LED Vf variances

[oharag]

I'm a mechanical by nature trying to design an LED circuit. I'm going to have 22 parallel-series circuits with 7 LEDS in each series.

I'm designing for a 24 constant voltage PS. The amperage is 3.3 A (22 parallel chains each requiring .150 A -> 22*.150 = 3.3 A).

Each LED is nominal 3.2 Vf - so 7*3.2 = 22.4 volts required per series circuit. Each LED requires .150A.

Calculate resistor to place in line:

24 volts input - 22.4 volts required = 1.6 volts to disappate.

V=IR -> 1.6=.150A*R -> R=10.67 ohms.

My question is:

What happens if my LEDS come in with a higher Vf? With only 24 volts supplied - and with some being disappated by the resistor will be there be a possibiliy that a chain will not light? There is a maximum Vf on these chips of 3.6 volts. I ordered these to a specific color temp - but no binning on Vf.

I appreciate your responses. I believe I'm being overly concerned. But I need some assurances before sending this out to a board house.

 

[CKOD]

Just double checking, it appears that youre putting the 10.x ohm resistor in series with each chain of LEDs? If so, thats probably the best layout to use. You shouldnt have any strings which dont light, as the Vf is a value at a test current, if a LED has a higher then normal Vf, then with your nominal 24V applied, then less current will flow.

Since your chains are decently long (7 diodes) that should provide some averaging, so a single diode with a high Vf shouldnt hose stuff up too bad. And if youre ordering the diodes all at once, they should be from the same batch, so hopefully the Vf is similar between the devices.

Worst case, you get a string with a Vf lower then the rest, and that string is brighter then the rest of the strings. As long as the current isnt excessive, it shouldnt be a problem other then not looking nice. Vice-versa for a higher Vf.

 

[Freeze_XJ]

You've already done most of the math, all you need to do now is connect the dots.
Assume that you have a faulty LED that only takes 2.2 V (or 0.0, as in : shorted), then your chain suddenly gets 2.6 V / 10.6 Ohm = 243 milliamp, which might be enough to one. With a sudden short, you'd get 450 mA, which should burn most non-power LEDS. However, since you're chaining LEDs, variation will average out (with the square root of the number of elements in your chain, in case you wondered), and most LEDs are binned quite tightly on Vf, so i think you're OK.
If you really want to play it safe, take a bigger resistor, and waste more power during operations... Or just check with a multimeter in the loop, but that probably is over the top. I'd just go for it, the little LEDs are cheap enough to waste some in case of errors.

ps. in case of higher Vf, the string would just go a bit dimmer. A multimeter could tell you what's happening then, and you can swap some LEDs with other strings, or adapt the resistor at will.

 

[CzarDestructo]

As it's already been stated, although most companies won't give you Vf binning, for the most part LED come pretty tightly packed around nominal. However keep in the back of your mind worst case is the upper region of the voltage bin X 7 which could exceed your power supply voltage. Also Vf increases with current but DECREASES with temperature. The reason I bring this up is since you're only using a resistor to balance the current, if one of the many strings you have has significantly less forward voltage it will sink most of the current, self heat, further lower its voltage and you'll end up with thermal run-away until something pops.

Read here:
http://www.ledsmagazine.com/features/4/8/1

So to indirectly answer your question you should be more concerned with the lower voltage string. In the case of the higher voltage string, lets say the LEDs worst case use 23V, that only leaves 1V at the resistor which will stifle that string (1V divided by the resistance). On the other hand the BEST string will begin eating all that extra current*.

I'd recommend some constant current circuitry in this system for current balancing if you value the LEDs -or- some adequate heatsinking and the understanding that you might not get extremely uniform light output.

Godspeed.

 

[MikeAusC]

. . . .On the other hand the BEST string will begin eating all that extra current*. . . . .

Why ??? If he's using a constant-voltage supply, the other six strings will be totally unaffected by current variations in one string.

 

[smpowell]

......design an LED circuit. I'm going to have 22 parallel-series circuits with 7 LEDS in each series.

I'm designing for a 24 constant voltage PS. The amperage is 3.3 A (22 parallel chains each requiring .150 A -> 22*.150 = 3.3 A).

Each LED is nominal 3.2 Vf - so 7*3.2 = 22.4 volts required per series circuit. Each LED requires .150A.

Calculate resistor to place in line:

24 volts input - 22.4 volts required = 1.6 volts to disappate.

V=IR -> 1.6=.150A*R -> R=10.67 ohms.

My question is:

What happens if my LEDS come in with a higher Vf? With only 24 volts supplied - and with some being disappated by the resistor will be there be a possibiliy that a chain will not light? There is a maximum Vf on these chips of 3.6 volts. I......

I appreciate your responses. I believe I'm being overly concerned......

You have a poor design that depends on luck to work properly.

First, the Vf of an LED isn't fixed, will vary with current and the temperature of the LEDs. Lets do a couple of quick rough calculations:

If all the LEDs have a Vf of 3.6 volts at 150 mA, then seven is series will have a Vf of 25.2 volts. Since you only have 24 volts, the current through the string will drop until the sum of the 8 Vfs equal 24 volts. So the string will be dim.

But what happen if you get a shipment of LEDs with Vfs of 3.0? Seven of them in series will have a Vf of 21 volts at 150 mA. But to get a 3 volt voltage drop across an 11 ohm resistor you need 273 mA. (Since Vf increases with current and drops with increased temperature, it's probably not going to be that exactly)

If the LEDs, power supply, and application can tolerate that kind of current variation, then fine. If not, you need to redesign.

For example, if you used 6 LEDs in each string and a high value resistor, the potential current variation would be a lot less. And if you put 6 LEDs in each string with a constant current driver instead of a resistor, it would be a lot more stable.

 

[kethd]

This simple design could work. But you need good heatsinking on LEDs to prevent thermal run-away. It would be wise to put a fast-blow fuse in each series string in case an LED fails short, in which case you will get domino failure.

If this were my design, I would never just populate with random parts. If you must populate the LEDs without pre-testing for Vf, I would definitely start with a high series resistance, and then custom-tune the series resistance in each string -- not that much extra work.

Be aware that a "24V" power supply is not usually exactly 24 volts. There are many variables!

 

[AnAppleSnail]

Why ??? If he's using a constant-voltage supply, the other six strings will be totally unaffected by current variations in one string.

That's a big IF. And the Vf of all the strings will equalize by changing the current through each string.

 

[CKOD]

That's a big IF. And the Vf of all the strings will equalize by changing the current through each string.

I'm designing for a 24 constant voltage PS

not so much.

 

[MikeAusC]

The interior lights in my car work this way - each string has 3 LEDs in series plus a resistor, with several of these strings in parallel.

The supply is far from regulated - 12.5 to 14.5 volts - and lots of spikes no doubt. The temperature varies from freezing cold to painfully hot.

They just keep working - no noticeable variation in brightness between strings or within strings.

These are budget lights from DX - I doubt that any matching gets done - though I assume all LEDs in a light are from the same batch.

 

[CzarDestructo]

Why ??? If he's using a constant-voltage supply, the other six strings will be totally unaffected by current variations in one string.

Because if you have 24V across all the strings of LEDs and one string only wants 15V then 9V is across the resistor which is essentially the only current controlling item in the circuit, therefore the current gets bumped on that string.

edit: Oops, you're right that quoted fragment had me confusing CV and CC. Either way you'll have wildly different current draws on each string and still have the potential for thermal run away due to Vf. Resistors controlling a CV system isn't fantastic but it can get the job done sometimes.