LED Blinker rate - Toyota Nation Forum : Toyota Car and Truck Forums

meicalnissyen · 02-14-2005, 09:38 PM

Well, I wish this made sense but here goes.

I have a 99 rola and I wanted to put leds on it got the back end done, stop turn and that little running lamp and the blinker ran fast, not unexpected.

put the fronts in and lit up the fuse.
put the standard lamps back in the front

now some guys sell like 5watt resistors to fool the flasher which defeats the low consumption of the LED but I, at this point, could go for it.
the other deal is a no load flasher. I tried one from azone and had to make little jumper leads (literally a round peg in a square hole) WHOAAA that was somethin else whole instrument cluster blinkin, flasher hot, fuses poppin. the leads were correct

HELP PLEASE

TekPhobia · 02-14-2005, 09:50 PM

What resistors are you using for your rears?

-Tim

meicalnissyen · 07-08-2005, 02:44 PM

None at the rear, tried it w/o the resistors

flex-appeal · 07-09-2005, 03:08 AM

Seriously, resistors are like 20c each. Why dont you just stick with that? Remember to put them in series with the LEDs, just select a resistor that has an impedance equal to the difference between the impedance of the old bulbs and new LEDs.
While you're at it you could just put a variable resistor in there so that you can get the flashing speed exactly the way you want it

***Eye8Pussies*** · 07-09-2005, 04:08 AM

^ werd

and you get leds to have brighter lights, not so they take less power on the car???

TakeABao · 07-09-2005, 04:14 AM

Or if you're bold, do an LED taillight retrofit.

ZXLNT · 07-09-2005, 04:59 AM

Quote:

Originally Posted by flex-appeal

Seriously, resistors are like 20c each. Why dont you just stick with that? Remember to put them in series with the LEDs, just select a resistor that has an impedance equal to the difference between the impedance of the old bulbs and new LEDs.
While you're at it you could just put a variable resistor in there so that you can get the flashing speed exactly the way you want it

Thats not correct. The LEDS have a higher resistance to the flasher than the stock bulbs do. In order to correct the load the flasher sees you need to decrease the total circuit resistance to increase the circuit current. You do that by adding parallel resistance to the flasher circuit. In other words you need wire the resistors in parallel with the leds, NOT IN SERIES. That will only make it worse as your adding more resistance and decreasing the circuit current. Series resistance always increases total circuit resistance, parallel resistance decreases total circuit resistance. When you put resistors inline the current has to flow thru each resistor one after the other, its like multiple restrictions in the flow of current. When you have resistors in parallel, the current has multiple paths to follow, the more paths to follow the easier the current can flow, hence increase current flow and decreased circuit resistance. That's basic ohms law.

Also you will want to use at the very least 10watt or 20 watt resistors if possible, as the current flow thru them will heat them up quickly. Try about a 4 ohm resistor to start with.

Figuring 14VDC and 4 ohms resistance you get 49 watts of power that 10-20 watt resistor needs to dissipate.

flex-appeal · 07-09-2005, 02:06 PM

I'm trying to figure out what you are trying to say here. Thanks for the revision on 'basic ohms law' ...
In a series circuit the current is simply V/R , as anyone can derive from V=IR. With parallel resistance, the value for R is given by (R1R2)/(R1+R2).
I agree, in parallel the total circuit resistance is always lower than the leg with the lowest total R, so your technique is correct for lowering the overall circuit resistance, and therefore increasing the current flowing through the circuit.
What i can't fathom is why you would want to do that? Firstly, your average high intensity LEDs wont generally tolerate a continuous forward current higher than 50mA, i.e without something to impede current flow , it will soon overload and burn out.
Secondly, the flashers work by saturating a capacitor with power and then releasing it, less resistance means more current which means the capacitor will saturate quicker and the flash rate will be quicker.
Or is it that the flasher units arent constructed in the way that i thought? Maybe someone can shed some _light_ onto how they are built

I'm actually surprised that i remember my electronics, uni was a long time ago hehe

ZXLNT · 07-09-2005, 02:48 PM

Quote:

Originally Posted by flex-appeal

I'm trying to figure out what you are trying to say here. Thanks for the revision on 'basic ohms law' ...
In a series circuit the current is simply V/R , as anyone can derive from V=IR. With parallel resistance, the value for R is given by (R1R2)/(R1+R2).
I agree, in parallel the total circuit resistance is always lower than the leg with the lowest total R, so your technique is correct for lowering the overall circuit resistance, and therefore increasing the current flowing through the circuit.
What i can't fathom is why you would want to do that? Firstly, your average high intensity LEDs wont generally tolerate a continuous forward current higher than 50mA, i.e without something to impede current flow , it will soon overload and burn out.
Secondly, the flashers work by saturating a capacitor with power and then releasing it, less resistance means more current which means the capacitor will saturate quicker and the flash rate will be quicker.
Or is it that the flasher units arent constructed in the way that i thought? Maybe someone can shed some _light_ onto how they are built

I'm actually surprised that i remember my electronics, uni was a long time ago hehe

Last I knew they were a pretty simple bi-metallic strip that heats up with current flow. The more current flow the hotter the bi-metal strip gets and the longer it takes to cool down, therefore the slower the flash rate.. U say you can't fathom why I would want to increase the total circuit current, and it sounds like you believe that extra current would flow thru the LEDS. Thats not the case. The total amount of circuit current is divided amongst each resistance in the circuit. That being the LED and the resistor, were ignoring any other circuit resistance at this time. Each time you would add a resistance in parallel the total circuit current would increase, yes, but there would be an additional path for that current to flow thru.

Your saying they use an RC type circuit to determine flash right and not the plain old simple bi-metal flasher??

Now... if they are using an RC circuit which is basically what your describing, a capacitor in parallel, maybe with the resistor I can see where your coming from on most of it..

The LED should have an internal current limit resitor anyway. So unless you supplied them with a lot more than 12V that should protect the LED from over current. Only way to overcurrent them is to over supply them with voltage.

Back to you. Interesting discussion...

flex-appeal · 07-09-2005, 10:21 PM

I hadn't considered that a bimetallic may be being used, if thats the case then what you were saying is totally right. I had a hunt around just now and couldnt really find anything definitive (i think we need the electrical service manual).
Although I did come across one resource which said that if a bimetallic strip is in use, the flash rate will slow down if a bulb is removed or fused. If a transistor controlled flasher is in use, then the rate will increase when a bulb is removed or fused. Based on that i guess both are being used depending on the vehicle.

As for the internal current limit resistor, it's possible - if the thing is being sold as a unit with a number of LEDs packed into a bulb shaped housing. Whenever i have used LEDs, it has just been the actual component and i have mounted them onto a PCB in the required layout. Hence the need to be careful with the amount of current that you allow through them.

As a sidenote - how much do you want to bet that the guy with the original question about his LEDs has been totally confused by all this?

ZXLNT · 07-10-2005, 01:08 AM

Well I cant say for sure if its the bi-metal type or an electronic type with a relay output. I would guess bi-metal type as you can sure hear the relay clicking when the flasher is on. I was just going by what works on the motorycles I have messed with. They all use the bi-metal type, and lots of people integrate turn signals and use leds for the brake lights and signal lights combined. They end up either using an electronic flasher unit or "tricking" the stock flasher with load resistors..

And yeah I am sure this all sounds confusing..

V6LE_CAMRY · 07-12-2005, 01:05 AM

i think that the lead bulbs use way less power and is a less of a load so thats why your bliker is going crazy fast. its just like having a burned out bulb. you understand what im saying?

ZXLNT · 07-12-2005, 05:20 AM

That pretty much covers it...

88 LE · 07-12-2005, 05:53 AM

Quote:

Originally Posted by ZXLNT

The LEDS have a higher resistance to the flasher than the stock bulbs do. In order to correct the load the flasher sees you need to decrease the total circuit resistance to increase the circuit current. You do that by adding parallel resistance to the flasher circuit. In other words you need wire the resistors in parallel with the leds, NOT IN SERIES.

Correct.

LEDs also draw a lot less current then a bulb.

1 led = ~20mA

1157 bulb = ~2A

It would take 100 LEDs to equal the current draw of a single 1157 bulb.

Resistors:

http://spectrumled.com/equalizers.htm

http://spectrumled.com/images/leds/installation.jpg

Hajoca · 07-12-2005, 06:07 AM

Also, if your handy, wire in a LM 7812 type voltage regulator (looks like a transistor). I had LED license lamps, but they burned out sooner than expected from voltage spikes. Look up the part # on the web.

LED's will last longer, if the current is not exceeded their design capacity.
So any over voltage, will lead to more current being pushed through, and that will shorten the LED's life considerably.