EXPERIMENT NO.4

EXPERIMENT NO. 4

Component : 1 * resistor, 1 * 5v1 400W Zener diode,  1* Diode 1N4007 .

Exercise:  Vs = 10& 15v, =1K ohms.

10 volt                  15V

Volt drop V1:         4.30v                    6.45

Volt drop V2:         0.70v                    0.70V

Volt drop V3:         5v                         7.35V

Volt drop V4:         5V                        7.25V

Calculated current A :

I = V/R = 10/1000 = 0.01 AMPS (1 A)       

I = V/R = 14.80/1000 = 0.0148 AMPS (1.48 A)

In this series circuit V1 – Zener diode in reverse bias and takes about  5v to turn it on (gates open) and the switcing diode(V2) is in forward base. In there should voltage for V2 is ranged 0.6v-0.7v to turn it on, v3 is the total of (V1 + V2) voltage drop and the rest of the voltage is consumed in V4 which is giving us the total voltage supply.

V1 + V2 = V3 and V3 + V4 = SV

Experiment no.3

Experiment no:3

Zener diode

A zener diode is a special type of diode which allows current to flow in forward direction as normal, but it also allows it to flow in reverse direction, when the voltage above certain value- the brakedown voltage known as zener voltage

How it works?

The zener voltage of standard diode is high, but if a reverse current above that value is allowed to pass through then the diode is permanently damaged.

Zener diodes are designed so that their zener voltage is much lower

If zener diode is connected in forward direction it behaves exactly the same as a standard diode.example; small voltage drop 0.3 to 0.7 volts with current flowing through pretty much. In the reverse direction their is a small leakage current between ov and the zener voltage.

Example;  just tiny amount of current is able to flow. Then, when the reach the voltage(vz) suddenly current can flow freely through it

Uses of zener diodes. Zener diodes are specially used to regulate the voltage in circuits. Using resistors to ensure that the current passing through the zener diode is at leas 5 MA (0.005 amps)

Excersise

Components: 2 x resistors, 1 x 5V1 400mW zener diode (Zd)

For R= 100 ohms and RL= 100 ohms, Vs= 12V

I=V/R IL=5V/100 W = 50 MA

IS = IZ + IL, IS = 80 mA + 50 mA = 130 mA

R = V/I = 12-5 V/130 = 0.0538 ohms

What is the value of Vz? 5v

Vary Vs from 10 V to 15? 13.27V

What is the value of Vz?5.07V

What could this circuit be used for? Voltage regulator

Reverse the polarity of the zener diode

What is the value of Vz? Make short comment why you had that reading.

In the reverse direction is very small leakage current between 0.3 to 0.7 volts. In this circuit my meter reading is 0.3V and there was higher voltage drop so it means it needs higher voltage to push through, where as in the forward base. There is alot less voltage drop so it needs less voltage for current to go through

this is a picture of my experiment

Experiment no.2 diode

Experiment NO:2

DIODS

First I would like to explain about diodes . Diode is an electrical device that allowing current flue in one direction with far greater easy than other. The most common type of diodes in modern circuit design is semi conductor diode.And also diode an insulator to

current flow the opposite direction. Diodes are used for regulate voltage, switching applications, current guides, tuning device, circuit mixing can say protect the circuit. There are several types of diodes we have, some of semi conductor diode zener

diode. Dioode Light-emitting diode (LED) and silicon controlled

diode rectifier which we use in a upcoming experiment.
Forward voltage drop.The electricity use up a little average pushing its way through the diode. There is a small voltage across a conducting diode, it’s called forward voltage drop in normal diode about 0.7v.Reverse voltage is applied perfect diode

does not conduct but real diodes leak very tiny current a few ua or less. This can characteristic of silicon diode in the most circuit it would be very small amount of current. In the forward direction most diode have maximum reverse voltage (50v) or more. If this exceeded the diode will fail and a large

current pass in the reverse direction, this is called break down.

Testing Diodes
We can test a diode with a multi meter.Diode must be connected the correct way anod (a) and cathode (k) the cathode is marked by line printed on the body anod (+) and cathode (-). Diodes are level with their code number. In

the next experiment i used rectifier diode (large current) which are used in power supply to convert (AC) to (DC) in alternator also used in circuit where a large current must pass through the device.

Component : 1*diode, 1*LED

	Voltage drop in forward biased direction.	Voltage drop in reverse biased direction.
LED	1.8 v	oL
Diode	0.67 v	0.0L

Explain how you could identify the cathode without a multimetr.

Without meter the cathode (K) side of diode is colure with silver strip.

The LED cathode side is short lead and slight flat on the body of round leads. Inside the LED the cathode is the larger electrode

Component: 1* resistor , 1* diode. 1 LED.

Exercise: For Vs= 5v, R= 1Kohms, D =14700 build the following circuit on a breadboard.

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

Show working

Calculated                                                                                 Measured

I = V/R = 5-0.7 = 4.3/1000 Ohms = 4.3 mA                                   4.3 mA

The reading is which I expected, because the calculation and the reading is the same , at this experiment i follow the ohms law.

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

Calculated                                                                                  Measured

V= I*R= 4.3 mA * 1k Ohms = 4.3                                                   0.65 V

VD = VS - V1 = 5 - 4.3 mA = 0.7 V

The maximum current that can flow through the given diode is 1.0 Amps .

For R= 1Kohms the maximum value of VS so that diode operates in safe region is 

VS = IR      1 * 1000 = 1000 = R = 1K Ohms

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

Calculated                                                                                  Measured

A = VR = 5 - 1.8 / 1000 = 3.1 / 1000 = 0.0031 Amps                     3.1 mA

The voltage across diode is less than the voltage drop across LED, so LED has more current flowing then diode.

experiment no.1

identifying, testing and combining resistors

Resistors produce voltage across its terminal that is proportional to the electrical current. It is a two terminal electronic component.

v = IR

the first two or three bands to write down. the next band is multiple to added the number. gold multiplier makes one decimal place small. silver makes two decimal point smaller. the last band to right may be tolerance values.in this expirement im going to obtain 6 resistors of different values

im using the colour code to calculate the value of resistors
this including the maximum and minimum tolerence value of each resistor
the value of resistor measuring with multi meter

value (colour code) value (multimeter)

1. 100 ohms 5% 98.3k ohms
2. 560000 ohms 5% 555k ohms
3. 330000 ohms 5% 330k ohms
4. 10000 ohms 5% 9.9k ohms
5. 10000 ohms 5% 99.6k ohms
6. 560000 ohms 5% 566K ohms

Now im choosing two resistors to record their individual ohm resistance value measure with multimeter.

resistor 1: 98.4 ohms resistor 2: 258k ohms

And then i put these two resistors in series.

calculated value 1 and 2 in series: RT=R1 + R2 = 98.4 ohms + 258K ohms = 258.984k ohms

And then i put these 2 resistors in parallel

calculated value of 1 and 2 in parallel: 98.36 ohms

measured value 1 and 2 in parallel: 98.4 ohms

RT= R1 x R2/R1 + R2

what principles of electricity have you demonstrated with this? explain:

in series circuit we added up both resistors resistance to give us the total resistance. Resistance has increased in , but in parallel circuit the total resistance has decreased, an electricity rule in parallal circuit the total resistance must be less than the small resistor value in the circuit.

ie: RT = R1 x R2/ R1 + R2

resistors connected in series, when they are chained together in single line. all current flowing through the first resistor has no other to go out, it must pass through the second resistor

ie: RT = R1+ R2