Assuning a source Voltge of 0.8V, Determing the curretn in a 50mW LED(Light emitting diode)
If anyone can solve this, can you please explain any formula used to solve alonge with the answer?
THank you
2006-11-29 01:47:32 · 4 answers · asked by jeep10455 2 in Science & Mathematics ➔ Engineering
Well we are given voltage, and power. We are asked to find the current. Well the formula we have to use here is P=VI.
Rearranging this formula we get; I=P/V
However; there is only one thing wrong with this problem.
We do not know what the voltage is correctly.
If this diode is a ideal diode. An ideal diode acts like a short circuit with no voltage drop across it. Then we would use .8V as the Voltage Source across the diode.
Ideal Diode:
I = .05 W/.8V
I = 62.5mA
If it is not an ideal diode and there is a .75V drop across the diode then we would use .05V as the Voltage Source across the diode.
I = .05W / (.8V - .75Vdiode drop)
I = .05W / (.05V)
I = 1A
Here this might give your more information!
http://en.wikipedia.org/wiki/Diode_modelling
2006-11-29 01:57:56 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The LED converts electric current to light and heat. The voltage drop times the current is the amount of power consumed.
2016-03-29 15:43:35 · answer #2 · answered by Anonymous · 0⤊ 0⤋
If this is a real world problem, the current would be Zero. Most LEDs require at least 1.6Vf to conduct. Is this a trick question?
2006-11-29 18:31:12 · answer #3 · answered by charley128 5 · 0⤊ 0⤋
Power = voltage X current -- so
.05 watts = .8 volts X current.
I'm sure you can go from there.
2006-11-29 01:50:17 · answer #4 · answered by Gene 7 · 0⤊ 0⤋