Noobie Questions

[musatsu]

I have a few questions that I haven't found a suitable answer for. Can led heat dissipation be compared to halogen heat dissipation? For example, if one has a 6W halogen would the same amount of heatsinking be adequate for a 6W led? Is there a linear relationship to heat generation vs. amperage?
If you have an emitter rated for 80 lumen per watt at 350ma, and it is driven at 350ma would it need to dissipate 1 watt of heat? And if this same emitter was driven at 700ma would it need to dissipate 2 watts?
How much extra heat dissipation would be necessary for a driver circuit? What type of driver circuit is most efficient for a medium size flashlight that would retain the ability to run below the rated voltage (insurance if caught out with low batteries)?
Do typical constant current (buck?) drivers output the appropriate voltage to the emitter or does the input voltage to the driver need to equal the input voltage to the emitter?
While I really like tinkering and it would be fun to just buy a bunch of stuff and try different things, my hobby money is limited. So, I'm trying to learn as much as I can in an attempt to keep things from going
Also, what are some favorite sites for ordering emitters and drivers?

 

[Lumencraft (Matt)]

Welcome to CPF musatsu.

I have a few questions that I haven't found a suitable answer for. Can led heat dissipation be compared to halogen heat dissipation? For example, if one has a 6W halogen would the same amount of heatsinking be adequate for a 6W led?

This is just my opinion but no. The heatsinking for an led is more important than the heatsinking to an incan bulb. Improper heatsinking can quickly destroy and led, where with a halogen bulb the risk is only in melting other nearby parts or burning your hand. A 6 watt halogen is going to generate a very minimal amount of heat, but a 6 watt led driven at max will quickly produce enough heat to destroy itself, however with an led the heat is not radiated out in the same was as the the halogen so the risk to other parts of the light (ie reflector, switch, lens) is nearly non existent.

Is there a linear relationship to heat generation vs. amperage?

In short yes, more amp = more heat.

How much extra heat dissipation would be necessary for a driver circuit?

Depends on the driver. Several of the popular drivers designed for use with 3 watt and other leds are able to run without heatsinking, where some are more fragile (or sometimes just cheap).

What type of driver circuit is most efficient for a medium size flashlight that would retain the ability to run below the rated voltage (insurance if caught out with low batteries)?

I think what you would want is a driver with whats called "moon mode" this means that the light will still operate below rated voltage, but at a diminished output. A driver with a 7135 type chipset is one example I can think of. After the voltage drops below a certain point ( I think for this driver 3.8V) the light falls out of regulation, but will remain on, but getting dimmer until the batteries are depleted. There are many examples of driver that will do this that may suit your needs better though.

Do typical constant current (buck?) drivers output the appropriate voltage to the emitter or does the input voltage to the driver need to equal the input voltage to the emitter?

A "buck" converter commonly will need to have at least 1.25 ish volts input higher than what the emitter requires. An example would be, if your led required 3.5 volts input, and your battery pack was 7.4 volts the driver would "buck" the extra voltage. Drivers that output the exact same voltage that is input to them are referred to as "linear"

Also, what are some favorite sites for ordering emitters and drivers?

For the cheap drivers DX and KD are a couple of options. For some better stuff Shiningbeam The Sandwitch Shoppe and Taskled are some others.

Hope this helps, and Have fun :wave:

 

[Conte]

Wow, those are some pretty technical questions coming from a "noobie".

Let's call you and "Advanced Noobie".

:welcome:

 

[musatsu]

vesture,

Thanks for the quick reply. So, judging by your responce, led transmit heat backward into the heatsink where as typical incandecent transmit forward to the reflector and lens. Correct?

Are leds rated the same way as incandecent bulbs? A 6w bulb uses 6w of power at 6v pulling 1 amp, therefor a 6w led rated for 3.8v would need 1500ma. Or are they rated in equivalent light output like flourescent lights are (where a 60w bulb actually only draws 13w)?

Following W=V*A and that leds are only about 20% efficient, I would assume excess wattage would be dissipated as heat. Does this relationship hold true or do variations in voltage and amperage lead to nonideal operating perameraters for the led? I guess this is more of what I should have asked instead of asking about linear relationships.

With the buck driver, does one set the output voltage via a resistor or do you just purchase one with the proper output?


Conte,

I was thinking noobie in a different way. I'm new to flashlight and light tinkering and modification. Usually, I'd just go to the store and buy a light in my price range, typically the two pack of rayovac 2D lights. The only led light I have is a watchcell powered direct drive keychain light I got at Wally World for a buck. I used to do testing and repair on computers, am a certified Audi mechanic, and am in school for mechanical engineering. I got started on leds because I recieved a free dive light with a bad bulb and a broken bezel. The next lower dive light in my series has an led replacement but mine doesn't, and I'd really like to have an led light. Also, I'd eventually like to make led tail lights for both of my vehicles ('01 CVPI and '80 Spitfire) for looks and safety. I do appreciate the warm welcome and hopefully will get enough knowledge and parts together to get some beamshots. :twothumbs

 

[AnAppleSnail]

vesture,

Thanks for the quick reply. So, judging by your responce, led transmit heat backward into the heatsink where as typical incandecent transmit forward to the reflector and lens. Correct?

Are leds rated the same way as incandecent bulbs? A 6w bulb uses 6w of power at 6v pulling 1 amp, therefor a 6w led rated for 3.8v would need 1500ma. Or are they rated in equivalent light output like flourescent lights are (where a 60w bulb actually only draws 13w)?

Following W=V*A and that leds are only about 20% efficient, I would assume excess wattage would be dissipated as heat. Does this relationship hold true or do variations in voltage and amperage lead to nonideal operating perameraters for the led? I guess this is more of what I should have asked instead of asking about linear relationships.

With the buck driver, does one set the output voltage via a resistor or do you just purchase one with the proper output?


Conte,

I was thinking noobie in a different way. I'm new to flashlight and light tinkering and modification. Usually, I'd just go to the store and buy a light in my price range, typically the two pack of rayovac 2D lights. The only led light I have is a watchcell powered direct drive keychain light I got at Wally World for a buck. I used to do testing and repair on computers, am a certified Audi mechanic, and am in school for mechanical engineering. I got started on leds because I recieved a free dive light with a bad bulb and a broken bezel. The next lower dive light in my series has an led replacement but mine doesn't, and I'd really like to have an led light. Also, I'd eventually like to make led tail lights for both of my vehicles ('01 CVPI and '80 Spitfire) for looks and safety. I do appreciate the warm welcome and hopefully will get enough knowledge and parts together to get some beamshots. :twothumbs

LEDs essentially generate their heat inside the chip, right against the heatsink. And if that chip gets above (something like) 160* C, it will quickly cause trouble with the LED. You've seen how quickly a 10-watt soldering iron will get hot - now imagine that same heat inside a 3-millimeter square. LEDs need more heatsinking than incandescent lights, at least because they break at much lower temperatures. My headlights will first experience thermal failure when the plastic reflector melts or discolors (at around 200*C, an inch or more away from the hot filament).

LEDs are rated by their power. The forward voltage of most white LEDs is 3.2v, and the ones we use these days are generally rated at 1 amp, so 3-watt LEDs are common, while a 5-watt one takes a little over 1 amp. Technically speaking, LEDs (and all diodes) have a 'forward voltage' for a given current. At a lower current, they take lower voltage, and at higher current, more voltage is required.

LEDs become less efficient at higher drive currents. You can look at a datasheet to find specifics - there's a minimum lighting current (10 mA or so), and then efficiency peaks and then falls. The rated current is often below the peak efficiency, and exceeding the rated current quickly decreases efficiency. Note that this will also heat up the LED - and hot LEDs are also dimmer than cold LEDs. Ideally, lights cut back the current supplied when the LED gets hot, preserving the LED and battery life, but reducing output slightly. Just like with other lights, the next lumen is less noticeable and more expensive.

As far as I know, constant-current drivers magically determine the correct voltage to supply metered current at.

Details on the current/heat/efficiency of getting more light from one LED: Take this with a grain of salt in case I'm horribly wrong.

If you put more current through an LED to get more power, you also drive it at a higher voltage - take the new Cree XP-G. At 350mA, it has a given luminous flux (132 lm/Watt, I believe). Doubling the current to 700 mA gives about 175% of the output at 350 mA, and at 1 amp you'll have 250% of the 350 mA output. But you've not only increased the current by a factor of 2.85, you've increased the voltage. This Voltage to Amperage is an exponential increase - small voltage increases will dramatically increase the current, while increasing current will affect the voltage less and less.

At 350 mA, the XP-G takes 3 volts to light. At 750 mA, it's more like 3.2v, and at 1000 mA, you're around 3.3v. This means that you have the following wattages and outputs, approximately:

.35A * 3v = 1.05W, 132 lm, 125 lm/Watt
.75A * 3.2v = 2.4W, 231 lm, 96 lm/Watt
1A * 3.3v = 3.3W, 330 lm, 100 lm/Watt

Now we talk about heat. Almost linearly, increasing heat decreases the luminous output of the LED. Running them at higher power can decrease the output if there isn't enough heatsinking to keep it cool. I hope this helps, and is correct.

 

[Lumencraft (Matt)]

Since AnAppleSnail addressed most of the other concerns ( better than I would have I think) I have only a couple of things to add.

vesture,

With the buck driver, does one set the output voltage via a resistor or do you just purchase one with the proper output?

Conte,

It varies from driver to driver, but from what I understand AppleSnails explanation of "magic detection" is fairly accurate. I cant remember any case where I have read of someone using a resistor in combination with a driver. I have only seen resistors used in direct drive situations to lower the voltage from a battery to the led. This driver is just plug and play, and I think it detects the VF of the led (not positive) I measured 3.05 v to the emitter off of a 7.4v battery pack, and the amp output is pre-set. This is also an example of a driver that has heat issues. I don't know if they have improved any but the early version would only run for a moment on high without heat sinking.

There are also drivers that you can set current output, I could be mistaken but I think voltage is still detected by the driver ( I don't have this driver). This driver is user configurable from 1-2.8 amp.

 

[Th232]

How drivers keep a constant current can be done in several ways, one way I know of is the use of a sense resistor, where the driver measures the voltage drop across the resistor, and hence knows the current flowing across it, and can adjust the voltage as required to increase/decrease the current.

Note that this is a feedback loop, and not just dissipating the excess power as heat.

Given the I vs Vf curve, you don't set both variables independently, to set one is to set the other. It's just that given variance in Vf and how steep the curve is, a constant current driver has a wider range of usefulness than a constant voltage driver. It's also safer when things heat up and thermal runaway becomes an issue. For a driver, the only two cases where I consider Vf are when I'm calculating efficiency, and when I'm checking to see if the total Vf of the LEDs is lower than the maximum voltage of the driver.

 

[musatsu]

I believe that this is exactly what I was looking for. So picking a driver really is just matching the desired output current and the desired input voltage. Approximately how efficient are drivers? Just a general conservative approximation.

Also, how does one go about predicting the heat dissipation needs of an led? From what it sounds like this will have a mildly exponential curve due to the increased inefficiency at higher driving voltages. This information would be particularly helpful for designing retrofit modules.

And to add a more traditional noobish question. What does the star after Flashaholic mean?

 

[gcbryan]

I believe that this is exactly what I was looking for. So picking a driver really is just matching the desired output current and the desired input voltage. Approximately how efficient are drivers? Just a general conservative approximation.

Also, how does one go about predicting the heat dissipation needs of an led? From what it sounds like this will have a mildly exponential curve due to the increased inefficiency at higher driving voltages. This information would be particularly helpful for designing retrofit modules.

And to add a more traditional noobish question. What does the star after Flashaholic mean?

I'm no expert but a smaller light (body) would require better heatsinking than a larger light.

Regarding the efficiency of the emitter look at data sheets for that emitter.

Just as an example for the SST-50 Vf is around 3.2 V at 1amp producing around 500 lumens. You can get 1200 or so lumens from this emitter but it then requires a vf of 3.6 V and draws 5 amps. Obviously, in the latter case a lot of heat is generated and needs dissipation.

For a dive light once you get thermal contact between the heatsink and the body of the light then the whole body becomes a heat sink and the water takes care of the dissipation.

I'm rather new to this myself and will stand corrected if I've misspoken or misinterpreted your question.

 

[AnAppleSnail]

Also, how does one go about predicting the heat dissipation needs of an led?

And to add a more traditional noobish question. What does the star after Flashaholic mean?

From PDF of Cree XP-G Datasheet is the information I saw. Also, it turns out that you should expect around 370 lumens at 1 amp, so adjust the other figures as necessary. To predict heat dissipation needs:

Decide on a ballpark drive current based on the luminous Flux [pg 5]. At 1000 mA, expect 350-370 lumens. It's suggested to start with a light level, because we're building a light for a need. Note that there really isn't a visible difference between 300 lumens and 350 lumens... except that one uses 15% less power, and makes 15% less heat. So we have our drive current! Let's say we go for 1000 mA because I like easy math.

Read across the [pg 4] Forward Voltage vs Forward Current graph to get an approximate voltage. I can't tell you how to -build- a light to dissipate (3.3v * 1000 mA) = 3.3 watts, but that shows you how much heat to expect for a chosen light level (lots, in my example).

One thing to keep in mind is that the luminous output decreases with temperature. Keeping the LED's innards at 100*C instead of 150*C gives you about 10% more light. That also means that if your design will make the LED sit at 100*, and you bump the power up 10%, bringing the temperature up 30*, you haven't really gained much. Now I'll stop so the people who make these things can help you!

 

[Th232]

I believe that this is exactly what I was looking for. So picking a driver really is just matching the desired output current and the desired input voltage. Approximately how efficient are drivers? Just a general conservative approximation.

You've pretty much got the driver selection in that, although you also need to check the maximum voltage the driver can put out, e.g. if you want to drive a string of 4 LEDs in series with a Vf of ~3.3 V at whatever drive level you've chosen, a driver that can put out only 9 V probably won't work to any degree of satisfaction, if at all.

Without knowing more about what your setup is going to be, I think a driver efficiency at around 80-85% would be a fairly conservative estimate. Driver efficiencies vary based on several factors. In general:

* The closer Vin is to Vout, the more efficient the driver
* In order of least to most efficient driver types, I think it goes: buck-boost, boost, buck. Note that this one is definitely general, and YMMV.

Generally you can find data on efficiency somewhere on the maker's website (or from the datasheets if you wish to make your own driver using a dedicated IC).

And to add a more traditional noobish question. What does the star after Flashaholic mean?

It's just an indicator of how many posts you have, I think it goes:

Unenlightened
Enlightened
Flashaholic
Flashaholic*
*Flashaholic*

Can't remember the post counts required though. Think it's 50 for Enlightened, 100 for Flashaholic and 500 for Flashaholic*?

 

[Lumencraft (Matt)]

musatsu,

You may already know this since you have some computer repair exp. but I will mention it anyway. One of the most important parts of heatisnking a good LED is making sure the layer of thermal compound between the LED and heatisink is very THIN. I use Arctic Silver, or Arctic Alumina depending on the project, but in either case if the layer is too thick its not going to make any difference if you have a 3" thick block of aluminum for a HS, the heat will not transfer, and in about 2 seconds you will get . It cost me 40$ to learn this, but I figured I would mention it for free.

I had not noticed that * next to the Flashaholic before, nor did I know what it meant. I am glad you asked about it, now we both know .

 

[musatsu]

Thanks guys! I'm going to do some searching and try to put together a total curcuit. Would it be more proper to start a new thread with prospective designs or continue in this one?

 

[Th232]

I think either way would be fine, I'd just keep it in this thread though, since it's a continuation of your past questions.

 

[musatsu]

What do you guys think about the 500-600 bin SST-50 running about 1.2-1.4A with 5 or 6 C cells? I say 5 or 6 because I could add extra heatsinking or reflector depth if necessary.

The host is a SL6 (6 C in series arranged in a 3 X 2 fashion). It utilizes a slider type switch, a rim contact, and a fairly thick plastic body. The original lamp was an 8W halogen intergral in a solid aluminum reflector with a spring on the bottom. The lense bezel is rubber coated plastic and in the original design press fit the combined bulb reflector assembly into the housing. I believe that the assembly will be most stable and easier to build with 5 cells. The Vi will be lower so a driver will be easier to find, and there will be more room to dissipate heat. The intent is to use it as an underwater light, so as long as heat can be transfered through the plastic the light should remain cool.

I'm also looking for suggestions for a reflector. I'd like something that is going to throw well with little spill. I'm not too sure what dimensions would be best (the housing would accept up to a 35 mm reflector).

I have absolutely no idea how much current the stock housing could handle, and it's not replaceable. Via math, I can deduce that it is designed to withstand at least 900ma. Everything is connected with a 2mm X .25mm copper strap.

I chose the SST-50 because of the possiblity of 400 lumen output, and chose 1.2-1.4A to get as near as possible to this number with the lumen ranging in the bin. This would yeild approximately double the lumen output of the original bulb. I'm not quite sure how to calculate runtime with a driver, but I'm hoping this setup will deliver atleast 1.5 hours.

Opinions/Comments/Suggestions?

 

[musatsu]

Hoping that a subtle bump will arouse interest from some of the previous posters (or new ones ).

 

[Th232]

Sorry about the late reply, I figured someone with SST-50 experience would've chimed in by now. Can't help you too much since I haven't used SST-50s (and never needed to look at optics for them).

That said, if you're wanting throw, you may be better off with a Cree XR-E, the higher surface brightness will translate into more throw.

For good throw with little spill, you might want to consider the use of an aspheric lens instead of a reflector, if the casing has enough space. You'll sacrifice OTF lumens and beam quality, but an XR-E + aspheric is one of the best combinations for an LED thrower if range is all you want. If you want greater efficiency than an aspheric, a Polymer Optics #207 reflector/collimator should work quite nicely.

 

[Lumencraft (Matt)]

Hoping that a subtle bump will arouse interest from some of the previous posters (or new ones ).

Sorry about the late reply, I figured someone with SST-50 experience would've chimed in by now. Can't help you too much since I haven't used SST-50s (and never needed to look at optics for them).

+1

I saw your post before, but I dont have any SST lights either. I have been following a thread about one. Its not exactly what you are talking about, but maybe it will give you some ideas. Here, and here is another. The second probly has more of the type of info you are looking for.