Consider the following schematic:

LEDs and schottky in series reverse biased

simulate this circuit – Schematic created using CircuitLab

Main question:

1) I am accustomed to use Schottky diodes to protect my circuits against reverse polarity. However, in my current scenario, user may apply voltage in reverse polarity to LEDs directly. After some research, I started to suspect that every Schottky diode may not be suitable, because while their reverse voltage may be rated enough the Schottky to survive, but their reverse current may exceed of LEDs'.

Is it correct?

Other questions:

2) How is the reverse voltage distributed on components? Is it equal, or completely random, dependent on LED/Schottky characteristics?

3) Consider the reverse voltage to be 20 Volts, and shunt the Schottky. Will each LED handle the reverse voltage (approximately) equally, and they will stay functional?


1 Answer 1


I do not think that there will be any issue as the reverse current of the Schottky diode is very likely far to small to cause any damage to the LEDs. Due to the small leakage current not much power can dissipate in the LEDs so I do not expect that any damage can occur.

However, the reverse current of a Schottky diode is very temperature dependent, at higher temperatures it could become in the order of 10 mA! (Source: datasheet of IN5817, a 1N4001 silicon diode has a reverse current of only 50 uA at 100C) Even that might not be enough to damage the LEDs though.

Fortunately there are simple measures we can take to prevent all issues. Just add another diode in parallel with the LEDs such that this diode conducts the reverse current of the other diode:


simulate this circuit – Schematic created using CircuitLab

Note that I copied the Schottky diode, D9 does not need to be a Schottky diode, a Silicon diode (like 1N4007) will do the job just as well. The same is true for D8, there is little need to make that a Schottky diode, you can use a 1N4001 as well as long as the current is less than 1 A.

D9 will be in forward mode when the voltage from V1 is negative. That means that the reverse voltage across the LEDs is limited to less than 1 Volt so nothing will happen to the LEDs.

In your circuit however I do miss a resistor to limit the current through the LEDs. Using LEDs without a current limiting resistor is not advised as the current through a LED is heavily dependent on the voltage, type, aging, temperature. So in my circuit I added R1.

  • \$\begingroup\$ Also remember that schottky diode has usually lower reverse breakdown voltage than normal silicon diode. I don't know that the OP really intended to use 50V or not. 1N5187 schottky has 20V maximum reverse breakdown voltage while 1N4001 sillicon has 50V \$\endgroup\$
    – Unknown123
    Apr 3, 2019 at 9:18
  • \$\begingroup\$ @Unknown123 Also remember that schottky diode has usually lower reverse breakdown voltage than normal silicon diode. Not if you compare a 50 V silicon diode to a 50 V Schottky diode. I mean, both will have a 50 V breakdown. Also a 20 V silicon diode will have a lower breakdown than a 50 V Schottky diode. Your remark does not make sense to me. The IN5817 is just an example. \$\endgroup\$ Apr 3, 2019 at 9:23
  • \$\begingroup\$ Thank you for the answer & schematic. I will be happy to accept as answer if you have an opinion about questions 2 and 3. \$\endgroup\$
    – C K
    Apr 3, 2019 at 9:26
  • \$\begingroup\$ Yes, I mean usually, really it's relative, what my point is that, you need to check the diode datasheet to make sure everything is fine. \$\endgroup\$
    – Unknown123
    Apr 3, 2019 at 9:26
  • \$\begingroup\$ @CK Regarding questions 2 and 3: that will depend on the LEDs, often very little is specified in the datasheet regarding reveres voltages of LEDs. So it is better to avoid these unknown conditions which can be done easily by following my suggestion of an extra diode. \$\endgroup\$ Apr 3, 2019 at 9:28

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