# Circuit Design Questions - Voltage Drop

Hi there, I'm working on a circuit which will include a 3v, .7Amp motor and three 20mA LED chips, with a maximum forward voltage of 3.8v. The whole circuit is intended to run off a 6v battery.In calculating resistance for voltage drop, I've figured out that resistance for the motor is 2.1 Ohms and then (by Ohm's Law Resistance = Voltage / Current) the resistance of each LED chip is 190 Ohms. Is this correct? It seems astronomically high.?

active| newest | oldestSay you want 20mA to flow throught an LED who's forward voltage drop is 3.8V (and that seems a bit high), then you need a resistor in series with the diode to drop the remaining 2.2V. i.e. : R = (6V - 3.8V)/20mA , therefore R=110Ohms

So, just to be sure I understand what you're telling me, once the LED has taken up 3.8v to glow, the remaining 2.2v needs to be prevented from reaching the power source as 2.2v? Otherwise the LED will burn out from excess voltage, correct?

With no current limit the LED will gobble more current while raising it's

voltage only a little and really heating up.

This causes a visible change to the color output and eventually burns

the phosphors that change the blue to red and to green.

First blue should start and then the dark red glow or if there is enough power to a bang-crack dark !

It's worth pushing a fwe LEDs to see how they suffer over power

but not in a

NO SMOKING Zone ;-)the semiconductor once saturated (at 3.8v) acts as a short circuit, allowing almost as much current wants to flow as the circuit will allow. Adding the resistor knocks down the excess voltage and forces a more predictable ohmic current response. Over-Current kills semiconductors.