Permanent Latch Circuit - Page 1
Forum section: electronics / projects

author: Pentoad
author_rank: Regular Contributor
timestamp: March 23, 2020, 08:02:53 pm
content: Hello all,This circuit here switches on a relay when a trigger voltage is applied and keeps the relay on permanently until power is reset.What I need though is a system where the relay switches on when the circuit is powered up and switches off permanently when a trigger voltage is applied and stays off until power is reset. How would I change the circuit to allow this?

author: jbb
author_rank: Super Contributor
timestamp: March 23, 2020, 09:38:58 pm
content: Use the normally closed relay contact?

author: Pentoad
author_rank: Regular Contributor
timestamp: March 23, 2020, 10:19:13 pm
content: That can't work for my application as it would allow the load to run even if there is no power going to the circuit or the circuit suffers some kind of failure.

author: Prehistoricman
author_rank: Regular Contributor
timestamp: March 23, 2020, 11:58:58 pm
content: Here's something I just threw together in simulation.Here's what the circuit does, in chronologic order:This generator on the left is set up to deliver a power-up type slope over 5ms (to make sure the system performs with a realistic turn-on).R1 initially turns T1 on, so that the relay turns on at power-up.C1 makes sure that T2 doesn't turn on before T1 turns onSW1 is set up to close after 10ms to simulate some external input going highR3 limits the current into the base of T2 (this resistor must be smaller than R2)T2 pulls the base of T1 low to turn off the relaySW1 opens at 15ms to simulate the external signal turning offR2 provides feedback to keep T2 turned on (the current flows through the relay coil)I've attached a zip of the TINA-TI (that's the simulator I use) schematic, if you want to play with it.Improvements that should be considered:A resistor to discharge C1Diode on the input, so that a low state cannot be asserted by the external signal

author: Zero999
author_rank: Super Contributor
timestamp: March 24, 2020, 10:39:40 am
content: The circuit posted above is an example of a bistable multivibrator.https://en.wikipedia.org/wiki/Bistable_circuitC1 should discharge via the transistor's base when the power is remove, so it shouldn't be an issue.The circuit can be made less sensitive to the input being forced low, by increasing the value of R3.

author: Pentoad
author_rank: Regular Contributor
timestamp: March 24, 2020, 08:32:22 pm
content: Quote from: Prehistoricman on March 23, 2020, 11:58:58 pmYour solution was brilliantly simple and did exactly what I wanted. I built it on a breadboard and it worked first time. Many thanks!

author: Pentoad
author_rank: Regular Contributor
timestamp: March 30, 2020, 08:27:36 pm
content: I know I said this was working perfectly and that was using the BC547 transistors I had lying around. I tried 2N3904 when they arrived and the circuit just oscillates and a tone comes from the relay no matter what state the input is at which is strange as it worked in the simulation.

author: Prehistoricman
author_rank: Regular Contributor
timestamp: March 31, 2020, 11:08:03 am
content: I don't really know how that could happen.Try increasing C1 and changing the value of R2. Or perhaps it's transistor-to-transistor variance?

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 31, 2020, 12:06:54 pm
content: i guess solenoid is an inductor (L), R2 and C1 forms resonant circuit, hence oscillation. from the original circuit in OP, try adding capacitor (maybe 0.1uF) from 9V to input trigger, and resistor (maybe 1Kohm) from input trigger to ground. reduce R1 to say 1Kohm or 100 ohm. during startup, the capacitor should momentarily giving initiating pulse to turn the thing ON and applying trigger LO later on to deactivate relay (OFF state). it may work it may not depending on the leakage. and maybe also reduce R4 to between 1-10Kohm. ymmv.