Too much dissipated power?

Thread Starter

stoopkid

Joined Mar 3, 2011
146
I'm going to be powering 33 LEDs in series of 3 from a 12v battery that may range from 11-14v. The LEDs are 2.6v, 70mA. That's 7.8v at 770mA.

Here is the LED datasheet. HPWT-DH00.... Page 5:
www.philipslumileds.com/uploads/3/DS05-pdf

Here is the Linear Regulator I am looking at, .5v drop out, 1A:
http://www.newark.com/national-semiconductor/lm2940ct-9-0-nopb/voltage-regulator-ic/dp/41K4578

That's 2.31 watts of dissipated energy on the regulator and about 1 watt over the 11 resistors.

Is that too much dissipated power? I don't really have any idea what to base it off of.

I'd like to use 4 in each series and only have about a tenth of a watt of dissipated energy but I don't see how to get a constant 12v from a 11-14v range. I can't find any switching regulators that preform such magic.

This is not for a vehicle.

Thanks
 

Thread Starter

stoopkid

Joined Mar 3, 2011
146
I just often times see people talk about dissipated power for LEDs and simply say, "that's too much..."

I suppose I just want to keep temperature within acceptable limit without the use of particularly large heat sinks. I have this heat sink:
http://www.newark.com/jsp/search/productdetail.jsp?SKU=18M8193

That's about as large of a sink as I could probably fit. I guess I just want it to not get hot enough to melt any plastic or other components or enough to make it work particularly inefficiently.
 

#12

Joined Nov 30, 2010
18,224
It seems to me that each LED dissipates 182 milliwatts. ANY led can do that, even tiny little surface mount LEDS.

The whole pile (everything) can make 11 watts of heat. Think about the size of a 10 watt resistor. If you're going to use up more space than that will give you an idea of how hot things will get.

I'm really quite good with thermodynamics but you aren't giving me much to work with.
 

Thread Starter

stoopkid

Joined Mar 3, 2011
146
Sorry, I just really don't know enough about this.

Basically I just want to know if this is a wise setup, or if I should try to make something more efficient. Would you use a 9v regulator from a 11-14v source for 770mA?
 

#12

Joined Nov 30, 2010
18,224
No. That's too little voltage leftover to allow for variance in the LEDs from batch to batch. The resistance will be too small to give dependable results.

If you want to get more efficient, you should use a switchmode constant current circuit. Look at romanblack.com
http://www.romanblack.com/
 

Thread Starter

stoopkid

Joined Mar 3, 2011
146
So you are saying that the 9v to 7.8v drop needs to be greater? Would it be better to have a higher regulated voltage or to use fewer LEDs in each series?
 

WBahn

Joined Mar 31, 2012
30,088
For your power dissipation calculations, be sure that you use the worst case circumstances. If this is a 9V regulator, then the worst case is going to be when the supply voltage is 14V (yielding 5V across the regulator) which, at 770mA, is going to be nearly 4W.

If you are using the TO-220 cased part, then the thermal resistance from the junction to ambient air is 60C/W, meaning that for each watt of power dissipation, the junction temperature will be 60 degrees C above ambiate. At 4W, that would be a 240C temperature rise. The maximum junction temperature is 125C, so your part would have to be operating in ambient air that is about -115C, which I'm guessing is not the case. In ambient are at 25C, then you are limited to a tempurature rise of only 100C (and you want to avoid operating that high if you can) which means a power dissipation of 1.67W. Best to stay below 1W.

Now, this is for packages that are not heat sinked. In order to use a passive heat sink with 4W of dissipated power, you need to have a junction-to-ambient thermal resistance of less than 100C/4W=25C/W. You have 4C/W just getting from the junction to the TO-220 case, which means you need the sum of the case-heatsink and the heatsink-ambient to be less than 21C/W and, again, you don't want to push this, so you would want to look for something in the 10C/W range.

Shifting gears: Just using what you gave in your first post (i.e., not looking at the details in the data sheets), you say that each LED has 2.6V at 70mA. Well, four of those in series would be 10.4V and then you have 0.5V minimum dropout for your regulator and so your minimum supply voltage would, nominally, be 10.9V. Thus, if you configure a regulator to deliver 10.5V you would, in theory, be good. The problem here is that your current limiting resistor for each string would have to be R=V/I=0.1V/70mA=1.4Ω. The problem here is that a change in the regulator output voltage of just 0.1V (i.e., less than 1%) would either double or halve your LED currents. Not good.

You have a lot of options, depending on what it important to you. You could use switching regulator to output something comfortably in excess of 2.6V*33=85.8V. So say you chose 100V. Then you could run all of your LEDs in a single string with 15.4V across a 220Ω resistor (which would be dissipating just north of 1W). I'm not recommending this, for a number of reasons, but just using it to illustrate the kind of options available to you.

I think your approach is on a reasonable track. Use 11 3-LED strings. Each string has about 7.8V 70mA requiring the current limiting resistor to drop 1.2V. But note that the output voltage is only spec'ed to be within about 0.5V (over the entire temperature range), so you could see your LED current vary by as much as 40%. If that's tolerable (and keeping in mind that it will usually be less than this), the you can proceed. The current in the limiting resistors would nominally be 80mA with 1.2V (96mW), but could be as high as 1.7V with 113mA (193mW). So you would want to use 1/4 W resistors. I'm not too pleased with only using 15Ω current limiting resistors, because it doesn't take much variation in the 2.6V LED voltage to throw the numbers off (or, for that matter, for the tolerance in the resistors to make a significant contribution).

As for the power dumped in the regulator, there are a couple of ways to deal with this. One way is to move some of that power into a power resistor by placing a resistor between the supply and the input to the regulator. Unfortunately, your numbers don't let this technique really buy you much. Over the temperature range, you can have a dropout voltage as high as 1V and, with a minimum supply voltage of only 11V, that means that you can only permit 1V to drop across that resistor. To give yourself some margin, you probably don't want to use anything more than 1Ω, which would let you dump about 800mW in that resistor.

Another trick would be to use an NPN transistor with the emitter tied to the Vin of the regulator and the base connected to the Vout of the regulator through two or three silicon diodes (or, better yet, a 1.5V zener). The collector and base are also tied to the supply voltage through appropriate resistors (about 1Ω for the collector and something like 100Ω for the base). This will nail the voltage dropped across the regulator at about 1V so that it is only dissipating about 800mW, which would allow you to run a TO-220 case without heatsinking.
 

#12

Joined Nov 30, 2010
18,224
My rule of thumb is, "Waste at least 15% of the string voltage in the resistor, and 20% is better.
If 7.8 volts = 85% of Vmax
Vmax is equal to or greater than 9.18 volts.

If 7.8 V = .8 Vmax
Vmax is equal to or greater than 9.75 volts

Therefore, I'm saying use at least 9.18 volts for the resistor and LED string, but if you have 9.75 volts available, use that. You have 9.75 volts. In fact you have 10.5 volts. Use it.

There is a way to do this by calculating the variability of LED batches, increased LED voltage caused by increased current, number of LEDs per string, and change in LED voltage because of temperature, but it would require a page or two of math.

Edit: Yeah, what he said, only less wordy.
 
Last edited:

WBahn

Joined Mar 31, 2012
30,088
Don't overlook the option of going with no regulator and accepting the variance. By regulating it down to 9V we've seen that variability of something on the order of 50% is to be expected. So how much worse is it to just go direct?

With 3 LEDs/string, you have nominally 7.8V and a supply voltage of between 11V and 14V. Go for the midpoint of 12.5V, giving you a 4.7V drop with 70mA for a 68Ω resistor. At 11V, you would have 47mA and at 14V you would have 91mA, for a variance of roughly 30%. Of course, there's additional variable due to resistor variance, but that was true with the regulator, too. Your worst case power in a resistor is now 6W. This is one thing that the regulator helped out on quite a bit, but even then not as much as it might appear at first glance. The good news is that you can spread this heat out among many resistors much more easily and cheaply and you can also get away with thermally abusing resistors a lot more than you can most ICs.
 

Thread Starter

stoopkid

Joined Mar 3, 2011
146
This is getting a bit beyond me, even after a few reads.

I don't want to do anything that you guys think would not be preferable, I certainly do not have my heart set on the 9v regulator.

I guess I'm at the point where I just need suggestions on what exactly you guys might do, as much as I hate for that to be the case. My supposed understanding of the matter is out the window.
 

WBahn

Joined Mar 31, 2012
30,088
As you've already seen, there are a number of ways to proceed, each with pros and cons. What we would do would be highly dependent on why we were doing it and what was important. Right now, all we know is that, for some reason, you want to get 70mA flowing through each of 33 LEDs using a DC power source that can vary between 11V and 14V. Is this for a one-of-a-kind project and what's really important is just to get something working quickly? Is it a prototype for something that you are hoping to eventually market? How important is the 70mA? How much variation is acceptable? How important is it to have each of the strings have nearly the same current, whatever it is? How important is total cost? Is it important that some of the strings continue to work even if one or more of the other strings fail?

Very often, engineering comes down not to finding the best way to do something, but considering all of the many ways to do something and then eliminating the ones that, for whatever reasons, aren't acceptable. Then you choose from what's left (if any are left at all).

The more you can fill in our knowledge of why you are doing this and what is important, the better the suggestions we can make will be.
 

Thread Starter

stoopkid

Joined Mar 3, 2011
146
It is a fixed sign. It is just a one time project. Cost isn't a big deal but should still be reasonable, <$10 for the power supply would be good but it's flexible. Several tens of dollars might be too expensive. Exactly 70mA is not necessary but it shouldn't be variable to a point where it will be particularly noticeable, and I don't want to strain the LEDs to the point where their life will be significantly shortened. All of the sets of LEDs should be the same brightness without a great deal of noticeable variation. It would be preferable for an LED failure to not kill the whole thing. Space is limited but flexible, no computer sized PSU but anything that would fit on 2 or 3 square inches of PCB would be fine. I can not rely on mains power, so this 12v battery that is already available where my sign will be is my most reliable power source.

That's really all there is to it. It should stay plenty bright and variation between LEDs should simply remain less than awkward looking.

I'm sorry that I'm making this difficult. I really appreciate all your help so far.
 

#12

Joined Nov 30, 2010
18,224
Trying to sort this into short, choppy, sentences:

Run the regulator as high (in voltage) as you can, which is 10.5 volts, or it will have serious problems with heat. Using 10.5 volts distributes as much heat as possible across 11 resistors (@39 ohms and 1/2 watt size each) and 33 LEDs. The heat sink you posted will work.

dataman just posted a good page about a regulator
 
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