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Topic: 10W RGB LED constant current drivers (Read 8169 times) previous topic - next topic

Riva

I bought some 10W high power RGB LED's and now need to build 3x suitable drivers cheaply. Doing a web search gave me this interesting article http://www.tbideas.com/blog/build-an-arduino-shield-to-drive-high-power-rgb-led/ but the LED's are different ratings and the MOSFET and transistor will also be different as I want to get all components from one supplier (http://cpc.farnell.com/) if possible to reduce P&P charges.
Schematic...


LED's are...
VF = 6 - 6.6 (Red)
VF = 9.6 - 10.2 (Green/Blue)
IF = 350mA

In the circuit I wish to substitute the 2N5088 with BC635 and the BUZ71 with IRFZ24NPbF. I hope to drive it all with this http://cpc.farnell.com/jsp/search/productdetail.jsp?sku=PW02701 PSU running through 3x LM350 to step voltage down to something like 7V for Red & 10V for Green and Blue LED's. Using these values I calculate R2 in the circuit to be 1.43 ohms. Can someone please check my assumptions and calculations or offer alternate circuit to drive the RGB LED please.

terryking228

Hi, an off-the-shelf solution would be a Power FET driver board with optical isolation like this: http://goo.gl/lXCXP

Some how-to here: http://arduino-info.wikispaces.com/Brick-4ChannelPowerFetSwitch


DISCLAIMER: Mentioned stuff from my own shop...
Regards, Terry King terry@yourduino.com  - Check great prices, devices and Arduino-related boards at http://YourDuino.com
HOW-TO: http://ArduinoInfo.Info

Grumpy_Mike

Quote
PSU running through 3x LM350 to step voltage down to something like 7V for Red & 10V for Green and Blue LED's.

No need to use a voltage regulator let alone three of them. That is the job of the constant current supply. However, the heat will be dissipated in the circuit's FET so you need to look at the power dissipation. This depends on your input voltage.

Docedison

Simple 10W/3V = 3.334 A. so the Mosfet Should be at least 10A Idmax and have a heat-sink capable of 40W dissipation because you should figure that there 'might' not be the heat transfer to ambient that you expect, and there are 3 of them... The Circuit will not work as shown however as you are missing a 10K pullup to the gate of the mosfetas shown. The NPN keeps a constant .63... Volt base bias by controlling the gate to source voltage of the mosfet and you can putt a 500 to 1K ohm resistor in the emitter of the NPN and adjust the current limit too.

Doc
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MarkT

Nice little circuit, like current limiting in a series regulator.  Wondering if stability could be an issue with large FETs (high gate capacitance)?
[ I won't respond to messages, use the forum please ]

Riva


Quote
PSU running through 3x LM350 to step voltage down to something like 7V for Red & 10V for Green and Blue LED's.

No need to use a voltage regulator let alone three of them. That is the job of the constant current supply. However, the heat will be dissipated in the circuit's FET so you need to look at the power dissipation. This depends on your input voltage.

I was hoping to use a 12V supply so R dissipation would be about 2W, G & B would be about 0.5W each?

Docedison

The Mosfet current limiter is likely to be less expensive than trying to use 3 X LM350 as current limiters both circuits can be controlled with a PWM control easily. There is another consideration that should be looked at and that is the sense resistor. For a Bi-polar sensor like the 2N5088 the resistor needs to develop .6 volts drop across it to begin current limiting and the LM350 requires 1.25V which is going to double the required power dissipation of the sense resistors. On the surface the LM350 is an attractive device as it becomes a 2 part limiter but with the discrete's (BUZ71 & 2N5088) there is one additional part and a sense resistor of 1/2 the value and thus 1/2 the required dissipation. Both can be PWM controlled by bringing the sense feedback point to ground. The sense feedback is either the gate of the mosfet or the Adj terminal of the voltage regulator.. For sense resistors I would use metal clad resistors that can be mounted to the heat-sink for good thermal management.

Doc
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard

Riva

#7
Sep 22, 2012, 10:53 am Last Edit: Sep 23, 2012, 11:15 am by Riva Reason: 1
I have decided on the component values I want to use and have revised the drawing to add R1 as suggested in one post but is it really needed? Won't it just default to turn the LED on if no PWM signal present?

Edited to replace image with cropped version.

Docedison

I would my self change the pull-up and the control resistors by a factor of 100. 1K and 10K for base and collector resistors. 10K and 1M would work But there exists the issue that with a 1.2 uA Max collector current for the 2N5088 the circuit might have a limited and possibly non linear control range and the extra 88 uA control current would hardly be missed by the 12V power supply.

Doc
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard

BenF


I have decided on the component values I want to use and have revised the drawing to add R1 as suggested in one post but is it really needed? Won't it just default to turn the LED on if no PWM signal present?

A pull-up to 12V will exceed max voltage for the PWM pin and you don't want that (eliminate R1 from your circuit).

Generally we want FET's to switch hard and fast and this dictates a low gate resistor (say 220 when driven from a AtMega PWM pin at 5V). Fot this constant current device (Q1 will never saturate), it's not cirtical and 10k will work equally well. The only limitation is that combined on/off switch time is at least an order of magnitude or so less than PWM cycle time (switch time will be around 40us with 10k and IRF520). R2 at 2.2 will limit current (through LED, Q1 and R2) to about 300mA . A 1W resistor for R2 will then be suffcient.

Docedison

Great, you have an expert advising you... it's a shame he didn't read the data sheets first,

Doc
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard

Grumpy_Mike

Quote
Generally we want FET's to switch hard and fast

Yes we do but not in this circuit. The FET is running in linear mode and is dropping the excess voltage.

Riva

Quote
A pull-up to 12V will exceed max voltage for the PWM pin and you don't want that (eliminate R1 from your circuit).

Good point never thought of that. I prefer the LED's to be powered off by default and only turn on with HIGH PWM. Question is will it work okay without R1.

I have reduced the 10K to 1K as suggested and removed R1. Will it work as I want?


BenF


Question is will it work okay without R1.

Yes it will. You can also keep the gate resistor at 10k if you like as it really doesn't matter for this circuit. Note however that your FET willl drop close to 3W and so will neat a heat-sink.

Riva

Finally got the parts and the time at work to build the drivers.
Calculations did not work out quite right or slight component variations from datasheet mean the calculations figures were wrong.
I was aiming for about 330-350mA but final circuit is more like 240mA. I may change R2 if it's a bit dim when fitted in the light frame. B & G drivers Mosfet does not get to hot without a heatsink but R definatly needs one.
http://www.youtube.com/watch?v=bI8p4E3tFjs

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