EXPERIMENT No.2 DIODES

EXPERIMENT No. 2

DIODES

Fig 3 - Diode Symbol & Physical component

Fig 4 – Diode symbol and P.N. junction

A diode has the characteristics of:

·       An insulator when current tries to flow in one direction.

·       A conductor when current flows in the other direction.

  

Components: 1 x diode, 1 x LED

Exercise: Using a multimeter, identify the anode and cathode of the diode and the LED.

	Voltage drop in forward Biased Direction.	Voltage drop in reverse biased direction
LED	1.784v	0v
Diode	0.568v	0v

Explain how you could identify the cathode without a multimeter

To identify the cathode side of a LED without a multimeter look on one side and it should be flat. The flat side identifies the cathode. This is a good way to identify the cathode side because the leads could be cut. The anode lead is longer than the cathode. Below is picture I saved of the internet showing an LED and its flat side.

(Refference) http://elecrom.wordpress.com/2008/03/02/anode-and-cathode-of-led/

On a normal Diode the grey band represents the cathode side. The anode lead is longer than the cathode. Below is a picture showing a diode and its grey band showing the cathode side.

(Reference) http://camhardyreidttec4841.blogspot.com/2011/05/diodes.html

Table 1: Data sheet of 1N4007 is as follows

Absolute Maximum Ratings, TA = 25OC
Symbol	Parameter	Value	Units
IO	Average rectified current @ TA = 75oC	1.0	A
PD	Total device dissipation Derate above 25oC	2.5 20	W mW/OC
	Thermal resistance, Junction to Ambient	50	OC/W
	Storage Temperature Range	-55 to + 175	OC
	Operating Temperature Range	-55 to + 150	OC
VRRM  (PIV)	Peak repetitive reverse voltage	1000	V

Components: 1 x resistor, 1 x diode. 1 x LED

Exercise: For Vs=5V, R= 1KΩ, D= 1N4007 build the following circuit on a breadboard.

Fig 5

Calculate first the value of current flowing through the diode, now measure and check your answer?

Show your working

Calculated                                                                          Measured

I = V/R = 5 – 0.6 = 4.4/1000W = 4.4mA                           4.4mA                                                 

Is the reading as you expected; explain why or why not?

The reading is as I expected because I calculated the current using ohms law and then I measured the current flow with a meter and compared answers. 

Calculate the voltage drop across the diode, now measure and check your answer?

Calculated                                                                          Measured

V = I X R = 4.4 X 1000 = 4.4mA                                         0.6VD

VD = Vs – V1 = 5 – 4.4 = 0.6v                                                                                  

Using the data sheet given in Table 1 above,

What is the maximum value of the current that can flow through the given diode?

The average rectified current that can flow through the given diode is 1.0A @ 75°C

For R = 1KΩ.  What is the maximum value of Vs so that the diode operates in a safe region?

The maximum value of Vs so that the diode operates in a safe region is 1000 V.

Replace the diode by an LED & calculate the current, then measure and check your answer?

Calculated                                                                          Measured

I = V/R = 5 – 1.8 = 3.2 / 1000 = 0.0032A                           3.2mA

What do you observe? Explain briefly.

Looking at the LED and Diode, the LED requires a higher potential difference to move electrons through the boundary layer and turn it on than the Diode. electric charge flows through the Diode at a less voltage to open the gate therefore more electrons can flow freely.