How do I use a pull-up resistor to trigger a relay?

I have a 12v automotive electrical system that has a pressure switch that switches at 40psi. It is normally closed to ground (low pressure), and goes open when switched (high pressure). The switch only has one wire, the body of it is grounded. On the coil side of the relay, I assumed I could ground one side and connect the switch wire to the other side with a pullup resistor to give it 12v when it goes open. The low pressure state is grounded, making both sides of the coil ground and no relay actuation. When the pressure goes high, the switch opens and one side of the coil is pulled up to 12v, actuating the relay. (This is a normal Bosch style automotive relay, with a coil current of ~150-200ma)

After some testing, I have found that the relay will not trigger through a resistor greater than about 100 ohms (Ohm's law calculator says 60-80ohm pullup is needed). Shortly after, the resistor goes up in smoke- so I'm assuming it is shorting to the ground.

This diagram is roughly what I was thinking, but I may be way off in understanding how the pullup works. I need to trigger the relay on an open signal, and I thought pulling up was the way to accomplish that. How is a switched ground that is pulled up not a short?

• Was the relay wired like this from the start, (less the resistor), or is this all your own circuit? Are you simply trying to get VCC on the output when S1 is opened? Is the relay actually a SPDT type.
– Nedd
Feb 1 at 5:33
• Can you swap the NC and NO contacts of the relay? If so, just do that and put the relay coil where your pull-up resistor is. This will mean the relay pulls its 150 mA coil current when the switch is closed, instead of when it's open, so it might waste more power--but it will waste less power than a suitable resistor would. Feb 1 at 13:38
• This is a custom circuit, everything but the inputs can be modified. Input 1 (pressure switch) is normally ground, open when switched. Input 2 (ecu trigger) is normally open, ground when switched. My end goal is to energize a solenoid when IN1 is open and IN2 is ground. I am not concerned with the ~2w dissipation, the coil side would only be energized when the vehicle is running. Feb 1 at 13:59

If you must use a relay and switch arrangement like this you could use your original circuit but the resistor would need to be a 5, 10, or higher wattage type. With the resistor value range you listed (60-80 Ω) the power dissipation would be 2.4 W to 1.8 W. Do note that even if you use a 5 or 10 W resistor it will still get fairly warm. Also, if the switch is normally closed most of the time you will be wasting all that power most of the time, so you wouldn't want this circuit powered if your car was in the off condition.

One way to handle this would require a second relay, thus:

simulate this circuit – Schematic created using CircuitLab

Or you could use a couple of transistors:

simulate this circuit (resistor values may require some adjustment)

If you try to use a pull-up resistor, you will need a fairly low value to provide the current required to operate the relay when the switch is open. When the switch is closed, the full 12 volts is across the pull-up resistor, resulting in enough current through the resistor to destroy it.