Reverse Polarity Protection 1N5817 or 1N4001

[anotherjim]

Definitely after the diode - the diode would block the capacitor if it were before the diode.

[Rob Strand]

Where do you buy your diodes?

Often I hear that the Schottky is more expensive,

The price certainly depends on your sources.
Some small vendors want near $1.00 each and people's local store in some countries might not even keep Schottky's.

The small-time hobbyist and one-time builders have different limitations to people doing production.

My post stands from a technical perspective but there are of course many other factors.  Like a design might *have to* use 1N4004s in one location so in order to minimize part you might choose an 1N4004 in another location.  When you use surface mount parts you might not want to buy 4k reels of two parts.  Reducing the number of parts can help reduce supply problems as well.

I only wanted to give a short answer!

[PRR]

> The only limit on how much current flows in the diode is the diode's forward resistance, which is deliberately as small as it can reasonably be made, the resistance of the lead wires, and the resistances of the transformer windings. These are all, in fact, as small as they can reasonably be made.

For clarity of theory, or for folks doing smoke-tests at home......

"as small as ...reasonab[le]" can be read "as small as you PAID for". You pay for V times A, and get just low enough resistance to get what you paid for.

Transformers in general can output 10 times their rated current, into a short, for a short time.(*) For the "1 Amp" tranny mentioned, 10 Amps.

Transformers are massive, take a long time to burn up, minutes. OTOH, as R.G. says, diodes heat quickly.

So, as R.G. says, a "1A" AC wall wart can usually burn-up a "1A" diode. And (as said) melted diodes often go "short" both ways, at least for a while. Sometimes they eventually burn open, sometimes they don't (depending maybe on just how much current flows).

(*) "10 times" is a starting point for thinking/guessing. In practice, small transformers are cheap, in every way, including excess winding resistance. If you smoke a few, you may find that ~1A warts may "only" deliver 5X rated current into a short. OTOH, big transformers are more about efficiency. The 100 Amp street transformer feeding my house can deliver 50X rated current. Partly for low loss at nominal load, and also so a short will burn-up a drop-wire quickly and save the transformer.

[Rob Strand]

Transformers in general can output 10 times their rated current for a short time.(*) For the "1 Amp" tranny mentioned, 10 Amps.

Transformers are massive, take a long time to burn up, minutes. OTOH, as R.G. says, diodes heat quickly.

So, as R.G. says, a "1A" AC wall wart can usually burn-up a "1A" diode. And (as said) melted diodes often go "short" both ways, at least for a while. Sometimes they eventually burn open, sometimes they don't (depending maybe on just how much current flows).

Very true.

I have a very old plug pack rated at 300mA, with known poor regulation and hence lowish short circuit current. The short circuit current is 1A.   Once the output got shorted for many hours.  It got freakin hot..  However because it was from the bad old days it didn't have the (one shot, non-replaceable) thermal fuse you see in modern units.  It survived.

The short circuit current/rated current ratio goes up disproportionally as you increase the rating.

[amptramp]

I have several plug-in doorbell transformers rated from 10 to 14 volts.  They are distinguished by the fact that they have enough series leakage inductance that they can operate with a shorted output because the current does not rise to a value they cannot tolerate.

These could be a good choice for a pedal power supply since you could have the supply (either AC or DC) plugged into a pedal with the centre pin grounded then have the barrel of the output grounded accidentally, maybe for some time.  It would also be good for parallel protection since the input current is strictly limited and could be set to a value a diode could carry permanently.  Some of them plug directly into a wall outlet and have screw terminals on the back for the output so they are like wall warts from before the days of wall warts.

[merlinb]

The series diode method (low-drop diode such as 1N5817 in series with power supply) is essentially perfect except for the voltage drop you seem very worried about. But the fact of the matter is, that the 0.2-0.4V or so are rather small compared to the actual voltage difference between different supplies.

+1
I used to worry about the series diode drop when I was a beginner. Since then I've matured, series Schottky is fine and gives best value for money IMO. Worrying about <0.5V drop on the rail is misdirected attention; it's bugger all and you can always claw back more than that by swapping to rail-to-rail opamps. Except nobody bothers with that because <0.5V is bugger all!

[Rob Strand]

Worrying about <0.5V drop

It does waste battery capacity. (Not that I use batteries much.)
With the cost of 9V batteries these days the % loss adds up quite quickly.
The more elaborate solutions give you payback quite quickly!

[Fancy Lime]

The more elaborate solutions give you payback quite quickly!

Does it, though? For a low-draw fuzz or something similar, batteries can have years of life in them and I for one would always use cheap zinc-carbon batteries for that. For a Flanger or whatever, that draws a lot of current, battery operation makes no sens anyway, economically, ecologically or just plain practically.

Andy

[bool]

"Elaborate solutions" - and schottkys - can both give you a slight un-ease. When you think about it, and simplify, you have to deal with two major nuissances: the reverse leakage thing, and the failure mode. Both can (and most probably will) damage the circuit they are supposed to protect, in case "something goes wrong" and exploit precisely those two weaknesses.

When you think about it some more, and you seek a "stupidly simple & reliable" - and possibly affordable solution, you really can't ignore the exactly "wrong" thing to do: two (or more) ordinary Si rectifiers in parallel (!!) It's cheap, and lower leakage compared to schottkys, withstands significantly more reverse voltage etc. etc.

If you proportion the circuit current consumption to fall within the lower forward drop knee of the diode, you can of course minimize the said drop.

In short: SM4004's and SM5404's are a cost effective and sturdy solution that wastes a little portion of your PCB real estate.

You can beat around the bush as much as you want, but that's the way it is.

[Fancy Lime]

Hi bool,

I'm afraid I don't quite follow.

Reverse breakdown voltage (which I assume you mean with "failure mode"?) is 20V for a 1N5817 or 40V for a 1N5819. So enough for someone plugging in a 9-18V universal PSU with the polarity switch in the wrong position. And if someone insists on hooking their stompbox up directly to 220V AC power, there is not much hope of reliable protection anyway. But that will hardly happen by accident. Thats why vastly different voltages and power ratings use different connectors.

I could not find reverse leakage data quickly but I would be surprised if enough voltage came through to destroy even a sensitive circuit. Definitely not for anything discrete transistor based and almost certainly not for most common modern op-amps. If it does, there is something else wrong with the circuit.

If you have two diodes in parallel, wouldn't just one open? The voltage drop across one diode is exactly what the other diode of the same type would nominally need to open. So unless they are exactly identical, only the one with the slightly lower drop will open, leaving you with the same 0.7V as before. Things change when the forward drop increases due to high power draw, then you can open both. But then again, if your stompbox eats 3 amperes, you have a whole other sort of problems.

I see your solution work and be relevant for a space heater, kitchen stove, dryer or other high power appliances, but for a 9V stompbox, I don't see how this would be useful. So, to beat around the bush some more: no, I really don't think that thats the way it is.

Respectfully,
Andy

[bool]

Think whatever you like... that's your constitutional right

What it is that makes the described configuration work is the fact that diode junction resistance somewhat varies with the passing current; and if diodes are over-specified in that regard just enough (i.e. for a couple milliAmps or couple tens of milliAmps), said diodes will act somewhat as paralleled resistors; sharing the passing currents.

By reading the datasheets, it won't be too difficult to locate (and extrapolate) the combined V drop at a specified passing I ...

And the combined V drop will be a little lower for paralleled diodes than for a single diode.

[Phoenix]

Think whatever you like... that's your constitutional right

What it is that makes the described configuration work is the fact that diode junction resistance somewhat varies with the passing current; and if diodes are over-specified in that regard just enough (i.e. for a couple milliAmps or couple tens of milliAmps), said diodes will act somewhat as paralleled resistors; sharing the passing currents.

By reading the datasheets, it won't be too difficult to locate (and extrapolate) the combined V drop at a specified passing I ...

And the combined V drop will be a little lower for paralleled diodes than for a single diode.

You can't simply parallel diodes like that though - that's NOT how they work. Diodes are like any other semiconducter material, there are always slight variances, regardless of if the diodes have come off the same reel, been made at the same factory, made from the same bool of silicon. There will be slight variations in their doping, and therefore their forward voltage, it's just the nature of the beast. This is why high power diodes are made, and we don't just use lots of little diodes in parallel.

Now, in high power stuff, doides often ARE used in parallel, however, they MUST have ballast resistors in series with them (so the diodes themselves aren't actually in parallel, it's a series diode-resistor pair in parallel with others), because otherwise, whichever diode has the lowest forward drop will conduct ALL of the current, while the others in parallel sit there and do basically nothing, until the first one burns up, and then the one with the next lowest forward voltage steps up to the plate to sacrifice itself in a vain attempt to save the others. The difference in forward voltage only has to be microvolts, it doesn't make a difference.
However, if you have some ballast resistors in series with each diode, they will help to balance the current between the diodes, because if any one diode is conducting more current than another, the voltage drop across the resistor will increase, leading to some self-balancing.

Andy's quite correct, the forward voltage will always be the same, the only time that you use parallel diodes (with ballast resistors) is when you can't get one with the current ratings you require for a price you're willing to pay, or in a form-factor that's suitable for your application. Yes, forward voltage increases with increased current, but even for high-current digital pedals that's frankly a non-issue, any of your typical rectifier diodes that can be had for $0.005 each in 100's of quantity are nowhere near their knee at the currents we demand of them. Seems you have a fundamental misunderstanding of how diodes work.

[bool]

That's correct whent diodes run at their rated pass currents.
Their junction parasitic R will increase when ran at significantly smaller currents compared to their rated current; and consequently this internal parasitic R will accomplish what otherwise externall added ballast resistors would.

(JK: If it wasn't so; diode bridge compressors wouldn't exist).

[duck_arse]

[pardon]

....  made from the same bool of silicon. There will be .......

ohhh, well played.
[/pardon]

[bool]

Well played indeed. But it is Mr. Phoenix who is coming along as a bit too smart for what he is preaching. (I would just quietly move along as if I didn't notice it if you didn't push it like that).

What I described above is a mechanism that is a basis of operation of some very expensive (with a reason) studio compressors - diode bridge compressors. So guys, learn more before you serve generic text-book answers.

Exact same parasitic characteristics of diodes can be used to drop the forward drop a little (usually a couple tens of millivolts) cheaply. The answer to what I described is staring at you right from the datasheet.

F.e, download here
www.vishay.com/docs/88503/1n4001.pdf
Look at Fig.4; and extrapolate. Take f.e. a passing current of 20mA.

The current-sharing balance between several un-padded diodes (as will be the case with crude paralleling) won't be ideal, but just from the graph, it is obvious that the voltage drop is going to be lower than compared to a single diode. A soon as you move current-wise into the lower portion of the fw. characteristics knee, what I described is just how it works.

It's a no-brainer really.

[Fancy Lime]

@bool

I and I'm sure many others before me have had your idea as well and wondered: Why does no-one use this obvious and perfect solution? General rule of thumb that I and I'm sure many others before me have learned along the way: If it's obvious and perfect and nobody has ever thought of it, it's almost certainly not going to work. Not to discourage anyone of course... but if someone tries to tell me that they invented a perpetuum mobile, cold fusion, the levitation machine or that a Nigerian Prince wants to give me half his inheritance for a small service fee, I have to make a conscious effort keeping to take them seriously.

I don't see what Fig.4 has to do with this. It shows the instantaneous forward current vs instantaneous forward voltage. What you are describing has to do with resistance vs forward current (like in a diode bridge compressor).

What I described above is a mechanism that is a basis of operation of some very expensive (with a reason) studio compressors - diode bridge compressors. So guys, learn more before you serve generic text-book answers.

Well, partly but not quite. The poblem is, that you are missing a different, much more important factor. Let me try to explain:

The solution you propose relies on the parasitic resistance increasing with increasing current. For this to balance the currents between two diodes, this relationship would have to be stronger than the dependence of the breakdown on the voltage. And that is simply not the case in anything that can reasonably be defined as a diode. It is not like the effect you describe does not exist, its just that there is another much stronger effect canceling it out. What you describe assumes that a diode behaves mostly like a resistor. But it doesn't. It behaves a little bit like a current dependent (or voltage dependent because those two are coupled as well) resistor but it also behave a lot like a diode. In a diode bridge compressor that does not matter because there are no identical parallel paths and we can use the little bit of resistor behavior without the diode behavior messing things up. I won't say hat there are no diodes with which your technique would work. But those would be diodes that are terrible at being diodes. With a modern Si power diode, I don't see this happening. Just try and measure it. If you parallel several diodes you will get a drop that is slightly below the drop of a single one because one diode is open but still has some resistance and the other is closed but lets a tiny bit of voltage through due to the small parallel resistance it sees.

Hope that was comprehensible,
Andy

[bool]

Let me put it that way: have you ever seen a circuit that uses a BAV70/65 or a BAS40-05/06 that has both halves in parallel? It happens, I'm afraid.

Otoh, you have already given yourself half of the answer when pondering on the Fig.4. Just think a little deeper. It's simple.

Disclaimer: just don't ever do this to "up" the diode current ratings on the cheap.

[POTL]

Hi. I don't want to clutter up this forum with regular silly topics, on this I will revive the old one.
I was looking for a replacement for the 1N5817 in a compact smd package.
And found this diodes MBR130LSFT1 & MBR120VLSFT1
I looked at the specifications and it seems to me that it is even better, but I am not an engineer and could have missed something.
It will be used in conjunction with icl7660s / max1044 to double the voltage.
what do you think?
https://www.onsemi.com/pub/Collateral/MBR120VLSFT1-D.PDF
https://www.onsemi.com/pub/Collateral/MBR130LSFT1-D.PDF
and classic
https://www.onsemi.com/pub/Collateral/1N5817-D.PDF

[POTL]

nobody?

[vigilante397]

I have some SMD 5817 diodes that I use, give me a second and I'll find the part number for you. They absolutely exist.

Try out the SM5817 or B5817. Same thing as 1N5817, SMD package.

https://www.mouser.com/ProductDetail/Micro-Commercial-Components-MCC/SM5817PL-TP?qs=sGAEpiMZZMtQ8nqTKtFS%2FEo19YTfnJP1ysRBdx90Swg%3D