Im trying to make a LED circuit for my E bike which is a 48v-55v supply voltage. Im not used to working with such a high voltage. I tried looking at the original LED circuit that was there and dont know how they are powering 3 cheap amber 5mm LEDs from 55v input . The resistors on it are only 1/4 watt resistors and they dont get hot or anything . There is a reversed glass type diode(guessing maybe a Zener) of the input before the LEDs which must be whats bringing the voltage down but how does that work .Heres the schematic for the circuit below

Can someone help me figure this out .Thanks

 

Welcome to AAC!

Why such high voltage. The LEDs shown would need around 12V, max.

You could do this much better with a buck boost converter, with much higher energy efficiencies.

 

The Ebikes battery is a 48v - 55v max. I have built tons of 12v LED circuits which is simple but iv never worked with a 48v supply before

 

There are classes of electronic circuits that convert, instead of dropping voltage.

 

Thanks .I know a switching regulator would work but i was more interested in how the original circuit above works and so i can put something together quickly with radioshack parts for now.

 

can you get any info off of the zener diode that is in the circuit? It looks as if they have it reverse connected and uses its breakdown voltage to drop the voltage across the LED's.

 

Truth, I wouldn't even consider the listed circuit. About 90%+ of the energy from the batteries is pure waste, and that zener is going to get very, very hot.

I agree they are using the zener as a voltage drop. Problem is, when zener fails it shorts, not always, but it is a common failure mode. Pop go the LEDs.

A resistor, on the other hand, open, so all the parts are saved.

 

Im guessing they used the Zener because its a turn signal and not on for more then a second between blinks. The headlight circuit on the bike is just two resistors and no zener.

 

I calculate about 33 volts for the zener and .65 watts for the zener...if that helps at all.
33 volt, 1 watt, zeners are probably not Radio Shack stock.

 

I calculate about 33 volts for the zener and .65 watts for the zener...if that helps at all.
33 volt, 1 watt, zeners are probably not Radio Shack stock.

So the LEDs are seeing 33 volts and the 750 ohm resistor is resisting it down to 6ish v ?

How do you determine what that 1k resistor is for ?


Also if i were to use a regulator does anyone know what regulator i could use . I know Linear regulators are pretty much out unless i use a 10lb heatsink.

Btw thank you for all the help everyone .I really appropriate it.

 

Guessing about 3.5 volts for a yellow LED, the 1k resistor has 7.5 volts across it for 7.5 ma added to the 4th LED.
Guessing that the LED maximum current is 20ma, the three LED string has 20ma-7.5ma = 12.5ma.
.0125 amps times 750 ohms is 9.375 volts
9.375 + 7.5+2.5 = 19.375 volts
55-19.375 = 35.6 volts OR MORE on the Zener
if the zener is 36 volts, P= .02 x 36= .72 watts

There. That's the math. If I guessed wrong about anything you have the method to figure out the truth.
I don't know squat about making a 20 ma switching regulator.
You should probably just use (3) 12 volt, 1/2 watt, zeners in series.

 

So the LEDs are seeing 33 volts and the 750 ohm resistor is resisting it down to 6ish v ?
...

As others have said the zener is what drops the voltage, so the 3 LEDs and resistor get a much lower voltage which sets the current as it nromally would (say on a 12v battery).

...
How do you determine what that 1k resistor is for ?

The 1k resistor is typical (although i would have used about 4k7) and holds the LEDs firmly OFF until the zener is doing it's job. It is important for stopping transients and button sparking etc that can occur with a 55v button to a turn signal.

...
Also if i were to use a regulator does anyone know what regulator i could use . I know Linear regulators are pretty much out unless i use a 10lb heatsink.

Considering you have limited electronics knowlege maybe the best system would just be a nice chunky 5W resistor and the 3 LEDs in series. That will be simple, reliable and easy for you to test and repair if needed. It will have less change in brightness too as the battery voltage drops than the zener circuit. It will waste exactly the same amount of heat as the zener circuit so no loss there.

If you go with that idea, I would also suggest a cap in parallel with the 3 LEDs, a 10uF cap or even a 0.1uF greencap or similar.

 

Guessing about 3.5 volts for a yellow LED, the 1k resistor has 7.5 volts across it for 7.5 ma added to the 4th LED.
Guessing that the LED maximum current is 20ma, the three LED string has 20ma-7.5ma = 12.5ma.
.0125 amps times 750 ohms is 9.375 volts
9.375 + 7.5+2.5 = 19.375 volts
55-19.375 = 35.6 volts OR MORE on the Zener
if the zener is 36 volts, P= .02 x 36= .72 watts

There. That's the math. If I guessed wrong about anything you have the method to figure out the truth.
I don't know squat about making a 20 ma switching regulator.
You should probably just use (3) 12 volt, 1/2 watt, zeners in series.

The weird thing with that 1k resistor is im still getting the full voltage across it when testing with the DMM.It would be so much easy to find a way to run the circuit if my adjustable bench supplys went up to 55v but they only go up to 20v

 

I will try the 2-3 12v zeners in series with a 4k7 off of it and see how it goes. I tried a single .5w 12v zener yesterday and the zener was starting to burn up.