# voltage drop equation

Discussion in 'Homework Help' started by jonisonvespa, Mar 30, 2012.

1. ### jonisonvespa Thread Starter Member

Mar 25, 2012
49
0
hi
im currently trying to make a project using a ardunio uno, the outputs are 5v and the leds i want to use are surface mount 2v max 30ma current, what equation can i use to drop the voltage to 2v, 30ma?
thank you

2. ### MrChips Moderator

Oct 2, 2009
12,414
3,353
Ohm's Law is I = V/R

hence,

R = V/I

3. ### jonisonvespa Thread Starter Member

Mar 25, 2012
49
0
hi mr chips im confused with the v value, is that the supply or the max voltage of the led? (supply)
ok thanks

Last edited: Mar 30, 2012
4. ### Jess_88 Member

Apr 29, 2011
174
1
The way I would look at it is;
current in a series circuit is the same, while a voltage drop can be achieved across one resistor and the led in parallel with the second (same voltage in parallel).

IR1 + IR2 = 5V
IR2 = 5v-3v

...whats how I understand it anyway

5. ### Jess_88 Member

Apr 29, 2011
174
1
wait... sorry, that can be right. current would change in the LED

3v = 30mA x R

with just a sires circuit

6. ### MrChips Moderator

Oct 2, 2009
12,414
3,353
Vr = 5V - 2V = 3V
R = 3V/30mA = 100 ohm

30mA for LED is too much. I would reduce it to 10mA
R = 3V/10mA = 300 ohm

Even 3mA may be bright enough
R = 3V/3mA = 1K ohm

7. ### jonisonvespa Thread Starter Member

Mar 25, 2012
49
0
thanks mr chips will try some different values when i get the leds