Polarity protection diode substitution

[Ale_jellybelly]

I was trying to desolder components from my old Speaker Cranker perfboard build and I broke the 1n4001 polarity protection diode in half
I only have 1n914s at the moment, can I use one of those? Or can I just jumper it since I will be using it with my power supply?

 

[Robert]

Does it go to ground? If you KNOW you aren't going to plug in the wrong supply you can just omit it.

Or, what I would do is omit it and replace the positive wire from your DC jack with the 1N914, striped end pointed at the PCB.

 

[Brett]

Don’t jump it if it goes to ground.

 

[Ale_jellybelly]

It does go to ground
I'm following this Effectslayout schematic:

So I could add a 1n914 between +9v and the + of the DC jack?

 

[Robert]

So I could add a 1n914 between +9v and the + of the DC jack?

Yep. Anode at the jack, Cathode (stripe) at the PCB.

 

[Ale_jellybelly]

Thanks, it works but... If I leave it plugged in with no power and activate it, for some reason some signal passes through, and it sounds like a very voltage starved distortion... Where is power coming from if it's not connected? Through the guitar cables?

 

[jimilee]

Sounds like a solder bridge, but it works, so…audio probe the footswitch to see how it’s sending the signal.

 

[Robert]

With a simple circuit like this the output from your pickups might be enough to cause the transistor to conduct in not-so-musical ways.

 

[Feral Feline]

Dave doesn't give the schematic on his layout page.

I suppose you're following the schematic for his PCB?

 

[Ale_jellybelly]

Sounds like a solder bridge, but it works, so…audio probe the footswitch to see how it’s sending the signal.

I'll take another look inside

With a simple circuit like this the output from your pickups might be enough to cause the transistor to conduct in not-so-musical ways.

Even without powering the circuit? I want to check with my little practice amp and with a buffer upfront
Quick question, the diode I added on the 9v is protecting the polarity but on the + instead of the ground right?

Dave doesn't give the schematic on his layout page.

I suppose you're following the schematic for his PCB?

View attachment 60990

Yeah looking at the layout and at the schem, it's the same

 

[Robert]

Even without powering the circuit?

Technically you are powering it with a very weak AC signal applied to the wrong place in the circuit. (The input)

Quick question, the diode I added on the 9v is protecting the polarity but on the + instead of the ground right?

It was always protecting the + input, just using a different method.

Before moving the diode you had parallel "crowbar" polarity protection:

If reverse polarity is applied the diode shorts the voltage/current to ground. With enough current the diode can short, effectively breaking the pedal. Sure, the rest of the circuitry is protected (until the diode burns open) but to the average user the pedal is still broken and the diode needs to be replaced.

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What you have now is series polarity protection.

If reverse polarity is applied the diode does not conduct. The circuit is protected, the diode doesn't short, the pedal simply does not work unless the correct polarity is applied.