I need to detect SLA battery voltage & when the battery comes closer to 10.5V the relay must turn ON.The below circuit works ok, but the hysteresis gap is pretty closer, when I touch the circuit the relay jitter fastly.

The transistor arrangement allows only 2-3mA current to flow which cannot drive a 400 Ohms coil, That's why I used an opto-coupler to drive the relay.

I need to remove the opto-coupler part & needs to drive the relay from the transistors with hysteresys. How to do that?


simulate this circuit – Schematic created using CircuitLab

  • why does the relay have to activate when the voltage drops .... would it not make more sense to deactivate the relay when the voltage drops so that there is less drain on the battery when voltage is low? – jsotola Nov 23 at 9:30
  • I want other side, When Battery drains it must turn ON the relay. – Joseph143 Nov 23 at 9:38
  • 1
    it is unclear what you mean by turn ON the relay ...... does it mean energize the relay? .... or does it mean close the relay contacts? ...... those two things are not the same – jsotola Nov 23 at 9:42
  • Hi, Sorry for the mistake. Turn ON the Relay = Energize the Relay.I hope now its clear. – Joseph143 Nov 23 at 9:45
  • Just use a relay with normally closed contacts that gets turned off when the battery voltage drops. That way it will stay off when the battery dies altogether. – Finbarr Nov 23 at 9:45

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