EXPERMENT NO. 2 DIODES

Componint: 1 diode, 1 LED
Exercise: Using a multimeter, identyfy the anode and cathode of the diode& LED.

	Voltage drop in forward Biased direction.	Voltage drop reverse biased direction
LED	1.77 V	OL
Diode	0.65 V	OL

I have set my multimeter to "diode check function , then first connected mutimeter (+) led to the anode and (-) led to cathode . I have correctly identetyfy the anode and cathode the meter reading wich ihave got at forward biase in diiode is 0.65 V . the reverse biase i hade connect the multimeter (+) led to cathode and the (-) led to anode wich the reading is OL it is contintue no current pass on revers biase.
A diode is valve that means electrical current flow in one diriction.
For the LED i have used the same prosseger . LED has two termenal Anode & Cathode, the cathode has short lead and the anode is has biger wich connected to (+). The reading is forward biase 1.77 V and reverse biase OL.

We can identify the diode without multimetr anode and cathode:
LED has two legs the cathode terminal is shoter then anode, and also we can see two leads go inside the LED, we can notice that each leads has tringular node attach to it the anode is physically bigger node.
Thediode has silver strip around that is cathode the black side is anode.

Table 1: Data sheet of 1N4007 

Components; 1 resistor, 1diode , 1 LED
For Vs=5V, R= 1K ohms, D= 1N4007 Ihave build the follwing

First the valve of current throgh the diode
calculated;
A= V/R
VS=5V, R=1Kohms
Id= (5V-0.6DV)/1000 ohms=00.0435 (4.35 MA)

Measured
4.5MA
Is the ading as you expected; explain why or why not?
If we know ohms law and the valve of the component such as resistor, volt supplay and diode and voltage drop theen we can gets easey to find the current in the circuit. that expirement my calculation and measurment the same and correct i expected that.

Calculate the voltage drop acoss the diiode and measure.

Calculated
V= R*A
1000 ohms * 0.00435A= 4.35V +0.65DV= 5V
OR Vs=5V-0.65DV=4.35V
Measured : the measur you can see on the picture wich show 0.63V .

Using the data sheet given the the table 1 above
What is the maximum valve of the current that can flow throgh the given diode?  From the data sheet 1.0 A.

For R= 1Kohms. what is the maximum valve of Vs so that diod operate in safe reagion?
From data sheet VRRM (PIV) 1000V.
Avarage rectiied current @ TA= 75 c 1.0 A.
For 1Kohms we can flow ohms law to find it.
R= V/I
1000V/1.0 A= 1000 ohms.
For 1K ohms  1000V> Vs that diode operate in safe reagion.

Reflace the diode by an LED & calclate the current that measure and check your answer?

Calculated
I= Vs- V LED / R
5Vs- 1.8 V LED / 1000ohms R = 0.0032 (3.2 MA)
I= 3.2 MA

Measured  3.1 MA .
Sorrey i have delated the photo for this circuit by mistake.

I have noticed diffrence of the current flow between LED and diode in this circuit indiode more currnt flow then LED, because the Vd of diiode 0.6 V & Vd of LED is 1.8 V. That way high voltage more current flow & low voltage less current flow throgh the circuit. I happy withe my expermit cause the calculation and measurment was smame.

EXPERRIMENT NO. 1 Identyfing Testing and trubleshooting Semiconducter componenta

Identyfying , Testing and combining Reseators

The first two or thre colour bands have valve we can write the number.
The third or next number is the multiplier numbers (how many zero can add to the number).
The 3rd band also has two new colours where Gold=.1 and selver  = .01
The last band is the resistor tolerance valve Gold = 5% and Selver = 10 % and no band whatever = 20%

Ihave choosed 6 resistors diffrent colours bands including tolarance vaalve below chart i have record the valve of colours code and valve of multimeter.

Colour of bands	Valve (colour cods)	Valve (multimeter)
Green,violet,grey,gold	570000000+-5%	5.68 M ohms
Brown, black, orange, gold	10000+-5%	10 K ohms
Brown, black, red, gold	1000+-5%	1 K ohms
Red, violet, green, gold	2700000+-5%	2.7 M ohms
Orange, orange, black, gold	330+-5%	33 ohms
Red, red,green, gold	2200000+-5%	2.190M ohms

I have  Choose two resistors and recorded their individual ohm resistance value measured with a multimeter:
resistor 1: 1.93 ohms                Resistor 2: 33 ohms
I have put these resistors together in series and then calculated and measured their
calculated value 1 and 2 in series: RT= R1+R2 (R1:993 ohms + R2: 33 ohms = RT 1026 ohms
measured value 1 and 2 in series: 1.26 K ohms

And then i put these two resistors together in parallel
Calculated value 1 and 2 in parallel
RT = (R1 x R2/R1+R2)  993 x 33 = 32769/1026 = 31.93 ohms

What principles of electricity have you demonstrated with this? explain.
In series circuit the total resistance equal to resistor 1 and resistor 2 (RT = R1+R2+R3) . In this experiment i have put two resistors in series and the resistance for both were equal.
If two resistors are connected in parallel the total resistance can be found using the formula RT = (R1 x R2/R1+R2)
The total resistance in parallel circuit must be less than the smallest resistance in the circuit.