I am a beginner in electronics and would like to request comments and critics for the following circuit I have designed: Bistable relay driver The objective is to drive a two coils 12V bistable relay upon the state of a relay input provided by a BMS hooked on LiFePO4 battery:

  • input relay closed: short pulse applied to the ON side of the bistable relay
  • input relay open: short pulse applied to the OFF side of the bistable relay

Operating voltage range from 11.5 to 14.8V. As the battery is charged by an alternator, electro magnetic environment can be quite perturbed (similar to what is observed in a car).

CD4093 has been selected as it is able to operate up to 20V. Anti bounce has been added to the input command. FET protection is provided by a zener and a TVS.

Thanks in advance,

Best regards,

  • \$\begingroup\$ What those two series caps for and how are they supposed to work? \$\endgroup\$
    – Lundin
    Commented Aug 17, 2022 at 9:51
  • 1
    \$\begingroup\$ You should consider simulating things to see that the pulses lengths are sufficient for the relay coils. \$\endgroup\$
    – Andy aka
    Commented Aug 17, 2022 at 10:02
  • \$\begingroup\$ R6,7 wont do too much as the mosfet gate is always actively driven by 4093. If it were a microcontroller that it would be a different story. D1,3 aren’t required unless you have a really poor pcb layout. You’re switching low current slowly so spikes due to source inductance should be non existent. D4,5 are unnecessary unless you expect to have long wiring between the mosfet and the relay. Better to put transzorbs on the incoming power supply. Being automotive, add protection for load dumps. \$\endgroup\$
    – Kartman
    Commented Aug 17, 2022 at 10:32
  • 1
    \$\begingroup\$ At first glance, it's excessively complicated, though I've currently no opportunity to sraw a smaller one and prove it. \$\endgroup\$
    – TonyM
    Commented Aug 17, 2022 at 12:35
  • \$\begingroup\$ Decoupling capacitors? \$\endgroup\$
    – winny
    Commented Aug 17, 2022 at 16:07

2 Answers 2


Besides the good suggestions by Kartman, I also recommend the following.

Use a non-polarized cap. The slight negative voltage the polarized cap will see is probably OK, but it can be avoided. 4.7 uF ceramic is fairly common now. Increase the resistor accordingly.

Add diodes so the IC won't have to absorb the rebound pulses outside the power range.

Add resistors in series with the IC input to limit the current from whatever the diodes don't clamp.

If you have an MCU in your system, I would use it to make the pulses.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ 10uF is also common for ceramics nowadays, though it depends a lot on what voltages that's expected. \$\endgroup\$
    – Lundin
    Commented Aug 17, 2022 at 12:57

Thanks a lot for the very valuable comments. My apologies for the long delay replying, I have been away for some time.

@Kartman and Mattman944, I have modified the schematic according to your suggestions. I will test the circuit on a breadboard before designing a PCB and will pay attention to PCB layout ; the latching relay will be soldered on the PCB to reduce the length of the wiring between the FET and the relay. I have located a transzorb on the incoming power supply. Updated latching relay driver

I have found the way to simulate the circuit with KiCad (ngspice). Here follows the results : Model schematic ON side OFF side The first graph focus on the ON side and shows in red the output of the first Schmitt trigger and blue the potential behind the capacitor. Indeed, a negative spike was present in the original proposal and can be reduced with the addition of the protection diode D1. The second graph aims to the OFF side. The positive spike in here once again clamped thanks to the diode D3.

The last graph shows the current in the protection resistors R7,R8 that limit indeed the current for the part that is not clamped from the diodes: enter image description here It is ranging from -250uA to 280uA (providing the model for 4093 is adequate...) and seems to be within the specs of the 4093.

Thanks again to everyone, the helped a lot to sort out various aspects of the circuit. Still a lot to learn...

Best regards,


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