protection diode

[glinjik]

hi can someone tell me an alternative for this protection diode please

[mariush]

1n5401  - max 100v voltage, 3A diode.    Forward Voltage is most likely irrelevant (it's big)

You can use higher voltage rating, you can use higher current rating, you can probably use slightly less current rating but I'm not sure because I don't know the rest of the circuit ....(or you could parallel 3 1A diodes to get a 3A diode, sort of)

So some options  (sorted by amount in stock, not by price, but lots of generic cheap diodes there) : https://www.digikey.com/short/zn1zwq9p

[bob91343]

1N4007 might do.

[fourfathom]

1N4007 is 1000V, 1A.  Probably not a good replacement for a 3A diode.

[bob91343]

Maybe not, but still would do the job.  The 3A rating is somewhat arbitrary.  For nearly all its useful life, the diode will very likely never conduct.

[MK14]

Maybe not, but still would do the job.  The 3A rating is somewhat arbitrary.  For nearly all its useful life, the diode will very likely never conduct.

I don't want to be rude. But at the moment, we have very little idea as to the overall circuit, and what it does. Maybe it conducts nearly 3 Amps, nearly all the time, through that diode. The OP doesn't seem to have provided extra details.

So you can't say a 1 Amp diode will be ok, probably.

[fourfathom]

I could be wrong, but I assume that this is part of a power supply or regulator circuit and the diode is there to protect the pass transistor should the input be shorted while the output filter capacitor is still charged. (reverse breakdown voltage).  This diode will seldom if ever see any current, but in this fault condition it could pass many amps -- depends on the short, and on the capacitor.

[floobydust]

Note the diode doesn't actually give full protection- in the case the transistor has voltage at its emitter
you can still get backfeed across the E-B junction and damage the base-drive circuit.
An example of this is a DC power supply charging 12V battery.
If mains is turned off, there is still 12V at the emitter and the transistor backfeeds with ~7V avalanche to the drive circuit, which usually causes damage. OP's diode will just keep the filter capacitor charged.

[fourfathom]

Note the diode doesn't actually give full protection- in the case the transistor has voltage at its emitter
you can still get backfeed across the E-B junction and damage the base-drive circuit.

No doubt there are still fault current paths, but when the input is grounded the diode in the OP's circuit will discharge the cap on the transistor emitter, so I don't see where the emitter voltage comes from to drive the E-B current.  If there is charge on the base circuit there could be B-E and B-C current, but the base circuit is usually at a high enough impedance to limit the base current.  But I'm imagining the rest of the circuit, so I could be wrong.

An example of this is a DC power supply charging 12V battery.
If mains is turned off, there is still 12V at the emitter and the transistor backfeeds with ~7V avalanche to the drive circuit, which usually causes damage. OP's diode will just keep the filter capacitor charged.

Which filter capacitor?  The invisible one on the transistor collector?  Yes it will, but the discharging cap I'm talking about is the invisible one attached to the emitter.  I guess there ought to be a series diode at the output of your battery charger.

Anyway, back to the OP's question:  It's better to match or exceed the current rating of the original diode -- it was probably selected for a reason.