I require a SPDT relay (changeover) that will switch up to 600Vdc, which presents some problems because on the major suppliers such as Digikey, such a product is not readily available at such a high voltage.

However, there are SPST relays available and I was thinking of using two of those. However, I am concerned because I need the contact to the first line to disengage before connecting to the second line (otherwise I might cause a short).

To make things a little more difficult, I have only one digital control signal available, and would like the switch to connect to either one line or the other based on whether I am outputting a logic 0 or logic 1.

Anyways, all this to ask: does anyone have any recommendations for introducing this switching delay via hardware components? I was looking around for a "break-before-make" design but wasn't finding much luck.

Any help or direction would be appreciated!

Edit: Based on the accepted answer here is the schematic I came up with, for reference. Thanks everyone!:


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ There is always a good chance that one relay will be faster and make contact before the other breaks contact. \$\endgroup\$
    – PlasmaHH
    Nov 27, 2015 at 15:25
  • \$\begingroup\$ how many poles you need? \$\endgroup\$ Nov 27, 2015 at 15:47
  • \$\begingroup\$ @MarkoBuršič SPDT implies single pole \$\endgroup\$
    – Andy aka
    Nov 27, 2015 at 15:48
  • \$\begingroup\$ look here, something similar? schneider-electric.com/en/product-range-selector/664-tesys-d/… \$\endgroup\$ Nov 27, 2015 at 15:57
  • 1
    \$\begingroup\$ looks kind of dangerous. why not use diodes to cover the gap instead of shorting the two supplies together. \$\endgroup\$ Nov 27, 2015 at 20:04

2 Answers 2


Study this circuit: -

enter image description here

There is a single input called PWM but it can be any on-off circuit from any old logic device. It produces two outputs and notice the little bit of deadband due to the RC network.

For your application you can invert the OR gate output to guarantee it never rises until the AND gate output has gone low for the delay incurred by the RC network AND, importantly the AND output will never go high until the (N)OR output has fallen to low for the same period.

1 input, two outputs with shoot-thru (aka deadband) protection.

As an aside I'd probably use schmitt trigger input logic gates.

  • \$\begingroup\$ This is exactly the type of thing I was looking for--very useful and clever design. Thank you very much! \$\endgroup\$
    – jssoe
    Nov 27, 2015 at 16:31
  • \$\begingroup\$ I'm pretty sure that won't work since, if the relays are SPSTNO, the circuit's outputs should be out of phase. \$\endgroup\$
    – EM Fields
    Nov 27, 2015 at 16:33
  • \$\begingroup\$ @EMFields I think that's why he mentioned I might want to invert the OR gate output to be a NOR, which would allow them to work with SPST-NO \$\endgroup\$
    – jssoe
    Nov 27, 2015 at 16:41
  • \$\begingroup\$ @jssoe: Well, the terminology was a bit spotty, in that it's not a question of if you want to use a NOR, since for your application, the NOR isn't optional and that should be reflected by the schematic. \$\endgroup\$
    – EM Fields
    Nov 27, 2015 at 17:02
  • 1
    \$\begingroup\$ Don't forget you need to make sure your logic gates have Schmitt trigger inputs, so they're OK with the RC input. Alternatively, you could buffer the RC filter input. \$\endgroup\$ Nov 27, 2015 at 18:00


Here's Andy aka's circuit, but with the OR converted to a NOR, the AND converted to its DeMorgan equivalent in order to provide a one-chip solution, and a simulation run and plotted using LTspice..

His is a nice circuit, and a caveat with these kinds of circuits is that the RC delay must be substantially longer than either relay's release time in order to let things settle once the made contacts have opened.

enter image description here

  • \$\begingroup\$ Coolio squared LOL! \$\endgroup\$
    – Andy aka
    Nov 27, 2015 at 17:53
  • \$\begingroup\$ BTW what sim are you using? \$\endgroup\$
    – Andy aka
    Nov 27, 2015 at 20:33
  • \$\begingroup\$ @Andy aka: LTspice, available at and a TTL (74HC actually) available at \$\endgroup\$
    – EM Fields
    Nov 28, 2015 at 6:13

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