Thoughts on a simple circuit failure

Thread Starter

Jwooky

Joined Mar 16, 2019
4
hello newby here with a little problem I’m scratching my head over.

I am wiring up a custom set of LED lights. The intent here is to have a running voltage that is lower than when the Brake switch is closed and then would apply full 12 volt battery power.

Everything was working fine till I repeatedly cycled this switch and then the circled resister blew. (Literally caught fire).

It seems to me that when the switch is closed, I would have equal 12v on either side of this resister, so why would it have an issue?

I am thinking I will add a diode before the resistor so the 12v Brake voltage will not have a chance to flow back through it, but I guess I’m not sure why I have to.

Thanks for your thoughts.

2D175CD0-AC0D-4B52-AD1D-EE8F301B9429.jpeg
 

jbeng

Joined Sep 10, 2006
72
Sound to me like the resistor was physically too small. What wattage device were you using? Also, since it's not shown in your diagram, you do have current limiting in place for the LEDs, right?
 

dendad

Joined Feb 20, 2016
3,426
Something like this (L/R switches not shown)
BrakeLight.jpg

Calculate R2 for standing current and then R1 for (full curent - standing current).
The diodes stop the current feeding back.

LEDs are current devices, not voltage. They do have a aprox 2V drop for each LED but you MUST include current control , like a suitable resistor or a constant current supply.

This will give around 20mA current for the LED.
 

Thread Starter

Jwooky

Joined Mar 16, 2019
4
First to directly answer your questions. This is a 1/4 W resistor.

The led strip I am using has 3 3528 SMD LED’s in series with a resistor strung together. I have ~ 10-15ft of them, which is ~500 diodes.

I just ran a quick test, the resister is getting extremely hot, to the point you can’t touch it, without activating the switch.

It is very apparently undersized.

Recommendations on sizing it? Doing some quick checks, 10 ohms gives me the brightness I want.
 

Tonyr1084

Joined Sep 24, 2015
4,369
Assuming automotive voltage of 13.8, which is typical, a 47Ω resistor means a current of 294 mA (0.294 amps). That times 13.8 = more than 4 watts. It's no wonder a quarter watt resistor would burn out. Since you don't tell us the forward voltage of your LED's I can only guess you're still very close to 4 watts.

13.8 v ÷ 47Ω = 0.294 A
0.294 A x 13.8 v = 4.05 watts.

4 watts on a 1/4 watt resistor = POOF!

AT THE VERY LEAST use a 5 watt resistor. Better still, use a 10 watt resistor and you won't have issues with heat. No matter how you look at it you're still generating a lot of heat. With a larger resistor you have greater surface area to dissipate that heat.
 

mvas

Joined Jun 19, 2017
537
Assuming automotive voltage of 13.8, which is typical, a 47Ω resistor means a current of 294 mA (0.294 amps). That times 13.8 = more than 4 watts. It's no wonder a quarter watt resistor would burn out. Since you don't tell us the forward voltage of your LED's I can only guess you're still very close to 4 watts.

13.8 v ÷ 47Ω = 0.294 A
0.294 A x 13.8 v = 4.05 watts.

4 watts on a 1/4 watt resistor = POOF!

AT THE VERY LEAST use a 5 watt resistor. Better still, use a 10 watt resistor and you won't have issues with heat. No matter how you look at it you're still generating a lot of heat. With a larger resistor you have greater surface area to dissipate that heat.
These are 12 Volt LED's
3 @ LED's in series + the built-in in Current Limiting Resistor of ~100 Ohms
So, don't you need to subtract the Voltage Drop across each LED Module =
the sum of ( 3 @ LED's in series + the built-in in Current Limiting Resistor ) ?
12 Feet is about 72 modules in parallel.

I am going to "guestimate" ~5 Volts across the latest 10 Ohm Resistor ( per message #6 ) = 2.5 Watts of heat in the 10 Ohm resistor.
 

dendad

Joined Feb 20, 2016
3,426
One problem I see is the lack of blocking diodes so you may be back feeding to the supply circuits.
If the LEDs have built in resistors, the 47R in not needed.

But you need to make sure the brake light only feeds the lamps and LEDs, and not the "IGN" circuitry.
Likewise, the IGN does not try to power the brake. Add diodes to feed the LEDs. Don't just hook the IGN and BRAKE together as you are doing.

And, use higher current diodes that the 1A ones in my circuit.
 
Last edited:

Tonyr1084

Joined Sep 24, 2015
4,369
Whatever the sort of LED's, the TS wants them to be dim like driving lights but brighten when you step on the brakes. Whether it's 2.5 watts or 4 (or more - or less) the quarter watt resistor is going to burn up.

I don't know how they do it on cars that have LED's from the factory. Perhaps they use some form of PWM to dim them. Perhaps for full brightness they overpower them for a microsecond (again, I don't know) to gain brightness.

Extra resistance should work, but again, at a substantially higher wattage for resistance.

1/4 watt resistance is futile.
 

Thread Starter

Jwooky

Joined Mar 16, 2019
4
I measured the voltage after the 10 ohm resistor and it is 8.4V.

How would I properly calculate the wattage needed?

I have replaced with a 5W to try and it’s not getting hot at least.
 

dendad

Joined Feb 20, 2016
3,426
Current = Voltage/Resistance
I=E/R
I=(13.8-8.4)/10
I=5.4/10
I=540mA

then P=E*I
P=5.4*.54
P=2.916Watts.

Also, you can use..
P=(E*E)/R
P=(5.4*5.4)/10
P=29.16/10
P=2.916Watts.
 

Thread Starter

Jwooky

Joined Mar 16, 2019
4
Ah ok.

I started to go down this road and was questioning which voltage to use.

I stated with 12V/10 ohms=1.2 A

1.2Ax12V=24W that just seemed rediculously high.

Guess I’m still not clear on why. I think what I am missing is the resistance on the LED array which also has resistance.

Having said that....


R=V/I =12/.54=22ohms

Therefore the LED’s are 22ohms-10ohms=12ohms.

Correct?

Note...(I do realize we are using 13.8V and 12V for system voltage). The car is not running when I measured so it was at 12V. It could be up to 13.5-14V when running.
 

dendad

Joined Feb 20, 2016
3,426
You measured 8.4V on the resistor, and assuming it was 13.8V on the other side, there is 5.4V across tge resistor, not 12V (or 13.8V).
You do not use the 12V (or 13.8V for a charger car battery) as that is not what is across the resistor. You only use the supply volts total when the load resistor is across the full supply, like from 12V to 0V. In this case, the resistor is in series with other stuff, so only use the voltage that appears across the resistor. You can use your meter to measure the volts across the resistor directly. Just put one probe on each end of the resistor.
 

shortbus

Joined Sep 30, 2009
7,599
I don't know how they do it on cars that have LED's from the factory. Perhaps they use some form of PWM to dim them. Perhaps for full brightness they overpower them for a microsecond (again, I don't know) to gain brightness.
Pretty sure it is a PWM from the BCM, body control module.
 
Top