Of these two diodes which is the better polarity protector?

[lowvolt]

It's Monday morning, I'm shooting for "dumb question of the week" right off the bat here.   

I'm adding polarity protection to a DIY distortion pedal someone gave to me that isn't outfitted with it.  I have these two diodes on hand ...

1N914

1N4001

Which of the two makes a better polarity protector?  9vdc application.  Can I just put it across the DC jack terminals?

And just so I know for future reference, can either of them be used for protection with 18vdc power?

Can an LED be used for that purpose?

Thank you .... brain operating on LOWVOLTage .. per normal.   

[Mark Hammer]

They should both work.  It's hard to imagine a pedal that would draw that much current, but if you know you will be drawing a lot of current, use a 1N4001.

[R.G.]

Mark's correct - they should both work if you use them in series with the input voltage.

But I saw "across the DC jack" in the post. If what you have in mind is putting the diode across the power supply voltage so it's normally reverse biased, but forward biased when the power is reversed, I would use the 1N4001. That's because the current out of a 9V battery or wall-powered DC adapter can be an ampere or more, and the 1N914 (VERY similar to the 1N4148, BTW) is only rated for 100ma. In the shunt protection use, the 914 is likely to die protecting the circuit, then the circuit will be damaged.

The 4001 is rated for 50V and the 914 for 100V, so both work the same as noted above for an 18V supply. I'd still use the 4001.

An LED can't be used for protection either way in a 9V circuit because their reverse breakdown is only several volts, depending on the type/color.

[Kesh]

With a reversed diode across the DC, is its purpose to blow the wall wart or its fuse, or melt the battery, rather than hurt the pedal?

[slacker]

The cynical would say it's there so that if you connect the wrong supply it will blow the diode, there by proving you've used the wrong supply and voiding your warranty.
With a 9 volt battery it will protect the pedal if you accidental try and connect the battery the wrong way round, I don't think it's really designed to protect against plugging in the wrong power supply even though people use it for that.

[R.G.]

With a reversed diode across the DC, is its purpose to blow the wall wart or its fuse, or melt the battery, rather than hurt the pedal?

I think the purpose is to clamp the reverse voltage across the circuit to one diode drop, which protects the circuit.

I *don't* think much additional thought went into what happens to the thing supplying the current. I believe that the origin of this design was primarily intended for batteries, not wall wart power. Batteries have a limited ability to supply current, and are replaced normally in practice, so this worked well. In addition, batteries are usually sagging lower than the nominal voltage, so losing a little voltage to a series diode is a problem, worse with batteries than a wall supply.

Probably the originator didn't intend to melt batteries, but I doubt he worried about whether it happened or not.

Like true bypass, it's a solution to a problem that no longer is prevalent in modern pedal practice.

In other words, yes, what he said:

I don't think it's really designed to protect against plugging in the wrong power supply even though people use it for that.

[lowvolt]

Wow ... learned a ton here.  I've been sleeping poorly the last several nights so my brain is really tired.  I tried to study-up on this but I just couldn't wrap my head around it all conceptually.  I'm normally a bit quicker study than that.  So thanks for the "PSU 101" stuff folks.

Don't let me put a stop to the conversation, by all means carry on.    I'm learning a lot and it's making other semi-related branch issues more clear as well.  Thanks again!

LV.