Microchips for LED Lighting

Now days there are plenty of companies that produce very efficient and cheap constant current microchips specifically designed for use in LED lighting. Gone are the days of needing to piece together LM350 regulators and 2N3055 transistors mounted on a brick of a heatsink.

Try looking at the websites for the following companies for numerous LED lighting control microchip options (many of which should be available from your local electronic component stockist).

National Semiconductor

ON Semiconductor

Fairchild Semiconductor

Linear Technology

Microsemi Corporation

Philips Semiconductors

Power Integrations

STMicroelectronics

Texas Instruments

Analog Devices Home

Alternatively try a google search for "led lighting driver"

Hi, Urbain,

This is a school project? What kind of school?!

Your list of requirements sounds like some kind of bizarre joke - please register, and tell us more.

Thermal protection will be achieved by keeping the average drive current (per LED) within spec.

What's all this about sine wave and blockwave frequencies?

What possible lighting application could need that range? 5 orders of magnitude covers effectively total darkness to blinding.

And "90% attack speed over 0.5us" ???

Also puzzled by your mention of "100LEDs" - does that mean "hundreds of LEDs" ? If so, how are they connected? (Parallel, series, combination or what?).

Please reply & make some sense of all this.

Hi guys,

thank you all for all your comments. But still haven't got a solution.

Let me put the question again in this form may be it could be well understood. On a board I have 100LEDs, 10 series strings and 10 parallel strings. On the same board I have the led driver, made of a constant current source. In the current source I have a comparator which the reference input is either, sine wave range 0-100Hz, square wave range of 0-10Hz, ramp signal, etc.

You all know LEDs produce lots of heat during power dissipation. With less than no time the board could burn my fingers because of too much heat and when it stays a little longer some LEDs get damage.

Now my question is why does LEDs dissipate so much heat? what can be done to reduce or eliminate the heat? how can the bandwidth of the current source be improved? is changing the current source be of any help to the heat reduction.

I look forward to hear from you guys, thanks in advance, and Andy is a semester 7 project at Technical University Eindhoven, so is serious and not a joke.

Greetings

Urbain

Now my question is why does LEDs dissipate so much heat? what can be done to reduce or eliminate the heat? how can the bandwidth of the current source be improved? is changing the current source be of any help to the heat reduction.

They don't generally produce enough heat to burn the user (or the printed circuit board they are mounted to). From what you are describing you are driving the LEDs too hard (exceeding voltage or current rating of the individual LEDs), or the LEDs are not sharing the current (some strings are drawing more current than others exceeding the rating of the LEDs). How are you balancing the current thru the different strings of LEDs (are you using resistors?). They need to be balanced, you wouldn't connect 100 batteries or capacitors up like this and expect each one to share the voltage and current without help would you (remember each device has slightly different electrical characteristics even thou it is the same device).

Since this is a school project I would suggest looking at the LED data sheet for the specific LEDs you are using, and measuring the current in each individual string (and the voltage across numerous LEDs in the strings) to see if the current is in fact balanced and the voltage across the individual LEDs is not exceeding the LEDs rated voltage. Also since you are using a waveform signal you really need to look at the waveforms with an oscilloscope to check for voltage spikes or waveform distortion that a standard voltmeter or multimeter will NOT pickup.

Let the learning begin!

Check out the application notes for National Semiconductor's Simple Switcher® voltage regulators. A common technique is to feed a string of LEDs from the output through a resistor (R(LED)) connected between the regulator feedback input and ground. The LED current is then Vref/R(LED). You probably don't want to drive all 100 LEDs in one string (160-350V, depending on LED forward voltage), so you have the regulator measure the current through one string (e. g, 5 LEDs in series), and use the regulator output to drive 20 strings (the one measured plus 19 more strings of 5 LEDs, each through a resistor equal in size to R(LED)).

You may not find a Simple Switcher that could drop into the circuit and do the job, but a switch-mode regulator with an external transistor could be adapted.

If the instructor asks about the comparator, it's inside the voltage regulator IC... Some regulator chips have built-in thermal protection or, if they are designed to drive external power transistors, they may have temperature sensor input. Your biggest challenge is to get the regulator to operate stably over such a wide dynamic range. One possibility is to vary the reference voltage for fine control, and switch in different sizes of R(LED) for order-of-magnitude control.

You've already noticed that LEDs don't like heat. The colder they are, the better they perform, so pay some attention to the thermal design of your system. At full power, you could have several Watts of heat generated; don't trap it in your enclosure.

Please register, the benefits are large and cost nothing.

More members will help if you are registered.

Hello Urbain,

I assume you are using common T 1-3/4 LEDs. These usually have a maximum current of 25 mA and are usually operated at 5-20 mA. If your LEDs are overheating, you are obviously exceeding the maximum rated current. (In my first post, I had thought the LEDs were being cooked by heat from the current source.) As I mentioned in my earlier post, you should have a resistor in series with each series string of 10 LEDs, otherwise the string with the lowest voltage drop will draw more than its share of current. This is the same principle behind the use of emitter current sharing resistors when transistors are connected in parallel in audio power amplifiers.

I would be amazed if you could get 100,000 Lux from 100 conventionally sized LEDs. A typical "super-bright" 5 mm. white LED emits 0.7 lumen at its rated current, so even if you crammed all of them into a 6 cm x 6 cm square and measured the illuminance right on top of them, the best you could get is about 19 kLux. (A 230V / 100W incandescent bulb emits 1300 lumens.) Perhaps you should verify this specification with your instructor, including the measurement conditions. See this link:

http://en.wikipedia.org/wiki/Lux

Paul

Hi Paul,

Thank you so much, I think you are getting it right. The previous circuit had 10 resistors of 1ohm each, I thought about it that way also and decided to change to 10ohms resistors, but things didn't get better. For now my current source produces 1.5A for each series string. I am not sure of the type of LED, I will check later.

I hope you come on with new ideas.

Greetings

Urbain

Hi Urbain,

1.5 A per string?! Wow! The only LEDs I know of that are rated for this much current are the CML Innovative Technologies CMDA51Cxxxxxxx and CMDA52Cxxxxxxx high power series units. These have 9 mm square metal surface mount packages made for soldering to large areas of copper foil on the circuit board or are attached to factory supplied heat sinks. 5 Watts at 1.4 A produces 60 lumens. They aren't cheap - a set of 100 white LEDs without heat sinks will cost you $1,634 US from www.mouser.com. There are other high power LED assemblies that contain two or more LEDs in series to keep the current requirement down, but that's another topic.

Most likely, you need to turn your current source way down to about 200-250 mA TOTAL output to avoid frying your LEDs. (Most of the common LEDs are designed for 20 mA.) The maximum amount of current per string would be the same as the maximum allowable current for an individual LED.

To determine the current sharing resistor size, you could estimate the probable difference of voltage drop in each string from the manufacturer's data sheet listing for forward voltage drop at the typical operating current. For example, if Vf is 3.4 to 3.6 V at 0.02 A, then you may have as much as 2.0 V difference between strings of 10 LEDs. 2.0 V / 0.02 A = 100 Ω. If you match LED strings for a lower variation, you could use smaller resistors, but you need some resistance to account for different behavior over changes in current and temperature.

Paul

Hi Paul,

Once more thanks for staying in touch.

I have decided to put everything on a piece of paper for you to see and may be come with an idea. here is the link: http://www.sendspace.com/file/13bysv

I have also given the type of LED we are using and where you can see more about it.

Will be looking forward to hear from you.

Greetings

Urbain

This is NOT a standard simple LED lighting project using inexpensive LEDs, you are using the new generation (and expensive) Luxeon K2 series LEDs. When is an LED not an LED?, when it requires heat sinking to dissipate the excess heat generated when it is driven at 1A.

These LEDs have special requirements that you MUST take into account (you need to take them into account for standard LEDs, but standard LEDs are a LOT less expensive)

1) Read the data sheet completely thru, then do it again. It is very important that you don't overdrive these LEDs - excessive heating and damage will result.

2) Check your drive current - from the data-sheet 1500mA is the "ABSOLUTE MAXIMUM" rating (ie- before damage occurs), never run an electronic device at or near its absolute maximum. 1A is indicated in the data sheet, limit the maximum current to that to start with.

3) Check your voltage drop across each LED. From the data sheet it should be around 3.72V at 1000mA drive current.

4) Unlike standard LEDs, Luxeon K2 LEDs use and dissipate large quantities of heat (it is their biggest problem over standard LEDs). Heat sinks are absolutely vital and must be used to remove the excess heat. This I think is your biggest problem, 100 x K2 LEDs are going to dissipate an enormous amount of heat in a small area very quickly. The Luexon data sheets are a little vague with regard to the need for a heatsink, you haven't got one have you?

Here is an example heatsink sold for the K2 LED (38x35x16mm)

http://www.dotlight.de/products/en/LEDs/LED-High-Power/LED-HP-Heat-Sinks/Heat-Sink-for-P4-Rebel-K2.html

You will need one of these (or something similar) for EACH LED, and it will need to be exposed to room air directly (you cannot just mount a cover over the top to hide it otherwise the heat will just buildup underneath the cover). Since you are using 100 LEDs you will either need a large fan forced air-cooled heatsink (if space is a problem) or one or two much larger natural convection fin heat sinks to get rid of the heat. If it were me, I would use a compact fan forced air-cooled heatsink (with a set of small, quite fans).

You will also need to take care in designing and constructing the heatsink interface between the LEDs and the heatsink to ensure that all the LEDs transfer and dissipate heat uniformly (ie- you don't get a buildup of heat between a few LEDs resulting in a hotspot and premature LED failure).

For further information refer back to the Luxeon website or try a google search for "Luxeon K2 thermal design".

Hello Urbain,

I found the spec sheets for the LEDs. You are using high-power devices very similar to the CML types I mentioned - that changes the assumptions I was using.

If your current source is regulating properly and no string of LEDs is taking more than the rated maximum of 1.5 A, then your burnouts may be caused by a thermal management problem (I note that the manufacturer specifies its 50,000 hour lifetime figure for your LEDs at 1.0 A at a junction temperature below 120 °C). In particular, if you are using the gull-wing surface mount package, keep the printed circuit board traces as wide as possible so the copper foil can dissipate the heat. Use the thickest foil available for your board material. You may even want to attach metal fins to the foil to help carry heat away from the board. These can also serve as reflectors to help focus the light if properly arranged. Make sure you will have proper ventilation and heat dissipation in all mounting conditions. You may still need to use a temperature sensor to limit the maximum drive current if the LEDs get too hot. Some regulators have inputs for this purpose; otherwise you can apply it to the brightness control circuitry.

Another issue is lead and PCB trace inductance, particularly if your current source has a high bandwidth. Keep lead lengths short, especially between the regulator device and the current sense resistor. Use a wide-bandwidth oscilloscope to check for unwanted oscillation - this can exceed device instantaneous maximums and generate heat. The classic op-amp current source is very prone to oscillation if the output lead lengths are long - I've seen a power op-amp fry under these conditions.

It would be most efficient and predictable to use a regulated maximum current pulse-width modulated (PWM) supply (i. e., set the maximum current at 1.0 -1.5 A, then turn it on and off to regulate the brightness, as recommended by the manufacturer). This again would be a wide-band circuit, so you will have to keep leads short and take precautions against unwanted oscillations or RF emissions. You may have to have multiple current sources just because longer lead lengths from one source make the system unstable. If you must use wire between sections of your circuit, it should be either shielded cable or twisted pair. Single pieces of wire can start an oscillation just by vibrating in the magnetic fields from the wires carrying the drive current.

Circuit layout is so critical to the proper operation of high frequency switch-mode power regulators that manufacturers provide recommended circuit board patterns for some models. The ARRL Radio Amateurs Handbook is an excellent reference for high frequency construction techniques. Even though your control signals may be DC or very low frequency, PWM waveforms have high frequency harmonics, and parasitic components arising from the layout of current source feedback circuits can be significant.

Let me know if this helps.

Paul

Hi Paul,

I got a carnaval break and I'm now back to work. Thank you so much for the effort you take to see that I get it right.

To be honest I'm still a younger student eager to learn, you have given a number of solutions but I didn't get them well so if you don't mind I would like you to explain some few items. For instance

1- Looking at the schematic I posted last week, how can you determine the bandwidth of the current source or improve the bandwidth?

2-You talked of regulated maximum current PWM supply. Is PWM not more completed? Is PWM not digital? If I choose to use PWM will there not be a need for a DAC to bring back the signal and making it more difficult? If not how will the sine wave or triangle signal at input be conveyed?

3-Reading at the datasheet of Luxeon K2 LED as you said, do you think of any other LED that can generate high intensity of light like K2 but produces less heat?

4-I did test the system with 3 different values of resistors (3.3ohms, 6.8ohms and 10ohms) for each string, I noticed a little different in current(10mA) between each string but no string consumes more than 1.5A rated. Is this of any influence to the system? Don't you think that the larger the resistor value, the more power loss? of course using 10ohms reduces the deviation percent.

Once more thank you and hope to hear from you again.

Greetings

Urbain

Hello Urbain,

I had not seen your schematic before - it did not show in the simple text editor I first used to view your document. It is a basic analog design. Perhaps a better way of characterizing the current source would be its impulse response - how it behaves in response to a step or pulse input. However, as Jack of all Trades pointed out, it is not very efficient - you are dissipating about 200-300 Watts in the transistors at higher current levels, which is enough energy to run two standard household incandescent lamps. That's why you can get away with using relatively large (10 Ohm) current sharing resistors - they are just taking some of the heat for the transistors.

That's why we recommend pulse width modulation. With this technique, the drive transistor(s) is turned on and off at a fixed frequency. The amount of on time is controlled by comparing the output with the control voltage (input waveform). (The "on" time is the output analog quantity, so no DAC is needed.) The big advantage of this is that instead of converting a lot of power into heat, you can store it in reactive elements (inductor and capacitor) and feed it to the LEDs when the transistor is off. Also, since you aren't heating the transistors so much, you can get by with a much smaller heat sink. To make it even more efficient, you can use power MOSFETs so you don't need as much transistor drive current and can obtain a much smaller voltage drop across the transistor than you would with a bipolar device.

Another reason for using a switched, fixed current level is that LED color varies with drive current. A fixed level gives you more predictable results.

However, as you surmise, a PWM circuit is more complex. The most common circuit uses a reference voltage in a ramp generator circuit. The feedback voltage (taken from the LC filtered output or a load current sense resistor) is compared with the ramp. If it exceeds the ramp voltage, the output drive is turned off.

Fortunately, since this circuit is so commonly used in modern switch mode power supplies, there are plenty of integrated circuits available with all of the really complicated stuff inside them. That's why I recommended you look at the data sheets and application notes for some of these - so you can see how they would be used. The most complicated part is the output drive design. Following is a list of some of the design considerations.

1. It appears that you want to vary the LED current with an input voltage waveform. That means your PWM ramp frequency (usually called switching frequency) needs to be at least twice the highest frequency component of the input waveform to avoid unwanted low frequency variations (the Nyquist criterion). Are you familiar with the Fourier transform? This is used to mathematically convert a waveform into a sum of sine-wave components. Square-waves, sawtooth waves, etc. have a lot of high-frequency content. In practice, you would pick an upper frequency limit that will permit the circuit to do the job you specify, and filter out any input frequency content over that (you're already doing that to some extent with the first-order RC filter at the input of your current source). There is nothing wrong with having your switching frequency much greater than the Nyquist minimum; that would give you a better resolution of the waveform and more stable operation. The trade-off is that circuit layout and EMI reduction become more critical.

2. Your switch-mode regulator IC needs to have an external voltage reference pin (your input goes there), otherwise you would have to use an extra PWM circuit to gate the output using a digital control input (this was what I meant by modulating a regulated maximum current). In other words, you would be looking for an adjustable switching regulator rather than a fixed voltage unit.

3. There are a number of output circuit design topologies. You can use one of the simplest, the "buck" configuration. In this design, the output transistor switches current into an output capacitor through an inductor in series with the transistor drain. The inductor bucks the change in current from the higher input voltage, provided the transistor on-cycle time is not too long.

You can use a regulator IC with a supply voltage less than the LED drive voltage, but then you need a level shifting circuit so you can connect your transistor drain circuit to the higher voltage power supply. This is also available as an integrated circuit called a "high side driver," for example, the International Rectifier IRS2117 or IRS2118. https://ec.irf.com/v6/en/US/adirect/ir;jsessionid=8D81CA35DFC0F4DEAFB221A491F20631?cmd=catSearchFrame&domSendTo=byID&domProductQueryName=IRS2117PBF

4. Study the IC manufacturer's recommendations for the output circuit design. In particular, proper consideration of the inductor flyback voltage and diode reverse recovery current is critical to prevent blowing your output transistor (I recommend using a fast-blow fuse in your prototype to protect the LEDs!) or unnecessarily wasting energy.

http://www.national.com/pf/LM/LM3423.html is a good example of the kind of circuit I'm talking about in "2." above. This IC is designed specifically for driving the type of LED you have, and is rated for up to 75 Volt input, so no high side driver is needed. Note that you would need a separate PWM converter to convert your input waveform to its pulse equivalent. If your waveform is digitally generated, you can skip the DAC and do the PWM generation in software. The serial control signal can then be connected directly to the nDIM input, using an optical isolator if necessary.

A note on selecting inductors and capacitors: make sure they are rated for the current! If you use a capacitor not rated for the high-frequency ripple current you will run, even though it may otherwise be the right size, it could overheat or even explode. I mention this point because it is sometimes neglected in textbooks.

That should cover the circuitry for now. The next topic is thermal and reliability considerations. You should not run right at the maximum current of 1.5 A - keep it down to 1.2 A or less. The lifetime of a device goes down exponentially as its temperature rises, more current generates more heat, and the LED is less efficient at higher currents (see the data sheet graph of light output vs. current). If you are running at maximum current, a voltage surge could push you over the limit and destroy your expensive LED collection. Also, some high-power LEDs have internal back-to-back zener diodes for protection against ESD or other over-voltage; raising the drive power to the limit only wastes energy in the zener diodes instead of providing more light. (I didn't see that in the K2 data sheet, but similar products from other vendors have them.)

Now, to address your biggest concern: heat. The data sheet for the K2 LED states that the thermal resistance is 9 degrees C per watt., meaning for every watt dissipated in the device, its internal temperature rises 9 degrees Celsius. If at 1 A, the voltage drop is 3.72 V (from the K2 data sheet), your device junction could be 33.5 degrees warmer than the point at which it is mounted to its heat sink, assuming all of the power is converted to heat. This is in addition to the thermal resistance of the heat sink and case.

If you get an LED already mounted to a star heat sink, this is an additional 4 °C/W (that's from the 13 °C/W value given for the star package). If you mount a bare LED directly to a circuit board with narrow traces, the additional thermal resistance is several tens of degrees per watt, so the LED will be much hotter than a star plate mounted part. (Each material between the heat generator and the ambient environment is an added thermal resistance in series. Moving air has a lower thermal resistance than still air, which is why fans are commonly part of the solution.)

The star plate offered by Phillips is probably the easiest way to make electrical connections and still take care of the heat. The thermal resistance measurement of 4 °C/W (13 °C/W total) assumes that there is free airflow on both sides of the star, so you would either need to arrange for that or attach the backs of the stars to a larger heat sink. Even so, your LED will still be hot to the touch.

If you don't want to use a lot of space between LEDs to dissipate the heat, consider mounting the LEDs on a ceramic-on-metal substrate (instead of a fiberglass PCB) attached to a finned or liquid-cooled heat sink. I don't know if you have the resources to make this. Since tooling costs are quite high, this technology is rarely used for prototyping or limited runs. If money is of no concern, you could try liquid nitrogen cooling .

The web page at http://www.national.com/analog/led/high_brightness has a number resources for your project, including more on the material I have discussed. I hope I haven't overwhelmed you with detail or insulted you with stuff you already know; I tried to cover the background without knowing how far you have progressed in studies. I look forward to your next post.

Paul

Hi Paul and Jack of all trade,

Thank you guys for all your supports, I am really grateful and appreciate your time.

Paul, is like you are a professor you have lots of experience on this field, you have given lots of points but I must admit I still have a long way to go before I could understand most of them.

I will be trying to make good use of your ideas.

Paul, you said you have not seen my schematic, here is it: http://www.sendspace.com/file/2bkb14

Consider you were a young student like me and this above schematic is given to you that you should improve it or design another one. The problem with this schematic is that the LEDs get hot so fast and current source bandwidth need to be increased. What should you do?

a) In reply to your last post, the LM3423 is expensive for this project the best one cost $110.00

b) My waveform is produced by a function generator and is analogue.

c) I thought if PWM is used sending sine wave or saw tooth will be a problem since high frequency is required. The question still stands how will sine wave for instance be conveyed by PWM? If is possible can you send me one schematic example?

Once more thanks again.

Best regards

Urbain

You could use a cheap'n'cheerful PWM controller like the LM3524 - there's a link to the datasheet (with several application circuits) on the RS site, or you can easily find it by googling.

The PWM switching frequency can be much higher than your modulation frequencies, so you don't need to worry about that side of it.

I'll try to think it through a bit better at the weekend, if I can find some time.

Re the heating - if you've got those currents going through the LEDs - they will dissipate a lot of heat. The only thing you can do is get rid of it (heatsinks & maybe forced air cooling).

One point about driving LEDs with a pulsed supply - provided you keep the average current within the rated limit, you can (sometimes - consult manufacturers literature) get away with higher peak currents. This gives a greater perceived brightness for the same I_ave (and hence the same heating).

Hello Urbain,

Once I realized there was a schematic on your original file, I was able to view it with NeoOffice. Given the inefficient design you were provided, I would be inclined to start over with a switching-type circuit. That would be Part One of the project.

The price you were quoted for the LM3423 must be for the evaluation board, including all of the parts and the circuit board, plus import duty and other taxes. The chip itself is $4.15 US from http://www.digikey.com. As you noted, the PWM control input is not the easiest thing to use given your analog input signal.

I have another one for you to look at - the LM5088. This part costs less - $3.80 US. It is a general purpose bucking switch-mode regulator. Referring to the schematic on the front page of the manufacturer's datasheet (http://www.national.com/ds/LM/LM5088.pdf), one of your LED series strings would replace Rfb2, and its current sharing resistor would be Rfb1. The other strings would simply run from the output to ground. The current sharing resistors would be 1.205V/Imax, or 1.0 Ω at 1.205 Amps. You can control the output current up to the maximum current by applying your signal to the SS pin, provided that the signal is no more than 1.205V. If the control voltage waveform exceeds that level, your LEDs will simply run at the maximum current (a control voltage over 7 Volts will damage the regulator).

Note that the output transistor circuit has its own current sense resistor. This circuit is sized for the total output drive current. Also, the on-resistance of the transistor must be low enough so the source-drain voltage is less than 1 V at maximum current, otherwise the regulator's protection circuitry will shut down everything. Those are some particular items I picked up from a brief perusal of the data sheet.

The drawback of this design is that instead of pulsing the maximum set current to get the desired average, you are actually reducing the set current with the control voltage at the "soft-start" input. This may produce inconsistent color. The LM3423 and similar products are designed to get around this issue. It appears that what you would like to have is something like the LM3423 with an analog control input rather than a PWM input. You would be looking for something designed to drive an external transistor, and your input power supply voltage needs to be at least 40V. I've only looked at National Semiconductor parts; other manufacturers may have a better match. I like to use National Semiconductor parts as a starting point because they have some of the best (most detailed and informative) data sheets in the industry. I learned as much about electronic design from these as I did from my formal studies. Certain far east manufacturers would benefit greatly from their example...

The next question is the maximum frequency of the input control voltage. If the project is a light panel for special effects, I would not expect it to be much over 60 Hz, since the human eye does not readily discern flicker above that frequency, even in the relatively faster peripheral vision regions. Even animals with only monochrome vision would not notice flicker much above that frequency. That is why you can watch television with a 60 Hz vertical frequency - you don't perceive the individual frames or sweep lines; rather your eye and brain integrate the raster into a single time-continuous picture.

If I understand the application requirement correctly, then even at its slowest speed of 50 kHz, this regulator will have enough bandwidth for the task. Certainly at 1 MHz it should be able to keep up with variations below 20 kHz. 200 kHz is probably good enough, but you may want to check the radio spectrum usage in your region to make sure that the fundamental and first three harmonics won't interfere with anything.

Now for the question of a sine wave represented by PWM: if the PWM waveform is integrated, you will get the sine wave back again (plus possibly some harmonics and other results of errors in the process). The on-time fraction of each pulse is the ratio of the instantaneous input voltage to the reference voltage (assuming the input is less than the reference). Negative voltages won't work, so your waveforms will need a DC offset to avoid cut-off at the bottom, just as in your analog current source.

There are many ways to obtain this ratio. A relatively simple analog-based method is to sample the instantaneous voltage and compare it with a ramp waveform running at a frequency much greater than that of the waveform to be converted. When the ramp = sample voltage, the pulse is switched off, so it stays on longer for higher voltages. At the end of the ramp cycle, the input voltage is sampled again and the process is repeated. (Sampling: using a solid-state switch, connect the voltage to a small, very stable, low leakage capacitor. Open the sample switch, then do the ramp comparison with the voltage on the capacitor. See also "sample and hold amplifier.") For lesser precision and easier timing requirements, the input voltage is applied directly to the ramp comparator.

How are you coming with Part Two (thermal management)? The high-efficiency PWM circuit will get rid of most of the heat from the current source electronics, but you still have to deal with the thermal losses in the LEDs. The thermal power figure I calculated for the last post was worst case, assuming all of the input energy went directly to heating the LED. Fortunately, most of it should be going to light production instead. Even so, you have to keep the total thermal resistance to ambient as low as possible to avoid having to dim the light at high ambient temperature or frying the devices. See Figure 25 in the data sheet for the K2 LED - this gives you an idea of how much you would have to cut the current at high temperatures for various heat sink sizes (including the thermal resistance of the LED).

If you want automatic thermal protection for your LEDs, you could use a temperature sensor circuit to clamp or reduce the magnitude of the control voltage the temperature increases. The reduction rate would be set to match or exceed the appropriate derating curve in Fig. 25. The curve you can use depends on how effective your heat sink is. For example, if you have a good-sized heat sink with a thermal resistance of less than 2°C/W per LED (assuming you are taking the heat from star plates), you can use the top curve and will probably rarely if ever have the de-rating function kick in.

The tricky part with heat sink mounting is the connection to the LED string used for regulating the current. You want the top of the string to be close to the output inductor, and you want the bottom LED of the string to be close to the feedback /current sharing resistor, which must be close to the regulator IC. This string won't be a straight line since it would have to fold in the middle to get back to the regulator.

I recommend making sure your heat sink has a solid connection to the power ground at the regulator; this should prevent a lot of problems with noise and instability (it would be almost like a ground plane in a circuit board).

Paul

Hi guys,

Thank you all for all your supports, I will try to make use of all what I have gathered so far and if any question I will get back to you again.

Please if any one finds something later I will be grateful to have it while waiting for JohnDG to have some time over the weekend to find something.

I had a weekly meeting with my tutor today and I discussed with him lots of possibilities and some ideas I got from this forum. He told me PWM should not be an option because a continuous signal needs to modulated by the current source.

If PWM is to be used, the PWM should have a 1ms duty circle plus locking to V-pulse '20micro second'.

I found it difficult because the PWM will have to be locked with signal from the camera, I don't think this is my level.

I will be waiting for more ideas.

I wish you all a nice time and a nice weekend in advance.

Greetings

Urbain

Sorry, Urbain, been working all weekend - didn't have a chance to look at it.

Hello Urbain,

It appears your intended use is a light source for 1000 frame/second video or high-speed photography (1 ms duty cycle). If that is so, then a PWM output regulator would have to clock fast enough to provide the desired average light level on a frame-by-frame basis. (We're back to the light level modulation bandwidth issue.)

If you are using a digital camera that reads the light sensor pixel data instantaneously rather than using a store-and-read method, a minimally filtered PWM scheme would give you an interesting but undesirable pattern of dark and light stripes in the picture. So far as I know, however, CCD sensors are inherently store-and-read, that is, the light induced charge production takes place while the shutter is open, then the pixel charges are transferred to the readout ADC(s) after the exposure is complete. In that case, a PWM output should be OK. (By the way, film is a chemical-based store and read medium.)

This does not mean you can't use a PWM drive for your output transistor. The question is whether you must filter out on-off switching of the peak current with a capacitor. If you don't have to, then you can maintain consistent color since the LEDs will always have the same current whenever they are on - assuming that the current source can settle to that level during the shortest pulses.

On the other hand, if you have light level variations within a video frame or photograph, it is better to filter to the average current and let the photo editor take care of color shift in software. Note that an analog light source would have the same limitation - you may have to do a sample and hold of the input voltage gated with the V-sync pulse in order to maintain a constant level for the duration of the frame or at least greatly constrain the input frequency, but you would be unable to avoid the current-dependent color shift.

Some more detail on the intended application for the lighting appliance will go a long way to help us understand the requirements. In particular, what is the highest fundamental frequency component of the control signal? What is the minimum time for which the average output light level must be constant? How significant is color shift in the lighting for the intended application?

One possible configuration is to use an LM3524 in an open-loop configuration with the input signal connected to pin 9 (see the data sheet application comment on use for a variable speed motor drive.). This is only to convert the input to PWM form. The output is then connected in a totem pole configuration and used to drive the input of an LM3421 which is set to drive the LEDs at 1.2 A. The limitations of the LM3421 dimming function set the following constraints: The highest permissible frequency component of the control voltage is 50 kHz; the response may be non-linear at a very low light level (if the input PWM pulse is very short, the output circuit will not be on long enough to let the LED current rise to 1.2 A). Many cameras have color inaccuracies at low light levels, so an error in color output from the LEDs may not be significant in this regime.

There are a few other resources in this setup you can use. For example, the reference voltage pin on the LM3524 can be used to set a trip level for a thermal protection circuit. That is, compare a fraction of this voltage with that of a temperature sensor in the LED array, and use the comparator output to shutdown the enable pin on the LM3421 if the LEDs are too hot. If you do this, I suggest you have the comparator also turn on a red indicator LED to let you know what happened.

Note that this design calls for using one or two large MOSFETs driven at high speed. That means there will be a fair amount of gate current going through the regulator chip, so you need to pay attention to its thermal considerations. See Note 4 on page 6 on the data sheet for the LM3421 (even if you pick a different regulator).

Paul

Hi Paul,

Nice to hear from you again. I'm grateful for your concern

Unfortunataly the PWM is not to be used after a meeting with my tutor, the reason is that we need something analogue and PWM is like digital.

Yes you are correct the Light source is for camera testing purposes. The problem here is that we need a constant current source and this current will have to be modulated.

I am still looking for other possibilities to realize this.

Will be waiting for more supports and idears.

Thank you once more.

Best regards

Urbain

Hi Paul,

I must admit you have been very helpful and I am grateful.

For the current source, I have decided ten different current sources for the ten string, I will build that I see how it looks like.

Now on the Thermal management, my problem is how to manage the losses on the LED as you said. I have some few questions if you don't mind.

1) looking at the datasheet of the K2 LED, the current derating curve shows that when lighting the LED at 1000mA under 75degree C the heat of 15degree C/W would be good, in that case may be I can divide 15/100 to get the right heat sink for all the 100LEDs.

2)if I assume taking for instance the FISHER SK157 +/- 0.25C/W heatsink and use a fan or cooler, how can I determine the amount of airflow needed to reduce the C/W?

3)the datasheet of most LEDs are in lumen and my specification is in lux, where 1lux=lm/square meter. I don't know how the square meter is considered. is there another way of converting lumen to Lux?

4)Are there other ways of determining this thermal dynamic, because is causing me more problems.

Will be grateful to hear from you. Once more thanks

Best regards

Urbain

Hi Paul,

I must admit you have been very helpful and I am grateful.

For the current source, I have decided to use ten different current sources for the ten string of LED, I will build that now and see how it looks like.

Now on the Thermal management, my problem is how to manage the losses on the LED as you said. I have some few questions if you don't mind.

1) looking at the datasheet of the K2 LED, the current derating curve shows that when lighting the LED at 1000mA under 75degree C the heatsink of 15degree C/W would be good, in that case may be I can divide 15/100 to get the right heat sink for all the 100LEDs.

2)if I assume taking for instance the FISHER SK157 +/- 0.25C/W heatsink and use a fan or cooler, how can I determine the amount of airflow needed to reduce the C/W?

3)the datasheet of most LEDs are in lumen and my specification is in lux, where 1lux=lm/square meter. I don't know how the square meter is considered. is there another way of converting lumen to Lux?

4)Are there other ways of determining this thermal dynamic, because is causing me more problems.

Will be grateful to hear from you. Once more thanks

Best regards

Urbain

Hello again,

I missed your third question in my last reply. When measuring illumination in lux, you are looking at an area of a sphere with the light source at the center. The source intensity is rated in lumens, but the light on the illuminated object is in lux. The two units would be numerically identical if the light were integrated over a sphere with a surface area of one square meter, but if the sphere is twice the radius, the illumination would be one-fourth as much (four times the surface area). That's why different units are used.

You would have to work backwards from the geometry of your lighting problem to get the required lamp intensity. The radius r of the unit integrating sphere (1 m^2 area) is 1/√(4π) meters. Let R be the distance between the lamp and the target, and I be the intensity of the lamp in lumens. Then the illumination in lux is I*(r/R)^2, or I/(4πR^2).

Paul

This webinar that was sent to me on "LED Driver Selection Methodology" may be of interest to you (link below).

http://www.techonline.com/learning/livewebinar/212902264

Hello Urbain,

If you can afford the additional circuitry, driving each series string with its own current source has a number of advantages. The output device requirements are less stringent; it should be easier to minimize the effects of parasitic reactances; and you can control the individual LED strings with different inputs to get interesting shading effects.

As for the thermal design, a good reference, if you don't mind converting English units to metric, is "Introduction to Thermal Management", referred to as "ITM" in the following dissertation, available for free download at http://www.wakefield.com/pdf/thermal_tutorial.pdf

Don't forget that the thermal resistance of the bare LED is already 9°/W from the junction to the bottom of the case. If you order it with the star-shaped heat sink plate, the junction to plate thermal resistance is 13°/W. You may come close to this figure if you use the bare LED on a thin double-sided board with many small vias connecting the copper under the case to the bottom of the board to transport the heat through to the heat sink underneath. If you use the latter approach, you will need to put a silicone or mica spacer under each LED to keep the case slug electrically isolated (see Note 1 on page 14 in the data sheet under Mechanical Dimensions for the 4-lead Gullwing Form). The board would be mounted directly on the heat sink in this arrangement, using thermal grease or a heat transfer pad in between, and attaching the board using screws going through metal strips to ensure uniform clamping force. See http://www.fischerelektronik.de/index.php?id=72&L=0

If you use star-mounted LEDs attached to the heat sink, you could either do point-to-point wiring (200 connections plus a terminal block), or you could make a circuit board with holes cut out for the LEDs and plated vias over the connection pads for soldering. (This board would be held away from the heat sink by the LED star plates.) Use white soldermask on the board to reflect light from the LEDs outward.

Assuming you are using the star package and your target thermal resistance is 15°/W, your heat sink must be (15°/W - 13°/W)/100 = 0.02°/W. Using the graph relating heat sink volume to thermal resistance on page 2 of ITM, I estimate by extrapolation that you would need an airflow velocity of 30 m/s to keep the STK 157 sufficiently cool (assuming you are using a 300 mm square section). That would be loud and disruptive - think papers flying all over the room!

The ultimate (and expensive!) approach is to mill the mounting surface of the heat sink to make mounting studs for bare LEDs. Your circuit board would have a hole under each LED and each stud would be just high enough to contact its LED thermal slug through the corresponding hole in the board. This reduces the LED to heat sink resistance to about 1.2 °/W (with a silicone heat transfer pad on each stud - see page 6 of ITM), so your heat sink could be (15-10.2) °/W /100, that is, 0.048 °/W. The advantage of this scheme is that you can get closer spacing on the LEDs and use a shorter heat sink than the other methods require. However, that is still out of practical reach for the STK 157 (forced air velocity would need to be more than 10 m/s). The drawback is a risk of cracked solder connections due to differential thermal expansion (the studs would try to push the LEDs off of the board at high temperatures). Mounting the board with spring-loaded shoulder screws is one possible solution.

Instead of the ST 157, you may get better results with the Fischer LA HLV 3 or an equivalent. This should work with any of the suggested LED mounting schemes. It is designed for optimal cooling with fans mounted on one end (the fans are included in the price).

A suggestion: attach a temperature sensor to the heat sink and use it to control the fan speed. The voltage controls a PWM clocked at the zero crossing of the power line to generate a triac gate signal for the fan. (Be sure the fan motors are "impedance protected" or they may fry at low speed.) I did this to a large power supply years ago, and that made it much more civilized at low current. The sensor can also be used as a fault detector if a fan fails or the system otherwise overheats. However, don't bother doing this until you get the important parts working. You'll get your grade based on the performance of the light source - the fan control may be a nice project for another student.

I assume you will use another heat sink for your current source output devices to keep the LED cooling problem from getting more complicated than it is already.

Paul

Hi Paul,

Thank you so much for your time and your assistance.

You have just answered my question, I'm grateful. I have really learned much in this project. And please if you don't mind I will be coming back again for more helps.

As for now I will try to work on the information I have got from you.

Best regards

Urbain

Hi Paul,

It is a pleasure writing once more. Here I am again. During my research on LEDs I have come out with another LED the luxeon rebel. See http://www.philipslumileds.com/pdfs/DS56.pdf

This type of LED produces high brightness at low current 700mA and less voltage 3.15V. I have few problems and wish you could help.

1) I want my target to be the 15°C/W looking at the derating curve (page 18). How can I determine the heat sink? This LED thermal heat sink is 7K/W which make me confusing to calculate the thermal management based on the ITM at http://www.wakefield.com/pdf/thermal_tutorial.pdf. More over it does not have the star heat sink like the formal K2. Can you counsel me on this please?

2) At typical the intensity is 180lm; from the last formula you gave me I could get 1,432lux if I have a radius of 10cm. Now my question is how to determine the number of LEDs needed to get the 100Klux, since I could not just divide 100Klux by 1,432lux because they don't have a common source and the surface area will need to be taken into account. On page 20 the Lambertian angle can be of help may be to determine the intensity with respect to the radius. May be excel can do it but I don't know how to put my data.

Would be grateful to hear from you.

Best regards

Urbain

Hello Urbain,

1) Thermal design with this LED is much easier since the thermal pad is electrically isolated from the LED die. The double-sided FR4 circuit board is the primary thermal interface. See http://www.philipslumileds.com/pdfs/AB32.pdf for details on board design recommendations. The authors of this application note have done almost all of the thermal design work for you. It describes two board construction methods: the cheap and simple open plated-through via, and the more expensive but lower thermal resistance (3°/W) filled and capped via. In the latter method, the plated-through holes are filled with epoxy, then the entire surface is plated with more copper.

The thermal resistance from the LED junction to the thermal pad is 10°/W. If an open via board is used, total thermal resistance to the bottom of the board is 17°/W, and the LEDs must be separated by at least 4 mm. to ensure adequate heat dissipation area. If a closed and capped board can be used, then the total thermal resistance is 13°/W and the LEDs can be placed as close as 2 mm. These numbers assume that the board is no more than 0.8 mm thickness, the via plating thickness is at least 35 microns and the surface plating is 70 microns.

No specific mention is made in the application note of an additional heat sink, but I would seriously consider adding a pad to each bottom-side pattern for soldering to a right-angle or U-channel copper strip fin running the length of each row of LEDs. One reason is that these fins will help to stiffen the board, which, at 0.8 mm would be susceptible to mechanical vibration that could cause visible variation in the lighting. The other reason, as we've noted before, is that the LEDs are more efficient at lower temperatures and will provide more repeatable results. Alternatively, the board can be mounted on a finned heat sink using an appropriate thermal transfer medium as described in ITM and one of my earlier posts. If you use a heat sink, you should plate the heat dissipation areas on the bottom of the board with tin (or gold if you can afford it). This is to prevent oxidation of the copper, which would severely compromise thermal conductivity. Organic soldermask protects the metal, but has relatively poor thermal conductivity.

2) A general calculation of illumination gets into solid-angle geometry. To provide a complete answer, you need to know not only the distance to the illuminated surface (10 cm?), but the area that must be uniformly lit. Is the target surface a plane, curved, or random (i. e., an unspecified object)?

However, if you just want the answer at one point, you can use the illumination vs. angle chart on page 20 of the data sheet, plus plane geometry. Assume you are looking at a lighted point 10 cm directly over an LED in the center of the array. The illumination from that LED is 1. Now, find the angle from the point to the nearest neighbor LED and look up the illumination on the chart. If the LEDs are on a square grid, you can multiply that times four, divide by the square of r/10 cm (r is the distance from the neighbor LED to the lighted point) and add the number to the 1 for the first LED. Repeat the procedure for the next four nearest neighbors. Draw a side-view picture and an overhead view to help yourself visualize this.

At 10 cm from the LED mounting plane, you will have significant illumination contributions from LEDs 27.5 cm away from the first LED (distance = 10 cm * tan 70°, which is half of the 140° viewing angle on page 6 of the data sheet). You could go as far as 57 cm (tan 80°), but I don't think you plan to make your light panel that large! (You can fit 100 of these devices on a board about 8 cm square.) Now that you have the sum of the relative intensities, multiply it times the illuminance from the first LED.

It should not be too hard to turn this into a spreadsheet formula set, with each cell representing the contribution from one, four or eight LEDs (take advantage of the symmetry of the problem). The fun part is calculating two hypotenuses - one in the LED plane to the lighted point, which is used to find the other from the LED to the lighted point and converting the equation into a formula that can be filled into a group of spreadsheet cells. I usually do this by referring to a row and column containing the coordinate distances - these also serve as axes for 3-D graphs if desired.

Start with a 7 or 8 mm grid (4 mm between LED edges) to determine whether this will be adequate. If it is brighter than needed, you can either spread out the LEDs to get more area coverage or reduce the drive current for greater efficiency. If you need more light, then you would have to use a filled-via board to get tighter spacing or add more LEDs.

Paul