I have a joystick control for a bow thruster. When you push the start button it enables the joy stick and also provides -ve 12 volts on an auxilary wire to be used in other circuits that I dont have. This wire normally has 0v on it but becomes -ve 12V and 0 ohms to negative supply when turned on. If I connect a 12v light bulb from it to 12V positive it will not light the bulb. It will however power up a small circuit board that has two relays mounted on it but when the relays are turned on they just pick up and drop and will not latch, I think the circuitry (encapsulated) must have some kind of current limiting. Can I use this -ve input to trigger a transistor switch to then apply the switched 12v battery supply to the relay panel. Any research that I do does not give me a definitive answer about tying -ve to -ve for turning switch on. Sw1 in diagram would be the joystick turning on the auxilary wire. Thanks


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


Your circuit needs to limit the emitter-base current, to avoid damaging Q1. R1 can do that but it also limits the emitter-collector current i.e. the load current. A suitably low base current will be also much too low for your load, be it a light bulb or whatever.

I recommend you reduce R1, commensurate with your load, and add another resistor in series with the switch leading to -12V. I don't know your load current but the ST datasheet rates your transistor's Ice at 200 mA so let's allow for that. It also rates hFE at 30 for Ice = 100 mA so let's assume a lower hFE and draw a base current of 12 mA. That leads to a switch series resistor (R3?) of (24-0.6)/0.12 = 1950 ohms. Try R3 = 2K2.

R2 can be removed or changed to a pull-up between the base and emitter (+12 V). A high value can be used to bias Q1 off when the switch is released. I would use 47 K or 100 K.

  • \$\begingroup\$ Thanks: So if I have this correct, I remove R2, or replace it with a 47 or 100K, put a 2.2K resistor in series with switch and change R1 to be the same as the resistance of my load. In the 0ff state, will I have any current drain? \$\endgroup\$
    – PJM
    Feb 28, 2017 at 18:17
  • \$\begingroup\$ @PJM, not quite. R2 is moved as said above, not changed. Add R3. Reduce R1 to whatever's suitable for your load...0 R? Q1 will be off, providing you've moved R2. \$\endgroup\$
    – TonyM
    Feb 28, 2017 at 18:38

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