Diodes

^                                ^
    Cathode "C           Anode "A"

Task 1# Identifying the Anode and the Cathode of a Diode
by simply looking at the diode you can instantly tell which is the anode and the cathode, in this case the grey band on the diode indicate that that side is the cathode which leaves the othersode to be the anode.

another method to identify the anode and cathode is to use a multimeter on the ohms funtion. The task was to use a multimeter to identify the anode and cathode of a diode and an LED using a multimeter.
first i measured the voltage drop of the diode in the foward biased direction (anode to cathode) with the red lead on the anode side and the black lead on the cathode and got a measurement of 1.9m ohms, this shows that the side with the red test lead is the anode because voltage can pass through diode from anode to cathode not in opposite.

next i measured the LED the same way as the diode in the 1st test except a different reading of 1.1m ohms which also identifies that the anode is the side with the red test lead, in the reverse biased direction you shouldd get an infinite reading both for the LED and Diode.
also another way to identifying the anode and cathode on an LED is to remember the flat side is the cathode and another way i remember which side is positive and which is negative is to think of the longer leg of the LED is positively longer.

Next task was to take 1 1k resistor, 1 1N4007 diode and have a voltage supply of 5v and put it in this circuit on a breadboard

  R=1kohm   D=1N4007 Vs=5V

Now I had to work out the current:  Calculated= I=V/R                   Measured= 4.2mA

                                                                           =5-0.7/1000

                                                                           =4.3mA

By simply using ohms law i calculated the current by dividing the voltage by the resistance, to get the correct voltage i had to minus the voltage drop over the diode which was 0.7v because according to the data sheet of a 1N4007 the voltage drop across it is 0.7V.

to measur it i used a multimeter on the Amps function and made the multimeter part of the circuit to get the reading of 4.2mA. The calculated and the measured amperage readings are only 0.1 mA out so therefore i can assume that the circuit is working properly.

Next was to replace the diode with an LED and then work out the current

calculated= I=V/R                          Measured= 3.3mA

                   =5-1.7/1000

                   =3.3mA

Even though the diode was replaced with a LED the ohms law is still used to work out the current, the only thing which changed was the voltage drop over the LED which was 1.7v because in this case this is how much the LEd needs to emit light.

The 3rd experiment was to obtain a breadboard, and grab the following components and build them in the circuit below : 2 resistors, 1 5v1 400mW Zener diode

                                                

           R=100ohm  RL=100ohm  Vs=12V
First i measured the volt drop of the Zener diode(Vz) and got a reading of 4.9v and when i varied the supply voltage from 12 to 15v the value of Vz remained 4.9v, what is happening here is this circuit is being a voltage regulator. what is regulating the voltage is the zener diode because of its properties, the zener diode has a zener voltage of 5.1 voltage which means when the voltage gets above 5.1v the zener will open and alow the rest of the voltage to go to ground which therefore keeps the voltage at Vz around 5.1v. Tis circuit can be used as a voltage regulator.

When i reversed the polarity of the zener diode the value of Vz decreased to 0.8v, because it is in reverse it just acts as a a normal diode which only needs 0.6v-0.7v to break through the diode which explains the small voltage of Vz in this polarity.  

the 4th experiment

componets: breadboard, 1 resistor, 1 5v1 400mW Zener diode, 1 diode 1N4007

Vs=10 & 15v, R=1k ohms

Volt drops:  10V          15V

-V1=             4.6          4.7

-V2=             0.63        0.67

-V3=             5.26        5.48

-V4=             3.75        7.48

Calculated

Current (A)    6.2mA     9.2mA

the voltage drop across the zener (V1) is measuring the zener voltage which is the voltage which passes through at a certain voltage in this case 4.6v with a supply of 10v and 4.7v at 15v supply.

the voltage drop across the 1N4007 diode (V2) is measuring how much the diode uses to breakthrough which in this case is 0.63 for 10v and 0.63v for 15v supply.

V3 is the combine voltage drop of the zener and the 1N4007 diode which is the 4.6 going through the zener and the 0.6 through the other diode (for 10v supply).

as the supply voltage changes from 10v to 15v the amperage in the circuit increases because the resistance in the circuit increases.