Current running through kick-panel wires
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desperately needs a drop
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Current running through kick-panel wires
Been a while since I posted.
I have been pondering a project involving some lighting, only the LEDs would not be 12V, they'd be 1.7V.
The issue I'm running into is a pretty simple one, just one I think would be easiest answered on here.
Now, the lines in the kick-panel I am talking about dealing with are all 12V, so no issue there, I just need to know the current in the lines.
Do the wires deviate or are they all the same amperage?
Thanks.
I have been pondering a project involving some lighting, only the LEDs would not be 12V, they'd be 1.7V.
The issue I'm running into is a pretty simple one, just one I think would be easiest answered on here.
Now, the lines in the kick-panel I am talking about dealing with are all 12V, so no issue there, I just need to know the current in the lines.
Do the wires deviate or are they all the same amperage?
Thanks.
Thread Starter
desperately needs a drop
iTrader: (1)
Joined: Sep 2008
Posts: 1,127
Likes: 1
From: Dallas, Texas
Ya, basically I need there to be 1.7V at each LED (2 of them) from the 12V supply. But I need to know the current the be able to calculate the resistance needed.
Unless you've got an idea?
Unless you've got an idea?
You need to determine the operating/nominal current rating of your diodes. We'll call that Id. Don't use max current rating or they'll probably burn out fairly quickly. They will be brighter with higher current though.
Then compute the value of the resistor you need in series with each diode.
R = (12.8 - 1.7)Volts / Id.
Note I used 12.8 volts. It's going to be a little higher than 12, up to something like 14.5.
Each resistor would dissipate the following power in watts:
P = Id x Id x R
That's the product of the diode current squared times the resistor value you computed above.
And you should select a resistor that's capable of handling at least 2x that power result.
Then compute the value of the resistor you need in series with each diode.
R = (12.8 - 1.7)Volts / Id.
Note I used 12.8 volts. It's going to be a little higher than 12, up to something like 14.5.
Each resistor would dissipate the following power in watts:
P = Id x Id x R
That's the product of the diode current squared times the resistor value you computed above.
And you should select a resistor that's capable of handling at least 2x that power result.
LEDs require a constant current supply. (As you know) I dont think you're going to find anything like that in the kick panels.
Wouldn't it be easier to buy a cheap little LED power supply? Probably 5 bucks and the size of a car key...
Wouldn't it be easier to buy a cheap little LED power supply? Probably 5 bucks and the size of a car key...
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