Checking my electronics math...

[ianacole]

It's been a long time since I've done any electronics calculations, so I'd appreciate someone in the know confirming my calculations.

I purchased some LEDs that are rated at 3.2-3.6V at 20mA. They supplied 470ohm resistors to step down from 12v. However, my transformer supplies 17.5v. So, falling back to E=I*R, I come up with:

(17.5-3.2) = .02*R (working in series)->
14.2/.02 = R ->
715ohm = R

Correct?

Thanks!

Ian

 

[ikallio1]

Looks good to me. Just remember you need to pick a resistor value higher
than 715 ohm to keep the current under 20mA. A 1k resistor would probably
work just fine unless you absolutely need the maximal brightness.

-ik

 

[kenw]

are you sure the xfmr supplies DC? sometimes the aux output is AC, if so you'll need a FWBR (or at least a diode, in which case you'll need to redo your math).

 

[ianacole]

Yep, it's got both an A/C and D/C connection, and verified with the multimeter at just under 17.5VDC.

 

[ianacole]

Looks good to me. Just remember you need to pick a resistor value higher
than 715 ohm to keep the current under 20mA. A 1k resistor would probably
work just fine unless you absolutely need the maximal brightness.

-ik

Thanks! I don't need maximum brightness. I strung two 470s in series and it at least allowed for proof of concept of what I wanted to light up. I'll seek out some 1k resistors. RadioShack here I come!!

 

[Habbyguy]

Now you have to figure out what watt the resistor will be.
Take your 14.2V X 0.02mA = 0.284.
That is just over 1/4 watt so it would be safer to get 1/2 watt resistors.

 

[ianacole]

Now you have to figure out what watt the resistor will be.
Take your 14.2V X 0.02mA = 0.284.
That is just over 1/4 watt so it would be safer to get 1/2 watt resistors.

Ah, thank you... I forgot P=I*E as one of the requirements.