What is a good resistor value for my circuit?

Question

I'm trying to tap into my car's brake circuit to switch between receiving a wireless camera signal and powering my device. For this purpose, I connected a relay to the circuit via a fuse tap. It should connect the phone to the receiver (displaying the camera feed) when the brakes are applied and charge the phone otherwise. Unfortunately, I'm having some difficulty figuring out how to get the relay to switch properly.

I wired the relay to the circuit in the fuse box that powers the car's instrument lights and the brake lights. The current draw for the instrument lights (which are always on) is 0.7 amps. The brake lights draw 1.15 amps. Without adding additional resistance between the car battery and relay the constant 0.7 amp draw will always keep the relay in the ON position (connecting phone and receiver). Hence, the need for additional resistor between battery and relay. The relay's coil draw is .5 amps.

^{simulate this circuit – Schematic created using CircuitLab}

Here is my most likely flawed resistor calculation. Please point out any errors.

First I calculate the internal circuit resistance

V=IxR
12 = .7xR
R = 17.143 ohms // with the brake lights off

12 = 1.15xR 
R = 10.169 // with brake lights on

To keep relay in OFF position, I need to supply a current less than the relay's coil draw (.5 amps). Adding an 8 ohm resistor should do the trick:

R = 17.143 + 8 = 25.143 //brake lights off
I = V/R
I = 12/25.143
I = .477 < 0.5 // relay switch in off position

R = 10.169 + 8 = 18.169 // brake lights on
I = 12/18.169
I = .66 > 0.5 // relay switch in on position

It seems like 8 ohms is enough to keep switch OFF with 0.7 amp draw, but my relay always stays in the ON position, even after I increase resistance to 16 ohms. So what gives?

Answer 1

If the car is set up like this:

^{simulate this circuit – Schematic created using CircuitLab}

And you want to tap into the fuse and make it like this:

^{simulate this circuit}

Then it would make sense that once you reduced the voltage through the circuit enough, by dissipating the power through R1, then the relay would turn off.

The problem is probably that the Relay doesn't require 500mA to switch but uses 500mA at its operating voltage. As the voltage decreases to turn the relay off it probably switches at a voltage threshold much lower than its operating voltage and thus a much lower current draw.

In theory, it would work with a high enough value resistor. But the lights will be very dim, you will burn a lot of heat through the resistor, and the relay won't last as long being partially energized all the time.

You would be better trying to connect the relay in parallel with the rear brake light or at least in series with only the rear light. If connecting to the fuse is the only option then putting a current sensor in series with the fuse and triggering the relay with an MCU (which reads the current value) would be the best option.

Answer 2

Operate the relay from the signal/voltage that powers the brake lights. This gives 12V when operated and open circuit (= 0V) when off (OR ground when operated and open circuit when off in some less usual situations).

The 12V power line from the battery will remain at about 12V at all times under all usual loads except starting. The starter motor MAY load the 12V line down to 10V or less, but nothing else will.

Answer 3

In a brake light circuit the fuse is typically before the brake light switch, by tapping the voltage at the fuse box you won't get a useful answer.

Easisest is probably to tap the voltage after the brake light switch, on most cars this is somewhere near the brake pedal lever, but some use a pressure switch on the hydraulic brake line.

^{simulate this circuit – Schematic created using CircuitLab}