Question about dual CCS

I'm building some LED based lights for my work trike, intended as safety daytime running lights, but with the ability to be used as headlights at night.

Using a pair of LM317s in constant current configuration, with the input source switched between them.

Would a diode be needed on the output to prevent reverse current damage to the inactive current source, like this? Yes, I know a diode will add an additional voltage drop of ~1.7V to the circuit, but given the power is being supplied by an 8Ah SLA, I think I have the headroom available, since I am driving a trio of 1W Osram type white LEDs wired in series with a typical voltage drop of 3.2-3.6V

Low beam would be configured for 90mA and high beam for 250mA.

Circuit:

**broken link removed**

Thumbnail is clickable.

No prob. And a reasonable idea.

Yes, I know a diode will add an additional voltage drop of ~1.7V to the circuit

Click to expand...

What ever they drop, it won't matter in this configuration.

Thank you for that.

You're welcome.

View attachment 60751

This circuit should do the same thing with fewer parts.

R1 is you high value resistor. With the switch closed the resistance is value of the two resistors in parallel.

3v0, I guess that's why they pay you the big bucks.

Yes, much simpler, and easier to do.

I could use that for my tail/brake lights as well.

Thanks!

I set up a test bed for 3v0's circuit, and found that 60mA -->320mA gives me the effect I want.

Resistor values are 17ohm for 60mA, and 5ohm for 320mA.

This keeps me comfortably below the max constant current for my 1W LEDs on a freshly charged battery. This is with the ammeter in the circuit, so I would expect a very slight difference without it.

I don't think there is enough head room for 3 3.6 volt LEDs and a regulator.

Hmmm.

Voltage checks:

Battery - 12.74V

On low power:

V-in to adjust - 4.27V

V-out to adjust - 1.21V

On high power:

V-in to adjust - 2.94V

V-out to adjust - 1.14V

I'd say it has just a small deficit, but nothing I can't live with.