Experiment 2

---------------------Voltage drop in forward -------------Voltage drop in reverse
LED-----------------1.761 volts---------------------------0 volts
DIODE--------------0.541 volts---------------------------0 volts

LED cathode side got shorter wire and flat side, and diode cathode side marked with grey stripe.



R=1000 ohm __________ V=5.31 v _________ Vd=0.657v

Calculated amps____________________________Measured

5.31-0.657=4.653___________________________4.6 mili amps

I=4.653/1000=4.653 mili amps

Its a voltage drop of 0.657 volts at diode as it use this voltage to run.

Calculated voltage drop--------------------------Measured voltage drop

Va=5.31-----------------------------------------0.664 volts

I=4.653 ma

R=1000 ohm

V=IxR=4.563*1000=4.563 volts

Vd=Va-V=5.31-4.653=0.657 volts

Max value of current that can flow through diode

1 amp at 75 C degrees

For R=1000 ohms, what is max Vs

Vs=1amp*1000ohms=1000 volts

Now diode replaced with LED

Calculated current amps-----------------------------Mesured current amps

Vs=5.31 volts----------------------------------------3.3 mili amps

Vd=1.967

V=Vs-Vd=3.343

I=V/R=3.343/1000=3.343 mili amps

LED light voltage drop was hiher then diode, and as it takes more voltage to run, then less amps flow through LED.