Okay, here's the story. I bought the Christmas LEDs from a garage sale
with ten missing LEDs. The guy that sold it to me said his brother took
ten of the super bright LEDs out to make a project. So he gave me a deal
and sold a Christmas light that is half working. As in 30 LEDs are lit but
the other 20 are not because they are missing ten.
The original question asked for the resistor value to be put in place of the
missing 10 LEDs. Someone answered 45 megohms. So that means
I need 10, 45 megohms resistor, correct?
---
Not correct.
---
Now in order to get the other 20 LED working again I will cut and splice
the wire and insert just 1 resistor in series to the LEDs. I'm not sure what is
the correct value would be or even sure if the 120V will affect the 20
LEDs which now wants 80VAC. If the LEDs are current dependant,
what would be the correct resistor value?
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It would seem that the strings are wired in anti-parallel and that
with 40 in each string the voltage across any single LED would be
120V 3V
------ = -----
40LED LED
If that's the case, then the 10 LEDS in the non-working string would
drop:
3V
----- * 10 LED = 30V
LED
Assuming the LEDs are passing 20mA means that the resistance you
need to simulate the stolen LEDs is:
E 30V
R = --- = ------- = 1500 ohms
I 0.02A
and the power it will need to dissipate will be:
P = IE = 0.02A * 30V = 0.6 watts,
so you should use a 1500 ohm one watt resistor if you want to use a
single resistor. Or two, 3000 ohm 1/2 watt resistors in parallel. Or
two, 750 ohm 1/2 watt resistors in series. Those are all standard
5% values, so you should have no trouble finding them.
You could also bridge each socket with a single resistor per socket,
in which case the value would be:
E 3v
R = --- = ------- = 150 ohms
I 0.02A
and the power dissipation would be:
P = IE = 3V * 0.02A = 0.06 watts,
so standard 5%, 150 ohm, quarter watt carbon films would be fine.
I'd prefer that solution becasuse the resistors will barely get
warm.
Finally, since there will be no current flowing in the string with
the resistors when the other string is lit, the resistors will drop
no voltage and about 170V will appear across the 30 LEDs, which is :
Vt 170V
Vr = --- = ------- = 5.7V per LED
n 30LED
I don't know if that's higher than the spec for the lamps, but just
in case it is, I'D go ahead and put a 1N4005, 4006, or 4007 in
series with the resistors in that string, with the anode facing in
the same direction as the LED anodes.
Matter of fact, it wouldn't hurt to put one in each string to take
the reverse load off of _all+ the LED's, like this:
View in courier
MAINS>---+---------+
| |
|A |K
[1N4007] [IN4007]
| |K
[R1] [LED]
| |K
[R2] [LED]
| |
. .
. .
. .
|A |K
[LED] [LED]
| |
MAINS>---+---------+
