On Fri, 13 Sep 2002, Robert Wade wrote:

> 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..

Your LED has a forward voltage of 1.6V, meaning it needs at least 1.6V
supplid.  There will always be 1.6V difference between the cathode and the
anode side.

Now, let's assume you're going to connect a resistor and your LED in
series.  Voltage across the whole mess is the same as the battery or other
supply voltage, or in this case around 9V.  You know the LED will have
1.6V across it.  That means the resistor will see 7.4V.  Actually it will
be more than that since the 9V battery will be a little higher -- that's
9V *nominal*.  But worry about that later.

So we know the voltage at various points in the circuit.  What determines
the resistance you want is the current you want through the LED.  Let's
assume you want a 10mA current (.010A), which will light most LEDs up
fairly well, and is generally safe.  The current in a series circuit
will be the same at all points in the circuit, meaning you will need
10mA through the resistor as well.  Now you use Ohm's law to figure R
using R=E/I, where E = 7.4V and I = .010...  so R would be 740 Ohms.  720
is a common resistor and will probably work fine, giving you right around
10mA.

One key thing to remember: Voltage drop across the LED will always be
1.6V, regardless of the current.  Voltage drop across the resistor can
vary depending on the current.

> 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. now if I take the
> amps and multiply it by 40ohm I end up with 9 volts. Ok I think I got Ohms
> Law. Now how do I get the value of the Volts after the resistor is applied
> to the line ?

If it's just the battery and a resistor, it's 9V (or whatever the battery
voltage is).

> 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?

Really, for all practical purposes especially when learning, either is
fine.  It's a little more efficient in most cases to connect the LED and
resistor to the + supply and drive the PIC pin low to turn on the LED, but
it's really your choice.  Some PICs can sink more current than they can
source, some are the same either way.

Dale

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