Custom LED light and driver

richard.cs

Joined Mar 3, 2012
162
With this design you should thermally bond Q1 and Q2 (easiest way for you is probably to pick a suitable package for Q1 and mount it on the same heatsink). That way as the MOSFET heats up the forward voltage of the Q1 base-emitter junction drops slightly and your current falls. This is enough to ensure that the regulator negative temperature coefficient and can't go into thermal runaway. Without this the dissipation in Q2 rises with increasing temperature.
 

richard.cs

Joined Mar 3, 2012
162
What would you recommend for q1
Pretty much *any* cheap general-purpose silicon NPN with a max collector voltage greater than your supply voltage and a package that you can easily mount onto your heat sink. That latter requirement might push you into slightly higher power transistors than you would normally consider for this.
 

Thread Starter

goody1928

Joined Apr 19, 2014
27
Pretty much *any* cheap general-purpose silicon NPN with a max collector voltage greater than your supply voltage and a package that you can easily mount onto your heat sink. That latter requirement might push you into slightly higher power transistors than you would normally consider for this.

are you talking about something like this?

 

wayneh

Joined Sep 9, 2010
17,498
2.8 A is what I was looking for they are 10w LEDS but are most efficient at 2.8a. Also how big of a resistor were you recommending? Just something tiny to regulate like 50k 1/4w?
OK, so each string carries 2.8A, and the current-sense R3 carries 5.6A. You need to drop 0.6V across R3, so it needs to be V=IR 0.6=5.6R R=0.1Ω. The power it will dissipate is I^2•R 5.6^2•0.1=3.1W. So it needs to be rated to a bare minimum of 5W. I'd use a 10W or more rated resistor.

The current-limiting resistors for each parallel string need to drop about 0.5V in my opinion (I'd like to hear other ideas), to make sure the current balances between strings. Again using Ohm's law: V=IR 0.5=2.8R R=0.18Ω. I'd probably lean towards a larger standard value, maybe 0.2 or 0.22Ω. The power it will dissipate is I^2•R 2.8^2•0.2=1.6W. So it needs to be rated to a bare minimum of 2W. I'd use a 5W rated resistor or more.

That transistor is overkill but would work fine. You could choose one in a similar package (leads, screw-hole) that has a lower current rating, if you like. 100mA would be plenty, for instance.
 

ronv

Joined Nov 12, 2008
3,770
It just dawned on me we have never seen the specs on your LEDs. I'm not sure I have seen a single LED that can run at 10 watts. Most are groups of chip LEDs and run at like 10 volts. Do you have a link?
 

Thread Starter

goody1928

Joined Apr 19, 2014
27
OK, so each string carries 2.8A, and the current-sense R3 carries 5.6A. You need to drop 0.6V across R3, so it needs to be V=IR 0.6=5.6R R=0.1Ω. The power it will dissipate is I^2•R 5.6^2•0.1=3.1W. So it needs to be rated to a bare minimum of 5W. I'd use a 10W or more rated resistor.

The current-limiting resistors for each parallel string need to drop about 0.5V in my opinion (I'd like to hear other ideas), to make sure the current balances between strings. Again using Ohm's law: V=IR 0.5=2.8R R=0.18Ω. I'd probably lean towards a larger standard value, maybe 0.2 or 0.22Ω. The power it will dissipate is I^2•R 2.8^2•0.2=1.6W. So it needs to be rated to a bare minimum of 2W. I'd use a 5W rated resistor or more.

That transistor is overkill but would work fine. You could choose one in a similar package (leads, screw-hole) that has a lower current rating, if you like. 100mA would be plenty, for instance.
With the Size of that transistor i was just worried about thermal proporties there is this one that is only .23 cents i like that more lol but wiht no mounting hole i will just thermally adhere it to the heatsink


so ive got a Resistor now .12 ohm 7W for R3

also ive found some resistors for limitting they are .22 ohm 5W ill make up a new schematic now
 

Thread Starter

goody1928

Joined Apr 19, 2014
27
It just dawned on me we have never seen the specs on your LEDs. I'm not sure I have seen a single LED that can run at 10 watts. Most are groups of chip LEDs and run at like 10 volts. Do you have a link?
so very sorry about that here is the specs straight from website i will obviously be buying from elsewhere

http://www.cree.com/LED-Components-and-Modules/Products/XLamp/Discrete-Directional/XLamp-XML

i want the XML-T6 witch is one of the cool white variants. only reason i want that one is most lumens per $ the U3 is another couple dollars each for only like 50 more lumens
 

wayneh

Joined Sep 9, 2010
17,498
As previously noted, the LEDs in the diagram are backwards, and Q2 is not a LM350.

You want to use those LEDs at greater than 90% of their specified maximum current? I think you'll need a more precise circuit for current control, or to aim for a more conservative level, like 80%. Make sure you read their application notes about PCB mounting, heat sink requirements and so on.

Have you told us what Vdd is? That Q2 MOSFET might have to dissipate a lot of heat also.
 
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Thread Starter

goody1928

Joined Apr 19, 2014
27
As previously noted, the LEDs in the diagram are backwards, and Q2 is not a LM350.
there is no symbol in scheme it for an NFET. its the closest i could find

You want to use those LEDs at greater than 90% of their specified maximum current? I think you'll need a more precise circuit for current control, or to aim for a more conservative level, like 80%. Make sure you read their application notes about PCB mounting, heat sink requirements and so on.
more than 90% is perfectly fine they will still live for well over 10000 hours and i will never use them to that life. also as long as correctly heatsinked they will be fine. i have a T6 flashlight i built with a pre made 2.8A driver that can run for over an hour without a hiccup just gets around 100*F so not too bad.

i would like the LEDS to run anywhere between 2.4A - 3A and im happy they will put out a stupid amount of light and that will be fine with me so if its not at 2.8A exactly its fine these are blinding. reason i want to build this circuit is to heatsink all materials properly and its fun.

Have you told us what Vdd is? That Q2 MOSFET might have to dissipate a lot of heat also.
my understanding is that VDD is input source voltage mine being a 14.4 V from my alternator in my car

here is the new schematic with LEDS flipped for you lol
 

Bernard

Joined Aug 7, 2008
5,784
Getting closer. Now change Q2 from LM350T to something like N ch FET, IRF540, 28A, 100V, TO220 pkg. Rfd is not critical as it will be operating in linear range. There are hold down clamps for some high W LED's, maybe for Cree?
I like .22Ω for R4, R5.
Also, I would measure Vf of each LED to match strings Vf as close as possible.
 

wayneh

Joined Sep 9, 2010
17,498
...more than 90% is perfectly fine
I don't disagree, as long as you know what you are doing with heat and such.

The concern is that when you aim for 90%, some unforeseen variable may result in 101% at times. I don't have personal experience with the current control scheme you're using and how precise it is in usage, so it would make me nervous. It looks good, but doo-doo happens.

If the supply is 15V (sort of a worst case), the combined voltage drop for the LEDs, balancing resistors, and the sense resistor is about 10.5V, leaving 4.5V at 5.6A, or ~25W for the MOSFET to dissipate. That sounds like a lot. Better make sure that will work. You may want to add a bit to the value of the balancing resistors so that they are taking more of the heat.
 

ronv

Joined Nov 12, 2008
3,770
I'm guessing you bought some mounted to heat sink boards like this?
http://www.ebay.com/itm/New-Cree-XL...984?pt=LH_DefaultDomain_0&hash=item4ad08c6600

They have a very wide Vf of .5 volts. You would probably not get high ones and low ones all on one side (I would but probably not you):eek: so there could be a large variation in current between the 2 strings. If it were me I would use a circuit for each string.

I think people are right to start to worry about the size of the heat sink. Do you have a spec or dimensions for it?
 

wayneh

Joined Sep 9, 2010
17,498
If it were me I would use a circuit for each string.
I'm leaning the same way. The extra components are not so expensive, and they would help spread out heat dissipation to more surface area. The OP could eliminate the balancing resistors I suggested earlier, as long as the MOSFETs can take all the heat.
 
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ronv

Joined Nov 12, 2008
3,770
Hmm. I just did an eyeball on the heat sink and make it to be about 400 sq. in of surface area. I think that will make it about 1C rise per watt. So with the circuit attached the LED are about 27 watts each string and 5 watts in each FET. This puts the heat sink at about 90C. add another 25 for case to chip on the LED and your at 115C without anything for the 2 interfaces between the LED and the heat sink. Probably ok, but keep the kids away from them.;)
If you can keep the case for the little transistor in contact with the heat sink it would be good as the hotter it gets the lower it will make the LED current.
 

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