I have attached an example circuit I made, after reading confusing information from google searches. Is a simple circuit 5817 Shotkey Diode, and 4733 Zener diode. Can someone look this over, am I doing this right? I laid this out from what diodes I have in my parts box. If this is not ideal or can be improved, please advise, I'm not afraid to buy another 100 more diodes for $.99, as long as they're better than what I'm using.
I would use this circuit for general purpose to design my projects - usually an Arduino nano with some modules, normally consume 1 - 1.5 amps. I usually use a step-down module, from 12v. Since the diode drops the voltage, I compensate by dialing the potentiometer on the step-down supply until the multi-meter reads just below 5 volts on the running project, AFTER this reverse-polarity projection circuit.
Unless there's a fuse, the zener will fail first on over voltage, and not protect the load
(although it can handle the odd spike - presumably the power source is a cheap USB
power supply and prolonged overvoltage isn't expected)
A fast blow fuse and a chunky TVS in place of the zener is probably a lot more robust
as a general protection circuit.
[I should add that TVS diodes often have ratings like peak power in the 100's of watts
rating (for brief periods), ie they have a good chance of blowing a fuse first ]
There's something to be said for the good old fuse + SCR crowbar circuit for over
voltage protection - much sharper voltage threshold, definitely blows the fuse so you
know there's an issue - for a larger more expensive circuit that might be worth having
especially in an automotive application.
Actually another thought - make the series Schottky diode really puny and it will act as a very
fast fuse(!). Problem is it have a higher forward voltage drop then.
If you're just worried about reverse polarity, why not just put a reverse biased general purpose diode across the input? It will short out any supply that you connect backwards to it. Unless you are connecting batteries to it, it's okay because most modern supplies (all switching plug in adapters) will just sense the overload and shut down.
Your "step down" circuit should be a voltage regulator, then you don't have to worry about overvoltage. Perhaps you could just use a step down module with better voltage regulation...
I am using batteries, lithium cells. Sometimes a power supply, most of the other devices are battery power.
Aarg: Do you mean a 4001 or 4007 general purpose diode? I read that the Shotkey diode drained less power, which is why I did not put a 4001 or 4007. But if that's just going to blow up, then it defeats the purpose of putting a circuit in - I would just find a fuse to use.
I think the thing with fuses, I have to have a big box of them for different current ratings on each and every project, then hopefully have the same one and / or remember to replace it if any of my devices change - which is alot, when I'm testing and designing.
Max pin voltage is NOT simply 5volt!!!
It's VCC +0.5volt and VCC -0.5volt.
VCC of a Nano on USB power is 4.6volt, because of a USB backflow protection diode.
That makes a 5volt zener almost useless when the Nano is on, and fully useless when the Nano is off.
A 10k resistor between unknown voltage and pin could be a lot safer than zeners.
With that, you rely on the internal clamping diodes of the MCU.
Ok for the occasional 'mistake'.
If you're paranoid about protection, use a (1N5817) schottky diode between pin and ground (cathode to pin) and a schottky diode between pin and VCC (anode to pin).
You still need a resistor between source voltage and pin, but it can be a lower value.
If fault current could get higher than the Nano's current draw, a 5volt (standoff) TVS diode across VCC is also required.
Leo..
