PRR said, there is no need to sag the LED so you should place it before the pot control as shown here.



regards, Jack

[stonerbox]

Through careful processing, you have discovered a way to make a LEP - a Light Emitting Potentiomenter.

Sadly, the expected rated operational hours on these are not great. After a short half-life, they transmute into Darkness Emitting Diodes, or DEDs.

You are killing me R.G! 

Bobandy welcome to the forum! We have all lit up a potentiometer at one point or another!

By the way is that you Randy? 

[GibsonGM]

Well, Vigilante...what happens when your variable resistor's resistance approaches low levels??   Do the math...if you are at 100 ohms, you have:

9/100=90 mA

.09A*9 = .81W!   

I agree, that's why I don't think one of the pot's lugs should go to ground. By grounding one lug you remove the rest of the circuit and the power supply only "sees" the resistance of the pot. If you wire the pot up as a variable resistance then it just becomes an additional current limiting resistor "seen" in addition to the rest of the circuit.

Yes, very true - and then you're ok as long as the draw of the rest of the circuit doesn't exceed the pot's rating, or else when you drop your pot's R below the threshold that will allow excess current flow, and it will smoke the same as if one leg were to ground, because it is, thru the circuit     

So we need to know max current in that circuit leg to find out if we need a limiting resistor or not.  Most 9V transistor circuits shouldn't be drawing 1/2W of current tho...

Smoking pots is awesome.  Everyone has at least tried it once or twice.  It's part of the learning curve!   

[Bobandy]

Thanks for the help. I rewired it earlier without lug 3 to ground and it worked great.  It is a fuzz circuit by the way. Build problem one solved, now gotta troubleshoot the return.

[R.G.]

If you simply must smoke pots, don't inhale.

[Jdansti]

If you simply must smoke pots, don't inhale.

Ah yes, one of our former presidents was into electronics... and cigars... and...