[Empeg-forsale] Re: I WANT THE CLEAR BLUE BUTTONS

[mtempsch@empegbbs-noreply.merlins.org]

The two cases are not the same, as in the case with a single resistor the voltage across all LEDs will be forced to be identical, as the anode (+ end) are all tied together at the resistor and the cathode (- end) are tied together at  ground) but with the individual resistors it can differ as the anodes are separated from each other. 
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Say we have a 5V supply voltage, nominally 2V LEDs @ 10mA current through each LED.

Case 1, One single resistor

Total current through resistor is 40mA -< R = (5-2)/0.04 = 75 Ohm
All LEDs will have the same voltage over them 5-(75*I_total)  (75*I_total is the voltage dropped over the resistor)
Unfortunately 2 of the LEDs aren't 2V @ 10mA, instead one is 1.95V @ 10mA which might translate to 2V @ 18mA and the other is 2.05V @ 10mA which might translate to 2V @ 2mA (note: using canned numbers to simply total math, and due to being to lazy to dig up a real LED UI graph). 

Now, the LED with 18mA going through it will be noticeably brighter than the one with 2mA going through it.

Note: In reality differences might be too small to notice.


Case 2, individual resistors

Through each resistor we have a current of 10mA and we need to drop 3 (5-2) volts over it -< R = (5-2)/0.01 = 300 Ohms (which is 4x the value of case 1, surprise surprise...)

Now, for the 1.95V @ 10mA LED there will be a slighty larger voltage drop across the resistor (worst case: assume all the extra 0.05V ends up at the resistor; in reality it will be divided over LED and resistor)
I = U/R = 3.05 / 300 = 10.16mA

Which is a far slighter difference than we could get in case 1.

Same sort of calculation can be made for the 2.05V LED -< I = 9.83mA

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Another point is if one LED were to fail open - due to a bad solder joint or whatever.

In case 1 we need to drop 3V over a 75 Ohm load. That was accomplished with a 40mA current that now needs to go through 3 LEDs, ie 13.3mA (in reality slightly less, as the voltage over the LEDs would increase slightly) each instead of 10mA. Not that that's likely to hurt them in any practical way, but still.
In case 2 we'd just get a non-lit LED, the others would remain unaffected,

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All in all, it's a better design to use individual resistors, but LEDs of today are probably consistent enough for it not to matter... (famous last words... )

/Michael