Can someone help me with the function of this Diode?

F

frogger_88

New Member
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.


Thanks
Frogger.
 
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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.
 
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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.
 
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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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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.
 
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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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ZipZapOuch said:
T

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