Luxeon 3 Watt

Forum 2005-2010 (read only) Hardware Interfacing
#1

I'm trying to connect a 3 Watt luxeon to my Arduino, but the whole project is driven off of a 9 volt supply. I'm using a 7805 atmega168 and related hardware, but the LED supply will be 9V.

I've tried an NPN transistor with 9V to LED to resistor to collector, emitter to ground. But with the Arduino IO limited to 5V and the LED only dropping 3.6V, I can't get the transistor into full saturation and I can only pump about 300 mA through the LED.

Anyone have any suggestions? Until yesterday, I didn't know that the emitter-ground volatage was limited by the base-emitter voltage. I supposed it would work with a couple of 1N4004's in series with the LED to drop the voltage, but that just seems clunky to me.

Thanks.

#2

Use a Darlington?

Which transistor are you using?

#3

I'm using a TIP31C.

A Darlington isn't the answer because it's not an issue with gain. hfe for the TIP31C is pretty good and with the Arduino able to pump out 40mA, I can get away with an hfe of 20-30.

The problem is with the arduino output voltage so far below my Vcc, I can't put the transistor into saturation.

For what it's worth, I will need a response time a lot better than a microsecond, so relays aren't an option.

#4

Take a look at my instructable: http://www.instructables.com/id/Super-simple-high-power-LED-driver/. There is a comment, where I've posted a schematic.

#5

I didn't know that the emitter-ground volatage was limited by the base-emitter voltage.

Um, it shouldn't be! Or rather, the base-emitter voltage should not be an issue in the emitter-grounded NPN configuration you describe, as long as it's larger than about 0.7V. Where are you getting the idea that you can't drive the 9V LED from a 5V IO pin?

#6

The problem is with the arduino output voltage so far below my Vcc, I can't put the transistor into saturation.

That is no an issue, as long as the base voltage is above 0.7V you should be able to drive the transistor into saturation given enough gain in the transistor. What value of base resistor are you using? That could limit the current drive.

nd with the Arduino able to pump out 40mA,

Not strictly true, that is the absolute limit of source current to prevent damage to the Arduino, what limits are you putting on it (again base resistor). If this is nothing then it could be your problem.

#7

I'm using a 220 Ohm base resistor. I calculate this gives me a base current of about 19mA (assuming 5V from the arduino and dropping 0.7V across the base-emitter junction). With an hfe around 50, that should be good for an amp.

I'm only getting about 8 volts acoss the LED and its 20 ohm series resistor (3.6Vf for the LED makes the current 220mA). If the transistor were in saturation, I should read Vcc across the LED and resistor.

#8

Looking at the data sheet for that transistor the maximum hfe is 50 and the minimum is 25 (for 1A Ic). This means that any individual transistor can have a gain of between 25 and 50. It looks like you are getting a gain of about 11.

You should always design assuming the transistor you have has the minimum gain.

I suspect the difference between the apparent gain of 11 and the minimum of 25 is that the collector current is only being calculated and not measured and you are not getting 3.6V drop across the diode but something more. Could you measure that as well?

#9

Thanks for your help so far.

I'll have to wait until tonight to measure the drop across the LED. The collector current is "calculated" using ohms law based on the voltage drop measured across the series resistor. I think at's about as close to a measured value as it gets. But as you say, Ic could be low if the actual drop across the LED is higher than the datasheet for the LED says.

I see the 25 and 50 figures on the datasheet, but looking at the graphs on a later page, it looks like DC hfe should be about 120. The discrepency was confusing to me. If I want to get up to 750mA, with an hfe of 25, I'd need a minimum base current of 30mA and I may as well take the Arduino to it's 40mA limit, so I should aim for a base resistor around 110 ohms.

#10

Um, it shouldn't be! Or rather, the base-emitter voltage should not be an issue in the emitter-grounded NPN configuration you describe, as long as it's larger than about 0.7V. Where are you getting the idea that you can't drive the 9V LED from a 5V IO pin?

Someone I asked in person last night told me that. But at the time, I had the collector connected directly to Vcc and the LED/resistor between the emitter and ground. At that time I was getting about 60mA through the LED.

#11

Take a look at my instructable: http://www.instructables.com/id/Super-simple-high-power-LED-driver/. There is a comment, where I've posted a schematic.

It's not a bad idea to use a 317 as a constant current driver. I'm focused on the transistor section of what in the schematic in the comments and for that purpose, a simple power resistor will do to limit the current instead of a proper constant current regulator.

#12

Did you take into account the (rather large) Vcesat of the TIP31 when calculating the resistor between 9V and the transistor collector?

at the time, I had the collector connected directly to Vcc and the LED/resistor between the emitter and ground.

That would be an entirely different configuration; it doesn’t saturate because as the current goes up so does Ve, and Vbe In the configuration you describe in your original posting, Ve is 0 because the emitter is tied directly to GND, and the problem shouldn’t arise.

#13

Did you take into account the (rather large) Vcesat of the TIP31 when calculating the resistor between 9V and the transistor collector?

I didn't.

That would be an entirely different configuration; it doesn't saturate because as the current goes up so does Ve, and Vbe In the configuration you describe in your original posting, Ve is 0 because the emitter is tied directly to GND, and the problem shouldn't arise.

It was his suggestion to change the configuration to what I have now, but his explanation didn't really make sense to me. Yours does, so thank you for that.

I measured around 4.5 - 5 volts across the LED using my scope, I'm not sure why it's so much higher than the spec, but it looks like Grumpy_Mike was right. I replaced the base resistor with 20 ohms, I know it's way too high a current for the MCU, but I've relied on its internal protection in driving LEDs before so it's fine for testing. Now I'm getting about 600mA through the LED which is in the ballpark I'd expect based on the responses here.

So...when I power this off alkaline batteries, just how bad will will the brownouts to the MCU be? :)

To get it higher, I guess I'll need a different transistor with a higher hfe or a Darlington array after all. Thanks for your help everyone.

#14

Use a N-channel MOSFET instead of a bipolar transistor or darlington array. Then current from the digital pin becomes a non-issue. You want a logic-level MOSFET with a low on resistance (Rds).

Although overkill from a current handling perspective, the IPP14N03 would work well. It has a typical drain-source on resistance of only 13.9 milli-ohms with a gate-source voltage of 4.5V.

#15