# Series resistor

#### knoblame

Joined May 24, 2005
1
i am trying to figure out how ohms law will change in regards to voltage in a series, does it affect the resistance. for example. if i have a source voltage of X and the voltage drops at a (LED) to Y. I have a current value of Z amps. How would I figure out the value of resistor at LED To achieve my Z amps. Do I have to take in affect that there was a change in voltage at the LED.

R=V/I

R= V (would i need to figure out the difference of X and Y by subtracting or just use Y value if that is where i need to find the resistance) divided by I (Z amps)

#### pebe

Joined Oct 11, 2004
626
Originally posted by knoblame@May 24 2005, 03:15 PM
i am trying to figure out how ohms law will change in regards to voltage in a series, does it affect the resistance. for example. if i have a source voltage of X and the voltage drops at a (LED) to Y. I have a current value of Z amps. How would I figure out the value of resistor at LED To achieve my Z amps. Do I have to take in affect that there was a change in voltage at the LED.

R=V/I

R= V (would i need to figure out the difference of X and Y by subtracting or just use Y value if that is where i need to find the resistance) divided by I (Z amps)
[post=7895]Quoted post[/post]​
Ohms law will never change!

Your resistor will have across it a voltage equal to the source voltage X minus the LED voltage Y. The current through both the resistor and LED is Z amps.

So from Ohms law R = V/I then R = (X - Y)/I

#### Brandon

Joined Dec 14, 2004
306
Say you have a typical red led at with a on voltage of 0.7v and an ideal current of 20 mA to acheive the brightness you want and you hve a 5v source to use. Since the led can not control the current only a resistor can, we need to generate the current limiting resistor.

You need to set up the KVL loop equasion to represent the circuit first.

5v (+ terminal) - 0.7v (led Vdrop) - Vr (Vdrop over res) = 0v (- terminal)

Vr=4.3v

Since we want 20mA and V=IR we set up the equasion.equasion.

5v - 0.7v - Vr = 0
5v - 0.7v - Ir*Rr = 0
5v - 0.7v - 0.020A * Rr=0

Rr= - (0.7v - 5v) / 0.020A

Anytime you need to solve for a led, you follow a method like this.

#### n9xv

Joined Jan 18, 2005
329

#### Brandon

Joined Dec 14, 2004
306
Originally posted by n9xv@May 24 2005, 11:22 AM
The LED foward voltage would typically be between 1.2 and 3.5 volts though not .7 volts.

Here's a cool link that does the work for you!
LED current limiting resistor calculator
[post=7898]Quoted post[/post]​
oops.. thats right.

Concepts still the same, lol.