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I want to protect my circuit against reverse polarity, by simply adding two diodes in parallel between the + and - input cables. So if the polarity is reversed, the current will go through those diodes at high current and blow the fuse. Change the fuse and flip the polarity and everything should be good to go.

The SS34 diode's datasheet says: Peak forward surge current 8.3ms single half sine-wave superimposed on rated load (JEDEC Method): 80A. I'll be using two of these, so total amperage is then 160A.

That should blow the 20A automotive fuse rather quickly, but is it quick enough? Seems the diodes can only allow such high amperage for 8.3 ms, and I don't know if normal automotive fuses will break so quickly.

Will using two SS34 diodes work?

EDIT: Found this document that has time-current curves for some common fuse sizes.

https://www.littelfuse.com/~/media/automotive/catalogs/littelfuse_fuseology.pdf

Seems a 20A fuse will need a whole 0.2s to blow at 160A. And from what I can tell, normal short circuit current of motorcycle batteries are in this range.

If the fuses really are that slow, then it seems impractical to design a circuit that relies on blowing the fuse for the reverse polarity protection.

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  • \$\begingroup\$ "I want to protect my circuit against reverse polarity, by simply adding two diodes in parrallel between the + and - input cables" Please draw a schematic. "I'll be using two of these, so total apmerage is then 160A" Diodes are NTC so assume they won't share any current. \$\endgroup\$ – winny Oct 29 '19 at 16:28
  • \$\begingroup\$ What is the max current that your circuit will use? is it DC or AC? \$\endgroup\$ – Voltage Spike Oct 29 '19 at 16:32
  • \$\begingroup\$ @Voltage Spike 20A DC \$\endgroup\$ – K0ICHI Oct 29 '19 at 16:41
  • \$\begingroup\$ Not so easy to answer. It will likely depend on the fault current capability of your supply and series resistance limiting the peak fault current. If the source is something like a car battery, the peak current could be much higher than 160A. A series MOSFET may be a better approach to reverse polarity protection. \$\endgroup\$ – Spehro Pefhany Oct 29 '19 at 16:49
  • \$\begingroup\$ @Spehro Pefhany Yes, the source is a motorcycle battery. I'm aware of the more advanced reverse polarity protection circuits, but they are overkill for this application. Because of the mating of the connectors I will use, it is very unlikely reverse polarity would ever occur, so a diode solution that blows an external fuse is acceptable, and preferred for it's simplicity. \$\endgroup\$ – K0ICHI Oct 30 '19 at 4:17
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Two diodes will not be enough, the problem is the diode might be able to handle large amounts of current as it can only handle 3A because you'll be running the diode in a forward current mode. You would need to parallel at least 7 diodes (21A of forward current).

For reverse current protection only one diode is needed.

The other problem is heat, even if you were running 3A, with a 0.5V drop this would be 1.5W of power, the diode would get ~30C above the ambient temperature. The 0.5V drop also a waste of voltage/power.

I would not suggest using diodes at all for high current applications. I would use a pmosfet. Using a mosfet keeps the resistance low when the mosfet is switched on and the current is going in the right direction. In the reverse, make sure the mosfet Vds voltage is higher than the voltage you will be using in your circuit.

enter image description here
Source: https://hackaday.com/2011/12/06/reverse-voltage-protection-with-a-p-fet/

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  • \$\begingroup\$ You surmise I would use the diodes in forward current mode. It should be clear from my post that I will not. Care to elaborate on why you think 1 diode in reverse will work? \$\endgroup\$ – K0ICHI Oct 30 '19 at 4:15
  • \$\begingroup\$ Yeah, because if you used the diodes in reverse, they would soak up at least 20V before they would turn on and if you figure your current is 20A they would dissipate 400W * dutycycle. Diodes are not for blocking large currents, they just aren't. A schottky diode is just a diode in reverse with a few bells and whistles. \$\endgroup\$ – Voltage Spike Oct 30 '19 at 4:33
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I want to protect my circuit against reverse polarity, by simply adding two diodes in parallel between the + and - input cables. So if the polarity is reversed, the current will go through those diodes at high current and blow the fuse. Change the fuse and flip the polarity and everything should be good to go.

That kind of reverse polarity protection (i.e. based on blowing/replacing a fuse) should be used if it is "really" needed (For example, during 8 years of my automotive electronics hardware design experience, I have been asked to design such a protection only once, and it was a customer's "special request").

Anyway, you have noticed that it may not be enough to replace only the fuse. The diodes may get short even before the fuse to blow. To protect also the diodes, you can use a PTC-based Resettable Fuse. Resettable fuses heats up when a large current flows through them then they become a high-resistive element to block the current flow. They also help to reduce the inrush current drawn from the battery. RFs can be "reset" by letting them to cool down by cutting the power. Depending on the component, they can be easly cool down before you re-flip the polarity and apply again.

schematic

simulate this circuit – Schematic created using CircuitLab

Since the current drawn from the rest of circuit will flow trough the RT during the normal operation, it should be selected carefully. The environment is important, because the trip current of the RF will decrease at high-temp environments. You don't want it to block the current flow during normal operation, right?

Or...

You can use an easier and better solution as given in Voltage Spike's answer. That circuit can also be built with NMOSFETs. NMOSFETs are easier to find and cheaper.

schematic

simulate this circuit

When you first apply the supply voltage in correct polarity, the MOSFET is off, the MOSFETs gate and source voltages are undetermined. But since the current will flow through the MOSFET's forward-biased internal drain-source diode (a zener) first, the voltage at MOSFET's source terminal will be nearly zero. After this first cycle of current flow, the MOSFET's gate voltage will be higher than its source voltage. So the MOSFET will turn on and it will short the internal diode then the circuit will operate normally. That is how MOSFET allows the current to flow with correct-polarity supply. If you apply the supply voltage in reverse polarity, the internal drain-source diode will block the current to flow so the circuit will never operate. Please note that the MOSFET's breakdown voltage (VBRDSS) should be higher than battery voltage (e.g. a 40V MOSFET can be enough for your needs).

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