Experiment #2

 Experiment #2

Diodes

A diode has the characteristics of a conductor when current flows one way. They also have insulator characteristics when current tries to flow in the opposite direction. Basically diodes allow current (amps) to flow in only one direction, Anode to Cathode.

Identifying diode's and/or light emitting diode's (LED's) polarity without using a multimeter is simple. Diodes display cathode with a white stripe around the end. LEDs display this by using a smaller terminal for cathode and a slightly longer terminal for anode.

We were given a data sheet to identify the characteristics of a 1N4007 type diode. Data sheets are very handy to have at your disposal when you need them. An example is given using a 5v supply, a 1k/ohm resistor and a 1N4007 type diode. When calculating the current (I) across the diode (d) we use the formula:
Id = (Vs - Vd)/R
= (5 - 0.6)/1000
= 4.4/1000
=0.0044A (4.4mA)                                                           Measured: 4.6mA


So the answers are not identical, but they're not far off and the concept is there. To calculate the voltage drop across the diode we use the following formula:
Vd = Vs - (I x R)
= 5 - (0.0044 x 1000)
= 5 - 4.4
= 0.6v                                                                                 Measured: 0.6v


Bang on. Now the maximum value of current able to flow through the given diode is 1A and the maximum supply voltage used to keep the diode working in a safe region is 1000v.


Replacing the standard diode with a light emitting diode (LED) and calculating the current is done as follows:
I = (Vs - Vd)/R
= (5 - 1.7)/1000
= 3.3/1000
= 0.0033A (3.3mA)                                                          Measured: 5mA

This observation shows us that there is low current being used by the LED.