A question: the consequences of reversing power leads

[Mark Hammer]

I'm trying to repair a buddy's pedal.  It uses the Madbean Quasar board, that is a clone of the Xotic RC booster.  The original builder included a toggle to select between clipping diodes.

In yet another one of those boneheaded binary flip-flops, I hastily wired up a plug + battery-clip adaptor to test out pedals with an external 9v battery, and inadvertently connected the leads to the wrong points on the battery clip.  T'was only a battery that was applied to the power jack, and there IS a reverse-voltage protection diode on the board, but it worked before I plugged that battery in and now it doesn't.

I have cleaned off the flux from the board, put my magnifying headset on, and scrupulously gone over every inch of the board with a bright light shining through to nicely outline all the traces.  I've scraped away anything that even remotely resembled a possible bridge (like those little speckles you sometimes get on an etch), but my meter is still saying I have a short between V+ and ground.  I pulled the chip, but still no dice.

So my question is:  Other than a simple physical connection between V+ traces and ground, what unseen things might result in a short?

(I know it's not situated in the power plug since I unsoldered the V+ lead from the plug to the board, and there is no problem with a short between the tabs on the external power jack).  I get the short when simply measuring between the otherwise unattached ground and V+ leads to the board)

[LightSoundGeometry]

from what i have learned in EET classes , sounds like a resistor is gone. somehow it got too much current and is open or shorted.

[Cozybuilder]

Looking at the schematic, I suspect the IC and C12 are fried.

[R.G.]

ICs and electro caps are the most likely offenders.

Electro caps are made by forcing electricity through the electrolyte between two pieces of very pure, very clean aluminum. The action of the electricity causes oxygen from the electrolyte to combine with the aluminum on one piece of aluminum, and that eventually forms a layer of insulating aluminum oxide. Current stops flowing when the forming voltage can no longer force the insulating layer to grow thicker.

If you reverse an electro cap, you are actually doing two things. One is that metal oxides act like sloppy diodes. This is the origin of the copper-oxide rectifiers from the first AC line powered equipment, and later stacks of selenium making rectifiers. Aluminum does it too, but sloppier. So reversing the cap lets current flow. This current busily undoes the building of the electrolytically deposited insulator, and soon eats a hole somewhere. When a hole starts, the plates are shorted through the electrolyte. This ...may... be reversable, but also may not. Depends on how much current for how long.

ICs are built by taking a slab of N-doped silicon crystal and then diffusing everything else on top of it, insulated from the bottom N-type layer by a first diffusion of P-type. This forms a reverse biased PN junction to the substrate crystal as long as the substrate is the most-negative voltage on the chip. Generally the substrate is used for the most-negative power supply connection, forcing it to be most negative. This is by the way why the 78xx and 79xx voltage regulators have different pinouts. The substrate has to be attached to the thermal tab to get heat out, and it has to be the most negative point, so for 7900 regulators, this has to be Vin, not ground.

If you reverse polarity on an IC, the internal structure may allow large currents to flow from the now-positive substrate to the terminals held below the substrate potential. Whether this kills the chip or not depends on the details of the path for current through the chip, as well as how much for how long, as for the caps.

Reversing power from a 9V battery can be fatal. 9V's can supply upwards of an amp or so when shorted or fed through a single silicon junction. So it depends on the battery too.

It is entirely possible that the reverse polarity diode is not functional. Shunt reverse polarity protection like most pedals use rely on the diode being forward biased by a reversed voltage, and conducting so much current that there's none left for the circuit to die from. Sometimes the diode dies. This is why connecting 9V AC to a pedal reliably kills even shunt-protected circuits. First the current kills the diode, then it kills the circuit.

To save you some time ( because time is money! ) I suggest replacing all the electros and ICs. This can be quite a bit quicker than puzzling out what it dead and why. There may be reasons not to do this, but it can be a time saver and is not all that expensive.

[Mark Hammer]

Thanks.  I'll start by pulling the socketed chip and unsoldering the electros  If the short disappears, I'll check each of the electros, reinstall them one by one (if they measure OK out of circuit) and replace any that result in my meter beeping.

[Mark Hammer]

Solved.

The caps were good but the protection diode was bad.  All is well, now.

Thanks all.