The power supply used on the product, is a DC-DC buck controller 24 V in (3 - 5.5 V out).

I need some suggestions to help protect a load (LED Display) which draws roughly 5 - 15 A the power supply on the product outputs 5.5 V and is capable of delivering 18 A - 25 A, the IC on our PSU can fail from time to time and cause over-voltages that destroy the LED drivers and LED displays.

I've looked into Zener clamping circuits, SCR crowbar circuits and Varistors,

But I don't know how well and how to calculate whether they would survive with the high current flow on the output of the PSU.

Any ideas?

  • \$\begingroup\$ Do you have any fuses in the circuit? If not, why not? \$\endgroup\$ Commented Mar 6, 2019 at 15:21
  • \$\begingroup\$ Would you really want to fix the effect of a failing PSU? Why not fixing/improving the PSU itself? Or should this by the single-fault protection? \$\endgroup\$
    – Huisman
    Commented Mar 6, 2019 at 15:27
  • \$\begingroup\$ Have you considered getting a more reliable power supply? Do you have an actual question for us? \$\endgroup\$
    – Dave Tweed
    Commented Mar 6, 2019 at 15:29
  • \$\begingroup\$ The PSU is fine, the over-voltage transients are a result of the IC being touched by field engineers (while on!) stupid as it sounds, and there are no fuses on the PSU. \$\endgroup\$
    – Jin Kazama
    Commented Mar 6, 2019 at 16:30
  • \$\begingroup\$ In that case, maybe you should just install a shield over the IC in question. It still seems overly fragile for the application, though. \$\endgroup\$
    – Dave Tweed
    Commented Mar 6, 2019 at 17:37

3 Answers 3


I was thinking along the same lines as Huisman — an electronic circuit breaker for the input that's triggered by the output overvoltage — and came up with the following circuit.


The 555 is a good building block for this, because it contains a flip-flop and a couple of voltage comparators. Here's a brief circuit description:

  • R1, D1 and C1 establish the power supply and reference levels for the 555 at 5.1V. R1 can supply about 5mA.
  • R2 and C2 create a trigger pulse that makes sure that the 555 starts up in the "triggered" state.
  • M1 is the main power switch, and Q1, R3, R4 and R5 drive it. Q1 serves as a switchable 1-mA current sink, which develops 10V across R3 to drive M1's gate, independent of the actual supply voltage. When the 555's "discharge" pin is grounded, this circuit is switched off.
  • R6 and R7 establish the trip point for the output voltage. The 555 will shut off the output if the "threshold" pin goes above \$\frac{2}{3}5.1V = 3.4V\$, and with these resistor values, that means if the output voltage goes above 6.0V.
  • C4 provides a 22µs time constant to filter out any fast glitches that might cause "nuisance" trips.

To restart the circuit, either interrupt the input power, or temporarily ground the "trigger" pin of the 555 (e.g., with a momentary switch).


Make a "smart" device that senses the output voltage of the PSU and interrupts the 24V input of the PSU in case of an overvoltage at the output.


You have powerful PSU, so protective circuit have to deal with large current and power. Zener diode and varisor are clamping-type solutions. Being activated they will dissipate huge amount of power due to high (higher than nominal) voltage. That is why they are inapplicable.
SCR crowbar, on the over side, is very well suited for the purpose for several reasons:

  • low voltage drop in open state (low power dissipation even under large current), load will not suffer from overvoltage
  • SCRs are capable of carrying huge currents for short period of time (easy to select proper part).

Of course, there must be a fuse at the input of DC-DC converter, that will be burned by opened SCR in case of converter malfunction.


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