between led 3 and 4 and another between 1 and 6?You need a resistor in series with each parallel leg. Otherwise one leg can hog all the current. Depending on your voltage supply,
.075 ohm 5 wattOh, and you'll need the right value and power rating of R3, to adjust for double current.
I mean one between Vdd and L3, and another between Vdd and L4. Or those resistors could be placed between L1 and Q2, and L6 and Q2.between led 3 and 4 and another between 1 and 6?
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.What would you recommend for q1
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?What current are you planning to run through your LEDs?
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.
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.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?
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 heatsinkOK, 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^2R 5.6^20.1=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^2R 2.8^20.2=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.
so very sorry about that here is the specs straight from website i will obviously be buying from elsewhereIt 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?
there is no symbol in scheme it for an NFET. its the closest i could findAs previously noted, the LEDs in the diagram are backwards, and Q2 is not a LM350.
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.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.
my understanding is that VDD is input source voltage mine being a 14.4 V from my alternator in my carHave you told us what Vdd is? That Q2 MOSFET might have to dissipate a lot of heat also.
I don't disagree, as long as you know what you are doing with heat and such....more than 90% is perfectly fine
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.If it were me I would use a circuit for each string.