Running Luxeons on AC power?
- Thread starter hth
- Start date
grnamin
Enlightened
I had a similar question and this is one of the answers I got.
shankus
Flashlight Enthusiast
No AC on diodes, without another diode, or rectifier circuit to protect it from the reversal of current on the opposite cycle.
Why 12VAC, instead of 12VDC?
Why 12VAC, instead of 12VDC?
The_LED_Museum
*Retired*
The Enerleds MR16 retrofit "bulb" (using a 1.2W Luxeon) can run on 12VAC or 12VDC; I'm guessing there's a full-wave bridge rectifier and a resistor in there somewhere. I'd have to take a hammer and smash open my test unit to verify this though; but I do know it works regardless of the polarity, which is why I'm pretty sure there's a full-wave bridge in there.
You can run a Luxeon on 12VAC if you put a diode somewhere in the 12VAC line, in series with the resistor you'll need so the Luxeon doesn't let all of its magic smoke out. A better solution would be a full-wave bridge rectifier and a good sized electrolytic cap before the Luxeon's resistor so the Luxeon doesn't flicker too much.
You can run a Luxeon on 12VAC if you put a diode somewhere in the 12VAC line, in series with the resistor you'll need so the Luxeon doesn't let all of its magic smoke out. A better solution would be a full-wave bridge rectifier and a good sized electrolytic cap before the Luxeon's resistor so the Luxeon doesn't flicker too much.
Shankus, I have a 12VAC transformer that is currently used to power my garden lights. I thought of replacing the bulbs with Luxeons?
The LED Museum, is that all I need? A resistor and diode in series? Is this a 'patch' solution or will it be ok for say 8-10hrs on time at night? I have little electronics knowledge so a 'full wave bridge rectifier' is a bit out of my depth. /ubbthreads/images/graemlins/frown.gif
Thanks for your replies.
The LED Museum, is that all I need? A resistor and diode in series? Is this a 'patch' solution or will it be ok for say 8-10hrs on time at night? I have little electronics knowledge so a 'full wave bridge rectifier' is a bit out of my depth. /ubbthreads/images/graemlins/frown.gif
Thanks for your replies.
highlandsun
Enlightened
A full-wave bridge rectifier is just 4 diodes arranged in a diamond layout. The diodes and power are connected such that the incoming AC sine wave is converted to all positive - the negative half of the sine wave is flipped over the X axis. If you use a single diode, the negative half of the sine wave is zeroed out, so half of the AC power is lost. With the bridge layout, all of the energy is usable. In both cases, the power still rises and falls cyclically, and your LEDs will have a visible flicker pattern. Using a capacitor will flatten the output, eliminating the flickering.
It probably makes more sense in pictures than in words, but hopefully you get the idea, since I don't have an illustration handy.
It probably makes more sense in pictures than in words, but hopefully you get the idea, since I don't have an illustration handy.
Enerleds MR-16 available yet?
I've seen the specs, etc, but haven't found any place that sells it yet. Is it available and what's the cost?
In my office, we have three overhead tracks, each with 5 MR-16s and it gets a little hot at times. A couple have burned out and it would be cool to replace them with LED versions. Though my guess is the LED versions would cost something like $30, which is about 10x the original lights.
I've seen the specs, etc, but haven't found any place that sells it yet. Is it available and what's the cost?
In my office, we have three overhead tracks, each with 5 MR-16s and it gets a little hot at times. A couple have burned out and it would be cool to replace them with LED versions. Though my guess is the LED versions would cost something like $30, which is about 10x the original lights.
Burnt_Retinas
Enlightened
Run a LED of AC? Yes....but,
A full wave bridge (4 diodes) with a large electro cap will do the job. It can also be used for DC!. You could go half bridge (1 diode) but you'll need a bigger cap. Why the cap? This is because the LED has less persistence than phosphor on a fluoro or the heating/cooling of the glowing element of a incandescent ie you'll see flicker, particularly in your peripheral vision.
I've not tried rectified AC for a 1W, but I guess perhaps a 4700uF electro would do the job. If it flickers still, go larger until you don't notice it. Maybe less will do?
Don't forget the resistor between the rectified DC and the LED to limit the current!
Also remember that the capacitor will charge to the peak of the voltage. A 12V AC is the RMS. The peak will be more. The exact amount will depend on transformer and current being drawn. Measure the volts/current under load to tweak your resistor values.
Chris
A full wave bridge (4 diodes) with a large electro cap will do the job. It can also be used for DC!. You could go half bridge (1 diode) but you'll need a bigger cap. Why the cap? This is because the LED has less persistence than phosphor on a fluoro or the heating/cooling of the glowing element of a incandescent ie you'll see flicker, particularly in your peripheral vision.
I've not tried rectified AC for a 1W, but I guess perhaps a 4700uF electro would do the job. If it flickers still, go larger until you don't notice it. Maybe less will do?
Don't forget the resistor between the rectified DC and the LED to limit the current!
Also remember that the capacitor will charge to the peak of the voltage. A 12V AC is the RMS. The peak will be more. The exact amount will depend on transformer and current being drawn. Measure the volts/current under load to tweak your resistor values.
Chris
Burnt_Retinas
Enlightened
You could start with the full wave bridge, cap - about 1000uF to 4700uF at 16V, and a 33 ohm 5W resistor in series with the rectified DC to the LED. The resistor will get hot - very hot. Rectified 12V AC is a high voltage relative to what the LED needs hence the resistor getting hot having to drop so many volts at I what I assume to be about 350mA or so for a 1W LED (you didn't state the LED type - 1W or 5W). Less volts would be better and the resistor would need to be less, therefore less power will need to be dissipated in the resistor. If you have a choice on transformer volts, go less, perhaps 4 volts or so (for a 1W LED). A different resistor will be required though.
Good luck.
Chris
Good luck.
Chris
I would like to use 3 x 1w Luxeons in series with the 12VAC 7.5A transformer (I already have that transformer for the existing light) and overdriving the Luxeons to say 450mA (1.35A total). 1.2 Ohm resistor will do? So any 16V electrolytic between 1000uF and 4700uF will do? Changing the value of the capacitence will determine the blink/flash rate of the LEDs? Now, I just need to find out what a full wave bridge circuit is like? Will figure 3.1.2 do?
Jonathan
Enlightened
1) You need to understand the difference between series and parallel. When you connect devices in _series_, their voltage (or voltage requirement) adds up, but the _same_ current flows through each device, so the total current does not change. When you connect devices in parallel, then they _all_ see the supply voltage, but the current is divided up through various paths, so the supply currents add up but the voltage stays the same.
If you connect 3 Luxeons in series, then the total current will just be your target 450mA (not 1.35A), but the needed supply voltage would be about 12V.
2)You need to understand RMS voltage and how voltage is measured on AC supplies. The output of a 12VAC transformer is _not_ 12 Volts. Instead it is a constantly changing value which averages (for a particular type of average called root mean square) to 12V. The peak Voltage in this waveform will be nearly 17V.
3) You need to understand what happens when you rectify AC. You lose a bit of voltage in your diodes, and then the capacitor charging time, the capacitance, and the load, and the peak voltage all act together to determine the output voltage of the rectify and capacitor, and the output voltage ripple. In your case I would expect an output voltage of 15-17V, a bit below the 17V peak.
4) You need to understand Ohms law. E=I*R, voltage = current times resistance. With 12V of LEDs, and say 16VDC supply, you need to drop 4V in your series resistor. You want 450mA, so 4V = 0.45A * R
5) It is unclear which diagram you mean, but yes, there are full wave bridge circuits on the page that you reference.
-Jon
If you connect 3 Luxeons in series, then the total current will just be your target 450mA (not 1.35A), but the needed supply voltage would be about 12V.
2)You need to understand RMS voltage and how voltage is measured on AC supplies. The output of a 12VAC transformer is _not_ 12 Volts. Instead it is a constantly changing value which averages (for a particular type of average called root mean square) to 12V. The peak Voltage in this waveform will be nearly 17V.
3) You need to understand what happens when you rectify AC. You lose a bit of voltage in your diodes, and then the capacitor charging time, the capacitance, and the load, and the peak voltage all act together to determine the output voltage of the rectify and capacitor, and the output voltage ripple. In your case I would expect an output voltage of 15-17V, a bit below the 17V peak.
4) You need to understand Ohms law. E=I*R, voltage = current times resistance. With 12V of LEDs, and say 16VDC supply, you need to drop 4V in your series resistor. You want 450mA, so 4V = 0.45A * R
5) It is unclear which diagram you mean, but yes, there are full wave bridge circuits on the page that you reference.
-Jon
I think that using a three LED string depending on a series resistor to control current in such a poorly controlled system (from a voltage stability standpoint) as the yard lights I've dealt with is a mistake.
At least to start, I'd suggest a much simpler route. Go to 'Radio Shack' and buy a 'one amp bridge rectifier' for each LED. These come in lots of different voltages, anything above 50 is fine for us. RS 276-1152 is $1.39, Jameco has several under half a buck, thirty cents or less each for ten:
Bridges at Jameco
Anyway, you'll need a resistor as well. Assuming 12.6 Volts RMS (a nominal value, but you really should measure it *at the light, under load*) and 3.6 Vf for the LED (again, measuring might be in order, although variance here is sure to be slight relative to the transformer voltage (and wiring losses)). So we have 12.6 Volts less 3.6 (for the LED) less 1.4 (for the diode bridge) or 7.6 Volts across our resistor. We want to (overdrive) at .45 Amps so we need 7.6/.45 Ohms, 16.888... Ohms, twenty is probably close enough (and safer in that it has lower current). Power needs to be 7.6 Volts times .45 Amps or about 3.5 Watts, making 5 Watt parts marginal but probably OK.
You might consider two LED in series, but the lower drop across the 'ballast resistor' will make you even more subject to wide changes in LED current with relatively small changes in Voltage. Consider that with 'only' 7.6 Volts across our resistor, a change of a single Volt in delivered voltage (for whatever reason) is a 13% change in LED current (meaning over 500 mA). Adding another LED (with 3.6 Volt drop) increases overall efficiency (although we probably have 'energy to burn') but reduces the resistor drop to 4 Volts so now a one Volt change is 25% increase (562 mA).
I wouldn't worry about capacitors and flicker. Florescent lights also flicker at 120 Hz, does this bother you? Large value electrolytic capacitors are probably the least reliable component WRT to heat I know of, and yard lights get hot. And they're expensive.
The circuit is rock simple. Bridge AC inputs to the AC from the transformer, Bridge plus to one side of the resistor, other side of the resistor to the positive side of the LED, negative of the LED to the bridge negative.
Doug Owen
At least to start, I'd suggest a much simpler route. Go to 'Radio Shack' and buy a 'one amp bridge rectifier' for each LED. These come in lots of different voltages, anything above 50 is fine for us. RS 276-1152 is $1.39, Jameco has several under half a buck, thirty cents or less each for ten:
Bridges at Jameco
Anyway, you'll need a resistor as well. Assuming 12.6 Volts RMS (a nominal value, but you really should measure it *at the light, under load*) and 3.6 Vf for the LED (again, measuring might be in order, although variance here is sure to be slight relative to the transformer voltage (and wiring losses)). So we have 12.6 Volts less 3.6 (for the LED) less 1.4 (for the diode bridge) or 7.6 Volts across our resistor. We want to (overdrive) at .45 Amps so we need 7.6/.45 Ohms, 16.888... Ohms, twenty is probably close enough (and safer in that it has lower current). Power needs to be 7.6 Volts times .45 Amps or about 3.5 Watts, making 5 Watt parts marginal but probably OK.
You might consider two LED in series, but the lower drop across the 'ballast resistor' will make you even more subject to wide changes in LED current with relatively small changes in Voltage. Consider that with 'only' 7.6 Volts across our resistor, a change of a single Volt in delivered voltage (for whatever reason) is a 13% change in LED current (meaning over 500 mA). Adding another LED (with 3.6 Volt drop) increases overall efficiency (although we probably have 'energy to burn') but reduces the resistor drop to 4 Volts so now a one Volt change is 25% increase (562 mA).
I wouldn't worry about capacitors and flicker. Florescent lights also flicker at 120 Hz, does this bother you? Large value electrolytic capacitors are probably the least reliable component WRT to heat I know of, and yard lights get hot. And they're expensive.
The circuit is rock simple. Bridge AC inputs to the AC from the transformer, Bridge plus to one side of the resistor, other side of the resistor to the positive side of the LED, negative of the LED to the bridge negative.
Doug Owen
Johnathan, yes, I do need to understand current/voltage across parallel/series circuits. I also need to understand RMS vs peak voltage. I also need to understand AC rectification. I do understand Ohms Law but my misunderstanding of point 1 led me to the miscalculation of the resistance value. Thanks for your explanations
Doug Owen, many thanks for your response. Can I clarify that each Luxeon will need its own (50V+) doide bridge with its own (22ohm for 350mA drain, I understand Ohms Law /ubbthreads/images/graemlins/wink.gif ) resistor and that each Luxeon/bridge "component" can be run in parallel with the transformer output? As you know, garden lights are always wired up in parallel so that in case one bulb blows, the other lights are still lit.
Also, if I add say a second/third/forth Luxeon/bridge "component" to my garden light circuit, I would effectively have half/third/quartered the current flowing through each "component"... would I need to change the 22 ohm resistance in each "component" to reflect this or is that 22 ohm resistance for the desired current output to the Luxeon. Given that total current is lower now, surely I would need to change/lower that resistance? /ubbthreads/images/graemlins/confused.gif
You suggested 2 Luxeons in series, I presume for each "component"? You (if I read correctly) advised against it yourself with your increase efficiency statement with a caveat of a potential high increase in current due to voltage fluctuations. If I use 1 Luxeon in each "component", then your statement can be ignored (but noted)?
Your link to the part "RS 276-1152" brought back no results... session variable issue. Could you cut and paste the bridge information so i can search for it myself?
Eg.
"DIODE,BRIDGE,DF01M(DB102)
Jameco #103000
Mfg Ref # DF01M"
Ok regarding the capacitance. Thanks for the 'rock simple' circuit. /ubbthreads/images/graemlins/smile.gif
Doug Owen, many thanks for your response. Can I clarify that each Luxeon will need its own (50V+) doide bridge with its own (22ohm for 350mA drain, I understand Ohms Law /ubbthreads/images/graemlins/wink.gif ) resistor and that each Luxeon/bridge "component" can be run in parallel with the transformer output? As you know, garden lights are always wired up in parallel so that in case one bulb blows, the other lights are still lit.
Also, if I add say a second/third/forth Luxeon/bridge "component" to my garden light circuit, I would effectively have half/third/quartered the current flowing through each "component"... would I need to change the 22 ohm resistance in each "component" to reflect this or is that 22 ohm resistance for the desired current output to the Luxeon. Given that total current is lower now, surely I would need to change/lower that resistance? /ubbthreads/images/graemlins/confused.gif
You suggested 2 Luxeons in series, I presume for each "component"? You (if I read correctly) advised against it yourself with your increase efficiency statement with a caveat of a potential high increase in current due to voltage fluctuations. If I use 1 Luxeon in each "component", then your statement can be ignored (but noted)?
Your link to the part "RS 276-1152" brought back no results... session variable issue. Could you cut and paste the bridge information so i can search for it myself?
Eg.
"DIODE,BRIDGE,DF01M(DB102)
Jameco #103000
Mfg Ref # DF01M"
Ok regarding the capacitance. Thanks for the 'rock simple' circuit. /ubbthreads/images/graemlins/smile.gif
Ray_of_Light
Flashlight Enthusiast
Take into consideration that the diodes bridge will decrease the peak voltage output of about 1.4 Volt, and will dissipate heat accordingly.
The calculation of electrolitic cap is done by using the thumb's rule of 2200 uF per Amp of current (minimum). Higher is the value of the cap, longer it will last.
Single diode rectifier are not suggested for high currents since they produce a magnetization of the transformer core, with higher hysteresis and thus overheating. A transformer is usually rated for 20.000 hrs, while with a single diode rectifier, at full rated power, its average life decreases to 1000 hrs.
Transformers dies due to the lost of insulating property of the insulating lacquer on the copper wiring.
Anthony
The calculation of electrolitic cap is done by using the thumb's rule of 2200 uF per Amp of current (minimum). Higher is the value of the cap, longer it will last.
Single diode rectifier are not suggested for high currents since they produce a magnetization of the transformer core, with higher hysteresis and thus overheating. A transformer is usually rated for 20.000 hrs, while with a single diode rectifier, at full rated power, its average life decreases to 1000 hrs.
Transformers dies due to the lost of insulating property of the insulating lacquer on the copper wiring.
Anthony
[ QUOTE ]
hth said:
Doug Owen, many thanks for your response. Can I clarify that each Luxeon will need its own (50V+) doide bridge with its own (22ohm for 350mA drain, I understand Ohms Law /ubbthreads/images/graemlins/wink.gif ) resistor and that each Luxeon/bridge "component" can be run in parallel with the transformer output? As you know, garden lights are always wired up in parallel so that in case one bulb blows, the other lights are still lit.
Also, if I add say a second/third/forth Luxeon/bridge "component" to my garden light circuit, I would effectively have half/third/quartered the current flowing through each "component"... would I need to change the 22 ohm resistance in each "component" to reflect this or is that 22 ohm resistance for the desired current output to the Luxeon. Given that total current is lower now, surely I would need to change/lower that resistance? /ubbthreads/images/graemlins/confused.gif
You suggested 2 Luxeons in series, I presume for each "component"? You (if I read correctly) advised against it yourself with your increase efficiency statement with a caveat of a potential high increase in current due to voltage fluctuations. If I use 1 Luxeon in each "component", then your statement can be ignored (but noted)?
[/ QUOTE ]
That's right. I'm suggesting treating the bridge rectifier, luxeon and resistor as a single component. It goes in parallel in the system, just like the standard yard lights.
Adding a second Luxeon to 'the unit' puts it in series (at least as I was thinking) meaning the voltage across the resistor is lower (by the voltage across the second LED), so we'll need a lower value. We also enjoy a doubling of light output for the same total energy used. In this case, as in all series circuits, the current is the *same* in each component, voltage gets distributed (in parallel, voltage is the same for each component, current gets distributed). You can put a second LED in parallel with the first (and lower the resistor) but I wouldn't . I'd use a second resistor for the second LED in this case (series is another matter, of course). You do need a higher current bridge rectifier (although 1.5 Amp units are very common, more than enough). Vfs don't match all that well, I'm more comfortable with separate current limit resistors lest one LED get greedy (or the other gets lazy) and the magic smoke comes out. The remaining one then is double driven, and follows suit. A second resistor is cheap insurance (or go with two in series).
I used 'bridge rectifier' to search with, let me know if you have trouble.
Not to sweat too many of the fine points here. You can safely do the math with RMS values as they represent true heat delivered over time, which is what we're worried about, right?
Doug Owen
hth said:
Doug Owen, many thanks for your response. Can I clarify that each Luxeon will need its own (50V+) doide bridge with its own (22ohm for 350mA drain, I understand Ohms Law /ubbthreads/images/graemlins/wink.gif ) resistor and that each Luxeon/bridge "component" can be run in parallel with the transformer output? As you know, garden lights are always wired up in parallel so that in case one bulb blows, the other lights are still lit.
Also, if I add say a second/third/forth Luxeon/bridge "component" to my garden light circuit, I would effectively have half/third/quartered the current flowing through each "component"... would I need to change the 22 ohm resistance in each "component" to reflect this or is that 22 ohm resistance for the desired current output to the Luxeon. Given that total current is lower now, surely I would need to change/lower that resistance? /ubbthreads/images/graemlins/confused.gif
You suggested 2 Luxeons in series, I presume for each "component"? You (if I read correctly) advised against it yourself with your increase efficiency statement with a caveat of a potential high increase in current due to voltage fluctuations. If I use 1 Luxeon in each "component", then your statement can be ignored (but noted)?
[/ QUOTE ]
That's right. I'm suggesting treating the bridge rectifier, luxeon and resistor as a single component. It goes in parallel in the system, just like the standard yard lights.
Adding a second Luxeon to 'the unit' puts it in series (at least as I was thinking) meaning the voltage across the resistor is lower (by the voltage across the second LED), so we'll need a lower value. We also enjoy a doubling of light output for the same total energy used. In this case, as in all series circuits, the current is the *same* in each component, voltage gets distributed (in parallel, voltage is the same for each component, current gets distributed). You can put a second LED in parallel with the first (and lower the resistor) but I wouldn't . I'd use a second resistor for the second LED in this case (series is another matter, of course). You do need a higher current bridge rectifier (although 1.5 Amp units are very common, more than enough). Vfs don't match all that well, I'm more comfortable with separate current limit resistors lest one LED get greedy (or the other gets lazy) and the magic smoke comes out. The remaining one then is double driven, and follows suit. A second resistor is cheap insurance (or go with two in series).
I used 'bridge rectifier' to search with, let me know if you have trouble.
Not to sweat too many of the fine points here. You can safely do the math with RMS values as they represent true heat delivered over time, which is what we're worried about, right?
Doug Owen
Hello Doug, here is a list of bridge rectifiers available from my local electronics store. Giving that my garden transformer is 12VAC 7.5A and say I wish to run 5 sets of 1w Luxeons, which rectifier would I need?
W04 1.2A 400V Bridge Rectifier
DIL 1A 100V Bridge Rectifier
PO4 6A 400V Bridge Rectifier
BR106 10A 600V Bridge Rectifier
400V 35A Bridge Rectifier
2A 800V Inline Bridge Rectifier
6A 400V Inline Bridge Rectifier
Thanks
W04 1.2A 400V Bridge Rectifier
DIL 1A 100V Bridge Rectifier
PO4 6A 400V Bridge Rectifier
BR106 10A 600V Bridge Rectifier
400V 35A Bridge Rectifier
2A 800V Inline Bridge Rectifier
6A 400V Inline Bridge Rectifier
Thanks
