# 126 led lamp

#### alank878

Joined Feb 21, 2016
5
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 #### dl324

Joined Mar 30, 2015
16,206
Welcome to AAC!

Did you intend to post a picture and a schematic?

#### alank878

Joined Feb 21, 2016
5

#### ScottWang

Joined Aug 23, 2012
7,389
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.

#### alank878

Joined Feb 21, 2016
5
Hi scott,

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

#### ScottWang

Joined Aug 23, 2012
7,389
Hi scott,

maybe this is why the resistor is burned? so you say this resistor should be changed for a 60 ohm part?
The 60 Ω just guessing can calculation, do you remember how many hours do the leds can be light up continuing?

#### alank878

Joined Feb 21, 2016
5

#### ScottWang

Joined Aug 23, 2012
7,389
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?

#### alank878

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

#### ScottWang

Joined Aug 23, 2012
7,389
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%