Electrical fires

[spargo]

So, I just boxed up an effect and plugged it in.  Got a bit confused when the LED wasn't coming on while flipping the bypass.  I left the back cover off for testing, so I lifted it up off the table to look inside and smoke poured out.  Smelled a lot nastier than solder smoke, that's for sure...  Upon closer inspection, I seem to have accidentally crossed the +9V and GND connection at one point.

Now, there is a 1N4001 diode to GND in the power filtering.  Should this not have helped?  Or only specifically for reverse polarity protection?

Also, I inspected the components as much as I could from the top side of the board, and I see no visible damage of where the smoke was coming from - but there was a significant amount.  Safe to fix the short and flip it back on?

[spargo]

Fixed it, fired it back up...no smoke, LED lights.  No sound for bypass or effect.  What's most intriguing is that it does still seem to have one function: the volume pot varies the brightness of the LED. 

[LucifersTrip]

I believe smoking is usually from overheated resistors. I would check those first...

[wavley]

I'll tell you a secret about electronics.  It won't work if you let the magic smoke out shhh

[spargo]

Ok, wow. Some very noob mistakes. I fixed all of the incorrect output wiring and now the bypass works, and the LED is all good.  The effect does not work though, so I'm assuming something got fried.  The circuit has a fairly average power filtering section, resistor, caps, etc. as well as a transistor in the circuit.  Could the transistor be gone, or do you still suspect a resistor? (I'd assume the one in the power section?)

[freeride]

ah the magic smoke.  gotta keep that inside man.  A lot of things could have gotten fried here.  It could be a simple wire that burnt up, or an IC that fried...I'd check some bias points and see if you can determine where it happened.  If we could see a circuit and you could tell us where your wiring mistakes were (like the power/ground short) we could probably help you figure out better what got toasted.  Do you have anything else accidentally grounded...like the output?

[R.G.]

"LED" is an acronym for Light Emitting Diode.

It's not all that widely known, but if you hook them up backwards to a power source capable of enough power, the high reverse current inverts the action, and instead of emitting light from the junction, they actually suck photons out of the surroundings, hence the name Darkness Emitting Diodes.

Or "DED".

[oldschoolanalog]

RE: "Magic Smoke":
http://www.diystompboxes.com/smfforum/index.php?topic=83237.0

[freeride]

Hey R.G. I like the direction you went! You've clearly got way more knowledge on here than most everybody, but hey let's broaden the scope and not limit ourselves to LEDs.  In fact all diodes (or really PN junctions) work as even solar cells and photodetectors depending on your bias.  Light hits the PN junction and gets absorbed by an electron, giving it the energy to leave it's spot and get into the conduction band where an electric field moves it (moving electrons = current), and a "hole" is left behind where it was.  So these electron-hole pairs get created and then split apart in the depletion region of the PN junction (where the P and N material touch) and then get swept to either side of the diode by the built in electric field = current.  So in fact, we don't even need the big reverse bias (needed for effective photodetectors though) in order to work the opposite of LEDs and actually absorb photons, we just need light which gets absorbed to create those electron hole pairs and the subsequent reverse current to power something - all from just light!  Who's going to make the first solar powered effect pedal?  I bet it will work great in dark bars...

[R.G.]

Hey R.G. I like the direction you went! You've clearly got way more knowledge on here than most everybody, but hey let's broaden the scope and not limit ourselves to LEDs.  In fact all diodes (or really PN junctions) work as even solar cells and photodetectors depending on your bias.  Light hits the PN junction and gets absorbed by an electron, giving it the energy to leave it's spot and get into the conduction band where an electric field moves it (moving electrons = current), and a "hole" is left behind where it was.  So these electron-hole pairs get created and then split apart in the depletion region of the PN junction (where the P and N material touch) and then get swept to either side of the diode by the built in electric field = current.  So in fact, we don't even need the big reverse bias (needed for effective photodetectors though) in order to work the opposite of LEDs and actually absorb photons, we just need light which gets absorbed to create those electron hole pairs and the subsequent reverse current to power something - all from just light!  Who's going to make the first solar powered effect pedal?  I bet it will work great in dark bars...

Actually, it's been done. The smokey bar thing may be a problem though... 

[spargo]

...............which begs the question, why the heck is my mosfet boost getting a full 9V at the drain when everything else seems to be wired correctly?  No solder bridges visible, voltage divider is working correctly, etc.

[ayayay!]

...............which begs the question, why the heck is my mosfet boost getting a full 9V at the drain when everything else seems to be wired correctly?  No solder bridges visible, voltage divider is working correctly, etc.

You answered your own question:  It's getting 9V at the drain because it's wired up to the 9V supply.  The proper question would be what voltage(s) are you getting at the Source and Gate?  (And I'm assuming this is a 1 mosfet boost.  You didnt' provide a schem.)   

[spargo]

...............which begs the question, why the heck is my mosfet boost getting a full 9V at the drain when everything else seems to be wired correctly?  No solder bridges visible, voltage divider is working correctly, etc.

You answered your own question:  It's getting 9V at the drain because it's wired up to the 9V supply.  The proper question would be what voltage(s) are you getting at the Source and Gate?  (And I'm assuming this is a 1 mosfet boost.  You didnt' provide a schem.)   

http://www.muzique.com/schem/mosfet.htm

I'm getting about 5V at Vr and 9V at point A, with R4 being about 3.6k and a 100R filter on the 9V line.

[PRR]

> getting about 5V at Vr and 9V at point A

As Jono asked: "what voltage(s) are you getting at the Source...?"

In most amplifier stages there are _five_ points of interest:

Ground (is it really zero V? ground wires are often overlooked).

B+ power (nearly 9V? Or did that get lost somewhere?).

Plate/grid/cathode
Collector/base/emitter
Drain/gate/source

The common amplifying devices have _three_ pins. Even if you don't know their names, measure all three, maybe someone can sort it out.

Five numbers. Easier than a lottery ticket. Better odds too.

[amptramp]

...............which begs the question, why the heck is my mosfet boost getting a full 9V at the drain when everything else seems to be wired correctly?  No solder bridges visible, voltage divider is working correctly, etc.

You answered your own question:  It's getting 9V at the drain because it's wired up to the 9V supply.  The proper question would be what voltage(s) are you getting at the Source and Gate?  (And I'm assuming this is a 1 mosfet boost.  You didnt' provide a schem.)  

http://www.muzique.com/schem/mosfet.htm

I'm getting about 5V at Vr and 9V at point A, with R4 being about 3.6k and a 100R filter on the 9V line.

FET's are notorious for having cutoff voltages that vary all over the place.  It is obvious that the FET is not carrying any current or you would have a voltage drop across the drain resistor, R4.  You may need to increase R1 to turn the FET on.