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When I wired my boost gauge, I used the radioshack LED's, 5V @ 30ma. I used two in series, and assumed a 12V supply. Using an online calculator, I found that I needed a 68 ohm resistor. I ended up using 2 33 ohm resistors in series. The LED's only lasted about 6 months and went out. Reading through the ambient lighting thread, they used a single LED and a 470 ohm resistor. Doing the reverse math, this would mean they assumed an 18V supply. Is that right? What supply should I assume? Is the 18V just a safer supply to use for calculating the resistor?
 

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Josh, the voltage is probably closer to 14 volts (or more) than 12 most of the time (when the car is running). the LEDs burned out because there was slightly higher voltage (and therefore, more current) running thru them. it isn't the voltage that destroys these, tho - its the resultant current!
recommend you use the calculator and go with a 150 ohm resistor (that I saw when I ran thru the calcs on that website). I've seen voltage higher than 14 in a car - it can be as high as 14.4 - 14.5 volts.
 

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Even thought the diode is rated 5 volts at 30 ma you don't want to run them at 30 ma. They'll burn out fairly quickly. I would assrme 20 ma continuous for normal application. Thus, at 14.4 volts max, the total resistance you want is R = V/I or 14.4/.02 = 720 ohms. The diodes resistance at 5V, 30 ma = 5/.03 = 167 ohms. Thus, the resistor you need = 720 - (2*167) = 386 ohms. If you had one LED than the resistor would be about 553 ohms. Since they don't sell 553 ohm resistors the other guy bought the standard 470 ohm. For your application, you should be able to find a 330 ohm.
 

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For DC, the equation is E = (I)(R) where voltage is e, current, measured in amps is I, and resistance, measured in ohms is R. When kenblasko says (ma) he is referring to miliamps or thousandths of an amp.

For AC, things are a little more complicated, and imaginary numbers are required if there is any capacitance or inductance at all. If you want me to explain this in greater detail, let me know...
 

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citispot said:
For DC, the equation is E = (I)(R) where voltage is e, current, measured in amps is I, and resistance, measured in ohms is R. When kenblasko says (ma) he is referring to miliohms or thousandths of an ohm.

For AC, things are a little more complicated, and imaginary numbers are required if there is any capacitance or inductance at all. If you want me to explain this in greater detail, let me know...
a) ma = milliamps, NOT milliohms. :whistle:
b) automobile electrical systems coming from the alternator are DC. :poke:
c) this thread is from the middle of Feb. why resurrect it at all? :rolleyes:
 

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JayTheSnork said:
Josh, the voltage is probably closer to 14 volts (or more) than 12 most of the time (when the car is running). the LEDs burned out because there was slightly higher voltage (and therefore, more current) running thru them. it isn't the voltage that destroys these, tho - its the resultant current!
recommend you use the calculator and go with a 150 ohm resistor (that I saw when I ran thru the calcs on that website). I've seen voltage higher than 14 in a car - it can be as high as 14.4 - 14.5 volts.
Jay is right. The fuze box for me amp lead off of my battery has a LED readout of the the voltage. When the car is off it's at about 12.3 but when the engine is running it's up to 13.7 IIRC.
 

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JayTheSnork said:
a) ma = milliamps, NOT milliohms. :whistle:
b) automobile electrical systems coming from the alternator are DC. :poke:
c) this thread is from the middle of Feb. why resurrect it at all? :rolleyes:

Actually, all generators and alternators generate AC. In order to make a DC output, a commutator is required.
 

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citispot said:
Actually, all generators and alternators generate AC. In order to make a DC output, a commutator is required.
Don't confuse these guys with the big words. He's talking about the rectifier, which on our cars is built into the alternator but on high output units is outboard.
 

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citispot said:
Actually, all generators and alternators generate AC. In order to make a DC output, a commutator is required.
the diode block is built into the body of most modern alternators nowadays, and in order to make the DC output, the diode block is used as a rectifier. the commutator and associated brushes enables an alternator to make AC while it is revolving (ie, turned by the rotation of the engine with the serpentine belt). yes, the output of the alternator itself is AC. HOWEVER, the output you will see if you connected an oscilloscope to the back of ANY modern alternator will be "pulsating DC" because of the aforementioned diode block, and certainly NOT AC - the battery acts as a "sink" to help filter it to less change about the mean. it has an average value (RMS) that is somewhere north of 12 volts while the car is running (and the alternator is supplying power for the car electrical loads).

and what I said was, automobile electrical systems coming from the alternator is DC - re-read my post.
 
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