Hacking cars LEDs 12V 47R internal resistor to 100mA max current limit ...

[szan]

Hello,
I bought such 12V LEDs in automotive cars case and after quick teardown I've discovered inside this thing ... 47R Ohm resistor  instead of any kind of current limit 

Not too much place there since diameter of this thing is around 19mm, but it looks like maybe LM317 LZ  (100mA max) based current limiter circuit in small TO-92 package with 15R Ohm resistor (it gives around 83mA output current) we were able fit there instead of this bloody single resistor to have more stable light output eg. when powered from 3 x Li-on 4.2Vmax in series  ?



Any other ideas ?!

[Benta]

Did you calculate the power dissipation in the regulator? If not, please do.

[szan]

Did you calculate the power dissipation in the regulator?

Nope, because of I need to know forward voltage of those LEDs at ~100mA current and didn't connected oryginal circuit with 47R to 12V before its teardown 
I haven't got any specyfication of those yellow LEDs.

Anyway, I've managed to design circuit with LM317 LZ 100mA max, but during the tests plan is to use bigger LM317T capable to output 1.5A max and start with lower input voltages around 6V (2 x Li-on 4.2Vmax) instead of 12V

BTW, LM317 LZ in TO-92 package nice fits on top of this small round PCB with additional resistor and bypass diode   

[szan]

... power dissipation in the regulator ...

LM317LZ datasheet says its power dissipation (PD) is internally limited and in electrical characteristics of LM317L in its electrical characteristics Io(max) when voltage difference Vi-Vo is between 3V-13V Io(max) minimum 100mA, typical 200mA, which looks good when I use 15R 5% resistor in worst case scenario I have: Io:  1.25V/(15R*(1-0.05))+100*10^(-6)mA= ~88mA

UPDATE: When we take into account not typical Vref=1.3V instead of typical 1.25V we get ~91mA, so everything so far looks good (for me) - let me know if I missed something

Lets solder test circuit  and see  what happends 

 

[Benta]

The parameter you need to watch is the junction-ambient thermal resistance, which is 200 K/W for the LM317LZ. This means that for every W you dissipate in the part, junction temperature will rise 200 K above ambient.
In a temperate climate, ambient would probably be 40 C maximum.
Maximum junction temperature for the LM317LZ is 125 C.
125 - 40 is 85 K.
This means you can maximally dissipate 85 / 200 = 0.425 W in your regulator.
In a warmer climate it will be even less, and if it's a car left in the sun it can easily be 80 C inside. So even worse.

The "Internally limited" power dissipation just means that the regulator will shut down when it gets too hot = no LED light.

I think the 47 ohm resistor is not such a bad idea after all...

[szan]

I think the 47 ohm resistor is not such a bad idea after all...

Thank you very much for figure out those junction temperatures.

However, when we assume 3Vf of those LEDs (I didn't tested it yet) than at 12Vcc we have current I: (12Vcc-3Vf)/47R = ~191mA, so power dissipation in this 47R resistor will be as high as 1.7W ... but as we can see there is small 6mmx2.5mm 0.25W 47 Ohm 5% resistor ?! 
It is very bad idea or maybe simply my 3Vf estimation is too low at ~200mA currents in those LEDs 
That is why - when I saw this tiny 0.25W resistor inside this thing I decided to do not connect it to 12V battery at all 

[Benta]

I can well imagine that there's more than one LED chip in that thing (in series). Reconnect the 47 ohm resistor, wire it to 12 V and see what the voltage across resistor or the LED(s) is. Or use a 470 ohm resistor to be on the safe side. Easy.
Right now you're fumbling in the dark.

[DavidAlfa]

Smelling a little overthinking here.
A running car will work at a pretty much constant voltage, about 13.7-14.4V depending on the car, so a resistor will do a similar job being way more cheaper.

The lamp probably has 3 leds in series, each Led Vf=3.3V, total Vf=10V
For Vbat=13.7V,  (13.7-10)/47 = 79mA
For Vbat=14.4V, (14.4-10)/47 = 92mA.

Anyways those values look too high for that little resistor, with a power disipation of 0.3-0.4W.
Nevermind! Chinese overpowered electronics. For sure that little 0.15W resistor is working at 200% of its nominal power.

Not such great variation to justify adding a regulator, caps, and making the bulb to cost twice.
Also the resistor will tolerate much better high temperatures or voltage spikes, a regulator would make the bulb more prone to fail.

If you truly want to make it nice, remove teh resistor and add a constant buck current regulator instead a linear reg, will be a lot more efficient.

[szan]

The lamp probably has 3 leds in series, each Led Vf=3.3V, total Vf=10V

I've already finished soldering prototype PCB with LM317LZ for current limit ~83mA and resistor ~15R made of two (22R||47R) 0.25W in parallel (1/22R+1/47R = 1/14.99) 
Power dissipation in this 15R @ 83mA resistor 2x0.25W will be  around 0.1W  
I have also bypass diode 1N4007 between LM317 output and input just in case, as well as I've protected its input by second 1N4007 

[szan]

I can well imagine that there's more than one LED chip in that thing (in series).

Looks like there are 12 light sources in hexagonal pattern.
I've added also 470R resistor as current limit to prototype PCB, but no time today to play with this circuit and make experiments ...

[Terry Bites]

If you have one or can borrow one use a bench PSU. Put you power supply in cc mode. set to the current you need.
The psu voltmeter shows Vf.
Then you can calculate how much power will be sweated off by the LM317. Assume the LED will fail short at some time.

You can do better with a PWM current source- save this planet!

[szan]

The lamp probably has 3 leds in series, each Led Vf=3.3V, total Vf=10V

I've connected those LEDs to 12.3Vcc (3s Li-on battery pack fully charged to 4.1V each) and with 470R current limiting resistor we have 7.77Vf@10mA, while voltage drop on resistor in series was 4.52Vr 
Probably Vf for those LEDs will be higher at bigger currents ~100mA, while when we calculate now for 14V running car voltage with its oryginal design 47R:
(14.0V-7.77Vf)/47R ~132mA 
Looks like it is around ~1W power LEDs (no electric markings on this aka light bulb metal case)
However, power losses in 47R resistor are around 827mW but they put inside this thing something which looks like 0.25W resistor (maybe it could be 0.5W 6mmx2.5mm size) 

[strawberry]

carbon resistors fail when stressed

LED 2S 6P ~20mA per diode (I bet its more like 10mA LEDs)

[szan]

carbon resistors fail when stressed

It may also be a deliberate limitation of the lifetime of this LED lamp by the manufacturer    (it was quite cheap one) through the accelerated destruction of this resistor and, as a result, possibly also LEDs - car users are used to the fact that the light bulbs burn out 

I've connected those LEDs now again to 12.3Vcc battery, but by using 100R 0.25W resistor and we have slightly higher  8.13Vf@41mA on those LEDs while 4.1Vr voltage drop over this resistor 

However, no supprise because of when I've estimated before this test LEDs current for custom battery 3s Li-on 4.2Vmax charged up to 3.9V to extend lifespan of its cells, we get nice LEDs cuttoff even when voltage on single battery cell will drop to 2.6V:
(3*2.6-7.77)/100=0.1mA
while when fully charged:
 (3*3.9-8)/100=0.036A~40mA 

When used in a car where I have around 14V when running:
(14.0-8)/100= 60mA
so estimated power dissipation in 100R resistor will be: 0.06^2*100~0.4W
which looks much better then oryginal design.
LEDs will output then around: 8*0.060~0.5W 
which is not so bad, while even at 40mA light is very bright from this thing 

NOTE: While we know expected forvard voltage of those LEDs there is poblem in using LM317, since as I know it need input voltage +4.25V higher than regulated output, so it doesn't look good now even in the case of 12Vcc battery (especially my custom 3*3.9Vmax=11.7Vmax) , while maybe it could be fine in a running car at 14Vcc:
14Vcc-8V=6V 




 

[Benta]

car users are used to the fact that the light bulbs burn out

Yes... every 5...8 years in my experience.
It's still totally unclear what you're trying to achieve here.

[szan]

It's still totally unclear what you're trying to achieve here.

Make things like those LED lamps usable and lasting longer while looks like manufacturer shortened its lifespan 

Do not turn on - take it apart ! 
Hopefully, by adding 100R resitor to existing 47R one in series we can make this thing usable and lasting much longer by limiting current to a range  41mA-31mA at very low cost (+100R):
(14Vcar-8Vf)/(100R+47R)= 0.0408 ~41mA

0.04081632^2*147R= 0.245W ~0.25W

(12.6Vcar -8)/147 ~31mA ~0.14W

In my design based on LM317 I've decided also use lower current limit around ~60mA simply by changing 15R to 22R:
1.25/0.060=20.83R ~22R
1.25/22=0.0568 ~57mA ~60mA

[tooki]

The lamp probably has 3 leds in series, each Led Vf=3.3V, total Vf=10V

I've connected those LEDs to 12.3Vcc (3s Li-on battery pack fully charged to 4.1V each) and with 470R current limiting resistor we have 7.77Vf@10mA, while voltage drop on resistor in series was 4.52Vr 
Probably Vf for those LEDs will be higher at bigger currents ~100mA, while when we calculate now for 14V running car voltage with its oryginal design 47R:
(14.0V-7.77Vf)/47R ~132mA 
Looks like it is around ~1W power LEDs (no electric markings on this aka light bulb metal case)
However, power losses in 47R resistor are around 827mW but they put inside this thing something which looks like 0.25W resistor (maybe it could be 0.5W 6mmx2.5mm size) 

Total fool’s errand, it seems to me.

The 12 LEDs in the module are likely wired either as four parallel sets of 3 LEDs in series, or as 3 parallel sets of 4 LEDs in series. Given your 10mA test, I’m guessing it’s the first one. That means each LED was being run at 2.5mA, hence the very low Vf.

Why are you even worried? What is it you think is being overstressed by the 47R resistor? Your calculations are all wrong since you can’t use the Vf at 10mA to calculate the voltage drop at 100mA. As current rises, Vf across the LEDs rises, and consequently the voltage dropped across the resistor drops.

And since it’s multiple LED strings in parallel, it means each LED isn’t getting much current.

If we assume 3.3V per LED (a much more reasonable value), 3 in series, then the resistor would dissipate 85mW at 12V or 340mW at 14V. Nowhere near the 827mW fantasy value you came up with.

[szan]

Your calculations are all wrong since you can’t use the Vf at 10mA to calculate the voltage drop at 100mA.

Maybe you missed one of my tests when I've connected those LEDS to 12.3Vcc battery with 100R resistor in series ?! 
We now know that we have voltage drop 8.13Vf@41mA  
So, as we can see between 10mA and 41mA LEDs current there is only:
8.13Vf-7.77Vf=0.36V ~0.4V difference
When we assume 9Vf@100mA then... we get:
(14Vcar-9Vf)/47=106mA which gives ~0.5W power dissipation in a very small resistor 6mmx2.5mm which looks like 0.25W 
Show me better estimation of LEDs Vf since I wouldn't like to connect those LEDS via oryginal 47R current limiting resistor to runing car battery which is around ~14V in my car ? 

[tooki]

So you’re still basing your entire argument on assumptions, and don’t actually know the behavior of the LEDs as delivered. You haven’t explained how you even came to believe that 47 ohms is too low a value.

If you connect the LEDs to 12V and 14V via the 47 ohms, you can then measure the actual voltage drop and current. It’s not going to instantly fry anything, and it’s frankly not likely to fry anything in the long run.

I saw the 8.13Vf test, but that’s not what you used for your calculations…

The ~0.4V difference in Vf between 10 and 40mA doesn’t tell you much because the relationship is not linear. So you can’t use that to predict with any certainty what Vf will be at 100mA.

[pcprogrammer]

Why are you even worried? What is it you think is being overstressed by the 47R resistor? Your calculations are all wrong since you can’t use the Vf at 10mA to calculate the voltage drop at 100mA. As current rises, Vf across the LEDs rises, and consequently the voltage dropped across the resistor drops.

Your reasoning feels wrong to me. The resistor has linear behavior in voltage versus current, so at 10mA the voltage across the 47 Ohm resistor is 0.47V, and at 100mA it is 4.7V. So the voltage across the resistor will not drop when the current rises.

You are right about the LED forward voltage not being linear, and will not go up very much at the higher current. For this you need the specifications of the LED's. But when assumed to be 3.3V with 3 in series it is 9.9V, leaving 4.1V for the resistor when the supply voltage is 14V. This means a current of ~87mA and ~0.35W dissipated in the resistor.

[szan]

But when assumed to be 3.3V with 3 in series it is 9.9V, leaving 4.1V for the resistor when the supply voltage is 14V.

It is bad to assume 9.9Vf since this forward voltage depends on current.
So far I made some measurements of Vf of those LEDs in experiments with current limited to ~10mA, ~40mA but I will make also experiment with ~100mA, as well as for a short time I can use also oryginal 47R and we'll see what we get.
Using data from those experiments than I can make non linear approximation eg. by assuming:
Vf(I)~ a*(I)^b   , so we should be able better predict this forward voltage of those LEDs at higher current 

[pcprogrammer]

Take a look at this site http://lednique.com/current-voltage-relationships/iv-curves/ You might find it helpful.

And to make an example calculation it is not a crime to assume something. It was to show that the voltage across the resistor does not drop when the current increases. The whole relation between the current, the LED forward voltage and the resistor is that it will reach a balance based on input voltage.

[Zero999]

Regardless of whether a resistor or linear regulator is used, the power has to be dissipated as heat.

If it's overheating, use a higher power resistor, or a higher value, to reduce the current.

If you want to use a linear regulator, it will need to dissipate the same amount of power, as a resistor for a given current and voltage drop. The TO-92 LM317L won't be able to dissipate ate 0.6W, at high temperatures, without overheating and going into thermal shutdown.

Use the LM337 in the SOT-223 package, with the tab soldered to the lightbulb's metal can, which can act as a heat sink. The circuit is similar to the LM317, but input is negative and output positive. The reason I'm suggesting the negative voltage regulator is because the tab is negative, so it can be soldered to the can.
https://www.ti.com/lit/ds/snvs778e/snvs778e.pdf

Another option is a switched mode power supply, such as the PAM2861, AL8862, NCL30160 etc..

https://www.diodes.com/assets/Datasheets/PAM2861.pdf
https://www.mouser.co.uk/datasheet/2/115/AL8862-1274720.pdf
https://www.onsemi.com/download/data-sheet/pdf/ncl30160-d.pdf

[szan]

The TO-92 LM317L won't be able to dissipate ate 0.6W, at high temperatures, without overheating and going into thermal shutdown.

In the case of LM317L I've changed current limiting resistor from 15R to 22R, so now we have:
1.25V/22R~57mA 
current limit, so power dissipated in this resistor (0.25W) will be:
0.057^2*22~0.1W 
When those LEDs bulbs will be used in a car then:
(14Vcar-9Vf)~5V difference  which is good for LM317L since it needs +4.25V higher input voltage than regulated output as I know from its datasheet 
In this case we have 5Vdiff@57mA difference in LM317L regulator, so estimation of dissipated power in this thing probably will look like this:
5Vdiff*0.057A~0.28W 
I looks good,  especially when we'll take into account such thing like those LEDs bulbs has metal back  cover 14mm in diameter x cylinder 16mm high, so there is quite big heatsink for those LEDs car bulbs 9Vf*0.057A~0.5W   

[tooki]

Your reasoning feels wrong to me. The resistor has linear behavior in voltage versus current, so at 10mA the voltage across the 47 Ohm resistor is 0.47V, and at 100mA it is 4.7V. So the voltage across the resistor will not drop when the current rises.

You are right about the LED forward voltage not being linear, and will not go up very much at the higher current. For this you need the specifications of the LED's. But when assumed to be 3.3V with 3 in series it is 9.9V, leaving 4.1V for the resistor when the supply voltage is 14V. This means a current of ~87mA and ~0.35W dissipated in the resistor.

The voltage drop across the resistor is, by definition, the residual voltage after the LED’s voltage drop. The current of the LED + resistor system finds an equilibrium. LED Vf rises —> resistor voltage drops —> current drops —> LED Vf drops —> resistor voltage rises —> current rises —> LED Vf rises. (Additionally the LED’s temperature also comes into play but let’s ignore that for today.)

Suppose an LED that has a Vf of 2.7V at 1mA and 3.2V at 30mA, and a supply voltage of 5V.

1. At 1mA, the resistor needs to drop 2.3V. (P = 2.3mW)
2. At 30mA, the resistor needs to drop 1.8V. (P = 54mW)

Using the voltage at one current as the basis for calculating the voltage at a totally different current is, thus, nonsense.

Anyhow, the OP first got alarmed because they assumed a single LED Vf, such that a little 1/2W resistor would have to drop 9V, causing dissipation of nearly 2W. But since discovering that it’s multiple LEDs in series, they still are freaking out at the 47R value, when in fact it might be perfectly reasonable.

This whole thread is about finding a new resistor value (or equivalent regulator). So what I was trying to say is this: as you select a value to increase (or decrease) the current, the LED’s Vf will increase (decrease) and the residual voltage the resistor must drop will decrease (increase). So when OP inserted a larger resistor value, this necessarily means the resistor will have to drop a larger voltage than the original one would.