# Pre-wired led question

#### john2k

Joined Nov 14, 2019
146
I have some 3mm pre-wired LEDs for car projects. The LEDs come wired with a resistor and which is covered with a heat shrink. The specs for it says supply 12v for max brightness and 5v for dim. I want to run these LEDs in dim brightness from a 12v power source directly from the car. But ideally don't want to have to use some sort of 12v to 5v convertor. Is there a way to add another resistor to turn the input voltage from 12v to 5v?

#### peterdeco

Joined Oct 8, 2019
356
Take the heat shrink off and give us the value of the resistor.

#### ericgibbs

Joined Jan 29, 2010
12,884
hi,
You could add a series resistor, say a 1k for starters and try slightly different values around the 1k range.
E

#### SamR

Joined Mar 19, 2019
3,400
A 7805 regulator isn't that much larger.

#### MisterBill2

Joined Jan 23, 2018
8,723
You can measure the current drawn by the LED on 12 volts and then calculate the resistance to drop six volts at half that current. That should get you close. OR you can just put a 470 ohm resistor in series with the existing LED and see how close that comes to the amount of light you want. I am thinking it will be close.

#### john2k

Joined Nov 14, 2019
146
Thanks for the replies, what kind of resistor any particular wattage etc?

#### Tonyr1084

Joined Sep 24, 2015
5,966
Assuming a 12 volt automotive source - that can be as high as 14.5 volts. If the LED has a built in resistor, assume it's bright at 20mA. So, just going off a guess, the Vf (forward Voltage) of the LED will have a small effect. Assume it's Vf is 2.1 volts. 14.5 - 2.1 = 12.4V. 12.4V ÷ 20mA = 620Ω. If you double the resistance (in series, NOT parallel) then you will effectively cut the current in half. That is to say 10mA. 12.4V ÷ 10mA = 1240Ω. Increasing the resistor will reduce the amperage.

Wattage is calculated by multiplying the voltage times the current. So 14.5V x 10mA = 145 mW. For that you'd need a quarter watt resistor. (250mW).

You don't have to put the resistor under the existing shrink sleeve, you can put it wherever you make the connection. Just be sure to insulate the connection so it doesn't short out to something.

#### MisterBill2

Joined Jan 23, 2018
8,723
Really it should be at least a half watt resistor so that it can better handle the power AND be mechanically more rugged.

#### SamR

Joined Mar 19, 2019
3,400
AND be mechanically more rugged.
A bit of shrink wrap will help with that.

#### Tonyr1084

Joined Sep 24, 2015
5,966
Really it should be at least a half watt resistor so that it can better handle the power AND be mechanically more rugged.
145mW is just 20mW more than an eighth watt resistor. Quarter watt is almost double the actual wattage. And a higher Vf is going to mean even lower wattage; so a half watt resistor is just way overkill.

#### MisterBill2

Joined Jan 23, 2018
8,723
For the difference in cost it makes sense to use the bigger resistor to have a stronger package.

#### Tonyr1084

Joined Sep 24, 2015
5,966
Whatever. It's up to the TS to decide. Personally it doesn't move the cheese on my cracker.