Pulse width driver for Luxeon

Candle Power Forums

Help Support Candle Power:

[ QUOTE ]
6pOriginal said:
Has anyone tried this? It supposed to give you full bightness while using far less power and generates less heat. I am thinking about modding the brake light on my car with that.

[/ QUOTE ]
The product description at this link is full of bogus claims. Except in the case of extreme underdrive [<10% of rated] PWM is less efficient than continuous DC.
 
No it wouldn't.

LEDs are more efficient at lower drive currents. PWM pulsing an LED runs it at full current for a low duty cycle to achieve dimming. Thus, even though no power is "wasted" in a resistor, the LED is operating less efficiently during it's on time in a PWM dimmed setup.


Let's say you want to dim a Luxeon III to 100mA. At 100mA, a luxeon's Vf is around 3.0V. Let's also say you're using a power source of 4.5V (3 alkalines). To keep things simple, let's assume there is an input cap on the PWM setup, so voltage drop due to the battery's internal resistance is negligable.

Now, to drive the L3 at 100mA with a resistor requires a resistance value of 15 ohms. Power dissipated by the Luxeon is 300mW, power dissipated by the resistor is 150mW, for an efficiency of 67%

To drive the L3 at 100mA with a PWM requires a 1/7 duty cycle. The current for the luxeon would be 3.7V @ 700mA (1/7 duty cycle). You still also need a resistor (4.5V power supply). The resistor would be 1.1 ohms. During each pulse, power dissipated by the resistor would be 0.56W, power during each pulse for the luxeon, 2.6W/pulse, efficiency for each pulse would be 82%. Average power usage is 0.08W and 0.371W respectively.

Okay, first, look at those numbers. constant current vs. pwm. Just averageing to the same average current level, the resistored setup is dissipating 300mW, while the PWMed is dissipating 371mW. For the same average current usage (assuming right now, the same light output, but I'll address that in a second), the resistored luxeon is using about 20% less power.

However, you must look at the output vs. current curve to know exactly if you're really putting out the same amount of light in the PWM setup. The human eye does a good job of integrating high speed pulses into an average brightness level.

At 700mA, an L3 (white) will produce, on average, 65 lumens (per the datasheet). Heating is neglegable due to the low on time of each pulse and low average power dissipation. Averaging to 1/7 gives us about 9.3 lumens of light.

Now, running at 100mA, you get about 20% of the light as at 700mA. Heating at this power level is negligable (5C higher junction temperature), so output will be about 65*0.2 = 13 lumens.

So, we're getting more light for less power used at the LED. However, since all that really matters is what the power draw is from the batteries, we're getting more light (13 vs 8.6) for the same power input.

If we wanted to equal the power output, we have to increase the duty cycle to about 1/5. So now the PWM setup is dissipating, on average, 520mW in the LED and 112mW in the resistor (almost as much in the resistor as in the pure resistor setup). Average current from the batteries in the PWM setup is 140mA, 40% more than the simple resistor setup. So, to get the same amount of light from the PWM setup, requires 40% more power from the power supply.

[edit: updated per Doug's post to remove heat derating of the PWM setup]
 
Darin, very nice explaination. I have one very small quibble. It is the *average* power that determines the die temperature [assuming sufficiently short pulse widths] so the flux derating for die temperature is roughly the same for both the PWM and DC cases, subject only to the small difference in average power. BTW, I once [actually several times] measured the thermal time constant of the die itself and came up with values in the 5-20 msec range.
 
Ah - good point. For some reason, 1/7th duty cycle in my mind equated to 142mS on in one pulse. Of course, the length of a single on pulse depends on the frequency of the PWM driver, and at sufficiently high frequency, the pulses are easily under 20ms.

So the light output for the PWM is about 9.3 lm. The duty cycle should still be increased to 1/5 (due to rounding I did in the original post) to equal the output of the resistored solution.

I'll edit the post to reflect this.
 
As an Amazon Associate we earn from qualifying purchases. Product prices and availability are accurate as of the date/time indicated and are subject to change.
Please correct me if I'm wrong, but doesn't the data sheet for this device claim that it will run five 1 watt LEDs while consuming 500ma? If so I'm fairly sure that you will find that it is pulsing each of the 5 at 1/5 duty cycle. Somewhere in there it's listed as a "flasher".

It's easy to imagine pulsing a LUX at a frquency above 30 Hz and getting a fair output.

Daniel
 
So what is the most efficient dimming method for LEDs? If I want to be able to run the LEDs at either full power or be able to dim them, what method should I use? So PWM is ultimately not as efficient as reducing current with resistors. Is there a method of adjusting the current delivered to the LEDs but without the inherent inefficiencies of resistors/potentiometers? Basically, what is the most efficient LED dimming circuit?
Thanks, Eli
 
The best way would be with a buck converter with adjustable current output. One of georges80's drivers (uFlex, nFlex for buck; fatman for boost) would do the trick. The uFlex and nFlex have multiple brightness levels (I think through button presses), and the fatman allows an external potentiometer to be connected to the driver board to adjust the current.
 
gadget,

Based on the 1W datasheet, estimate each 500mA pulse is 1.22 times the output at 350mA, and 100mA is about 0.35 times the output at 350mA.

So 1.22 / 5 = 0.244 the light of a 1W at rated current (PWM), where the 100mA constant current gives 0.35 times.

The PWM luxeons will actually run hotter, since the Vf is substantially higher at 500mA pulses (~3V @ 100mA, vs ~3.5V @ 500mA), so the average power dissipation will be higher.

Basically, you'd be better off running all the LEDs at 100mA with seperate resistors.
 
If you're going to use an even number of red luxeons, you could use one of georges80's driver boards, configured for 700mA, and have two strings in parallel of 2-3 luxeons in series for each string. Break lights switch on and off a lot, so I'd ask georges80 about frequent switching of the boards on/off
 
[ QUOTE ]
6pOriginal said:
Thanks for everyone's input! What driver would you recommend if I want to drive 4-5 red 1 watt luxeon with 12v input? (again, going to be used on the brake light on my car)

[/ QUOTE ]

4 or 5 Luxeons series connected will not run from 12V with a buck (step down converter).

To simplify things, I'd recommend 4 luxeons - hooked up as 2 strings (2 series connected Leds per string) in parallel. So, you would have 350mA through one string and 350mA through the other from a nominal 700mA output driver. The 2 LED's in series would be running at around 6V (for Red Luxeons), so easy to get from a 12V buck converter.

4 Luxeons at 350mA will get quite hot - mount them on a decent heatsink. I assume you'll use Stars so that the Luxeons' Slugs are isolated from each other via the PCB on the Star.

george.
 
[ QUOTE ]
6pOriginal said:
Thanks for everyone's input! What driver would you recommend if I want to drive 4-5 red 1 watt luxeon with 12v input? (again, going to be used on the brake light on my car)

[/ QUOTE ]

If you can get away with 4 reds, just string them in series and add a resistor. IIRC the reds have about 2.5V, so with 4 in series you'd be at 10V, gives you ~80% efficiency _and_ high reliability (if you don't have a switcher, that switcher can't fail), which IMO is not to be underestimated in a brake light.

Bye
Markus
 
I would never use resistor to limit current in a car's electrical system. Voltage can vary from 13V (car off) to 15-16V (running), and there is lots of spikes, transients, and noise. You definitely need a regulator of some kind (lm317 is the simplest to set up).
 
It is easy to underestimate the extremely large electrical transients and temperature range that a "normally used" auto part is exposed to. In addition to costs, these extremes are one of the reasons that auto electronics are often set up to run on simple voltage control devices.

Georges80 - above, has done some products he has used in his truck, but he also lives in CA with me, so I am not sure if his design covers the total automotive temp range. He is pretty thorough, so if you pm him, he can provide more details.

An alternative is to look at a site like International Rectifier (and others like TI) and search through their auto applications. Perhaps putting an automotive grade LDO (basically a fancy self-varying resistor that outputs a constant voltage from a variable input) would be a place to start. As mentioned above, if you put 4 Red Luxeon's in series, you are at approx 10 Vf, so very close to the natural voltage found in cars. A small resistor value in series with this might do the trick. A diode in line is always useful. This may not be as efficient, but perhaps not that bad considering the more complex althernative.

You are likely to have more challenges with optics than electronics for this application, especially if you really want to be "in compliance".

Others please chime in if my logic is off.
 
[ QUOTE ]
markus_i said:
[ QUOTE ]
6pOriginal said:
Thanks for everyone's input! What driver would you recommend if I want to drive 4-5 red 1 watt luxeon with 12v input? (again, going to be used on the brake light on my car)

[/ QUOTE ]

If you can get away with 4 reds, just string them in series and add a resistor. IIRC the reds have about 2.5V, so with 4 in series you'd be at 10V, gives you ~80% efficiency _and_ high reliability (if you don't have a switcher, that switcher can't fail), which IMO is not to be underestimated in a brake light.

Bye
Markus

[/ QUOTE ]

I'm not so sure about this Markus. I've measured 13.8V car systems where I've seen healthy 80V spikes on the line, especially during load dump, and various other events. Of course this will vary from car to car.

Hint, 80V on a LED that was resistored for 13.8V gets a +2A hit.

Now the LED die bond wire turns into a nice expensive fuse.

There are other ways of addressing this if you have an adversion to LDO and Switching Regulators.


[ QUOTE ]
evan9162 said:
No it wouldn't.

LEDs are more efficient at lower drive currents. PWM pulsing an LED runs it at full current for a low duty cycle to achieve dimming. Thus, even though no power is "wasted" in a resistor, the LED is operating less efficiently during it's on time in a PWM dimmed setup.


Let's say you want to dim a Luxeon III to 100mA. At 100mA, a luxeon's Vf is around 3.0V. Let's also say you're using a power source of 4.5V (3 alkalines). To keep things simple, let's assume there is an input cap on the PWM setup, so voltage drop due to the battery's internal resistance is negligable.

Now, to drive the L3 at 100mA with a resistor requires a resistance value of 15 ohms. Power dissipated by the Luxeon is 300mW, power dissipated by the resistor is 150mW, for an efficiency of 67%

To drive the L3 at 100mA with a PWM requires a 1/7 duty cycle. The current for the luxeon would be 3.7V @ 700mA (1/7 duty cycle). You still also need a resistor (4.5V power supply). The resistor would be 1.1 ohms. During each pulse, power dissipated by the resistor would be 0.56W, power during each pulse for the luxeon, 2.6W/pulse, efficiency for each pulse would be 82%. Average power usage is 0.08W and 0.371W respectively.

Okay, first, look at those numbers. constant current vs. pwm. Just averageing to the same average current level, the resistored setup is dissipating 300mW, while the PWMed is dissipating 371mW. For the same average current usage (assuming right now, the same light output, but I'll address that in a second), the resistored luxeon is using about 20% less power.

However, you must look at the output vs. current curve to know exactly if you're really putting out the same amount of light in the PWM setup. The human eye does a good job of integrating high speed pulses into an average brightness level.

At 700mA, an L3 (white) will produce, on average, 65 lumens (per the datasheet). Heating is neglegable due to the low on time of each pulse and low average power dissipation. Averaging to 1/7 gives us about 9.3 lumens of light.

Now, running at 100mA, you get about 20% of the light as at 700mA. Heating at this power level is negligable (5C higher junction temperature), so output will be about 65*0.2 = 13 lumens.

So, we're getting more light for less power used at the LED. However, since all that really matters is what the power draw is from the batteries, we're getting more light (13 vs 8.6) for the same power input.

If we wanted to equal the power output, we have to increase the duty cycle to about 1/5. So now the PWM setup is dissipating, on average, 520mW in the LED and 112mW in the resistor (almost as much in the resistor as in the pure resistor setup). Average current from the batteries in the PWM setup is 140mA, 40% more than the simple resistor setup. So, to get the same amount of light from the PWM setup, requires 40% more power from the power supply.

[edit: updated per Doug's post to remove heat derating of the PWM setup]

[/ QUOTE ]

Take a look in this document, figure 3 & 4, note the 40% light output at 10% duty cycle, and also note the decreasing efficiency as you mentioned earlier at higher pulse currents:
http://www.lumileds.com/pdfs/LED_pocket_illuminator.PDF


Though, pulsing can alter the white point, and thus be used to tune the emitter to a certain "color" of white.
 
Back
Top