On Fri, Sep 13, 2002 at 11:03:21PM -0400, Robert Wade wrote:
> [EE]: section
>
> ok don't stone me for asking this question as I have been read Radioshack's
> books on Electronics. I have got to http://www.csgnetwork.com/ohmslaw.html
> and too here http://www.contractorcafe.com/ohms_course.htm and as well some
> 12th grade nasa education site.

Very good. Unlike a lot folks who simply ask a question from out of the blue
it's good to see that you actually did some research.

>
> I have an LED that needs 1.6Volts to light up. and I have a 9 volt battery.
> now I see every calulation to go from Volts, watts, resistance, amps, etc..
> What I have not seen is what the end result is. It I have a 9 volt battery
> and a 40ohm resistor then I end up with a amp of 0.225....

...Through the resistor presming that one end of the resistor is connected to
one terminal of the battery and the other end of the resistor is connected
to the other end of the battery.

> now if I take the
> amps and multiply it by 40ohm I end up with 9 volts. Ok I think I got Ohms
> Law.

That's it.

> Now how do I get the value of the Volts after the resistor is applied
> to the line ?

This question is poorly formulated. Let's talk about resistors in terms of LEDs.
First a few givens:

* LEDs require a certain amount of voltage to light up. In this instance 1.6V.
* LEDs must be current limited otherwise they explode.
* As you see from the discussion above that resistors will only allow a
  certain amount of current to pass through them.
* If you have multiple components connected in series across a voltage source,
  collectively the components must use (usually called dropping) all of the
  voltage from one end to the other.
* And here's the kicker: each component in the string will consume the same
  amount of current! It took me a long time to figure this one out.

So here's the game plan: feed the LED its 1.6V, use a resistor to drop the
remaining voltage, and pick a resistor value to limit the amount of current
through the LED to a reasonable value. And yes Ohm's law comes into play.

The one remaining value we need to do the calculation is the amount of current
the LED can accept. Typical values are 15-20ma for continuous use. A story
for another day is that many LEDs can accept huge amounts of current (1 amp
or more) for brief periods of time making them extremely bright! So for the
sake of this discussion let's say 20ma max current for the LED.

So let's rock: The LED will use 1.6V of the 9V, so the resistor must drop the
remaining 9V-1.6V -> 7.4V. We want the resistor to pass 0.02A of current at
this voltage. So apply Ohm's law 7.4V=0.02A * R -> R = 7.4V/0.02A -> 370 ohms.
Not a typical value so pick the next higher standard value which is 390.
So in the end the LED will get 7.4/390 -> 0.01897A which it should be perfectly
happy with.

One last aside is the wattage rating for the resistor. Use the equation P=VI
where P is the power in watts, and V and I are the typical voltage and current.
So for the resistor the wattage will be P=7.4V*0.01897A -> 0.14037W. So a
standard 1/4W resistor (0.250W) will do just fine.

BTW it doesn't matter where the resistor is in the string. By convention it
usually comes between the positive terminal and the anode of the resistor.

>
> [PIC]: section
>
> Ok now is it better to have a PICs chip connected to the ground pin of an
> LED or is it best to have the PIC control the voltage to the device? will
> either solution result in a voltage leak or drain?

Doesn't matter and no. PICs fortunately will source or sink the same amount
of current: 25ma. So it doesn't matter. Most designers will connect to the
anode by convention so that setting the port pin to 1 turns the LED on and
setting to 0 turns it off.

>
>
> As I know people would love for this to be a PIC only type of list I wanted
> to know if anyone could recommend commerial tutoring software on the topic
> of EE or basic electronics. This way I could learn with out buging others.

I believe that the recommendation to get the second edition of Horowitz and
Hill's "The Art of Electronics" has been proferred recently.

BAJ

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