R.G.'s che and good polarity protection

[kristoipher612]

hi there i have a few questions about this polarity protection article.  here's a link.

here's a quote from the article that brings up my first question.

The slightly slyer trick is to place a silicon diode cathode to + and anode to -. That means the diode does not conduct in normal operations, but shorts out the power when power is inadvertently reversed, protecting the circuit.

However, what happens when the power supply is not a 9V battery, but an AC power supply? The AC power supply sits there and cooks the diode until the diode shorts, and now you have a dead pedal. If the diode opens, the reverse voltage then kills the circuits on the effect.

so here's the question: why does it kill the diode?  I'm not questioning it, just want to know the why as well as the result on the situation.  

question #2:  does this protection need to be at the beginning of the circuit?  im going to point to the power filtering cap for the ggg big muff and how its not at the beginning of either the schematic or layout. here's that info.  anyone that wants to chime in with their knowledge can.  thanks in advance.
kris

[H.Manback]

stick a fuse in directly at the + or -

[kristoipher612]

just checked my original post, and the link to RG's article doesn't work.  you can get to it from the main geofex page.  sorry about the confusion.

[R.G.]

why does it kill the diode? I'm not questioning it, just want to know the why as well as the result on the situation.

It kills the diode by overheating it in many -  although not all -circumstances. The diode holds the voltage source down to 0.7V, so the voltage source responds by putting all the current it possibly can into the diode. If the voltage source is current limited, like from a 3-terminal regulator, the current is held down to maybe 1A, and the diode can live. If the power supply is not current limited, ...large... currents flow, the diode dissipates 0.7V times the DC current, which can easily be several amps, the diode overheats, the diode shorts. Now the pedal does not work at all because the protection diode is shorted, and none of the effect circuit is hurt at all. I've fixed a number of broken pedals with this exact problem.

question #2: does this protection need to be at the beginning of the circuit? im going to point to the power filtering cap for the ggg big muff and how its not at the beginning of either the schematic or layout.

Large power filter caps can usually be hung across the voltage rails almost anywhere they fit. I often place power filter caps on a PCB last, and find a place where there is physically room, then route the power and ground over from the closest place.  That works because ... large... currents are not involved.

It's different for the reversed diode protection. Since current only flows down the wires and traces in the shortest path to the diode that is protecting things, then large currents can flow through any conductors that are between the power supply wires and the diode. The currents may overheat those traces or even burn them out. Sometimes you can get away with it, but it is very advisable to put the diode right where the power comes onto the board.

Better yet, don't use the reverse diode method. Using a series schottky diode or one of the active methods is far, far better.

[onboard]

Better yet, don't use the reverse diode method. Using a series schottky diode or one of the active methods is far, far better.

Shoot. And I just popped parallel reversed 1N4001's in a couple of pedals and felt good about it. At least the circuit is still ok if the diode shorts.

Would the mosfet trick be the best way to go?

[R.G.]

Shoot. And I just popped parallel reversed 1N4001's in a couple of pedals and felt good about it. At least the circuit is still ok if the diode shorts.

Don't get too upset. The 4001 is a 1A rectifier diode that can probably withstand 50A pulses. If you have to do reverse diode protection, use a beefy one. Yours is modestly beefy.

I like series protection so I don't run high currents. But that's not the only way.