resister/s needed to end up with the proper power levels.

[StirCrazy]

ok for all you electronic brainers.. I have a transformer that take 120 AC and goes to 18VDC / 80 mA

for a LED I need to step that down to 4.5Vdc and 20mA. so I need to limit both the curent and the voltage. it is easy for me to to limit one.. but maby I am not looking at it right .. or I am having a brain fart but it is confusing me at the moment.

Steve

[StirCrazy]

this is what I am thinking....

I have a 18V supply and I have to drop it down to 4V so that means I have to drop 14 V with a resister.

so if I tak the Volts I have to drop and devide it by the curent of the LED I should get my resister rating.. so 14/0.02 = 700.. so I am thinking I need 700 Ohm resister.. so I will set up to the next highest to be safe..

Steve

[Reef_Ready]

I think its cerebral gas Steve. Lol

Yes I belive you are right. The voltage is dropped between the two resistances in series and the current remains the same.

My reasoning is somewhat different but close to the same answer

Total resistance = 18v/.02ma… = 900ohms
4.5v/18v = 25% therefore 900ohmsX 25%= 225 for the LED and 900-225= 675ohms for the drop down resistor

[reefburnaby]

Hi,

Sounds about right :)

If 18Vdc and you are applying to a diode (LED) that has a forward bias (i.e. turn on voltage) of 4.5V@20mA.

So, a first order approximation would be (18Vdc-4.5Vdc)/20mA = 675 ohms. The 680 ohm resistor should be close enough. It doesn't hurt to try something larger (like 1Kohm or 830 ohms) to start and fine tune later.

20mA*20mA*676 ohms = 0.27W. That means, you'll need a 1/2 W resistor.

BTW, besure to check which pin is the cathode and which is the anode. If you apply the wrong polarity, then there is a remote possiblility of damaging the LED.

Hope that helps.

- Victor.