AC led circuit

Hello everyone!

I am designing a simple LED circuit to run off ~117 VAC. I would like to ask if anyone can provide me with feedback on the attached circuit design.

TIA

ACled.png

Not a very efficient circuit. I suppose you would get some light out of it however. 63 mA average. What are the specs on those LEDs?

Won’t one diode see much higher voltage than the other?
Say L1 was 120 and L2 was 0. Than the junction would be
120V*900/(100+900) = 108V.

Then the current switches, L2 = 120 & L1 = 0 , the junction would be
120V*100/(900+100) = 12V.

I think you’re better off with equal resistors, something like this.
What is Vf of the LEDs?

117 x 0.6 = 70 watts of power for the LEDs to produce a nominal 5 watts of light

Not exactly an efficient lighting system or doesn't that matter

Thanks for the feedback.

Yes more efficiency would be nice, I am currently using a single resistor but 70w is a LOT of heat to burn off. That's why I am thinking about a voltage divider config. I am pretty new to this stuff as you probably guessed. :confused:

The LEDs specs are (actually 6 wired in parallel, 3 in each polarity):

Vf: 6.1v
Current: 220mA

A simple voltage divider like that will not give you regulation when you apply the 10 ohm resistor and LED load.

The LEDs specs are (actually 6 wired in parallel, 3 in each polarity):

Ouch!
Sounds like a recipe for desisted. You need a constant current supply to drive a high power LED.

Won't one diode see much higher voltage than the other?

Unless he's changed his schematic, no.

The voltage divider is unnecessary and actually makes it less efficient, and therefore burns up more power.

The equivalent circuit is R = 900//100 + 10 = 100 ohms
V = 117 * 100/(900+100) = 11.7V

Hm. Already we only have 117mA rms, only if it is full wave and only if you short the LEDs. Not matching your numbers.

In any case, all the current through the 100 ohm resistor is wasted heat in both the 100 ohm and 900 ohm resistor. I'd use one resistor, or just use a suitable power supply.

Oh, so many things wrong!

Firstly, you want your 6 LEDs wired as two counter-parallel strings of three in series. Thus you only have to supply 220 mA average.

However, if the LEDs are rated for 220 mA average, then you are only driving them half of the time from AC, so depending on their specification, you would actually like to use a higher current, probably not 440 mA but perhaps 330 mA. Alternatively, (in fact much better,) put all six in series and use a bridge rectifier to feed them at 220 mA.

Now what you are doing is to imitate one of the now-readily available LED bulbs for mains use. I could be wrong here, but almost all of these will use a capacitor to limit the current (which is why they are “not dimmable”).

So, let;'s use the bridge rectifier. The capacitor value must have a reactance of about 110/.22 = about 500 ohms. I am not taking fully into account the voltage drop of the LEDs as it is a complex matter. Also your quoted Vf of 6.1V is rather odd.

Nevertheless to proceed, the capacitive reactance is given by:

so about 5.3 µF. 4.7 µF would be appropriate, rated at 150VAC.

Now to limit the “inrush” current to the capacitor if this arrangement is switched on at the peak of the cycle, let’s say we limit this to 1 Amp, so the voltage being 165-37 = 128V - we can use a 120 ohm resistor in series with the capacitor, and with 220 mA passing through this under normal conditions it will need to be rated at 6.2W - so you still need a 10W resistor!

Interesting exercise! This is why the 6W LED light bulbs use not 6, but more like 24 or 36 LEDs running at a fraction of the current, or a switchmode converter similar to what you would find on eBay.

Sorry for taking so long to get back to this. Putting the LEDs in series is a revelation! (It makes perfect sense in retrospect, it also explains the LEDs I have seen on the market with Vf values of 24v and 48v.)

I think I will settle for the attached circuit. I don’t need to buy any new components and I would like to fixture to be dimmable. I can just clip the wires in my current parallel config to modify it.

Am I correct in subtracting the Vf from the source voltage to calculate wattage? i.e. 117-36.6=80.4*.250=20.1.

I understand that the circuit will only be on half the time so the wattage is actually half.

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Pokey:
I think I will settle for the attached circuit.

I wouldn't do that if I were you!

I have wondered if people ever use a capacitive divider to drive a couple of LEDs from
mains? Here the power involved is too high, but for a couple of indicator LEDs back to
back surely a series resistor fed from a mains-rated capacitive divider would reduce
the circuit losses to something completely reasonable?

Not a divider, no. A divider only wastes power. A single capacitor in series, with a resistor to limit the charge current in the case where you turn the power on right on a voltage peak.

But you need either a full wave bridge, or two LEDs in antiparallel (parallel but pointing in opposite directions, or an AC LED. Otherwise the reverse voltage will destroy the LEDs.

Keep in mind that you -must- build this so it is double insulated, as the circuitry is connected directly to the incoming AC.

Thanks to everyone for the information, it is very much appreciated!

I was in a hurry with my last schematic and a bit distracted by other things. (Yes of course I cannot have all of the LEDs in series without rectification! :blush: )

So here we go again—I have attached yet another drawing which, while it is not perfect, is definitely a big improvement over my original in wattage.

Screen Shot 2015-02-04 at 11.17.13.png

I suggest making that two 640 ohm resistors to separate the two strings.

Otherwise, slight differences in the LEDs and drift with temperature and time could have one string hogging more of the current.

Pokey:
Yes of course I cannot have all of the LEDs in series without rectification!

And that is precisely the correct thing to do! It makes the arrangement four times as efficient. Just add a bridge rectifier.

Can I add a bridge rectifier without a capacitor? I really need the light to be dimmable.

TNX

Yes, you can. The 100Hz or 120Hz flicker, some people find objectionable even though it is normally below the level of perception.

Your dimmer may or may not operate properly with such a load, however. I have some dimmable CFL, but my dimmer won't work without at least one incandescent plugged into the circuit.

Pokey:
Can I add a bridge rectifier without a capacitor? I really need the light to be dimmable.

Adding a bridge rectifier is no different to having the two chains of LEDs counter-paralleled. It is just a different - more efficient - way of connecting them. And it doubles the flicker frequency of the individual LEDs - which is of course the whole point - having all the LEDs light on both halves of the AC cycle instead of just one.

polymorph:
I have some dimmable CFL, but my dimmer won't work without at least one incandescent plugged into the circuit.

Which is of course due to the dimmer requiring a minimum load to operate and the fact that the CFL driver contains a capacitor which does not discharge fully in between the half cycles (it of course uses a bridge rectifier) so that the dimmer does not "see" the load until the AC cycle comes up to the residual voltage on the capacitor. This may also affect the dimming behaviour of your LED circuit.