TTEC 4842 exp 6

Identify the legs of your transistor with a multimeter.

Diode test (V) meter readings	Table Cell	Table Cell	Table Cell	Table Cell	Table Cell	Table Cell
Transistor number	Vbe	Veb	Vbc	Vcb	Vce	Vec
NPN	O.L	O.L	0.834	O.L	O.L	O.L
PNP	O.L	0.841	O.L	0.84	O.L	O.L

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With this experiment what we were trying to do was to figure out which way current will flow on an NPN transistor, and a PNP transistor. Our results show that an NPN transistor, current will flow from emitter to collector once a base current is supplied. The results for the PNP transistor show that current will flow from the collector to the emitter once a base current is supplied.

TTEC 4842 exp 5

Components: 1 x resistor, 1 x capacitor, 1 x O/N switch
Exercise: First, calculate how much time it would take to charge up the capacitor. Then, connect the circuit as shown above. Measure the time taken by the capacitor to reach the applied voltage on an oscilloscope. Fill in the chart below. Also draw the observed waveforms in the graphs below, filling the details on each one.





            

Circuit #	Capacitance (uF)	Resistance (Kohm)	Calculated time (ms)	Observed time (ms)
1	100	1	600ms	500ms
2	100	0.1	60ms	50ms
3	100	0.47	280ms	250ms
4	330	1	2.4s	1.75s

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circuit 1

circuit 2

circuit 3

                                           circuit 4

How does changes in the resistor affect the charging time?
When a resistor is changed, the amount of voltage supplied to the capacitor changes so therefore that effects the rate of charge in the capacitor.


How does changes in the capacitor affect the charging time?
This effects the charging time because a change in capacitor size means a change in storage size, the more storage space the longer it takes to fill the capacitor, and the same rule applies to a smaller capacitor except there is less storage so therefore it takes less time to fill. The capacitor also has a curve in the oscilloscope graph because a capacitor releases voltage gradually, the release time (graph curve) fully relys on the specific specifications of the capacitor being used, so the release time curve will vary for different capacitors

TTEC 4842 exp 4

                 10volts                             15volts
Vd V1:      4.63                                4.80
Vd V2       0.66                                0.68
Vd V3       5.29                                5.48
Vd V4       4.72                                9.50
Calculated current 10v: 10mA     Calculated current 15v: 14.98mA

Describe what is happening and why you are getting these readings:
When the voltage increases all the components use the required amount of voltage to power them which leaves the remainder of the voltage to be consumed by the resistor.

TTEC 4842 exp 3

For R= 100ohm and RL= 100ohm. Vs= 12v

What is the value of Vz?
Vd= 4.78
Vary Vs from 10v to 15v:
What is the value of Vz: 10v= 4.78v  15v= 5.13v

Explain what is happening here?
The voltage flowing through the diode increases when the supply voltage is increased.

What could this circuit be used for?
To regulate voltage in a circuit

Reverse the polarity of the zener diode.

What is the value of Vz? Make a short comment why you had that reading:
The reading we got at Vz in reverse polarity was 0.8v, it was 0.8v because when the diode is in reverse polarity it has more resistance than the resistor in this circuit, so therefore all the voltage will flow through the resistor because electricity will always flow through the path of least resistance.

TTEC 4842 exp 2 (2

Components:
1 x resistor, 1 x diode, 1 x L.E.D
Exercise:
For Vs= 5v, R= 1Kohm, D= 1N007 build the following circuit on a breadboard:


Calculate first the value of the current flowing through the diode, now Measure and check your answer?show your working:
Calculated:
                  4.3/1000  = 0.0043/4.3mA

Measured: 4.60mA

Is the reading as you expected; explain why or why not?
The reading is pretty much exact as to what I expected it to be, it was slightly different due to the tolerance/resistance involved in the individual components involved in this experiment.
Calculate the voltage drop across the diode, now Measure and check your answer?

Calculated: 0.7v

Measured: 0.4v

Using the data sheet given in Table 1,

What is the maximum value of the current that can flow through the given diode?
1A @ 75 degrees c
For R = 1Kohm. What is the maximum value of Vs so that the diode operates in a safe region?
1000v

Replace the diode with an L.E.D & calculate the current, then measure and check your answer?

Calculated: 3.1mA

Measured: 3.17mA

What do you observe? Explain briefly: V = Voltage
The resistor has a voltage drop because it is a consumer of V and is in series so the over all value of V is reduced after being partly consumed by the resistor. The L.E.D has a voltage drop of 1.83v because every L.E.D needs at least that amount of V to let current flow through the circuit and the drop is also due to resistance in both components.

TTEC 4842 exp 2 (1

Components:
1 x Diode, 1 x L.E.D (light emitting diode)


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

Explain how you could identify the cathode without a multimeter:
The easiest way to figure out which side of an L.E.D is the cathode is that it will always be the short piece of wire out of the two wires built into the L.E.D. For a rectifying diode the cathode will always be the side which has the strip on it.

TTEC 4824 exp 1

Exp number one:
Identifying, Testing and Combining Resistors.
1) obtain 6 resistors of different values. You are the going to determine their value two ways:
.Use the colour code to calculate the value of the resistor

.Include the maximum and minimum tolerance value of each resistor

.Then measure the resistor value with a multimeter.

results are shown in the graph bellow:

2) Choose two resistors and record their individual ohm resistance value measured with a multimeter:

Resistor 1: 467ohm    Resistor 2: 814ohm

Put these two resistors together in series calculate and then measure their combine value

Calculated resistor 1, 2 in series: 1281ohm
Measured value: 1283ohm

Put these two resistors together in parallel calculate and then measure their combine value

Calculated resistor 1, 2 in parallel: 467ohm
Measured value: 467ohm

What principles of electricity have you demonstrated with this? Explain:

If the two resistors are put together in series the resistance is what the individual value if each resistor is added together so therefore in series the resistance equals 1281ohm. If the resistors are connected in parallel the resistance of the two is equal to the resistance of the lowest resistor supplied