Driving 140x 3W LEDs

[Xenobius]

Hi, no objection at all. Im learning before I dive in.

Wouldn't the pcb heat up a lot at 3 watts each?

If I had to do all 70 in series, what voltage should it be?

Finally how do you calculate the value/wattage of the resistor for all 70 of them?

Thanks a lot

 

[betwixt]

If I had to do all 70 in series, what voltage should it be?

Take the worst case voltage (the highest) for the LEDs and multiply it by the number in the chain. This gives you the minimum voltage you must have available. Add maybe 5V to 10V to give a safety margin.
Then limit the current with a constant current circuit. Setting the voltage alone will not work but you do need a little excess to ensure the constant current circuit will work.

What you haven't grasped is that a constant current supply does just that, it adjusts it output voltage so the current flowing into the load stays the same. As LEDs are current driven devices, and all LEDs in the same string must carry the same current, the voltage across it will adapt according to the needs of the LEDs whatever their individual forward voltages are. As long as the supply voltage is higher than the LEDs need, (if it was less it would be impossible to make them fully conduct) a constant current will ensure they are all happy.

Lets say your LEDs need 3.6V maximum and you have 70 of them in series, that makes a worst case voltage requirement of (3.6 * 70) = 252V. So what you need is a supply capable of a bit more than that, maybe 260V and a current limiter at 700mA. By limiting the current, the voltage will automatically drop to 252V.

Now take the other extreme, all your LEDs need 3.2V so the voltage across them will be (3.2 * 70) = 224V, much less than the previous calculation. Now that same 260V power supply will still limit the current to 700mA and it's output voltage will drop to 224V so again all the LEDs are properly driven.

If you didn't use constant current and you applied 252V it might work if all the LEDs totaled exactly 252V but if they were anything less than that the current would increase and likely damage them.

Brian.

 

[c_mitra]

Wouldn't the pcb heat up a lot at 3 watts each?

Sure, it will be hot like hell!

The total will be more than 210W for each board. But that will be spread over a A4 size (roughly) board. For simplicity, consider an A4 size PCB has 600 cm2 area; so the heat will be spread over this area. You will have about 0.3W/cm2 which can be dissipated with a decent fan.

LED performance will be down if the temp becomes more than 60-70C; you should try to keep the LED temp below 50C. To dissipate by radiation alone, the temp will be around 200C; so you must have a good fan.

A desktop CPU uses about 100W but the heat produced is confined in a small volume /area. I guess you should be ok.

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If I had to do all 70 in series, what voltage should it be?

LEDs are current devices; the light produced depends on the recombination and the output is directly proportional to the current (within reason).

When you connect them in series, all of them are forced to have the same current. And they will produce the same light (roughly) on a per LED basis. Each will adjust to its own voltage.

But when you connect them in parallel, both of them see the same voltage and the LED with the lower forward drop will take more current and produce more heat (and light) compared to the LED with a higher forward drop.

Just consider two resistors in parallel: you put two 1R resistors in parallel and apply 1V. Each will take 1A current. But if one of them is 1R1 and the other is 0R9, they will not take the same current (one will be 0.9A and the other will take 1.1A). Heating will also be unequal.

Another point: when you put many (say more than 10) LEDs in series (that have individually different forward drops), the combination will act as if it has all the components (individual LEDs) have typical values of voltage drop.

If you are using a const current source, you can control the current with a simple potentiometer; a very reliable way to change the intensity (light output). In fact, you can have two current sources being controlled with one potentiometer.

If you are not bothered much about the intensity, you can use a traditional resistor (just like a common LED is being run).

 

[Xenobius]

This is really informative. I think if the voltage was around 230v I could just use a bridge rectifier and smoothing caps and work with mains, avoiding a powersupply altogether. I believe that a constant current circuit is a must and I will do some research on this department for sure.

So if I had to alter my design now, (and considering that 70 3W LEDS might be an over kill, I will do like so. If 230 is the max voltage I will get (without a power supply) (not taking into consideration that after smoothing the DC voltage will be higher than 230V)....... Using worse case I will have 230v - 10v to spare which leave me with 220VDC which can be split over 61 LEDs in series. (or maybe just 60 to have 10 rows x 6 columns per side)

As someone said above, with higher voltage its easier to control the current. We are talking about less than 1A per side? And no heat from power supply as its non existent, only the heat from the LEDs and the constant current circuit!

 

[c_mitra]

So if I had to alter my design now, (and considering that 70 3W LEDS might be an over kill, I will do like so. If 230 is the max voltage I will get (without a power supply)...

No really so; 220V AC after rectification and filtering can get you around 300VDC. If filtering is not good, the lamps will flicker.

You distribute the LEDs more or less uniformly over the board. The bottom of the LEDs must be soldered via a PTH hole on the PCB. Keep these pads isolated.

The LEDs will drop around 250V. You may have to use a dropping resistor for 50V and a current of 600-700mA. The resistor takes the same current as the LED. The wattage will be around 30-40W and resistance will be around 60-70 Ohm. You can use a suitable capacitor before the bridge rectifier. That will reduce the heat produced.

You can also fashion a one (or two) transistor constant current source - and choose one in which you can adjust the current. The final pass transistor will need a heat sink.

 

[KlausST]

Hi,

Connecting 70 (or so) LEDs in series with about 250V is causing a safety problem.
You need to properly protect everything in a way that nothing us touchable.
Hopefully you are familiar with high voltage and safety regulations.

Also consider: If one LED goes dead and getting high ohmic, then all (most) voltage will be across it's terminals.
...let's just consider the resistance goes up to 5kOhms and 50V across the dead LED then a current of 100mA causes 5W of heat.
In this situation the current limiting circuit will push all available voltage into the LED string.

Also consider: a capacitor after a bridge rectifier will be charged to about 325V.
In either case! ... unless there is a series circuit somewhere between mains and capacitor that prevents the capacitor to become fully charged to the mains peak voltage.
For sure, if the capacitor is small, then there will be voltage ripple ... reducing the average voltage.
But I don't think you want high voltage ripple.

With 220V across the LED string and 600mA LED current your current limiter produces up to 60W of heat.

All in all I recommend to use an isolated DC power supply (maybe 48V) and several LED strings with (simple) independent current limiters. In best case you just need an additional resistor and transistor for each string. (Only one complete current limiter circuit).

It all depends on your skills, knowledge and requirements.
Don't risk to hurt yourself or others.

Btw: maybe an overheat protection (thermostat) at the heatsink is a good idea to avoid fire.

Klaus

 

[Xenobius]

Hi, Ok so I've been looking around and so far its not easy to get a constant current driver capable of taking more then 60v input though I did find ones which output 45A. This is not my situation. Maybe multiple small ones that I can build would be ideal too. Initially this was my idea, using 24 / 48vdc but without the driver. Now I like the driver idea more and more.

As regards safety, you are correct and thank you for your concern because you have valid points. I am thorn between 2 good ideas. On one hand I could use mains directly, go back to the original 70 led design with the drawback of finding a suitable driver that takes 300+v input and output 600+mA (I could use 2 for each side of the board) while on the other hand I could use 48V with easier access to a driver, safer, but that will need an expensive power supply.

If I find a good driver I might go for mains >> smoothing >> driver >> leds but if not I will have no option

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Does it make sense to start a new thread just for the constant current driver? Anyways I found this page that explains how to build one, they are labeled with #numbers and the one that might be suitable for me is #3. However going on the assumption that if I just change the N-Mosfet to something more robust such as the STY60NM50 500V/60A and power dissipation of 560W I could have 2 for top/bottom sides. Thou I don't get a small paragraph as follows:

2) voltage. the "G" pin on Q2 is only rated for 20V, and with this simplest circuit that will limit the input voltage to 20V (lets say 18V to be safe). if you use a different NFET, make sure to check the "Vgs" rating.

Is it possible to build this circuit using this Mosfet?

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I think its not worth the hassle of trying to build one. I found the Mean Well LDD-600H which can take an input of up to 56V and outputs up to 52V at fixed 600mA. I will post more details later unless someone has a better idea

 

[volker@muehlhaus]

while on the other hand I could use 48V with easier access to a driver, safer, but that will need an expensive power supply.

Have you seen my post #13? That LCM-60 constant current supply is cheap because it is widely used for such LED applications. One supply per LED string of ~20 LED.

 

[Xenobius]

Hi volker,

Yes I did however I estimate for 140x leds I will need around 7 (not to mention that one string of 20 leds will be half on the bottom and half on top which is a problem since I want to switch them on separately) so perhaps even more then 7 and at 28eur each it would shoot my budget up too high. If not for anything however I did find out about Mean Well thanks to your post and I am considering the small buck ones but not yet sure how it will work for this application.

 

[c_mitra]

Safety first and safety last. ALWAYS pay attention to safety.

I found this page that explains how to build one, they are labeled with #numbers and the one that might be suitable for me is #3

I do not see why it cannot be used for your experiment /design.

If you want to have isolation, you need to design /look for a suitable one (power supply).

My personal opinion: cheap power supplies are also cheap and many are not safe. Good power supplies are unnecessarily complex and, you guessed it, unnecessarily expensive.

During testing etc., you should consider using a 1:1 isolation transformer for safety reasons.

If some 100Hz flickering is acceptable, you can use a capacitative limiter before the bridge rectifier and forget about any constant current driver.

 

[KlausST]

Hi,

its not easy to get a constant current driver capable of taking more then 60v input

A linear constant current circuit is simple: R, R, Zener, BJT....and it's easy to get 1000V if you want.

But in your case the constant current curcuit does not see the whole supply voltage.... just V_supply - V_LED.

Klaus

 

[c_mitra]

But in your case the constant current curcuit does not see the whole supply voltage.... just V_supply - V_LED.

That is true. But any const current driver has to work within the headroom it has.

He will be having about 40-50V headroom in this case and I guess that should be sufficient - if the filter capacitor is doing its job.

Because the BJT does not see the full voltage, 100V 1A BJT should be good for this experiment.

How much base drive will be needed? Anyway, the details are routine and can be worked out rather easily.

In reality, const current drivers need not be complex for the job in hand.

 

[KlausST]

Hi,

That is true. But any const current driver has to work within the headroom it has.

He will be having about 40-50V headroom in this case and I guess that should be sufficient - if the filter capacitor is doing its job.

Yes. But usually one wants the "headroom" to be low, to minimize loss.
50V headroom with 600mA current gives 30W of power dissipation in the bjt.

How much base drive will be needed? A

600mA / h_FE

There are options to optimize power loss in the base driving circuit:
* use a darlington
* use a MOSFET
* don't use the full power supply voltage but take it from the first couple LEDs that are connected next to the bjt

Klaus

 

[c_mitra]

Yes. But usually one wants the "headroom" to be low, to minimize loss.

True. But sometimes it cannot be helped.

If the power supply is 240V (many countries have that), the peak will be 340V; but with a 220V supply the peak will be 300V. If you have 110V, you may have to use a voltage doubler. I have no idea whether this is going to be a commercial unit.

But if the filter capacitor is not jumbo enough, it will drop to 250V in a few ms, and the headroom will become zero. If the filter cap is not sufficient, the lamp will turn off for the rest of the 10 ms time.

But with the load always present, the filter cap voltage perhaps will never reach 300V.

Perhaps the actual power dissipation will be about half of that (guess work, not calculated).

But anyway, I would recommend current sensing and a potentiometer to set the current (that will set the brightness) to be also used as a part of the design. With help from the community, it can be a robust design.

 

[KlausST]

Hi,

But sometimes it cannot be helped.

true.

, the lamp will turn off for the rest of the 10 ms time.

some designs do so. But it's not my way ... to desgn it flickering.

But with the load always present, the filter cap voltage perhaps will never reach 300V.

How can it be? Circuit? Simulation?

I still recommend the high quality and safe way to use a DC supply,

Klaus

 

[c_mitra]

How can it be? Circuit? Simulation?

Sorry, I misspoke. Getting old :sad:

It will always reach the peak voltage at least for a moment.

The droop will actually depend on the capacitor.

 

[Xenobius]

Hi again, so a quick recap about the circuit, I bought the 140x leds with a forward voltage of 3.2v - 3.6v and a current of 700ma. There will be 70 on each side of the UV Exposure Box and will need to be switched separately. From this thread it is clear that I need a constant current circuit and after some research I understand what was being told above by many nice folks

So I set out to find a constant current circuit and a common one is with the use of an LM317 however I have an issue. Firstly since the max voltage is around 30v, (worse case scenario) I will need (7 leds in series x 3.6v + 1.25v as needed by the LM317 + 3v headroom also from datasheet). That 29.45v so within the range of the LM317 however it also requires a resistor in series with the output with power rating of 29.45v x 700ma = 20.6W which considering is massive and I will need 14 of them. (NOTE: I hope I understood from the datasheet on pg 14 diagram 8.3.7 how to calculate the power rating of the resistor).

So with this in mind I continued to search for another solution and I found this circuit that does not use a resistor. So could anyone explain how it works please? https://www.youtube.com/watch?v=lnU5d-KBMLg&feature=youtu.be&t=6. Thanks

 

[KlausST]

Hi,

I see a lot of misunderstandings.

In the order of your post#37:

So I set out to find a constant current circuit and a common one is with the use of an LM317 however I have an issue

The aim of the constant current is not to provide a precision constant current. The aim is to limit the current to a safe value. 10% of deviation is no problem, no need for a 1% or better LM317
As already told there are simple circuits that use just Rs, zener, BJT.
But for sure you are free to use the LM317 circuit.

Firstly since the max voltage is around 30v, (worse case scenario) I will need (7 leds in series x 3.6v + 1.25v as needed by the LM317 + 3v headroom also from datasheet).

Again you misunderstood the voltages. Since LEDs and constant_current_circuit are in series there never is the full supply voltage across the constant current circuit.
Example: 70V clean DC supply:
Minimum voltage for the regulation circuit is 3V + 1.25V = 4.25 V
Leaving max 70V - 4.25V = 65.75V for the LEDs.
LED count: = 65.75V ÷ 3.6V = 18.26 --> 18 LEDs
Now to the other extremum:
Minumum voltage across the LEDs: 18 x 3.2V = 57.6V
Leaving max voltage for regulation: 70V - 57.6V = 12.4V
--> The LM317 circuit sees just about 4V ... 13V.
Don't go to the theoretical limits... you need to take care that power supply has some tolerance, overshot and so on. Also the voltage across the LEDs during low current may be lower than 3.2V

resistor in series with the output with power rating of 29.45v x 700ma = 20.6W which considering is massive and I will need 14 of them. (NOTE: I hope I understood from the datasheet on pg 14 diagram 8.3.7 how to calculate the power rating of the resistor).

Again a misunderstanding.
(Neither on page14 nor in section 8.3.7 I can find power calculations.)
Btw: you rather calculated power supply power requirement...but:
Power of the resistor (or any other device) is: voltage across_the_resistor x current_through_the_resistor.
In this circuit the voltage across the resistor is fix 1.25V during regulation. Thus the power dissipation is constant 1.25V x 0.7A = about 1W. Again here: dont go to the limit: maybe use a 2W rated one.

But you need to consider the power dissipation of the LM317:
Again here: power = voltage across LM317 x current through LM317: worst case roughly: 13V x 0.7A = about 10W. ( ten times of the resistor !!)

Power dissipation: Independent of what linear regulation circuit you use: The power dissipation is always the same.

Thus:
Power supply requirement (for each line): v_supply x LED current
For the LEDs: V_LEDs x LED current
For the regulation: (V_supply - V_LEDs) x LED current

No handstand will change this.

The only possible way to reduce regulation power loss is to go to "switching circuits" instead of linear circuit.
But keep in mind that LED power loss (= main power loss) will still be there, thus you need a heatsink, making the effort for a switching regulation circuit maybe useless).

Klaus

 

[c_mitra]

I do not think the 24R is god-given; you need to calculate the value based on your current requirement. The reference voltage is 1.25V=700mA *R; that makes R=1.8 Ohm; this should be rated at 1.25*0.7--> say about 1W. Big but manageable.

The LM317 will also dissipate about 3V (if that is the head room you are keeping) * 0.7A (the design current)= about 2W; you will need a small heat sink for the regulator.

You tube: they are going to work but I do not like a design that depends critically on the transistor hfe. If you are going to expose both sides, you must ensure that both sides have the same current and same light intensity and both sides should be controlled with one timer and one intensity control.

 

[Xenobius]

AHA that was the last bit I did not understand. Thanks very much Klaus so its possible! 1 or 2w is nothing. I was wrong thinking it was the complete voltage.... I should have known that.