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Is the function of 1N4001 before the 5V regulator reverse polarity protection?

enter image description here

Is this a fine method? And I sometimes see 1N4148. Can any diode(besides zener) be used for such function?

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  • \$\begingroup\$ No, it ensures the current will only flow in one direction \$\endgroup\$ – José Manuel Ramos Nov 20 '18 at 9:25
  • \$\begingroup\$ If you see the datasheet, the 1N4001 is capable to handle 1 Amp The 1N4148 only handles 500 mA. Take care of this. Take care of the voltage drop of each diode in the design stage of your schematic. For example, in this case, the regulator is ulseless if you don't use the analog ports to read values. A buck (or boost) could handle this situation better \$\endgroup\$ – José Manuel Ramos Nov 20 '18 at 9:27
  • \$\begingroup\$ Yes, this diode should prevent damage is the 12V input has swapped polarity \$\endgroup\$ – Andy Nov 20 '18 at 9:27
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    \$\begingroup\$ @JoséManuelRamos The 1N4148 only handles 500 mA That's peak current and not continuous current which is 300 mA. But personally I would only use 1N4148 when I'm sure less than 100 mA is flowing. Otherwise I'd use a 1N400x or a 1 A Schottky diode. \$\endgroup\$ – Bimpelrekkie Nov 20 '18 at 9:38
  • \$\begingroup\$ 1N4148 seems to be the default diode part number for some schematic drawing programs - so if you see 1N4148 used it may just mean the designer was too lazy to change the part number. \$\endgroup\$ – Peter Bennett Nov 20 '18 at 16:39
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Yes, this diode is probably for reverse polarity protection. It does its job but has the disadvantage that about 0.6 V is dropped across it. It needs to be rated for the current, both continuous and surge, as well as the reverse voltage it might encounter. 1N4148 is fine for lower current applications. A shunt diode with polyswitch/fuse is also an option. More elaborate schemes with MOSFETs are possible.

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any diode that can handle the current and voltage.

zeners have a low breakdown voltage and are thus unsuitable as you surmise. IN4148 can only handle a few hundered milliamperes continuous and so could be used in low current applications.

IN4001 can handle 1A, enough for many small devices.

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    \$\begingroup\$ zeners have a low breakdown voltage I disagree with that, it depends on what Zener diode you use. Zener diodes do have a more predictable breakdown voltage. Look at the Vishay BZT03 series zener diodes, they have breakdown (Zener) voltages up to 300 V, I do not call that "low". \$\endgroup\$ – Bimpelrekkie Nov 20 '18 at 9:34
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This is a correct reverse polarity circuit in any circumstance:

schematic

simulate this circuit – Schematic created using CircuitLab

This is completely different from the circuit you provide:

schematic

simulate this circuit

The main reason is the moment you stop: the inductances in your circuit will tend to conserve the current, and the circuit tries to achieve it from anywhere it want, harming your ICs, sensitive MOSFETs, sensors, or other components you have connected.

When you stop the circuit, the inductive equivalent of your circuit will try to retain the current, but your "protection" current diode does not allow to do that. Its mission consists in breaking the current flowing in the wrong direction.

The first circuit, instead, allows the current go trough the diode, and taking the diode a correct reverse polarity diode protection: all the current goes through this diode, breaking it if required.

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  • \$\begingroup\$ probably want something tougher than 1N4148 in the top picture \$\endgroup\$ – Jasen Nov 20 '18 at 10:41
  • \$\begingroup\$ It's only a schematic example. I hope it clarifies a bit about the differences between them \$\endgroup\$ – José Manuel Ramos Nov 20 '18 at 10:47
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    \$\begingroup\$ "This is a correct reverse polarity circuit in any circumstance:" No, that's wrong. That kind of solution can work reliably only if a fuse is put in series with the power source. A diode is not a fuse and it is not rated to be used as such and what you say is a real bad advice. If that 12V source were a car battery (e.g.), reversing the polarity could make that diode explode with flames. Moreover, if the fault current is big enough, even the explosion of the diode could not interrupt the circuit, but an electric arc between the terminals of the (now dead) diode could be generated. \$\endgroup\$ – Lorenzo Donati Nov 20 '18 at 11:02
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    \$\begingroup\$ The top circuit will only work if the diode is powerful enough to collapse the supply voltage. If the 12V supply were a car battery capable of supplying several hundred amps, then the result would be a blown diode, and the circuit wouldn't be protected at all. \$\endgroup\$ – Simon B Nov 20 '18 at 11:10
  • \$\begingroup\$ @LorenzoDonati The joule thief says hello to you. And... yes, you will need a fuse, too. \$\endgroup\$ – José Manuel Ramos Nov 20 '18 at 11:45

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