experiment 2

Experiment 2:

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	1.83V	2.5V(open circuit)
Diode	0.65V	2.5V(open circuit)

Explain how you could identify the cathode without a multimeter?

Diode: There is a line on the cathode side of a common rectifier diode.

LED: Shorter side of the LED is the cathode, the taller one is anode.



Table: Data sheet of 1N4007 is as follows


Calculate first the value of the current flowing through the diode, now measure and check your answer?
Calculated: Measured:
I = Vs – voltage drop ÷ R1
I = 5 - 0 .7 ÷ 1000
I = 0.0043 Amps or 4.3 mA
Is the reading as you expected; explain why or why not?
The answer is pretty close since we got the resistance and the voltage supply and voltage drop its quiet easy to calculate and the answer is close enough.


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

Calculated: 0.65 V Measure: 0.65 V
Using the data sheet
What is the maximum value of the current that can flow through the given diode?
1 Amp


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


Because of the gold strip at the end of the resistor it shows +5% or -5% of the 1000v by using ohms law we got the 1000V which is V= R * I .

Max Voltage = R * I max


Vmax=1000*1


Vmax=1000 Voltes


By using the table we get the max voltage that can be put in the reverse direction before destroying the resistor.


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

Calculated
I = Vs – VD ÷ R
I = 5 – 1.8 ÷ 1000
I = 0.0032 A or 3.2 mA


Measured
3.1 mA


What do you observe? Explain briefly


The current changes slightly due to the different voltage drop of the components. When we have the LED connected we have around 3.2 mA in the circuit but when the diode has less voltage drop then the current is a little bit higher as shown below.


I = Vs –VD ÷ R
I = 5- 0.7 ÷ 1000
I =0.0043 A or 4.3 mA