Experiment No. 7

Connect the multimeter between base and emitter.

Note the voltage reading and explain what this is indicating.

Vbe = 0.7 volts This is because the base and emitter are acting as a diode in forward bias which shows the transistor is operating.

Connect the multimeter between collector and emitter.

Note the voltage reading and explain what this is indicating.

Vce = 0.01 v Because the depletion area is saturated so this means the transistor is operating.

How dose a transistor work in these regions? Explain in detail :

At Vce 5 the Ib = approx 0.2 mA Ic 5mA the transistor starts to operate and as the Vce volts decrease

the Ic mA and Ib mA increase the transistor becomes fully operational at approx Vce 1v, Ic 30mA, Ib 1mA

What is the power dissipated by the transistor at Vce of 3 volts?

14mA x 3v = 42 mA

What is the Beta of this transistor at Vce 2,3 & 4 volts?

2 Vce = 20 Ic mA ÷ 0.8 Ib mA = 25 Beta

3 Vce = 14 Ic mA ÷ 0.5 Ib mA = 28 Beta

4 Vce = 5 Ic mA ÷ 0.2 Ib mA = 25 beta

EXPERIMENT No. 5 CAPACITORS

Exercise : First, calculate how much time it would take to charge up the capacitor. Then connect the curcuit as shown above. Measure the time taken by the capacitor to reach thr applied voltage on an oscilloscope.

curcuit 1 : 1 k Ω x 0.0001 x 5 = 0.5 ms

circuit 2 : 100 Ω x 0.0001 x 5 = 0.05 ms

curcuit 2

curcuit 3 : 470 Ω x 0.0001 x 5 = 0.235 ms

curcuit 3

 curcuit 4 : 1 k Ω x 0.00033 x 5 = 1.65 ms

curcuit 4

 How dose changes in the resistor affect the charging time?

The lower resistance the faster the capacitor charges in steps of 5 the first is rapid because the current is large but as the current decreases the charge builds up more slowly and the voltage increases more slowly.

How dose the changes in the capacitor affect charging time?

The higher the uf capacitor the longer it takes to charge.                                                       

EXPERIMENT No. 4

                       10 Volts                                                               15 Volts

volt drop V1 :  3.978 v                                                                   3.978 v

volt drop V2 :  0.743 v                                                                   0.745 v

volt drop V3 :  4.521 v                                                                   4.744 v

volt drop V4 :  5.763 v                                                                   10.53 v

calculated current A: 10 amps                                                         15 A

Describe what is happening and why you are getting these readings?

V1 = 3.978 because zener diode is reverse bias. V2 is 0.7 v because that is the knee voltage.
V3 = V1 + V2 = 4.521. V4 = 5.763 the zener in reverse bias in kneeds 5 v to get to break down voltage + the 0.7of the other diode = 5.763. With the 15 v curcuit the zener and other diode 1N4007 take 5.53 v so the resistor will use up the rest which is 10v.

                                                  

EXPERIMENT No.3

What could this circuit be used for?

To regulate voltage for some component.

Reverse the polarity of of the zener diode. What is the value Vz? Make a short comment why you had that reading.

Vz = 0.740 v This is because now the diode has forward bias it is using its knee voltage 0.7 v

 

Exercise Vs = 5v, R =1kohm, D =1N007 Breadboard Curcuit

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


Calculated value : 4.3 mA

Measured value  : 4.1 mA







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

Yes it is what I expected because the data sheet says average rectified current @ 75 degrees is 1.0 amp and my reading is 4.1 - 4.3 mA

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

amps x resistance =  volts = 4.3 mA = 0.0043 x 1000 ohms = 4.3 v =  vd = 0.7 v

measured vd = 0.663 v

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

1 amp at 75 degrees

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

1 amp x 1000 ohms = 1000 volts

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

5v - 3.102 = 1.898v devided by 1000 = 1.89 mA

Mesaured value = 1.59 ma

 

What do you observe? explain briefly.

The diode has a lower resistance so has higher milliamps going through and the LED has higher resistance so has lower milliamps going through.

Workings

Calculated value Rt in parallel : 19 ohms

Measured value Rt in parallel : 18.8 ohms




What principles of electricity have you demonstrated with this?

When two resistors are in series they are added together to get total resistance. When in parallel the total resistance is less than the lowest resistor.

Resistors

Choose 2 resistors and record treir individual ohm resistance measured with a multimeter:

Resistor 1 - 390 ohm Resistor 2 - 20 ohm

Put these two resistors together in series and calculate and then messure their combined value.

Calculated value in series R1 + R2 = 410 ohm

Measured value in series 409 ohm

Put the two resistors together in parallel calculate and then measure their combined value. Show workings.