2
\$\begingroup\$

I'm trying to find a way to have one relay lock its state if the other is on.

Example, 2 relays total:

If relay 1 is on then relay 2 can't change state, either turn on or off. Meaning if relay 2 is on when you turn relay 1 on then you CAN NOT turn relay 2 back off. And, if relay 2 is off and you turn relay 1 on then you CAN NOT turn it on. Both relays need to be able to work independently i.e. no flipflop action or toggle action.

Practical application = Relay one is a 32v supply. Relay 2 is a series of relays that can flip a transformer around (swap primary and secondary) to flip step-up step-down function. You can turn on the 32v with relay 1 without consequence regardless of relay 2's state. However, if you already have relay 1 on and flip transformer around (relay 2) then you will damage the circuit from the voltage spikes. So, if relay 1 is on and you want to change the state of relay 2 you must first turn off relay 1.

It seems like something in the SR latch world may help but I'm having a tough time actually figuring the circuit out. The switches to activate the relays can be momentary or latching, as I can make either work.

\$\endgroup\$
1
  • \$\begingroup\$ if relay 2 is a single coil latching relay this is trivially easy. - connect relay 2 coil in series with relay 1 NC contacts. \$\endgroup\$
    – Jasen
    Jun 10, 2016 at 3:07

2 Answers 2

Reset to default
1
\$\begingroup\$

This is simple, save for one big caveat:

schematic

simulate this circuit – Schematic created using CircuitLab

In the circuit above, when S1 is open, K1 is off, allowing S2 to control K2's coil through the NC contact on K1. Closing S1 energizes K1, which connects K2's coil to K2's own NO contact via K1's NO contact -- if K2 is energized at the time, it stays energized through that NO contact, and if K2 is de-energized at the time, it cannot be energized by S2 any longer. In order to prevent K2 from glitching and dropping out when K1 changes state, R1 and C1 slow the release of K2 down by limiting the rate of voltage decay -- the needed delay depends on the relay coil sensitivity, but for a common DPDT relay, $$t\approx2.3\tau$$ as a starting point, with 20-25ms being a reasonable amount of time to wait for K1's contacts to settle.

If you wish S1 and S2 to be momentary operation -- they can be made this way using other contacts on K1 and K2 to stick them up (hold them on).

\$\endgroup\$
8
  • \$\begingroup\$ R1,C1 values seem unrealistic, \$\endgroup\$
    – Jasen
    Jun 10, 2016 at 3:29
  • \$\begingroup\$ I kind of punted on the methodology for those...fixed them now -- (we're actually both using the same TC, just with different R and C -- I wonder where the 1/3rd Rcoil rule of thumb comes from for R?) \$\endgroup\$
    – ThreePhaseEel
    Jun 10, 2016 at 3:58
  • \$\begingroup\$ time constant includes the resistance of the relay coil, I guessed 400 ohms for a 24V relay. \$\endgroup\$
    – Jasen
    Jun 10, 2016 at 4:44
  • \$\begingroup\$ basically I punted. at about 1/3 voltage the relay will let go, so you need a resistor smaller than twice the relay coil resistance to get enough discharge current to hold the relay on for any time at all. and then it's a trade-off between voltage lost in the resistor,delay time gained by the resistor's additinal resistance, and inrush current . \$\endgroup\$
    – Jasen
    Jun 10, 2016 at 4:50
  • \$\begingroup\$ Thanks guys. What is the "one big caveat"? \$\endgroup\$
    – jackwarner
    Jun 10, 2016 at 4:56
2
\$\begingroup\$

perhaps like this?

schematic

simulate this circuit – Schematic created using CircuitLab

a capacitor-resistor snubber in parallel with RLY2 coil will also help with hold-on during RLY1 switching.

\$\endgroup\$
3
  • \$\begingroup\$ Almost there -- what happens while RLY1 is changing state, though? \$\endgroup\$
    – ThreePhaseEel
    Jun 10, 2016 at 3:18
  • \$\begingroup\$ put a capacitor with a resistor 1/3 the relay coil resistance in series arallel with RLY2 coil. time constant should be ball-park 50ms or so. \$\endgroup\$
    – Jasen
    Jun 10, 2016 at 3:22
  • \$\begingroup\$ Add that into your answer and then you should be golden. \$\endgroup\$
    – ThreePhaseEel
    Jun 10, 2016 at 3:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.