I built a standard relay driver circuit as shown in the first schematic here:

enter image description here

It works as expected, but I now realized my input is between the positive terminal instead of ground.

How can I adjust the circuit to handle the input that varies between positive (12V) instead of ground?

  • 2
    \$\begingroup\$ Why exactly is the circuit not doing what you want? The circuit shown will work with +12V/OV inputs, going ON/OFF. Is that not what you want, and if not, what exactly do you want? \$\endgroup\$ – Spehro Pefhany Mar 9 '15 at 19:15
  • \$\begingroup\$ Are you saying that your input circuit has a common positive power line with the relay driver circuit but not a common ground? \$\endgroup\$ – alexan_e Mar 9 '15 at 19:28
  • \$\begingroup\$ Yes, exactly, it doesn't have a common ground. \$\endgroup\$ – Kritz Mar 9 '15 at 19:29

Just turn the circuit upside down:

enter image description here

and change the transistor for a PNP one. Eg. BC327. Notice the supply rails, the diode and the transistor symbol.

A voltage near Vcc will turn the relay off.

| improve this answer | |
  • \$\begingroup\$ Thanks Jippie, I think that should work. I've already built the original circuit and I only have a npn transistor. Can you think of any way I can it it to work with a npn transistor? \$\endgroup\$ – Kritz Mar 9 '15 at 19:31
  • \$\begingroup\$ @Johan No can't think of a smart solution that uses NPN at the moment. You'd have to change architecture. \$\endgroup\$ – jippie Mar 9 '15 at 19:39
  • \$\begingroup\$ One other item, is the resistor between the base and ground really necessary? \$\endgroup\$ – Kritz Mar 9 '15 at 19:43
  • \$\begingroup\$ I don't see a resistor between base and ground. The resistor between base and Vcc is an improvement to the circuit. It is only required when the output of the controller end can be "high impedance" or not connected to anything. \$\endgroup\$ – jippie Mar 9 '15 at 19:55

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