# 126 led lamp

Discussion in 'General Electronics Chat' started by alank878, Feb 21, 2016.

1. ### alank878 Thread Starter New Member

Feb 21, 2016
5
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Hi Guys,

I have an under bonnet lamp for working on my car, it seems to have a burned resistor (larger black one) am i correct in saying there should be 2 for this type of circuit? can any one tell me which size it should be so i can get a replacement?

Specs are

Battery Type 3.7 V 4400mAh Lithium-Ion
Brightness 700 lumens
Charge Time, hours 5
Charger 110 V AC 50/60 Hz
Charging Voltage/Current 4.2 V/1 A
Description 128 LED Under Hood Light
Discharge Time, hours 3

Any help would be great

2. ### dl324 Distinguished Member

Mar 30, 2015
3,120
603
Welcome to AAC!

Did you intend to post a picture and a schematic?

Feb 21, 2016
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4. ### ScottWang Moderator

Aug 23, 2012
4,782
762
If the Leds are in parallel directly then it is not so good for the Leds, assuming that the spec of led is 3V/20mA then the total current of 128 Leds is :
I = 20 mA*128 = 2.56 mA,
R = (4.2V-3V)/20 mA = 1.2V/20 mA = 60 Ω,
W = V* I = 1.2V*2.56A = 3.072 W
So the values of W should be at least 3 times of calculation and that is 9W, but if using more than 5 times is better and that is 15W.

But I prefer to use 80% of rating current as 16mA, the product didn't shows the current, now we only can guessing.

5. ### alank878 Thread Starter New Member

Feb 21, 2016
5
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Hi scott,

maybe this is why the resistor is burned? so you say this resistor should be changed for a 60 ohm part?

6. ### ScottWang Moderator

Aug 23, 2012
4,782
762
The 60 Ω just guessing can calculation, do you remember how many hours do the leds can be light up continuing?

7. ### alank878 Thread Starter New Member

Feb 21, 2016
5
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hi its around 3-4 hours max

8. ### ScottWang Moderator

Aug 23, 2012
4,782
762
I = 4400 mA/3 hrs = 1467 mA = 1.467A
I = 4400 mA/4 hrs = 1100 mA = 1.1A

If the battery can be used for 3 hrs:
R = V/I = 1.2V/1.467A = 0.82 Ω

If the battery can be used for 4 hrs:
R = V/I = 1.2V/1.1A = 1.09 Ω

Can you recognize the value of resistor?

9. ### alank878 Thread Starter New Member

Feb 21, 2016
5
0
hi scott, the resistor is un recognisable as its very burned, i will take a photo today and show you

10. ### ScottWang Moderator

Aug 23, 2012
4,782
762
I checked the spec and it showed that Discharge Time, hours : 3.
I = 4400 mA/3 hrs = 1467 mA = 1.467 A/hrs
If the battery can be used for 3 hrs:
R = V/I = 1.2V/1.467A = 0.82 Ω
W = V*I = 1.2 V*1.467 A = 1.76W
As above calculation, I assuming that the Vf of LEDs is 3V.

I_1led = 1.467 mA/128 Leds = 11.46mA, each led draw current 11.46 mA.
If the assuming is right then maybe could choose 1Ω/10W when the 0.82 Ω/10W is not easy to buy.

For the safety reason, if we assuming that the Vf of LEDs is 2V, and then
If the battery can be used for 3 hrs:
R = V/I = (4.2V-2V)/1.467A = 1.5 Ω
W = V*I = 2.2 V*1.467 A = 3.23W, so it may choose 1.5 Ω/15W

HS15-1R5J -- Resistor: wire-wound with heatsink; screwed; 1.5Ω; 15W; ±5%