Messed up with LED driver calc, can't think, brain to mush mode....

[Rockingit]

Hi Troops,

not so much an electrical issue as an electronics one, just realised that I've dropped a hairy one with making some crimbo led display lights (for me, for home). LED driver I got isn't up to the job, and my brain has ceased to function (mixture of cider and soldering fumes).... 10 x varied 105-130mW 5mm leds @ 3.7vf - 4.2vf in series. What size driver/s both current and voltage?

R.

 

[spark 68]

You need to provide a bit more information, ie. what is the current requirements for the LED's ?

see this: LED series parallel array wizard

 

[topquark]

If we say average of 120mW (x10) @4V, that's 1.2W @ 4V = total current required of 0.3A. You'll need to adjust to fit the driver voltage but LEDs aren't too fussy.

 

[spark 68]

Hi TQ,

If he is putting them in series though, he will have to allow for the volts drop for each one, ie. if they have a VD of 3.6V each he will need to push at least 36 Volts through the string. (he has 10)

 

[topquark]

Hi TQ,

If he is putting them in series though, he will have to allow for the volts drop for each one, ie. if they have a VD of 3.6V each he will need to push at least 36 Volts through the string. (he has 10)

Quite right spark 68 (I've had a few beers ). As they're in series it's the VD that's important, the current will be the same through them all.

 

[topquark]

How about one of these guys: Buy External Power Supplies Power Supply,enclosed,IP67,36V,1A,36W RS LPV-35-36 online from RS for next day delivery. (suitable for outside use as well).

 

[spark 68]

Hi again,

he will need a current limiting resistor still, and some of his LED's have a VD of 4.2 Volts.

The formula for this is: R = (supply Voltage - Total VD)/ the required current.

Eg. for one LED on a 5v Supply, assuming a VD of 1.7V and requiring a 100mA is
R = (5-1.7)/0.01A = 3.3/0.01 = 330 so current limiting resistor = 330 ohms in this case.

He would have to add all of the VD's together and workout what his max mA is (this would be for the lowest rated LED) and then find a PSU which exceeds this margin by a couple of volts (at least) and then calculate the resistor (to the nearest preferred value), not forgetting it's wattage.

I hope this helps.

 

[Rockingit]

Cheers gents, mucho appreciated. I did use to know this stuff, a loong time ago, scary how you forget it if you don't use it.

The driver I got (maplins) is an auto sensing constant current driver, 3 - 36Vdc, output power 1-8W and output current of 350mA. The data sheet does show an exampe wiring set of both 3 and 8 leds wired directly in series, no resistors needed. But, is this making an assumption that you're using 350mW leds??? Have attached it.View attachment plc35xx_And_plc70xx_-03.pdf

Have I either grossly under calculated the demand, or do I just need to wire them as 'series - parallel pairs/triples' to make up my o/p current?

I've a suspicion that I may have accidentally fried a couple!

I've four leds which are:

DC forward voltage:	3.6V typ (4.0Vmax)
DC forward current	30mA
DC reverse voltage:	5V(max)
DC reverse current:	50 uA(max)
Power dissipation:	120mW
Optical output power:	6.0 mW typ.

two are:

Forward voltage at If=20mA:	4.0V
Forward current max.:	30mA
Reverse voltage max.:	5V
Power dissipation PT:	105W

four are:

If max = 100mA
Vf max = 3.6Vdc
Power dis = 100mW

 

[topquark]

Hi again,

he will need a current limiting resistor still, and some of his LED's have a VD of 4.2 Volts.

The formula for this is: R = (supply Voltage - Total VD)/ the required current.

Eg. for one LED on a 5v Supply, assuming a VD of 1.7V and requiring a 100mA is
R = (5-1.7)/0.01A = 3.3/0.01 = 330 so current limiting resistor = 330 ohms in this case.

He would have to add all of the VD's together and workout what his max mA is (this would be for the lowest rated LED) and then find a PSU which exceeds this margin by a couple of volts (at least) and then calculate the resistor (to the nearest preferred value), not forgetting it's wattage.

I hope this helps.

Hehe, thanks for the informative post. I did realise that.

 

[topquark]

Hi rockingit. Yes, that'll be making that assumption.

You'll first need to work out how you are going to wire them before you can work out what to drive it with (I missed the series specification in my drunken stupor). Are you happy to do the calcs for the limit resistor then or do you want it done (it's fairly trivial)?

 

[topquark]

Just totted that little lot up and you'd need 36.8V to drive them all in series

You could arrange them in to a number of parallel strings, so you'd then have a current limiting resistor for each string. It would depend on why you wanted them in series to start with I guess.

 

[spark 68]

Hi TQ,

Just totted that little lot up and you'd need 36.8V to drive them all in series

He would need and extra volt or two on top of this, as that (36.8V) is just the VDtotal he would need to overcome before any current can flow, plus the resistor will drop some.

 

[topquark]

Yes, I know. The issue remains the same, the transformer he has is not up to the job to drive them in series. Hence the suggestion to arrange them in (parallel) strings.

 

[spark 68]

Hence the suggestion to arrange them in (parallel) strings.

Yes agreed,

The problem here though is the uneven max. current ratings of the leds, we would need to balance each string series (where possible) because some of the Leds are only 30mA, and some are 100mA.

so 30mA is the absolute max for some strings, which would lead to uneven brightness between the strings, we would ideally have the same mix in each string, or alternately calc for 30mA for each string.

The Vd is taken care of by the calcs.

 

[Rockingit]

Having mis-interpreted the driver info, I had in mind to just run all ten in series, but it's arbitrary how they get done so long as they all are on and the same brightness. I've soldered extra-long flying leads on to them at this stage (being mounted in a line in some 50mm box conduit) so I can mix and match the wiring to suit - there's 4 x blue, 4 x white and 2 x indigo in a pattern.

I've only just got back in from a heating call-out and will probably leave it now until tomorrow as I'm feeling cold and grumpy(!), but what I think I might do is just go and get another two drivers (£12ish) and be done with it, that way each set can be running at optimum brightness. I like those particular drivers as they're 45mm wide which means they fit snug inside the trunking, tidy tidy.

Again, appreciate the advice given from all, many thanks.

 

[Rockingit]

OK, a couple of days later and I finally get back on it....now have a new problem! Having gotten my head around the principle and the formula, I'm now stuck again - because - the led drivers I have are constant current at 350mW AND will auto-sense the required voltage between 3 and 36V. So what do I use as my start values for the maths to derive my resistor value?!

If, for example, I've got 4 in series each at 4v / 100mA, can I just assume a supply value of 17v (needing an extra volt or two to get things moving) and therefore a 12ohm resistor? There's not much info with the drivers.

 

[topquark]

You'd use the maximum output voltage 36V. The problem you may get though (which is why I don't like them in series), is when one blows the circuit will fail (and you'll need a difference resistance to get it started again, having replaced). It's much easier to drive devices like this in parallel....with the same voltage across them (and possibly a small resistor if you're dealing with unequal forward resistance, just to make them all "appear" the same). They are often done in strings or arrays, to even out the voltage drop across them.

Say if you had four 4V ones and two 3.6V, then you'd arrange them as two strings of 2x4V and 1x2.6V with say 12V across them and a limiting resistor for each string. You would then base your calculations on using the 12V source.

What you can get with series is the problem when one blows they all blow as well (as the voltage drop across the first to fail will decrease during failure).

 

[Rockingit]

What you can get with series is the problem when one blows they all blow as well (as the voltage drop across the first to fail will decrease during failure).

Yep, just did. Beginning to wish I'd never started this ****ing thing now. Bah!! Humbug!!!

 

[topquark]

Unlucky, if I hadn't mentioned it, it prob wouldn't have happened (so blame me ). Stick with it! (a lot easier in parallel though, did I mention that? ).

[FONT=Verdana]6V _____________________________________[/FONT]
[FONT=Verdana]       |[/FONT]
[FONT=Verdana]       /[/FONT]
[FONT=Verdana]       \  <--- resistor value to match with the LED forward voltage to 6V, repeat in parallel for the next LED.[/FONT]
[FONT=Verdana]       /[/FONT]
[FONT=Verdana]       |[/FONT]
[FONT=Verdana]      LED[/FONT]
[FONT=Verdana]       |[/FONT]
[FONT=Verdana]0V__|___________________________________
[/FONT]

 

[Rockingit]

IT IS FINISHED!!
Can you tell I'm happy?!

And more's the point, Mrs Rockingit is pleased with the results, so I'm in with a chance.

Right you lot, Merry Crimbo to you and your's, I'm off to drown in various pleasant intoxicating liquids for a few days, see you when I surface!!

R.

 

[topquark]

Happy unwrapping