I was given a PCB which has had all the components removed and I want to resurrect it.

Essentially, it's a stop/tail light.

It uses 45 LEDs arranged in 3 groups of 15.

Like this...

(Full stops just to make it all line up. 12v input on left)

|--->--->--->--->--->--|

|. . . . . . . . . . . . . . . . .|

|--->--->--->--->--->--|---/\/\/\--- V-

|. . . . . . . . . . . . . . . . .|

|--->--->--->--->--->--|

12v supply, 2 v forward voltage and forward current 20mA, I know the resistor value for each group of LEDs would be 68 ohms. (this will be the stop light at full brightness)

But the PCB uses 1 resistor for the whole group of 15.

Anybody have the formula for calculating this resistor value, or could work it out please?

If you're feeling particularly generous, I also need a resistor value for the tail light (I'm guessing 1/3rd brightness)

Many thanks in advance.

Is your diagram one group of 15, arranged as 3 parallel strands of 5 diodes in series? With one resistor for the lot, on the right?

 

If so, I think your calculation should be R = (12 - (5 x 2)) / (3 x 0.02) = 2/0.06 = 33 Ohms. I get that by saying supply voltage minus 5 lots of Vf drop, all divided by 3 lots of 20mA, one for each strand.  It's not a nice arrangement, because there's not much to control how the current is shared between the 3 strands. 

 

The voltage feed will typically be nearer 14V if your charging system is working OK, so you may want to alter that sum to 4/0.06 or 66 Ohms, per group of 15. 

 

There won't be anything like a linear relation between resistor value and brightness, so choosing values for 1/3 brightness would be a matter of trial and error.

Be aware that - as it is a very basic system, as LEDs fail, it will put extra strain on the remaining LEDS, causing them to fail faster - until a cascade effect takes all the remaining lights out in quick succession.

 

You would be advised to AT LEAST fit one resistor per row, although 1 per LED would be best; and a zenner diode set up making sure the voltage fed to them was stable and not prone to any peaks.

Thanks both of you.

I knew someone would know the answers.

Wino, that's correct. 3 parallel strands of 5 LEDs in series.

Would it be better to play safe and go with a 100 ohm as I've used before with 5 LEDs in series?

GG, there are 2 Zener diodes marked on the board. I assumed they were to allow the stop and tail circuits to work independently.

They both use the same tracks and there are 2 12v inputs (1 for the tail light and 1 for stop light) and 1 v-

From what you say, they are also to provide a stable voltage.

What value would you advise?

Zenners are used to control voltage, they block upto a set limit, then allow current to pass. Anything wired in parallel to the zenner will "see" the stable voltage and excess voltage will be fed through the zenner and a load resister.

 

If the board already has them marked, you probably need 12V @ a suitable current rating; if you know the total current draw of the circuit, pick a zenner diode a bit higher, so if it draws 100mA, you want 120-150mA. Any higher current rating that will fit on the board is fine, but the voltage rating must be exact.

100 Ohms would be a good choice, I reckon. 

 

Got a photo of the board? Both sides ideally, so the tracking can be seen.

Not the best pics, but I hope you can see enough detail

I'm trying to rack my brains on how to work out the values needed. I'd suggest that on a car, the Zener voltage would be a lot lower than 12v, to allow for lower voltages.

Then the considerations are the two extremes -no load and full load.

Then first of all you need to calculate the maximum current that the Zener can safely carry, using the formula Watts /Volts ( both of zener ) = maximum current.

For no load conditions , start with something like at least 1/10 of the max. Then using Ohms law ,work out the value of the series resistor ( R= (Input volts - zener volts)/ current (in this case that of approx 1/10 of max). Don't forget that below a certain current, the zener effect is not noticed .

Then you must consider the voltage at the Zener input when circuit is drawing maximum load ( and hence voltage across zener is at minimum.Too low and Zener is not regulating). Then using the other power calculation ( Watts = current squared times resistor value) , you can calculate the power rating of the resistor. 

 

It's not that difficult, but may require a bit of component value juggling. Hope it helps.

Edited 10 May, 201511 yr by VWD

I was working on the basis that the zenners would regulate the voltage and use the resistor values to set how much the LED "saw". We dont know the exact voltage drop required for the LEDS, but assuming it is 1.5v, the supply needs to be about 7.5v for full power across 5 LEDs, plus whatever the resistor will use up.

Not the best pics, but I hope you can see enough detail

Need a better picture of the tracking really. I can't make out what's going on with the diodes from that, is there a position for a series resistor for each diode? It looks like R4 might be such for the North-South oriented diode (D2 I guess, but  can't make it out). The left hand end of D1 is all a blur to me, so I can't see what it's connecting to.

Here's a couple of links you may find useful ...

 

http://www.evilmadscientist.com/2012/resistors-for-leds/

 

http://www.instructables.com/id/Choosing-The-Resistor-To-Use-With-LEDs/

 

https://www.sparkfun.com/tutorials/219

Thanks all, for your input.

I think I have it sorted out now.

680 ohm resistor for each set of 15.

A 1.2 Kohm for the tail light, and a couple of 1N4007 ? Zener diodes.

Blimey, that's a much higher value resistor than I would have expected per 15; was it just too bright? Or did you mean 68?

 

1N4007 is just a rectifier diode, not a Zener, so no voltage regulating function? Reverse polarity protection maybe?

Wow.. Over my head is this...

 

Good luck CFB

Blimey, that's a much higher value resistor than I would have expected per 15; was it just too bright? Or did you mean 68?

 

1N4007 is just a rectifier diode, not a Zener, so no voltage regulating function? Reverse polarity protection maybe?

I haven't built it yet so don't know if the 680s are too much.

That value was worked out for me by a chap on an electronics forum.

I'd rather start with too high a value than have the whole thing go pop!

I think I'll breadboard a set of 15 to double check.

The diodes will be to stop reverse polarity as you suggest, as the stop and tail positive feeds are at different points on the PCB.

Trying to find a cheap source for reliable LEDs now.

Got a link to his working out?

 

Sensible to be cautious, but I think you'll find that pretty dim. 

When it's powered up/lit up, measure the voltage across the 680 Ohm resistor (i.e. put one meter probe at each end of it); divide that voltage by 680, and you'll have current in Amps. Divide that by 3 and you'll have current in amps per string of 5 LEDs. Compare that with datasheet ratings for the LED type you've used.

 

if you have 2V forward voltage per LED, then 5 in series would have a potential difference of 10V, with 20mA of constant current flowing through them.

 

you have 3 of these series networks. so across each branch of 5 LED's there is a voltage drop of 10V, but the total current going through them is 60mA (20mA in 3 paths), this 60mA would then also flow through the series resistor.

 

So if your powering it from 12V, you know where 10 of those 12 Volts are dissipated, so that leaves the remaining 2V to be dissipated at the series resistor.

 

Apply Ohms Law, Voltage/Current = Resistance. 

 

Therefore Desired Resistor is 2 Volts / 60 mA = 2 Volts / 0.06 Amps (use SI units) = 33.33 Ohms per each 15 LED group.

 

so a 33 Ohm resistor would be fine. Cheap ones tend to be +/- 10% so will be close enough for this.

 

 

Lower resistance will mean brighter but will burn out the LED's, Higher resistance will dim them.

or did i miss understand the question?

 

Cheap ones tend to be +/- 10% so will be close enough for this.

 

 

I havent seen a +/- 10% resistor since college 30 years ago; I am pretty sure the only place you will find them (other than colleges), is in junk yards - everywhere else 1% will be the MINIMUM, and many places selling 0.1%.

or did i miss understand the question?

I think you understood, calculated well, and explained your working out nicely.

 

Only thing I did differently (see post #2) was to allow for the fact that car '12V' is generally closer to 14 when all is well with the alternator and battery (and engine is running).

 

0.7V (guestimated) additional Vf drop across the polarity protection diodes would knock this down to 13-ish, so probably not worth quibbling about.

Got it slightly wrong. My helper actually quoted 600 ohm, not 680, but you've now got me questioning his calculations.

Quote from his workings is below..

Here you can see that, if V+ is 12V then as per Ohm's law you need R1-R3 to be 600 ohms each if you want only ~20mA to flow through the resistors. (Formula: R=V/I)

He doesn't take into account the forward voltages of the LEDs, and assumes 20mA current for the whole group of 15, rather than 20mA through each strand of 5 LEDs (your definition of the problem was slightly ambiguous in this respect, but 20mA is a pretty common choice of operating current for an LED).

 

Do you know exactly which LED type you'll be using?  I recommend something with a reasonably wide viewing angle.

Wino, I want to use 30* angle of view, superbright red.

They are 2v forward voltage, 20mA forward current.

I have done a bit more research and I reckon the guys calculations were way out.

My thinking...

5 LEDs in series. Voltage drop would be 12v - (2x5) = 2v

V/I 2v / 20mA gives a resistor value of 66.66 ohms

For the whole array of 15 LEDS...

Assuming the the voltage drop is the same for each series branch and current is also the same, as the 3 branches are in parallel then..

Total current would be 3x 20mA = 60mA

V/I 2v / 60mA gives a resistor value of 33.33 ohms

WTF? So which is correct? Logic says the above but the previous figures were given by a supposed electronics wizard.

2/0.02 = 100, not 66.66, but other than that, you've got it.

2/0.02 = 100, not 66.66, but other than that, you've got it.

Doh!