EXPERIMENT No. 1
Identifying, Testing and Troubleshooting Semiconductor Components.
Identifying, Testing and Combining Resistors.
- First two or three bands may be the numbers to write down
- Next band is the multiplier (how many zeros to add to the number)
- Gold multiplier makes one decimal place smaller, Silver makes two decimal places smaller
- Last band to right may be tolerance values
- Notice the examples on the right
- Brown, red, red = 1, 2, 100, 5% = 1200Ω 5%, or 1.2KΩ,1K2
Obtain 6 resistors of different values. You are then 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.
Record the values in the chart below.Value (colour codes ) | Value (multimeter) |
Choose two resistors and record their individual ohm resistance value measured with a multi-meter:
Resistor 1. 269.1 Ω Resistor 2. 471.9 Ω
Put these two resistors together in series (end to end, one right after another) calculate and then measure their combined value. Show workings:
Calculated value 1 and 2 in series: 741.7 Ω RT=R1 +R2
Measured value 1 and 2 in series: 739.00 Ω
Put these two resistors together in parallel (connect both ends when they are side-by-side). Calculate and then measure their combined value. Show workings:
Calculated value 1 and 2 in parallel: 171.3 Ω 1over RT = 1 over R1 + 1 over R2
Measured value 1 and 2 in parallel: 171.3 Ω
What principles of electricity have you demonstrated with this? Explain:
When resistors are connected in series the resistor values are just added together.
This circuit shows RT = R1 +R2 formula with its principles .
As a results total resistor values increases however in parallel circuit which uses R1 over RT=
1 over R1 + 1 over R2 formula and total resistance value is low than the lowest resistor value .
EXPERIMENT No. 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 | ||
Diode |
Explain how you could identify the cathode without a multimeter.
On LED the cathode side is smaller then the anode .and also there is
small edge cut off that represents the cathode side.
On diode the grey band represents the negative which is the cathode and some cases you cant find it...
On LED the cathode side is smaller then the anode .and also there is
small edge cut off that represents the cathode side.
On diode the grey band represents the negative which is the cathode and some cases you cant find it...
Table 1: Data sheet of 1N4007 is as follows
Absolute Maximum Ratings, TA = 25OC | |||
Symbol | Parameter | Value | Units |
IO | Average rectified current @ TA = 75oC | 1.0 | A |
PD | Total device dissipation Derate above 25oC | 2.5 20 | W mW/OC |
 | Thermal resistance, Junction to Ambient | 50 | OC/W |
 | -55 to + 175 | OC | |
 | Operating | -55 to + 150 | OC |
VRRM (PIV) | Peak repetitive reverse voltage | 1000 | V |
Components: 1 x resistor, 1 x diode. 1 x LED
Exercise: For Vs=5V, R= 1KΩ, D= 1N4007 build the following circuit on a breadboard.
Calculate first the value of current flowing through the diode, now measure and check your answer?
Calculated : I = V/R
I = 5/1000 omhs = o.005 A
Measured: 0.0055ma
Is the reading as you expected; explain why or why not?
Yes the reading we got is the same as expected because we calculated.
in series current flows one way throughout the circuit
and it also stays the same .
What is the maximum value of the current that can flow through the given diode?
Average rectified at TA = 75* a.00 A
For R = 1KΩ. What is the maximum value of Vs so that the diode operates in a safe region?
peak repetitive reverse voltage 1000V
Replace the diode by an LED & calculate the current, then measure and check your answer?
calculated I = V/R measured 0.003 A
What do you observe? Explain briefly.
The current changes a little due to different voltage drop of the components
when we have LED connected we have around 0.003A in the circuit
when the diode has less voltage drop then the current is a little bit higher.
EXPERIMENT No. 3
Components: 2 x resistors, 1 x 5V1 400mW Zener diode (ZD).
For R= 100Ω and RL= 100Ω, Vs= 12 V.
What is the value of Vz?
VZ= 4.96v
Vary Vs from 10V to 15 V: 10V = 4.77v 15V= 3.06V
What is the value of Vz: VZ = 4.77V VZ = 5.06 V
Explain what is happening here.
At low the value of VZ is 4.77 and as is increase the Vs to 15 the value of the vz
is 5.06V because there is more vs.
The voltage is staying at around 5v regardless of the supply voltage
increasing this is because the diode is acting as a regulator because
it is in reverse bios.
What could this circuit be used for?
Dis piking protection and stable voltageEXPERIMENT No. 4
Components: 1 x resistors, 1 x 5V1 400mW Zener diode, 1X Diode1N4007 .
Exercise: Obtain a breadboard, suitable components from your tutor and build the following circuit.
Vs=10 & 15v, R=1K ohms
10 Volts
Volt drop v1: 4.6vd
volt drop v2: 0.69vd
volt drop V3: 5.29vd
volt drop v4: 5.24vd
calculated current A: i =10v/1000 ohms = 0.01
i= V/R
15 volts
volt drop v1: 9.68 vd
volt drop v2: 5.74 vd
volt drop v3: 0.69vd
volt drop v4: 4.77vd
calculated current A: i = 15/1000 = 0.015 A
Describe what is happening and why you are getting these readings:
V4 value is changed at 9.5v under 15 v however the v1,v2 and v3 values are
similer between 10v and 15 volts because the Zener diode keeps 3 volts and
block less 5v.so the v4 of the resistor only changed.
The current value is at 0.01A all areas due to this circuit is connected in series.
EXPERIMENT No. 5
The Capacitor
The capacitor stores electric charge.
A capacitor consists of two metal plates very close together, separated by an insulator. When connected to a battery or power source electrons flow into the negative plates and charge up the capacitor. The charge remains there when the battery is removed. The charge stored depends on the “size” or capacitance of the capacitor, which is measured on Farads (F).
Identifying Capacitor “Size”
If the Farad “size” is not printed on the capacitor, you may find an EIA code listed. Use the table below to figure out the capacitance
μF | nF | pF | EIA Code |
0.00001* | 0.01 | 10 | 100 |
0.0001* | 0.1 | 100 | 101 |
0.001 | 1.0 (1n0) | 1,000 | 102 |
0.01 | 10 | 10,000* | 103 |
0.1 | 100 | 100,000* | 104 |
1.0 | 1000* | 1,000,00* | 105 |
10.0 | 10,000* | 10,000,00* | 106 |
RC Time Delay or “Charging Time”
Capacitors take time to charge. It doesn’t happen instantly. The charge time is dependent on the resistor in the circuit and the size of the capacitor. And it is expressed in the equation: R x C x 5 = T. This is the time it takes to charge up to the applied voltage.
For example, 1,000,000 Ω x 0.000001 F x 5 = 5 seconds to charge to applied voltage. This can also be expressed as 1 MΩ x 1 μF x 5 = 5 seconds.
Capacitors are often used for timing when events take place. And often the voltage only has to get up to about 2/3 the applied voltage, and this happens at about 1/5 the time of their charging. So this is why the 5 is built into the equation. The concept of “time constants” is used here, where whatever the time it takes for a capacitor to build up to the full charge, it takes about 1/5 of that time to build up close to 2/3 of the charge. So you can divide the charge time into 5 segments, and the first time segment is often the time you are interested in.
Fig 8-Capacitor Charging Circuit
Components: 1 x resistor, 1 x capacitor. 1 x pushbutton N/O 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 number | Capacitance (uF) | Resistance (KΩ) | Calculated Time (ms) | Observed Time (ms) |
1 | 100 | 1 | ||
2 | 100 | 0.1 | ||
3 | 100 | 0.47 | ||
4 | 330 | 1 |
capcaitance 100uf Resistance: 1k ohms
Capacitance: 100 uF Resistance: 0.1k ohms
Capactance : 100uF Resistance: 0.47 k ohms
Capacitance: 330uF Resistance 1K ohms
How does changes in the resistor affect the charging time?
The lower the resistance the more current getting through
Therefore quicker the charge time.
How does changes in the capacitor affect the charging time?
The more capacitance of the capaitor the more charge it can hold
therefore longer time it will take to charge up.
EXPERIMENT No. 6
Identify the legs of your transistor with a multimeter. For identifying and testing purposes, refer to the representation shown above.
Diode test (V) meter readings | ||||||
Transistor number | VBE | VEB | VBC | VCB | VCE | VEC |
NPN | ||||||
PNP | ||||||
Emitter has slightly higher voltage then the collector.
EXPERIMENT No. 7
Transistor as a switch
Components: 1 x Small Signal NPN transistor, 2 resistors.
Exercise: Connect the circuit as shown in Fig 12 and switch on the power supply.
Connect the multimeter between base and emitter.
Note the voltage reading and explain what this reading is indicating.
0.798 volts,This reading shows that the current is flowing from the base to the emitter
This means that the transistor is turned on and the current should
be flowing from the collector to the emitter.
Connect the multimeter between base and emitter.
Note the voltage reading and explain what this reading is indicating.
0.054 volte (VD) This reading shows thr voltage drop across the transistor
from thr collector to the emitter. this also shows that the
transistor is fully open or saturated and minimal voltage is required to
allow current to flow through the transistor.
In the plot given below what are the regions indicated by the arrows A & B?
How does a transistor work in these regions? Explain in detail:
This graph shows that all the area marked with A is the area when the transistor
is fully turned on or saturated this mean that no voltage is required to push
current through from the collector to the emitter .
Area B is the area where the transistor is off there is no current flowing from
the base to the emitter so no current can flow from the collector to the emitter.
What is the power dissipated by the transistor at Vce of 3 volts?
pd = p = v x i = 3 x 0.013 = 0.039 watts
What is the Beta of this transistor at Vce 2,3 & 4 volts?
Beta = gain B = 1/IB (base current)
2 volts = 21 /0.75 = 28
3volts = 14 / 0.5 = 28
4 volts = 7 / 0.2 = 35
EXPERIMENT No. 8
Summary: Vary the base resistor and measure changes in voltage and current for Vce, Vbe, Ic, and Ib. Then plot a load line.
Pick five resistors between 2K2 and 1M for Rb. You want a range of resistors that allow you to see Vce when the transistor is the saturated switch region and when it is in the active amplifier region. I used 47K, 220K, 270K, 330K and 1M, but this can vary depending on your transistor. Some may need to use 2K2. Put one resistor in place, and measure and record voltage drop across Vce and Vbe. Also measure and record the current for Ic and Ib. Then change the Rb resistor and do all the measurements and record the new readings. Do this for each of the resistor values above.
Record here:
Rb 4.39v Vbe 0.696v Vce 0.697v Ib 21.1 ua Ic 5.37mA
Rb 4.689v Vbe: 0.754v Vce: 47.4mv Ib: 461.6ua Ic: 6.92mA
Rb 4.821v Vbe: 0.668v Vce: 2.444v Ib: 8.3uA Ic: 2.16mA
Rb 4.635v Vbe: 0.757v Vce: 43.7mv Ib:562uA Ic: 6.65mA
Rb 4.716 Vbe: 0.673v Vce:2.217v Ib:10uA Ic:2.60mA
Your voltage drop measurements across Vce should vary from below 0.3 v (showing the transistor is in the saturated switch region) to above 2.0 v (showing the transistor is in the active amplifier region) If this is not the case, you may have to try a smaller or bigger resistor at Rb. Talk to your teacher to get a different size resistor, and redo your measurements.
Discuss what happened for Vce during this experiment. What change took place, and what caused the change?
The different resistance of the Rb meant there was more or less amperage flowing
from thr base to the emitter this will mean that the resistor is more of less
saturated or turned on .Although voltage from the base to the emitter is roughly the same
it is the amperage that determins how well the transistor is turned on and
how big or small of voltage drop the will be from the collector to the emitter.
Discuss what happened for Vbe during this experiment. What change took place if any, and what caused the change?
The voltage from the base to the emitter does not change too much as the control side
of the transistor only requires 0.6 - 0.7 volts switch on high power sides of the transistor from the collector to the emitter.
Discuss what happened for Ib during this experiment. What change took place, and what caused the change?
Current flow at base was affected by the transistor used if there was a large resistor
i.e higher resistance to current flow then current flow is reduced .
If smaller resistor used with less resistance to current flow then current flow is higher as
it easier for current to flow.
Discuss what happened for Ic during this experiment. What change took place, and what caused the change?
The resistors that changed at Rb meant there was more or less restriction to the current flow
through the circuit,this in turn affected how much current would flow
through the collector to the emitter.
Plot the points for Ic and Vce on the graph below to create a load line. Plan the values for so you use up the graph space. Use Ic as your vertical value, and Vce as your horizontal value.
Using Vbe on the Vce scale, plot the values of Ib so the finished graph looks similar to fig 13
Calculate the Beta (Hfe) of this transitor using the above graph.
B=ic/iB
Explain what the load line graph is telling you. Discuss the regions of the graph where the transistor is Saturated, Cut-off, or in the Active area.
The load line is telling us the relationship between current to the collector (IC)
and voltage drop from thr collector to the emitter.
(VCE) This can tell you when the transistor is fully saturated or turned on as the line
will show high current flow from the base to emitter meaning the transistor
is more turned on so there is small voltage drop which seen at VCE.
Reference
Google images
Unitec moodle
Not enough explanations, you need to do more , "theory" , "procces" & "refelction" writing not just what you did in class
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