| T O P I C R E V I E W |
| Arnak |
Posted - Apr 26 2009 : 6:03:34 PM
Hi,
I would like to use 32 gauge wire to connect up a standard LED at 12v, the length would be 2 mt's.
Would that be too thin?
Also would there be a significant voltage drop over that length?
The wire is copper with 8 strands, so 32 gauge is the overall size.
Thanks for any advice you can give.
Arnak |
| 4 L A T E S T R E P L I E S (Newest First) |
| Arnak |
Posted - Apr 28 2009 : 05:33:04 AM
Hi Audioguru,
Thanks for that info, I'm not that well up on electronics...
It must be my senior age.8-((
That makes it very clear thank you, I'll make a note of the formula in case of need again.
Arnak |
| audioguru |
Posted - Apr 27 2009 : 09:38:02 AM
It is easy.
32AWG wire has a resistance of 538 ohms for 1km. I got it from the first link in Google when I searched for Wire Gauge Table.
Then only 2m has a resistance of 538/500= 1.1 ohm but there are two wires so the total resistance of the wires is 2.2 ohms. It can safely pass up to 91mA continuously.
If you have a 3.3V blue LED and want it to pass 20mA then the current-limiting resistor value is (12V - 3.3V)/20mA= 435 ohms. If you use 470 ohms then the current is (12V - 3.3V)/470 ohms= 18.5mA. The resistor will dissipate 18.5mA squared x 470 ohms= 0.16W so use a 1/4W resistor.
The voltage drop in the 2.2 ohms wires is 18.5mA x 2.2 ohms= 0.041V. |
| Arnak |
Posted - Apr 27 2009 : 04:57:55 AM
Hi,
Thanks for the info.
I did try google but a lot of the tables didn't go down that far.8-((
Also it was difficult to determine about the dc voltage of 12v.
The charts I did find left me a bit confused...
Arnak |
| audioguru |
Posted - Apr 26 2009 : 6:25:24 PM
Simply look at the spec's on a Wire Gauge Table in Google to see that the voltage drop on the thin short wire is very small since the current is very low.
If you don't use a current-limiting resistor then the LED will blow up. |
