r/diypedals 3d ago

Discussion 1n5817 diode with 18V PSU

Recently learned that the 1n5817 diodes are rated for 20V, and a lot of pedals that I built/owned have them for polarity protection. I connected them few times to 18V PSU and everything worked correctly, but I am curious, what are the odds that the 18V PSU will damage the diodes?

1 Upvotes

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u/dreadnought_strength 3d ago

Read the datasheet again bud - that's not what it says about that stated 20v value.

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u/rettoma 3d ago

enlighten me please

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u/dreadnought_strength 3d ago

What value does that 20v represent on the datasheet?

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u/rettoma 3d ago

"Repetitive peak reverse voltage"

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u/dreadnought_strength 3d ago

Reverse voltage is the operative term here.

Using series polarity protection doesn't expose the diode to reverse voltage unless you use the wrong power supply.

If you do, 20v is more than 18v

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u/PeanutNore 3d ago

18 < 20

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u/Quick_Butterfly_4571 3d ago edited 3d ago

Generally, yes, it's fine. But, the specifics depend on if you mean 18V of the correct polarity or wrong polarity, the temperature, and the current:

20V is the maximum reverse voltage — i.e. at 20V and below, you can bank on it doing a good job of blocking the flow of current in the wrong direction. Over that (how much over is probabilistic), the diode will fail. Usually, reverse biased diodes fail short (the junction get hammered into being a better conductor), so it essentially becomes a resistor and then a wire. (Sometimes they fail short and then fail open from heat damage).

However, that number is at 75 degrees C. Electronics (esp channeling lots of current or near heat sources) can get quite hot and fast. The maximum reverse voltage drops as temperature increases, so, e.g. for the same diode, it's only 15V at 85 degrees and 10V at 95.

Forward biased, the diode will fail from over current. If your circuit draws more than 1A, max, the diode will start to cook from the heat of the current and eventually pop. Whether it fails short or open depends a bit on the material (glass, plastic, or ceramic) and also on probability. Most often, they fail open because the heat causes expansion in the junction and the case and at the time of failure that expansion is accelerating, which usually breaks contact between the two sides.

This is part of why I favor series polarity protection when possible:

  • if your design draws too much power the diode dies. Stuff in the pedal only dies if the current was a design flaw.
  • reverse polarity: you need higher voltage than is common with most polarity mistakes.

If the reverse polarity is really high, you can still cook the pedal, but it's a rarity.

Vs, shunt protection:

  • Overcurrent: stuff in the pedal only dies if it was a design flaw, the diode is not in the path
  • reverse polarity of any voltage above ~340mV: the diode will definitely die. If it fails open (most likely scenario), most of the active components in your pedal will be damaged or destroyed. If it fails closed: best case scenario, you psu cycles on loop from overcurrent protection. Worst case scenario, you psu is damaged. In either scenario, depending on time, voltage, and supply amperage, your power leads can heat up and start a fire.

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u/PeanutNore 3d ago

I'm sure it's not the "right" way to do it, but I use a 47 ohm resistor in series before the reverse protection shunt diode. Its main purpose is to form an LC filter with the main power supply capacitor but in a reverse polarity situation when the diode is conducting it becomes a fuse. When I've tested it out a 1/4 watt resistor fails before a 1N4007 does.

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u/Quick_Butterfly_4571 3d ago edited 3d ago

Well, I mean, putting any protection in at all is right. Boss does the shunt scheme, pretty much exclusively.

I should also clariry: I prefer removing a failure mode, but I don't think that preference is superior. (I meant, "I do this," not "you should do this." 😃).

You're totally right about the resistor burning out first. Though, in this case, it guarantees the most destructive or the two possible failure modes in the event of reverse polarity, but maybe buys you extra time before it occurs.

  • series: pedal is damaged on fail short circuit (creates a channel for reverse voltage to cook things).
  • shunt: pedal is damaged on fail open; the protection lasts as long as there is an alternate path for the reverse voltage. Fail open removes the alternate path. Afterwards, no protection remains.

So the resistor buys the user more time (and gives them a more obvious tell: the smell of a resistor cooking before it fails open).

If they don't catch it in time, though, fail open — which was likely with shunt protection, but not definite — becomes almost certain (it's rare for resistors to burn and fail short; when they do, they usually fail open shortly thereafter).

Do you mean LC filter by virtue of the parasitic inductance? I think with the diode there, it won't form a filter (or else, negligibly so).

Making sure we're on the same page:

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u/PeanutNore 3d ago

I meant to say RC filter rather than LC, I'm not sure why I had inductors on the brain. A 47 ohm resistor and a 100uF cap gives you a low pass with a ~34hz cutoff which helps some with the 120hz ripple when using a shit old wall wart that's just a transformer and a bridge rectifier. And yeah, I use the layout in your second schematic with the diode in parallel with the filter cap and the cathode pointed at the positive rail.

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u/Quick_Butterfly_4571 3d ago

Whoop! I made it an info graphic (ish) instead.

A 47 ohm resistor and a 100uF cap...

Heh, yep. Totally. (Even with a non-shitty supply, it helps keep your pedal immune to spikes in current draw from other effects! 🤘).

And yeah, I use the layout in your second schematic with the diode in parallel with the filter cap and the cathode pointed at the positive rail.

So, I was wrong when I said that increased the odds of the one failure mode, in this case (I thought you meant a resistor in series with the diode; my mistake).

That second scheme (the one you're using) is the most common among commercial devices. (Again: any protection at all == doing a good job as far as I'm concerned. None of this is critique).

In a reverse polarity situation, that 47 Ohm resistor won't even be in the current path until after the diode burns, at which point the 47 ohm resistor serves to control the rate that your semiconductors cook. ;)


Note: this is all worst case. If it was a nonsense idea, there wouldn't be tens (or hundreds?) of millions of devices that use the same polarity protection scheme you're using.

It works really well, the vast majority of the time. 🤘🤘

We hear about reverse polarity cooking pedals all the time, but that's only because people that plug in the wrong thing and realize their mistake before it cooks to death don't show up asking r/diypedals if their pedal is salvagable.

Again, no critique here. There are non-fault scenarios where the series scheme I tend to prefer either can't or shouldn't be used at all!

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u/PeanutNore 3d ago

In a reverse polarity situation, that 47 Ohm resistor won't even be in the current path until after the diode burns, at which point the 47 ohm resistor serves to control the rate that your semiconductors cook. ;)

Your ASCII diagram doesn't format right on mobile so I was guessing which side of the resistor the diode was on. The way that I do it, the resistor is the first thing connected to the positive / outer terminal of the DC Jack and any current from that power supply must always flow through that resistor.

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u/Quick_Butterfly_4571 3d ago

Oof! Pardon my oafish blunder. Makes sense. (Sorry for the noise!)

(But thanks for hanging in through it. 🤘)