# LEDs @ AC

Discussion in 'The Projects Forum' started by JMD, Dec 21, 2009.

1. ### JMD Thread Starter Member

Dec 9, 2009
96
0
Plan: Convert existing outdoor lamps from standard bulbs to LEDs.

Found a stack toroidal transformers, so im gonna use those for my xmas-project (since people dont like 230V ). Output is 15V@15VA and 15V/2VA.

Using a full wave bridge rectifier, the DC voltage will be in the area of 20V. I plan to use a capacitor to limit the steady state current, as well as a resistor to limit the turn-on surge.

Here goes:

Im not sure wether or not the capacitor is good enough to current limit in this setup - help? The above circuit should provide ~10mA half-wave, before entering the bridge rectifier. Only downside i see, is that the 200 Ohm resistor eats about 0.3W.

Suggestions are welcome.

Last edited: Dec 21, 2009
2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You don't need a rectifier. Use a capacitor in series with the LED to limit the current througt it nad place a diode in antiparallel with the LED as not to destroy it when the AC voltage goes negative.Another option is to use a capacitor to limit the current and two LEDs connected in antiparallel. Both LEDs will light. The problem with these solutions is that if the diode or one LED in the second case fails then the other LED will fail too.

If you want to use a full bridge rectifier, you don't need to use smoothing capacitors. Just connect a resistor in series with the LED.

3. ### JMD Thread Starter Member

Dec 9, 2009
96
0
That was the idea when i wanted to use 230V, but people went kinda balistic on me. But yea, youre right, i could just skip the bridge. I just like DC when its low voltage, dont know why

There we go.. should do the trick? (Small note: the capacitor is a 4uF, not 3)

4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You don't need the resistors, only the capacitor.

5. ### JMD Thread Starter Member

Dec 9, 2009
96
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Which do you refer to? There are two resistors in the circuit.

6. ### peranders Well-Known Member

May 21, 2007
87
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It's a good idea to have fullwave rectification since the LED's will flicker, not very pleasant.

(For 230 VAC) The circuit must have transient protection. The series resistor must have good properties in transient power handling. Three or four metal film 600 mW resistors will be enough or some wire wound type. The main cap should have a high ohmish discharge resistor in order to avoid surprises.

Last edited: Dec 21, 2009
7. ### peranders Well-Known Member

May 21, 2007
87
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YES! if you want the circuit to last. (if you drive it with 230 VAC)

Last edited: Dec 21, 2009
8. ### peranders Well-Known Member

May 21, 2007
87
0
May I say that my circuit is only suitable if you have the whole design properly insulated. If you have a transformer You will need a rectifier bridge and a smoothing cap plus a suitable series resistor for the LED. Having a cap in series and driving the LED's with AC is not very practical but the only advantage is less losses.

9. ### JMD Thread Starter Member

Dec 9, 2009
96
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mik3 --> Ah, you wrote "resistorS".. well.. the 1M is to bleed the cap - which might not be needed, since i dont run 230V anymore. But the 200 ohm is to protect against the high surge-current when the switch is turned on.

As mentioned earlier, people went balistic when i talked about using 230V, threads got deleted, locked etc. So.. lets keep the talk on 15V AC, okay?

The flickering was my first argument to use the bridge, but im not sure how visible the 50 Hz is. Might notice it if you wave your hand in the light or something.

If you look at the last circuit, whats your opinion then?

10. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Actually the LEDs will flicker with AC powering them, but it will happen so fast we'll never see it. I assume you want minimum components, and you do need a small resistor. If the timing is right and you turn power on with the AC waveform at max with a fully discharged capacitor it will surge the LED, slowly damaging it little by little. A simple resistor (a small one) will help prevent or slow it down dramatically.

The article in the AAC book explains the surge problem.

Basically eliminate R1 in your schematic and it is close (I haven't taken a calculator to the values).

I find it interesting that LEDs seem to favor capacitors as an analog to ballast, while florescent bulbs use coils (that is what a ballast is). Both are for the same reasons, a reactive device does not dissipate power, so it doesn't heat up, which leaves the device doing all the work (and heating).

Last edited: Dec 21, 2009
11. ### JMD Thread Starter Member

Dec 9, 2009
96
0
I did some testing with 230V (naughty me!), and there is some noticeble flickering when you move your hand in the light. Eventho its only 50 Hz, its definitely noticeble.

Okay, here's the new schematic:

- 18,85 mA full-wave (~9.5 mA half-wave).
- Max surge-current = 75 mA
- Power eaten by resistor = 283 mW

Anything else i should be aware of? (valuen might change a bit, depends on what parts if have in my drawers)

12. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Looks reasonable to me, have you had a chance to check out the high power LEDs yet? They can be pretty impressive.

13. ### JMD Thread Starter Member

Dec 9, 2009
96
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Sounds good - gonna try it out soon.

High-power LEDs - Ive been there, and beyond My latest project had a 35W HID

14. ### JMD Thread Starter Member

Dec 9, 2009
96
0
Tested it a few minutes ago, and all i can say is: GREAT SUCCESS !

Always when i work with 230V, i use my good old vario-trafo (if the circuit allows it ofc). In this case, it wasnt needed at all - worked first shot.

I didnt have any capacitors with no polarity (no idea of the english term for those), so i went with 2x electrolyte back to back.

Working range: 20-230VAC on the primary (= 1.3-15VVAC on the secondary)
Surge-resistors arent even warm (using two ~420 Ohms in parallel)

Just a small update - test phase is done, now i need to figure out the complete circuit.

Assuming my calculation for power wasted in the resistor is correct, its R*I^2, then im gonna loose quite a bit of juice @ 350 mA

Last edited: Dec 22, 2009
15. ### JMD Thread Starter Member

Dec 9, 2009
96
0
Gonna do the math, tell me if im off somewhere:

Input voltage = 15VAC
Surge-resistor (Rs) = 10 Ohm
Capacitor = 160 uF

Z = 1 / 2*pi*50*160*10^-6 = 19,90 Ohm
I = Vin / Z = 753,98 mA

Power in resistor the first few ms: Vin * (Vin / Rs) = 22,50 Watt

Continuous power in Rs: Rs*I^2 = 5,7 Watt

Any errors? Its getting a bit late, so id like some extra opinion on it!

16. ### JMD Thread Starter Member

Dec 9, 2009
96
0
No one at all ?

17. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You have to include the resistor value in the total impedance as to calculate the current.

18. ### JMD Thread Starter Member

Dec 9, 2009
96
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Ah yes, true! But i still get a huge waste of power in the surge resistor.
The math looks like this:

Loosing 11W in that surge-resistor simply isnt gonna work. Any ideas to what i can do, to eliminate that loss?

19. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Drop the resistance of the surge resistor. The whole idea of using one is to reduce the extreme currents to the merely high current surge, to make the LEDs last longer.

Wonder if it would be practical to use a inductor instead of the resistor for the same job? Basically the initial rush is probably measure in nano to microseconds, then the reactance of the capacitor takes over. With a small coil, you could prevent that initial rush much the same way, but then (if the value were low enough) it would practically be invisible to the AC.

It would be a series resonant circuit, but so what? The frequency is fixed (50 or 60Hz), the amplitude is fixed, it would still be a capacitive reactance.

20. ### zimbarak Active Member

Feb 8, 2009
56
0
you can just connect a resistor in series with the led and make the calculation
for 220v AC line the led need accros 1.5v to glow to 30 ma
if you follow the owm's law by deducting 220v from1.5 it will give 218.5and divide by 30 ma it will give you 7.1kohm u can use 8kohm instead since 7.1k cant be found
by calculating the heat consumption p=R*i square you can determine the power of the resistor