I thought I would share a bit more detailed information about load resistors that are needed if anyone is putting LED into their turn signals.
While in some cars you can put LED ready turn signal relay, in 2014+ Forte the blinking is controlled by BCM. Hence, you must add load to the circuit in order for the system to not think light bulb is burnt.
The problem here, when I read here and there, is a confusion with load resistors specification.
There are two parameters that they carry.
- resistance
- power dissipation
The former is given in Ohms (6 or 8 Ohm are most common) while the latter in Watts (25 W or 50 W).
Now, the resistance makes the load (not the power!). The resistance must be chosen based on "missing" power load from installing the LED. The power specified on the resistor is nothing more, but the MAXIMUM power it can handle. However, one should keep in mind that while this is the maximum power, the safe limit is half of that.
The confusion gets from reading the specs and comparing to your setup.
In short example.
Rear turn signal is light bulb model 1156. It is rated 27 W at 13.2 V. That gives you resistance of 6.5 Ohm at current of 2.05 A.
More detail
Resistance remains (basically within some limits) constant, so at 14.3 V (when engine is running) you will get 2.2 A and power will be 31.2 W (that's why they are righter when engine is running), or when battery is low at about 12.0 V power is 22.3 W and current 1.9 A.
Going back on track.
LED in this example will be 2 W. This MUST be measured. Do NOT believe in LED that are 15 W. If that was true 15 W, then it would need to have a big heatsink. You must use multimeter and measure current being drawn by this LED. Having current and battery voltage you will find out what is its power. Current x voltage = power (AxV=P).
So, you got 13.2 V, 0.15 A. power is 2 W.
Tungsten is 27 W, minus LED 2 W - missing 25 W.
Now, take 25 W and convert to Ohms!!
25 W at 13.2 V will create current flow of 1.9 A. 13.2 V/1.9 A = 7.0 Ohm.
You know now that the resistor must dissipate heat of 25 W and its resistance should be around 7 Ohm.
Common sizes are 6 Ohm and 8 Ohm... what's the difference?
6 Ohm will produce more power (will allow more current flow) and at 13.2 V will have 2.2 A resulting in 29 W.
8 Ohm at 13.2 V will pass 1.7 A and produce 22 W.
Either one is good.
If possible, 8 Ohm would be a better choice due to the fact at 14.4 V (engine running) it will be more about 26 W... so basically ideal load.
To keep it safe without extra heatsinks - the dissipated power must be doubled. Hence, load resistor should be of 50 W. This way it will not be glowing red when under load...
In very short.
In any LED setup, for one side, one LED bulb, you will need 6 Ohm or 8 Ohm resistor of total power 50 W (or more). If you choose 25 W, you MUST add heastinks.
While in some cars you can put LED ready turn signal relay, in 2014+ Forte the blinking is controlled by BCM. Hence, you must add load to the circuit in order for the system to not think light bulb is burnt.
The problem here, when I read here and there, is a confusion with load resistors specification.
There are two parameters that they carry.
- resistance
- power dissipation
The former is given in Ohms (6 or 8 Ohm are most common) while the latter in Watts (25 W or 50 W).
Now, the resistance makes the load (not the power!). The resistance must be chosen based on "missing" power load from installing the LED. The power specified on the resistor is nothing more, but the MAXIMUM power it can handle. However, one should keep in mind that while this is the maximum power, the safe limit is half of that.
The confusion gets from reading the specs and comparing to your setup.
In short example.
Rear turn signal is light bulb model 1156. It is rated 27 W at 13.2 V. That gives you resistance of 6.5 Ohm at current of 2.05 A.
More detail
Resistance remains (basically within some limits) constant, so at 14.3 V (when engine is running) you will get 2.2 A and power will be 31.2 W (that's why they are righter when engine is running), or when battery is low at about 12.0 V power is 22.3 W and current 1.9 A.
Going back on track.
LED in this example will be 2 W. This MUST be measured. Do NOT believe in LED that are 15 W. If that was true 15 W, then it would need to have a big heatsink. You must use multimeter and measure current being drawn by this LED. Having current and battery voltage you will find out what is its power. Current x voltage = power (AxV=P).
So, you got 13.2 V, 0.15 A. power is 2 W.
Tungsten is 27 W, minus LED 2 W - missing 25 W.
Now, take 25 W and convert to Ohms!!
25 W at 13.2 V will create current flow of 1.9 A. 13.2 V/1.9 A = 7.0 Ohm.
You know now that the resistor must dissipate heat of 25 W and its resistance should be around 7 Ohm.
Common sizes are 6 Ohm and 8 Ohm... what's the difference?
6 Ohm will produce more power (will allow more current flow) and at 13.2 V will have 2.2 A resulting in 29 W.
8 Ohm at 13.2 V will pass 1.7 A and produce 22 W.
Either one is good.
If possible, 8 Ohm would be a better choice due to the fact at 14.4 V (engine running) it will be more about 26 W... so basically ideal load.
To keep it safe without extra heatsinks - the dissipated power must be doubled. Hence, load resistor should be of 50 W. This way it will not be glowing red when under load...
In very short.
In any LED setup, for one side, one LED bulb, you will need 6 Ohm or 8 Ohm resistor of total power 50 W (or more). If you choose 25 W, you MUST add heastinks.