# Need to solve missing LED in Christmas light.

Discussion in 'Electronic Basics' started by Sam Nickaby, Feb 27, 2006.

1. ### Sam NickabyGuest

James has a 60 LED Christmas Lights made by Phillips. Each LED takes
4-Volts . They run on 120-Volt AC.

Question: If James brother steals 10 LEDs from the 60 LEDs, James now
has 50 LEDs. What is the value of the resistor in Watts and resistance
that James should put in place of the missing LEDs in order to prevent the
LED from burning out?

Thanks

2. ### Tim WilliamsGuest

Does the textbook mention the current drawn by the string?

You may be interested in Ohm's law, to finish this one.

Tim

45 megohms.

John

4. ### ehsjrGuest

Ahhhhh - but what value do you use if his *sister*
steals the bulbs?

Ed

6. ### PeteGGuest

60 leds x 4 volts is 240volts so they never worked.

If they're connected as two sets of 30 leds wired back to back so they
conduct on alternate halves of the AC cycle which you don't say but it's
kind of implied since 30 leds x 4 volts = 120 then you need to know did his
brother take 5 leds from each leg? If he didn't you need two resistors.

And whatever resistor(s) you put in it still won't work because there are 10
open connections were the leds have been removed

Pete

7. ### redbellyGuest

That wouldn't work either. A 120 Vrms signal has an amplitude of 170
V, which will fry the LED's.

Mark

8. ### PeteGGuest

Yeah I'd thought of that too although it doesn't say whether it's 120V peak,
or 120V rms so we assume rms because that's what we'd expect in the real
world. The whole scenario as he has presented it here is flawed.

I don't like contrived 'real world' type questions for this reason.

They'll probably cover AC peak and rms voltages in next weeks lesson.

Pete

9. ### James DouglasGuest

None, just go to WalMart and get another one for \$1.99. I know this was
a trick question.

10. ### redbellyGuest

The thing is, even a question for a basic electronics class homework
wouldn't be flawed like this. If it was a troll post, I guess I've
been hooked. Oh well.

Mark

11. ### Sam NickabyGuest

There are two series strings of LEDs driven at 20 mA should draw 40 mA
from the source, and 120V * 40 mA = 4.8W. Only half the LEDs are on at
once, and they are not lit for a full 50% duty cycle. 60 LEDs, 115VAC, 60Hz
and 40 mA appears to add up to two series strings of LEDs.

The two strings of 30 series LEDs connected in opposite polarity across
the AC source, so that only 30 of the 60 LEDs are lit at any given time.
Of the 30 LEDs, 10 are missing. What resistor value should be used to
keep the 30 LEDs from burning out?

12. ### Sam NickabyGuest

I thought they sell them for \$9-12. I got mine from a garage sale for
\$2. I'd just figured out how Phillips hooked up the lights. The approach
takes two strings of 30 series LEDs connected in opposite polarity
across the AC source, so that only 30 of the 60 LEDs are lit at any
given time. There's a certain elegance to the minimalism of that, but
it produces a perceptible flicker in the lights that look the Twinkle
Twinkle Little Star in the Wizard of Oz.

13. ### Sam NickabyGuest

What resistor value should be used to keep the *20* LEDs from burning out?

14. ### ehsjrGuest

You do not understand the question or the circuit.
The proper question is "What resistor value should be used
to keep the *30* LEDs from burning out?" You had it right
the previous time.

The 20 LEDs *cannot* burn out - no voltage is applied to
them, due to the missing LEDs.

If you tell us what you are really trying to learn,
perhaps we can help.

Ed

15. ### Sam NickabyGuest

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?

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?

16. ### John FieldsGuest

---
Not correct.
---

---
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>---+---------+

17. ### John FieldsGuest

---
Oops... I misread it as 40 LEDs in the working string.
Corrections follow:
120V 4V
------ = -----
30LED LED
4V
----- * 10 LED = 40V
LED
E 40V
R = --- = ------- = 2000 ohms
I 0.02A
so you should use a 2000 ohm two watt resistor if you want to use a
single resistor. Or two, 3900 ohm 1 watt resistors in parallel. Or
two, 1000 ohm 1 watt resistors in series. Those are all standard
5% values, so you should have no trouble finding them.
E 4v
R = --- = ------- = 200 ohms
I 0.02A
P = IE = 3V * 0.02A = 0.08 watts,
no voltage and about 170V will appear across the 20 LEDs, which is :

Vt 170V
Vr = --- = ------- = 8.5V per LED
n 20LED

I'm sure that'll be higher than the reverse voltage spec for the
lamps spec for the lamps, so what I'd do would be to 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_ of the LED's, like this:

18. ### redbellyGuest

No, that was not a serious answer. 45 M-ohm is way too high. That
poster suggested it to insure that the LED's do not burn out, which was
the question you were asking at that time. Unfortunately they will not
light up with 45 Meg.

Some information is still missing. There should already be a resistor
in the circuit, based on having 30 LED's. You need to know how much to
ADD to that resistance, NOT how much total resistance is needed for 20
LED's.

Mark