Solid State Flasher controlling other Relays
simulate this circuit &ndash; 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 &lt; 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.