# Question about Bill Marsden's "LEDs, 555s, Flashers, and Light Chasers"

Discussion in 'General Electronics Chat' started by skinner89, May 8, 2012.

1. ### skinner89 Thread Starter New Member

Jul 18, 2011
17
1

I am very new to electronics. Trying to go through the math myself I am already confused. Using Ohm's Law E = IR I try to find the resistance using the known voltage (9V) and desired current (20mA):

9 = .02 * R or 9/.02 = R, 9/.02 = 450

So why wouldn't you want to go with a common 470 Ω resistor in this circuit?

2. ### paulktreg AAC Fanatic!

Jun 2, 2008
767
167
You're forgetting the V drop of the LED @ 2.5V. (9V - 2.5V = 6.5V).

It's then R= 6.5V/20mA=325R.

3. ### skinner89 Thread Starter New Member

Jul 18, 2011
17
1
Alright so the LED is not current limiting like the resistor, however, it does reduce the voltage or pressure of the circuit. Going back to the water example would this be analogous to puncturing a small hole in the pipeline? The same amount of water can flow through the pipe, however, some of the pressure is being lost on a small stream of water spraying out the hole (being converted to photons)?

4. ### strantor AAC Fanatic!

Oct 3, 2010
5,025
3,116
sure, sound good to me.