I have the following sample schematic of a 12 volt automotive circuit to trigger about 60 seconds of power output to a latching relay on output pin 3 of the 555 Timer chip.

The schematic below is a just a sample and not the final circuit. The 12 volt automotive battery is driving the circuit. It requires 2 different types of input to the 555 Timer.


1) 1 second ground pulse coming from various Viper alarm outputs. It is a floating Viper model-5906V (-) 200mA 1 second OUTPUT. (This input should turn the relay on for 60 seconds, once it sees ground.)

2) Turning the 12 volt ignition on should turn on the relay (set) and leave on as long as the ignition is on.

3) Turning the 12 volt ignition off should trigger the 555 Timer. After 60 seconds the relay should turn off (reset).

4) Ideally if the ignition turns off and it has been 50 seconds (relay is still ON) and then the Viper alarm sends a 1 second ground pulse on the input, it should wait 60 seconds from the last input received. I am not sure this is possible however without adding a 2nd 555 Timer to the mix.

555 Timer to Latching Relay

The 555 Timer needs to be configured to turn on (set) the 12V Hella Latch Relay (Hella 933364027), specs are here: http://hellahd.com/index.php/default/electrics/relays/micro-iso-relays/electrics-product-12/ , when either of the above inputs are triggered. The relay should turn off (reset) after the 60 second Timer output expires, this should expire 60 seconds after the last ground pulse from the alarm trigger, or 60 seconds after the ignition is turned off, whichever is later.

The circuit is used to supply power thru the latching relay to power assessories after the ignition is off for 60 seconds, with a provision to allow for remote controlled activation via the alarm's 1 second ground pulse.

Need assitance in getting the outputs to fire the latching relay contacts briefly with the above timing. Currently the output signal to the Set (pin 1) contact is working but when I try to use that to power a transistor as in the example to the Reset (pin 2) it does not turn the transistor on.

How does this circuit need to be modified to set and reset the latching relay?

  • \$\begingroup\$ What is the relay coil resistance? What voltage rating (12V presumably.) Several simple circuits are possible. How fast does the 12V fall. Do you want the pulse AFTER the 12V is fully gone or when it starts to fall. If AFTER the 12V is gone the relay needs powering. \$\endgroup\$
    – Russell McMahon
    Commented Jan 20, 2020 at 2:23
  • \$\begingroup\$ I will be driving a 12V latching coil (Hella 933364027), specs are here: hellahd.com/index.php/default/electrics/relays/micro-iso-relays/… The coil resistance is 2 x 75 Ohm ± 10%. The 12V is the ignition wire in a vehicle so it should fall off fairly quickly. The igition wire will also power the coil so as soon as the ignition wire drops to 0V, it should turn on the 555 Timer. I shouldn't need to compensate for the ms between those changes because i will be using a latching relay. The output i did not show but it is taken care of already. \$\endgroup\$
    – Raptor
    Commented Jan 20, 2020 at 2:45
  • \$\begingroup\$ Definition of task now uncertain. Is the latching realy to emain on due to final pulse. Or is the pulse a delatching one. |Failure to properly descibe task wastes you time and ours. | Max data helps good answers. | Relay has a MAX pull in time of 10 mS. So your 1s spec is excessively hard.|| operating voltage (or a <6V not operating one). So you need > I = V/R = 7/75 = 40 mA. More is better. 80mA say x 12V x 10 mS = ~ 0.01 Joule = nicely small. Dumping a say 22o uF onto the coild ould probably do. 2 x small cheap transistors, cap, 2 r's. diode. Later ... \$\endgroup\$
    – Russell McMahon
    Commented Jan 20, 2020 at 2:53
  • \$\begingroup\$ Appoligizes for the output, I was trying to keep it simple. The output of the 555 Timer is driving a NPN transistor to power the relay off/on, its a standard automotive relay and that is working properly. Question is: I need to have the 12V to 0V change trigger a small momentary ground to the 555 Timer input to start its countdown. \$\endgroup\$
    – Raptor
    Commented Jan 20, 2020 at 3:02
  • \$\begingroup\$ Thanks. Simplifying doesn't always do so :-). | I'm confused though - The relay has a set/reset dual coil arrangement. How does the 555 do the two coil separate pulse arrangement? | Is the 15V present even when the 12V ignition voltage is removed? . | A "simple" description of the whole cycle to make sure I have it clear would help. [Things that ARE in fact quite simple once properly appreciated can sometimes be remarkably hard to convey]. \$\endgroup\$
    – Russell McMahon
    Commented Jan 20, 2020 at 4:37

1 Answer 1


Here is a circuit that may meet your requirement.
Inverters are a package of hex Schmitt trigger inverters.
MUST be Schmitt trigger. There are two spare inverters which can be used in various ways if desired.

You could convert this to a '555 based circuit with more components and less "properness". eg here the 1 second delays are driving a Schmitt triggered CMOS input.

D1 accepts 12V positive on ignition
D2 accepts a 12V positive going alarm pulse. As the contact is floating this should not be a problem. If desired you can invert either input sense with a spare inverter.

When either input is high, Q1 inverts this to low and D3 keeps C2 discharged.
An alarm pulse occurring during a R2C2 timeout resets the timeout.

When neither input is high R2 charges C2. A 1 minute delay could use eg about 10 uF / 4M7 . Values will need to be "select on test"ed.

R1C1 and R3C3 plus following inverters provide 1 second delays.
A negative going edge pulls the cap low and the resistor then charges it high.
When the alarm pulse activates C1R1 the output pulse lasts the lesser of the alarm pulse or C1Ra timeout. As the set/reset relay only needs a 10 ms pulse 'this is not a problem".


simulate this circuit – Schematic created using CircuitLab



Extra inputs eg floating/ground on/off

If you follow my circuit and see how it works it should be possible to add to it with the spare inverters or modify it.

"Floating" inputs are best pulled to non active state with a pull up/down. This can be behind an input diode (see eg existing Ignition-D1-R6 where D6 pulls the input to ground if Ignition is open) to ensure input idle state is defined.

If voltages spike too high, clamp them to relevant rail (12V or ground) with a diode.
eg in existing circuit the input to IC3 spikes to 2 x supply when IC4 out goes high.
A diode across R3 (Cathode/bar to V+) stops this.

  • \$\begingroup\$ Thank you @Russell McMahon I may need to order some more hex Schmitt trigger inverters, but will try it out! \$\endgroup\$
    – Raptor
    Commented Jan 29, 2020 at 16:41
  • \$\begingroup\$ the circuit works great! Now that it is built I have found 1 additional input needed. I need to have a constant ground act the same as an incoming pulse. I can use an inverter to make it high as well, the difference is it needs to act as a pulse to the input for about 1 second and then not trigger again unless ground is removed and then triggered again. I have it working with a RC circuit but must be missing the correct capacitor discharge because after ground is removed I cannot get it to trigger again. Any ideas? Thanks for the help. I appreciate it. \$\endgroup\$
    – Raptor
    Commented Feb 4, 2020 at 3:06
  • \$\begingroup\$ @Raptor My circuit should turn the relay on on a rising edge and off on a falling edge. Your added requirement is ambiguous - can you please spell it out in step by step detail. A constant ground inverted with a spare inverter and then fed in via D2 should act the same as ignition on D1. Either a detailed word picture or a sequence diagram is needed. \$\endgroup\$
    – Russell McMahon
    Commented Feb 4, 2020 at 9:39
  • \$\begingroup\$ - this input is floating input that supplies a constant ground when activated. It can remain at ground for 1 second or 24 hours. The requirement is when ground is received it should activate the relay for 60 seconds and then turn it off. Simlar to the 1 second input pulse, but because this input is constant the 60 second timer should start as soon as the ground signal is received. The Ignition signal does not start the countdown 60 second timer until the ignition turns off. \$\endgroup\$
    – Raptor
    Commented Feb 4, 2020 at 20:06
  • \$\begingroup\$ While this input needs to start the 60 second off timer as soon as the relay is activated and, then turn off the relay in 60 seconds if no other input has activated it. Ignition would continue to keep the relay on if the ignition was on after this signal was received. Ultimated just need to change this constant ground into a 1 second pulse and it could be fed into IC1. \$\endgroup\$
    – Raptor
    Commented Feb 4, 2020 at 20:08

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