First post here, please excuse my electronic ignorance.

I am trying to design a 12V circuit to open and close the nitrous bottle on my race car. Here are the specifics of the design: It needs to

  • open the bottle (CW motor) when the SPST switch is in the on position, stopping when it reaches a limit switch to prevent the motor from burning up.
  • close the bottle (CCW motor) when switched to "off", again stopping on a limit switch.

The reason for the SPST switch is that the rest of the nitrous controls on the panel are the same type of switch and I don't want to have one oddball switch that doesn't match. I would prefer to use relays to prevent running the motor current through the switch(es), and have obtained 2 automotive grade SPDT relays for this purpose. I found circuits for a SPDT activation switch, and another for continuous automatic reversing, but I need it to stop when the bottle is open, and not reverse until the switch is turned off. I don't mind adding components as long as they are automotive-rated. The no-load current for the gear-driven motor is 230 mA.
Am I overbuilding it to use relays?

  • \$\begingroup\$ Electric windows on cars use relays and pretty much the same circuit it seems. \$\endgroup\$
    – Andy aka
    Jan 13, 2020 at 9:35
  • \$\begingroup\$ Windows just stall the motor when they reach the end of the travel, and if you hold the switch it will overheat and destroy the motor. They also use a 3 position momentary switch. I need it to automatically stop the voltage to the motor when it reaches the travel limit, without manipulating the switch. The bottle is in the back of the car and the car is VERY loud, so I cannot hear the actuator motor to know when it stalls out at either end of its travel \$\endgroup\$ Jan 13, 2020 at 10:24
  • \$\begingroup\$ You asked: Am I overbuilding it to use relays? and the point of my comment was to remind you that electric windows use relays in pretty much a similar circuit. \$\endgroup\$
    – Andy aka
    Jan 13, 2020 at 10:29
  • \$\begingroup\$ Ah, now I understand what you meant. Thanks \$\endgroup\$ Jan 13, 2020 at 10:42
  • \$\begingroup\$ Start current typically is 100x no load current and 10x rated current under full load and needs reverse diode to opposite rail to suppress arc voltage or EMI on open current switch to snub current. Current sense with an RC delay can ignore normal surge but detect end-stop to cut-off motor .Use 50mV stall current drop R (23A?) or measure motor DCR to compute ground side 50mV rise to comparator to shutoff motor (if 2 milliohm must be low inductance and filtered to ~ 100ms) Then 20A failsafe PTC or same as fuse \$\endgroup\$ Jan 13, 2020 at 13:06

1 Answer 1


To get the reversing action you want, you need a DPDT switch somewhere, either on your panel or in a relay. I'd guess wiring is easier if it's a relay, and you have the advantage of your panel switches all being the same, as you desire.

Would this circuit suit your application?

enter image description here

Explanation: with switch S1 open, relay is unenergised, and with motor in mid-position limit switches S2 and S3 are closed. Thus current flows from +ve supply through relay contacts P1 through S2, through motor, P2 and GND: Motor turns counterclockwise and eventually hits limit switch S2 which breaks the circuit. When S1 is activated, relay moves to energised state and current flows from +ve supply through P1 to S3, bottom of motor, P2, then GND: motor turns clockwise until limit switch S3 is engaged.

Note that for inductive loads such as the relay and motor, you need to pay attention to whether you need flyback diodes and whether any sparking is entirely contained, as obviously sparks and your various car fuels are explosive. Modified race cars are definitely at your own risk!

  • \$\begingroup\$ That looks like it would work. I was somewhat concerned about putting the limit switches in the main (load) part of the circuit, but they are rated for 5A so I don't think they will cook. The relays and the motor are sealed for Automotive applications, so I am not concerned about sparks. Could you please explain what a flyback diode is? I am not electronic lingo-fluent \$\endgroup\$ Jan 13, 2020 at 10:35
  • \$\begingroup\$ When an inductive load is disconnected, electricity is generated in the other direction (by the change in magnetism on the coil). The voltage can be much higher than the voltage that charged the coil up in the first place, and often produces a visible spark. A flyback diode absorbs this energy. Very often, relays have them built in. en.wikipedia.org/wiki/Flyback_diode \$\endgroup\$
    – jonathanjo
    Jan 13, 2020 at 10:40
  • \$\begingroup\$ Thanks for the info about the diodes. One other thing I am a little confused about- What is the purpose of RLY1? Unless I am seeing it wrong, wouldn't the on/off switch energize/ de-energize P1 and P2 in the same way? \$\endgroup\$ Jan 13, 2020 at 10:45
  • \$\begingroup\$ "concerned about putting the limit switches in the load" circuit: well, you need either a limit switch or a relay contact, which are pretty much the same thing, and you just get ones rated for appropriate current. If you're worried about this, put a fuse in line, rated below the stall current of the motor. \$\endgroup\$
    – jonathanjo
    Jan 13, 2020 at 10:45
  • \$\begingroup\$ RLY1 is the coil of the relay; P1 and P2 are the contacts of the same relay. \$\endgroup\$
    – jonathanjo
    Jan 13, 2020 at 10:46

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