Driving common cathode LEDs from bridge rectifiers

[galaxywide]

I'm trying to put together a circuit to drive an LED module with high and low inputs and a common ground from an AC supply with one common leg and one leg ("low") that is switched to a 3rd wire ("high"). The "low" input is always on, and there is a switch that connects the "high" leg to the "low" leg. (it's a tail/brake light for a vehicle with an AC electrical system, originally it was an incandescent bulb with 2 filaments).

What I'm running into is that there does not seem to be a simple way for the high LED to be off until the switch is pressed, and then come on in full wave rectification. What I got on my first attempt was simply switching the high LED from half wave to full wave, which is certainly obvious in hindsight but was quite disappointing when I discovered it. The circuit I tried first is thus:



The best way I can find to solve this issue is with a DPDT relay, like this:



I'm not opposed to using a relay, but I'm very space constrained in this application and finding a very small one with an AC coil that's of reasonable cost seems problematic based on my limited searching. I'm by no means an EE, so I'm hoping that there is some obvious solution that I'm missing here, or a fairly simple circuit design that I could get in a small custom PCB.


The specs are as follows:

Low LED draw: 90mA
High LED draw: 450mA
Nominal voltage of both AC and DC systems is 12V
AC frequency is variable with RPM, as it is generate by a stator.

Thanks in advance for any suggestions!


Edit: Solved the image issue by slightly cheating with forum image urls.

[Ian.M]

You cant do that with bridge rectifiers.  Generally, if you have more than one bridge rectifier, *EITHER* you can join their AC input terminals *OR*  their DC output terminals, not both.   Consequences depend on the exact connections and range from one being ineffective, to a half or full-wave short across the supply through the rectifier diodes.

In your second circuit the righthand bridge rectifier is redundant (provided the always on circuit can supply the total current for both LEDs).  Simply use one pair of relay contacts to switch rectified positive to the high LED.

If you want to eliminate the relay, it gets tricky.  Probably the best option would be an AC input optocoupler driving the gate of a P-MOSFET as a high side switch powered by the positive DC output of the left rectifier, to feed the high LED.  If it isn't convenient to obtain an AC input optocoupler, use an ordinary dual one with its LEDs in anti-parallel and its phototransistors in parallel.  Don't forget a current limiting resistor in series with the opto's LEDs and a gate pullup resistor for the MOSFET to ensure it turns off properly.

[galaxywide]

Thanks! I definitely found out the hard way that I couldn't do what I wanted, I was just a bit stuck as to where to go next.

You're very right about the righthand rectifier being redundant in the second circuit, can't believe I didn't see that!

I really like the idea of an AC input optocoupler, and it looks like they're quite inexpensive and small. I managed to simulate a circuit that works, though I had to use 2 DC optos as falstad.com seems rather limited in it's selection of parts.

Now I just need to select some components, it looks like the LTV-814S should work fine. The rectifiers I already have, they are marked "KBP206" and seem to work well. I also have a bunch of 1mF 25V electrolytic caps that I'd like to use for filtering, any reason to get something different?

One part I have not looked into yet is finding a suitable P-MOSFET - there seem to be a bewildering number of options, are there any that are more common or recommended? Ideally surface mount, but large enough to be hand soldered. A max current of 2A should be more than enough based on the load I have.


Thanks again for the help, that was just what I needed! Now to remember how to use my old version of Eagle...