10 ohm resistor keeps getting hot on pcb for tap tempo tremolo pedal.

[acehobojoe]

So, I have already asked about this in the forum for this pcb, but maybe someone will see it here. I was wondering if anyone knew what could possibly cause a resistor to burn out.. It's the first one in the circuit, right after 9v. I've looked at the schematic for this build and I remain confused.  The resistor just gets hot when I plug it in. No idea the cause, I can send pictures. The build documentation is over at http://musicpcb.com/pcbs/tap-tempo-tremolo/

Hope you guys got some tips!

[WhiskeyMadeMeDoIt]

Surely that's wrong.  Quick off the top of the head says that's 8w. Maybe it should be 100ohm that would be closer to 1w

[amptramp]

D1 in backwards would do it.  Shorted C11, C12, C14 would do it.  That would be the first place to look.  Loads after the 78L05 would have some protection from the current limiter in the regulator.

[acehobojoe]

Hmm, you mean I should put a 100ohm? I believe I put it the diode in the right way. I'll check those other caps. One of them must be shorted. You think it is most likely somehow soldered across to something it shouldn't be?

[WhiskeyMadeMeDoIt]

Follow amptramps advice and look for shorts.  I shouldn't have  commented without looking at your link.   

[acehobojoe]

Ok, thanks. Tomorrow I'll let you guys know the results.

[R.G.]

I was wondering if anyone knew what could possibly cause a resistor to burn out..

It's burning out because (drum roll... ) it's dissipating too much heat.

That is, power P >> rated power. If you're using normal pedal style resistors, this is probably 1/4W, 0.25W. 

We know power can be calculated as P = I *V (that is the voltage across it times the current through it).

Ohm's Law says V = I*R or equivalently, I = V/R or R = V/I.

So P = I*I *R or P = V*V/R.

This means that there is too much current through it or two much voltage across it - those are exactly equivalent statements.

For a 1/4 W resistor, the maximum voltage it can have across it without burning out is V = SQRT(0.25W*R) = 1.58V, or I = SQRT(0.25W/10 ohms) = 0.158A.

Interestingly, V/I = 1.58V/0.158A = 10 ohms. Ohm was right.

So you have something past the resistor that is attempting to pull more than 0.158A through it by lowering the voltage on the downstream side by more than 1.58V.

As noted, anything downstream of the resistor that is shorted to ground, reversed in polarity and trying to leak more than 158ma to ground, or breaking over and trying to force more than 1.58V across the 10 oihm resistor will do it. And the advice you got was right: reversed diode, shorted or reversed capacitors; or shorted PCB/perfboard/stripboard, solder threads, ANYTHING which connects from the downstream side of the 10 ohm to ground, perhaps through a couple of other components.

As an alternate to easter-egging for "whatever could possibly cause a resistor to burn out??", a short session following the advice of "what to do when it doesn't work" would turn up the cause in one or two posts.

[duck_arse]

connecting an AC adaptor would also do the burnings.

[acehobojoe]

I thought that might be the problem, but this is what the adapter says.

[Seljer]

Something is definitely shorting out when it shouldn't be.

Use a multimeter the measure the resistance between the pedal side of the burnt 10ohm resistor and ground. Any low value should be of concern.



That or the reversed polarity protection diode mentioned above. (or even a reversed power supply for that matter with the polarity protection diode functioning as it should)

[Jdansti]

Check the polarity at your PS plug. Make sure that neg is going to ground on your board and pos to 9V.

[acehobojoe]

Alright, will do. Thank you everyone. As always, you guys are the greatest!