When reading application notes, I've came with the suggestion of adding reverse-based diodes across each of power supplies connected in series.

A diode in reverse bias must be added across output terminals of each power supply. This is to prevent -V voltage being applied to other power supply in fault conditions such as short circuit across load. During a short circuit, -V1 & +V1 will be connected across +V2 & -V2 which means the 2 power supplies output will be connected in opposite polarity and will cause damage to the power supplies.


Source: Mouser - Delta Electronics, How to Operate Parallel and Series Connection by Tassanai Cheangban

Considering these are isolated and "floating" power supplies, that is they have no common ground, how could we damage them if we have a short circuit across the load? Even in such case, the direction of current won't change.

  1. Are diodes are indeed required?
  2. How can we have situation here where current will go in the wrong direction?

I understand I'm missing something fundemental here, so curious to understand. Not to mention power supply has OCP for that.

  • \$\begingroup\$ Those power supplies are in parallel. Why do you think they are in series? Or is this a translation problem? \$\endgroup\$
    – Solar Mike
    Commented Sep 12, 2023 at 18:37
  • 4
    \$\begingroup\$ How are they in parallel if negative of first supply connected to positive side of another? \$\endgroup\$
    – seeker
    Commented Sep 12, 2023 at 18:39
  • \$\begingroup\$ Inductive loads can generate negative spikes \$\endgroup\$
    – Grabul
    Commented Sep 12, 2023 at 18:52
  • \$\begingroup\$ A hint. Try to examine what will happen during turn-on and turn-off period. For example, during Turn-ON the PSU 2 voltage rises much faster than PSU 1 ( e.g. PSU2 2 = 5V and PSU1 = 0V). \$\endgroup\$
    – G36
    Commented Sep 12, 2023 at 19:22
  • \$\begingroup\$ @G36, are you suggesting that one power supply (PS2) will feed another? Even if that's the case - that rise time is very slow. Would it be possible to damage power supply for such a short period? \$\endgroup\$
    – seeker
    Commented Sep 12, 2023 at 19:47

3 Answers 3


The diodes are needed to prevent reverse voltage across one power supply should the other one start first.

enter image description here

  • 1
    \$\begingroup\$ Thanks for ilustration. How did we get 1V in second case (with diodes)? Is that diode voltage drop? \$\endgroup\$
    – seeker
    Commented Sep 12, 2023 at 20:37
  • 3
    \$\begingroup\$ > Is that diode voltage drop? --- Yes. \$\endgroup\$ Commented Sep 12, 2023 at 20:58
  • 3
    \$\begingroup\$ The diodes also help at power down, or when the load is shorted. More generally, the diodes prevent damage to the electrolytic output capacitor in a PSU that's off, or provides less current than the other. \$\endgroup\$
    – fgrieu
    Commented Sep 13, 2023 at 7:11

There is no common ground, but the wiring clearly shows a common point of connection, PSU 1 (-) and PSU 2 (+) being connected.

The second connection, through the load, can be shorted, or could (if the load has capacitors, or batteries) be powered. So, there's a wide variety of possible fault conditions, and the manufacturer of the PSU probably decided to suggest a fault outside the PSU parts, because that's good public relations even if it's bad logic.

A 24V capacitor in the load could (if one PSU abruptly shuts down for whatever reason) reverse bias the output of a PSU, and the depicted diode would keep that scenario from being destructive.

  • \$\begingroup\$ A charged capacitor on the load wouldn't reverse-bias a power supply. The problematic scenario would be when one supply is putting out power and the load is passing current. In that scenario, current would flow out the other supply's positive lead until its potential is equal and opposite the potential produced by other supplies, or until it's sufficient to push current through. Adding a reverse-biased diode will allow current to flow without the potential having to exceed about -0.7 volts. \$\endgroup\$
    – supercat
    Commented Sep 13, 2023 at 16:20

Those reversed biased diodes prevent "backfeeding" the power supplies under fault conditions. Under normal operating conditions, those diodes don't conduct. But under faults (short-circuits inductor EMF faults) which would reverse-bias the output of the supplies, current gets shunted through the diodes instead of destroying the power supply. The fundamental idea is to keep the output positive rail always above the negative rail. You will find the same diode arrangement across solar cells in series for the same reasons. Also, if you look at the datasheets of most integrated circuits, you see the same reverse-biased diodes "clamping" pins to within one diode drop in the wrong direction before shunting current from the sensitive circuitry.


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