Can someone help me with the function of this Diode?

Reverse protection, to blow the input fuse if you connect power the wrong way. It's reverse connected between input and ground.

It looks like an afterthought for somewhere who has accidentally reversed the power supply connections in the past. It may have been better to put the diode in series with the input to get instant blockage of the reversed power input and not required any fuse replacement.

Cheers Nigel.

Sweet, its doing what I had guessed. I do like to confirm my guesses though, so thanks.

ZipZapOuch said:

It looks like an afterthought for somewhere who has accidentally reversed the power supply connections in the past. It may have been better to put the diode in series with the input to get instant blockage of the reversed power input and not required any fuse replacement.

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Yeah agree, I might end up replacing it in series anyway, as that does seem more logical to me.

My Father in-law had actually put a slow blow fuse in, and smoke came out of one of those components.

But from testing the regulator on my 12V supply, it was putting out 4.99V consistently, so I think that's fine. I'll throw a new cap and diode on for good measure.

The only reason to have the diode in parallel is if you can't afford the voltage drop from putting it in series. If you have a 12 V supply and an 7805, the 7805 needs about 7.5 V to work. With the diode that gets to about 8.2 V, so on a 12 V battery it's fine.

Having the diode in parallel you are relying on the fuse blowing before the diode. That is really dodgy as you should consider all voltages up to the maximum supply voltage. You can get the situation where the diode is OK with a good battery, but not with a flat battery, where you could get a low current for a long time.

Also, if the device is drawing a lot of current (I assume so from the heat sink), you can put 5 diodes in series between the 12v supply and the 5v regulator to distribute the heat from the (linear voltage drop x current) power. 5 diodes in series would be almost 3.5v off of your 12v supply and greatly reduce the heat generated by the voltage drop. 7watts at 1Amp without the diodes and 3.5watts with the 5 diodes.

Make sure each of your diodes can easily handle 0.7w (or so) required.

T

Diver300 said:

The only reason to have the diode in parallel is if you can't afford the voltage drop from putting it in series. If you have a 12 V supply and an 7805, the 7805 needs about 7.5 V to work. With the diode that gets to about 8.2 V, so on a 12 V battery it's fine.

Having the diode in parallel you are relying on the fuse blowing before the diode. That is really dodgy as you should consider all voltages up to the maximum supply voltage. You can get the situation where the diode is OK with a good battery, but not with a flat battery, where you could get a low current for a long time.

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I'm sure you didn't mean that a single diode would drop the supply from 12v to 8.2v.

ZipZapOuch said:

T

I'm sure you didn't mean that a single diode would drop the supply from 12v to 8.2v.

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As he didn't say any such thing, obviously not.

ZipZapOuch said:

T

I'm sure you didn't mean that a single diode would drop the supply from 12v to 8.2v.

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Yeah I read it as the supply needs to be at least 8.2v to ensure that there's at least 7.5v left for the reg to work.

Thanks for the help everyone, I feel a little more knowledgeable dangerous now

ZipZapOuch said:

T

I'm sure you didn't mean that a single diode would drop the supply from 12v to 8.2v.

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I didn't mean that, and I don't think that I implied it either.

I said that the 7805 plus diode would need 8.2 V or more to work, so that is a good margin below the lowest voltage that a 12 V battery will produce in normal use, so it's fine.

frogger_88 said:

Hi everyone.
I'm pretty unskilled in the world of electronics, but I've been doing some testing on an old control unit for a telescope, that has been blowing the power fuse.

What I'm not terrible sure of, is the purpose of the diode in this power regulator circuit. Would love to have someone explain it to me.

View attachment 147100
Thanks
Frogger.

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Hi,

The reason, as Nigel pointed out already, is to protect the input from a reverse voltage connection on the input. The diode draws too much current with a reversed input, the fuse blows. The diode may blow also, but it usually will blow shorted out. That has to be replaced, but at least it protects the IC and anything else downstream from it which could be something much more expensive.

Anyway, sometimes you can get away with putting a diode in series with the input as some others have mentioned, but sometimes you can't. That's because a diode in series with the input means you lose some overhead voltage, which means your minimum input voltage goes up by an amount equal to the diode voltage drop at full load.
I think the 7805 minimum input is 7 volts, but with a regular diode with a 0.7v drop at 1 amps in series with the input, the minimum input goes up to 7.7 volts, which could make or break a circuit. Most of the time we don't want that, but sometimes it works ok. It all depends how much input voltage you will always be able to supply.

Also, sometimes you might see BOTH a series diode AND a parallel diode. That's because the series diode has an equivalent parallel capacitance, and that can cause conduction for a short time even with a reversed input. The series diode prevents high power dissipation in the event of a reversed connection, and the antiparallel diode (the one there now) prevents short term input surges. These short-term input surges are very, very small, but they could be destructive anyway over time.

It might be interesting to talk about the output protection too. Series diodes on the OUTPUT (rather than the INPUT like this one has) do not work to prevent a reverse connection on the output such as with a battery charger. If the battery is reversed, current still flows through the series diode almost as if it was not even there. That means we must use an antiparallel diode and a fuse. If we use just an antiparallel diode then we assume that the diode will blow as a short circuit.

As a final note, I have to wonder where that circuit came from. That's not the right way to connect components for long term use. I can't imagine any company shipping something like that. I can imagine a hobby project like that though, and I've done stuff like that myself. I'd never ship that to anyone though even if it was just for a hobby project for them.

MrAl said:

Anyway, sometimes you can get away with putting a diode in series with the input as some others have mentioned, but sometimes you can't. That's because a diode in series with the input means you lose some overhead voltage, which means your minimum input voltage goes up by an amount equal to the diode voltage drop at full load.

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If a diode has too much drop, you can use an P-MOSFET to block a reverse voltage from a positive supply.
With that, the drop is just the MOSFET on-resistance times the load current.

You connect the drain to the +supply, the gate to ground, and the source to the regulator.
The P-MOSFET then normally conducts in its reverse direction (in the direction of the substrate diode), but that's okay since a MOSFET conducts equally well in either direction when biased on.

A negative supply voltage is blocked since, although that's the normal voltage connection for a P-MOSFET, its Vgs is 0V so it does not conduct.

Ideal diodes for minimal diode drop V :


Schottky Diodes :



Regards, Dana.

crutschow said:

If a diode has too much drop, you can use an P-MOSFET to block a reverse voltage from a positive supply.
With that, the drop is just the MOSFET on-resistance times the load current.

You connect the drain to the +supply, the gate to ground, and the source to the regulator.
The P-MOSFET then normally conducts in its reverse direction (in the direction of the substrate diode), but that's okay since a MOSFET conducts equally well in either direction when biased on.

A negative supply voltage is blocked since, although that's the normal voltage connection for a P-MOSFET, its Vgs is 0V so it does not conduct.

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Hi,

Yes sure, and this has been known and done for a long time now, just like Schottky diodes have lower voltage drop but lower reverse resistance. You could reply to the thread starter to provide this option although a diode is more common.
The cost difference would come into play in the production realm also, but in the hobby world that may not be as significant. For my personal projects I like to use overrated parts to get really good performance even though a little pricier.