If headlamps are tied to 12V with low side switches, the LEDs must also be tied high with common anode instead. The lamp resistance is powering the LEDs. Nice job on MGB. I had a '67 B when they had a positive ground system.
Note in your schematic, headlights are tied to +12 and switched to Gnd via relays. Thus when connected to LEDs with Common Cathode (CC) tied to ground via a resistor , they will stay on due lamp resistance to V+.
Therefore replace CC LED to gnd with CA LED to 12V and pull down LEDs with lamp relays to make this work.
Edit
I just realized, that I misunderstood the modifications added two more Relays rather than being wired to existing Relay sockets. With this in mind, I suspect there is a lot of EMI on the wires which is sufficient to power the LEDs. This noise can be suppressed with >=0.1uF capacitor across each LED. Are these HID lamps? That may do it.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Simple on-off control.
It's not clear why you think you need relays. They will probably add a nice clicking sound but you should be able to obtain complete isolation with the switches only.
OK. That's not going to work due to the leakage.
simulate this circuit
Figure 2. Adding a pair of diodes eliminates one of the relays.
How it works:
- Switching on LEFT will cause the relay to turn on when current flows from the flasher through the coil and D1. D2 prevents the right LEDs from turning on simultaneously.
The problem is that a 60 V relay will require little current to energise it so you may find that the relay stays on and now you have LEDs at full brightness.
If you supply all the requested info and some details on the LED specifications there may be a workaround.
LED current Voltage drop across flasher
On 80 mA 2 V
"Off" 20 mA 36 V
From those numbers appears that your flasher passes about 20 mA when off. This is required to power the internal circuitry and on a 3 A load is < 1% so it wouldn't be noticed. On your 80 mA load it is passing 25% of the "on" current.
Solutions:
- Buy a better flasher. A three-terminal one would have its own direct path to ground so that the output could be fully off.
- Try your relay solution.
simulate this circuit
Figure 3. (a) A common loading resistor. (b) Individual loading resistors.
- Add a load resistor. As shown in Figure 3 a common load resistor or individual resistors per side will "shunt" some of the current around the LEDs. Since you have about 32 V across the LEDs and 20 mA flowing through them they are acting like a resistance of \$ R = \frac {V}{I} = \frac {36}{0.02} = 1800 \ \Omega \$. Therefore, if we put 1800 Ω in parallel we will shunt about half that current away from the LEDs. (Because the LEDs do not behave like resistors the voltage and current will not split exactly as shown above.)
We now need to check the power rating of the resistor when the flasher is on. Power is given by \$ P = \frac {V^2}{R} \$ so let's say we want the LEDs very dim when off so we add 1k in parallel and we'll have 58 V across the resistor when on. \$ P = \frac {V^2}{R} = \frac {60^2}{1000} = 3.6 \ \mathrm W \$. Obviously a 0.25 W resistor isn't going to be good enough.
Work with the numbers and your available parts to find something that will work.
Best Answer
It's okay if you use a car relay as long as it handles more current than your load, you'll have the advantage of having terminals to easily connect all the wiring but it will cost you some more current when the relay is on, car relays can draw up to 2amps from your battery, so that is something you should keep in mind considering your load only draws 600mA.
Using pcb relays will work as well and draw much less current, but they're hard to connect if they're not soldered on a PCB