EXPERIMENT No.1
Identifying,Testing and Troubleshooting Semiconductor Components
Resistors resist the flow of current.The higher the resistance,the lower the current.Resistors are colour coded,and have a tolerance band.
The voltage gets divided according to the resistance
Voltage works to push the current through the circuit.
picture of a resistor
resistor colour code chart
Reference:chart from elexp.com
Obtain 6 resistors of different values.You are then going to determine their value two ways:
Brown, Black, Red, Gold
1000 Ohms - 5% = 0.982Ohms
950M - 1050M
2090000 - 3210000 Ohms
Red, Red, Green, Gold = 2.168 m Ohms
2200000 Ohms - 5%
105000
Brown, Black, Yellow, Gold = 098.7K Ohms
100000 Ohms - 5%
14.25 - 15.75
Brown, Green, Black, Gold = 015.2 Ohms
15 Ohms - 5%
Orange, Orange, Yellow, Gold = 329.2 K Ohms
330000 Ohms 5%
131500 - 346500 Ohms
Yellow, Purple, Black, Black, Gold = 470.4 Ohms
5%
446.5 - 493.5 Ohms
= 23.5
Choose two resistors and record their individuals ohm resistance value measured with a multi-meter: Brown, Green, Black, Gold
Resistor 1: 15.1 Ohms
Resistor 2: 981 K Ohms
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: 1015 Ohms
Measured Value 1 and 2 in series: 0.996 K Ohms
Put these two resistors together in parallel (connect both ends when they are trying are side-by-side)
Calculate and then measure their combined value. Show workings:
Calculated value 1 and 2 in parallel: 14.7788325 Ohms
Measured Value 1 and 2 in parallel: 014.8 Ohms
What principles of electricity have you demonstrated with this?Explain:
The resistance is high in a series circuit and very low in a parallel circuit
In a series circuit, the resistance is very low. There fore there is more current flowing through the circuit to all the required components and units.
This is very good for high cranking of the engine
For example,Four wheel drives or extreme cold weathers require higher cranking of the engine to start.
EXPERIMENT No.2
DIODES:
A diode is an electronic version of a check valve.It allows electric current to pass in one direction and blocks the current from returning back.
Forward Bias (switched on),P-type is connected with the positive terminal and the N-type is connected with the negative terminal.
Reverse Bias usually refers to how a diode is used in a circuit.If the diode is reverse biased,the voltage at the cathode is higher than the voltage at the anode,therefore,no current will flow until the diode breaks down.
Voltage drop in forward Biased Direction Voltage Drop in Reverse Biased Direction
LED 1.807 V 0
Diode 0.579 V 0
Example how you could identify the cathode without a multimeter:
The cathode is where the band is on the diode.
The anode of the diode has the silver band.
1N4007 Data sheet
Components:1 x resistor,1 x diode. 1 x LED
Exercise:For Vs=5V, R=1K Ohm, D=1N4007 Build the following circuit on a breadboard.
Calculate first the value of current flowing through the diode,mow measure and check your answer?
Show your working.
Calculated: Measured:
V=I X R 4.4 mA
=4.4/1000 Ohms
=0.0044 Amps
Using the data sheet given in Table 1 above,
What is the maximum value of the current that can flow through the given diode?
1 A
For R = 1K Ohm.What is the maximum value of Vs so that the diode operates in a safe region?
1000 V
Replace the diode by an LED & calculate the current,then measure and check your answer.
Calculated: Measured:
I = V/R 0.003 A
= 3.2/1000
= 0.0032 A
= 3.2 mA
What do you observe?Explain briefly.
When the LED circuit is switched ON,the voltage consumed is higher then a normal diode.
The component that is consuming the most voltage is the 1 kilo ohm resistor.
If the resistor is not connected,there would be much higher current flowing through the diode.
The diode cannot handle large currents.
So to prevent any damage to the diode,the resistor is used.
EXPERIMENT No.3
Components:2 x resistors, 1 x 5v1 400mW Zener diode (ZD)
Exercise:Obtain a breadboard,suitable components from your tutor and build the following circuit.
Fig 6
For R =100 Ohms and RL = 100 Ohms,Vs = 12 v
What is the value of Vz?
4.97v
Vary Vs from 10v to 15v
What is the value of Vz
10v : 4.72v 15v : 5.1v
Explain what is happening here
The value of Vz has not changed much between 10 v and 15 v (Vs).This is because the Zener diode requires 5 v to operate.
What could this circuit be used for?
Circuit can be used to protect sensitive equipment from voltage spikes.
Also for a stable voltage
Reverse the polarity of the zener diode.
What is the value of Vz?Make a short comment why you had hat reading.
The value was 0.8 V.
When the zener diode was reversed,it had the same function as the normal diode
EXPERIMENT No.4
Components:1 x resistor, 1 x 5V1 400 mW Zener diode,1 x Diode 1N4007.
Exercise:Obtain a breadboard,suitable components from your tutor and build the following circuit.
Vs=10 & 15v, R= 1 K ohms
Fig.7
10 Volts 15 Volts
Volt drop V1: 4.64 V 4.82 V
Volt drop V2: 0.67V 0.7 V
Volt drop V3: 5.31 V 5.5 V
Volt drop V4: 4.74 V 9.5 V
Calculated current A: 0.0047 A 4.7mA 0.0095 A 9.5mA
Describe what is happening and why are you getting these readings:
The value of the current in this circuit is 0.01 Amps throughout the circuit.The value of the current is not changing because the circuit is connected in series.
The value of V4 has changed to 9.5v from 15 v.
The V1,V2 and V3 values are similar between 10v and 15 v
The Zener diode consumes 5v.and anything under that voltage is blocked.
EXPERIMENT No.5
The Capacitor
The capacitor stores electric charge.
Few different types of capacitors.
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 the capacitance of the capacitor,which is measured in Farads (F).
Fig 8-Capacitor Charging Circuit
Components:1 x resistor,1 x capacitor. 1 x push button N/O switch
Exercise:First,calculate how much time it would take to charge up a capacitor.Then,connect the circuit as shown above.Measure the time taken by the capacitor to reach the applied voltage on an oscilloscope.
.................................................................................................................................................
Circuit # Capacitance(uF) Resistance(K Ohms) Calculated Time(ms) Observed Time (ms)
.............................................................................................................................................
1. 100 1 500ms 500ms
2. 100 0.1 50ms 30ms
3. 100 0.47 235ms 200ms
4. 330 1 1.655ms 1.5ms
.............................................................................................................................................
How does changes in the resistor affect the charging time?
As the resistor's value increases,the charging time also increases.
How does changes in the capacitor affect the charging time?
When the capacitance is higher in the same resistors,the charging time goes up as well.
EXPERIMENT No.6
Meter check of a transistor:
Diode test (V) meter readings
Transistor Number Vbe Veb Vbc Vcb Vce Vec
NPN C547 0.708V 0V 0.706V 0V 0V 0V
PNP C557 0V 0.705V 0V 0.703V 0V 0V
EXPERIMENT No.7
Transistor as a switch
Transistors amplify current.It can be used as a switch (either fully on with maximum current or fully off with no current).It can be used as an amplifier (always partly on)
Components:1 x Small Signal NPN transistor,2 x resistors
Exercise:Connect the circuit as shown in Fig 12 and switch on the power.
Connect the multimeter between base and emitter.
Note the voltage reading and explain what this reading is indicating.
The reading with the multimeter was 0.8v
The current flows from base to emitter.
Connect the multimeter between collector and emitter.
Note the voltage reading and explain what is the reading is indicating.
The reading with the multimeter was 0.6v.
The gate is opened between collector and emitter.
So a higher current flow from the collector to the emitter.
In the diagram below what are the regions indicated by the arrows A and B?
How does a transistor work in these regions?Explain in detail:
When the transistor is between saturation and cut-off.it operates in the active region.
Cut -off region is when there is no collector current.It is completely off.
Base current is too low or zero.
Saturation region is when the transistor is turned on at the maximum level.
What is the power dissipated by the transistor at Vce of 3 volts?
P = Ic x Vce Ic = 13 mA
= 0.013 x 3 = 0.013 A
= 0.039 Watts
What is the Beta of this transistor at Vce 2,3,& 4 volts?
Beta=Ic/Ib Vce=2v Vce=3v Vce=4v
Ic=20mA Ic=13mA Ic=5mA
Ib=0.8mA Ib=0.5mA Ib=0.2mA
B2=20/0.8 B3=13/0.5 B4=5/0.2
=25 =26 =25
EXPERIMENT No.8
Summary:Vary the base resistor and measure changes in voltage and current for Vce,Vbe,Ic, and lb.Then plot a load line.
Set up the following circuit on a bread board.Use a 470R fot Rc and a BC547 NPN transistor.
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 across Vce and Vbe.Also measure and record the current for lc and lb.Then change the Rb resistor and do all the measurements and record the new readings.Do this for each of the resistor values above.
Rb:47K Vbe:0.7V Vce:0.09V lb:0.11mA lc:6.17mA
Rb:220K Vbe:0.68V Vce:0.16V lb:0.03mA lc:6.02mA
Rb:270K Vbe:0.68V Vce:0.20V lb:15.9uA lc:5.71mA
Rb:330K Vbe:0.68V Vce:0.32V lb:13.1uA lc:5.71mA
Rb:1M Vbe:0.65V Vce:2.16V lb:4.2uA lc:2.02mA
Your voltage drop measurements across Vce should vary from below 0.3v (showing the transistor is in the saturated switch region) to above 2.0v (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?
During the experiment when the Rb value was increased,the Vce value also went up.
Discuss what happened for Vbe during this experiment.What change took place if any,and what caused the change?
During the experiment when the resistance was increased,the Vbe value dropped.
The Vbe has a constant 0.6V operating voltage.
Discuss what happened for lb during this experiment.What change took place,and what caused the change?
During the experiment ,when the Rb values were increased,the lb values decreased.
Discuss what happened for Ic during this experiment.What change took place,and what caused the change?
During the experiment when the Rb value was increased,the Ic value dropped.
The Ic values are higher than the Ib values.
Plot the points for lc and Vce on the graph below to create a load line.Plan the values for you so use up the graph space.Use lc as your vertical value,and Vce as your horizontal value.
Using Vbe on the Vce scale,plot the values of lb so the finished graph looks similar to fig 13.
Calculate the Beta (Hfe) of this transistor using the above graph.
B = Ic/Ib
47K B = 6.17/0.11 220 K B=6.02/0.03
= 56.09 = 200.7
270K B=5.95mA/0.0159 330 K B=5.71/0.0131
= 374 = 435.9
1 M B=2.02/0.0042
= 480.9
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 transistor is cut-off when the Vce is very large on the load line graph,and Ib will be nearly zero,and no collector current will flow.
On the other end of the load line,the value of lb increases with the maximum current passing through is 0.11 mA.Then Vce will be nearly zero,and the transistor is saturated.
In both cut-off and saturation.the minmum power is consumed by the transistor.
the transistor can be operated correctly in the active area with different values of lb.
Identifying,Testing and Troubleshooting Semiconductor Components
Resistors resist the flow of current.The higher the resistance,the lower the current.Resistors are colour coded,and have a tolerance band.
The voltage gets divided according to the resistance
Voltage works to push the current through the circuit.
- 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 place smaller
- Last band to right may be tolerance values
Reference:chart from elexp.com
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
Brown, Black, Red, Gold
1000 Ohms - 5% = 0.982Ohms
950M - 1050M
2090000 - 3210000 Ohms
Red, Red, Green, Gold = 2.168 m Ohms
2200000 Ohms - 5%
105000
Brown, Black, Yellow, Gold = 098.7K Ohms
100000 Ohms - 5%
14.25 - 15.75
Brown, Green, Black, Gold = 015.2 Ohms
15 Ohms - 5%
Orange, Orange, Yellow, Gold = 329.2 K Ohms
330000 Ohms 5%
131500 - 346500 Ohms
Yellow, Purple, Black, Black, Gold = 470.4 Ohms
5%
446.5 - 493.5 Ohms
= 23.5
Choose two resistors and record their individuals ohm resistance value measured with a multi-meter: Brown, Green, Black, Gold
Resistor 1: 15.1 Ohms
Resistor 2: 981 K Ohms
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: 1015 Ohms
Measured Value 1 and 2 in series: 0.996 K Ohms
Put these two resistors together in parallel (connect both ends when they are trying are side-by-side)
Calculate and then measure their combined value. Show workings:
Calculated value 1 and 2 in parallel: 14.7788325 Ohms
Measured Value 1 and 2 in parallel: 014.8 Ohms
What principles of electricity have you demonstrated with this?Explain:
The resistance is high in a series circuit and very low in a parallel circuit
In a series circuit, the resistance is very low. There fore there is more current flowing through the circuit to all the required components and units.
This is very good for high cranking of the engine
For example,Four wheel drives or extreme cold weathers require higher cranking of the engine to start.
EXPERIMENT No.2
DIODES:
A diode is an electronic version of a check valve.It allows electric current to pass in one direction and blocks the current from returning back.
Forward Bias (switched on),P-type is connected with the positive terminal and the N-type is connected with the negative terminal.
Reverse Bias usually refers to how a diode is used in a circuit.If the diode is reverse biased,the voltage at the cathode is higher than the voltage at the anode,therefore,no current will flow until the diode breaks down.
Voltage drop in forward Biased Direction Voltage Drop in Reverse Biased Direction
LED 1.807 V 0
Diode 0.579 V 0
Example how you could identify the cathode without a multimeter:
The cathode is where the band is on the diode.
The anode of the diode has the silver band.
Components:1 x resistor,1 x diode. 1 x LED
Exercise:For Vs=5V, R=1K Ohm, D=1N4007 Build the following circuit on a breadboard.
Calculate first the value of current flowing through the diode,mow measure and check your answer?
Show your working.
Calculated: Measured:
V=I X R 4.4 mA
=4.4/1000 Ohms
=0.0044 Amps
Using the data sheet given in Table 1 above,
What is the maximum value of the current that can flow through the given diode?
1 A
For R = 1K Ohm.What is the maximum value of Vs so that the diode operates in a safe region?
1000 V
Replace the diode by an LED & calculate the current,then measure and check your answer.
Calculated: Measured:
I = V/R 0.003 A
= 3.2/1000
= 0.0032 A
= 3.2 mA
What do you observe?Explain briefly.
When the LED circuit is switched ON,the voltage consumed is higher then a normal diode.
The component that is consuming the most voltage is the 1 kilo ohm resistor.
If the resistor is not connected,there would be much higher current flowing through the diode.
The diode cannot handle large currents.
So to prevent any damage to the diode,the resistor is used.
EXPERIMENT No.3
Components:2 x resistors, 1 x 5v1 400mW Zener diode (ZD)
Exercise:Obtain a breadboard,suitable components from your tutor and build the following circuit.
For R =100 Ohms and RL = 100 Ohms,Vs = 12 v
What is the value of Vz?
4.97v
Vary Vs from 10v to 15v
What is the value of Vz
10v : 4.72v 15v : 5.1v
Explain what is happening here
The value of Vz has not changed much between 10 v and 15 v (Vs).This is because the Zener diode requires 5 v to operate.
What could this circuit be used for?
Circuit can be used to protect sensitive equipment from voltage spikes.
Also for a stable voltage
Reverse the polarity of the zener diode.
What is the value of Vz?Make a short comment why you had hat reading.
The value was 0.8 V.
When the zener diode was reversed,it had the same function as the normal diode
EXPERIMENT No.4
Components:1 x resistor, 1 x 5V1 400 mW Zener diode,1 x Diode 1N4007.
Exercise:Obtain a breadboard,suitable components from your tutor and build the following circuit.
Vs=10 & 15v, R= 1 K ohms
10 Volts 15 Volts
Volt drop V1: 4.64 V 4.82 V
Volt drop V2: 0.67V 0.7 V
Volt drop V3: 5.31 V 5.5 V
Volt drop V4: 4.74 V 9.5 V
Calculated current A: 0.0047 A 4.7mA 0.0095 A 9.5mA
Describe what is happening and why are you getting these readings:
The value of the current in this circuit is 0.01 Amps throughout the circuit.The value of the current is not changing because the circuit is connected in series.
The value of V4 has changed to 9.5v from 15 v.
The V1,V2 and V3 values are similar between 10v and 15 v
The Zener diode consumes 5v.and anything under that voltage is blocked.
EXPERIMENT No.5
The Capacitor
The capacitor stores electric charge.
Few different types of capacitors.
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 the capacitance of the capacitor,which is measured in Farads (F).
Components:1 x resistor,1 x capacitor. 1 x push button N/O switch
Exercise:First,calculate how much time it would take to charge up a capacitor.Then,connect the circuit as shown above.Measure the time taken by the capacitor to reach the applied voltage on an oscilloscope.
.................................................................................................................................................
Circuit # Capacitance(uF) Resistance(K Ohms) Calculated Time(ms) Observed Time (ms)
.............................................................................................................................................
1. 100 1 500ms 500ms
2. 100 0.1 50ms 30ms
3. 100 0.47 235ms 200ms
4. 330 1 1.655ms 1.5ms
.............................................................................................................................................
How does changes in the resistor affect the charging time?
As the resistor's value increases,the charging time also increases.
How does changes in the capacitor affect the charging time?
When the capacitance is higher in the same resistors,the charging time goes up as well.
EXPERIMENT No.6
Meter check of a transistor:
Diode test (V) meter readings
Transistor Number Vbe Veb Vbc Vcb Vce Vec
NPN C547 0.708V 0V 0.706V 0V 0V 0V
PNP C557 0V 0.705V 0V 0.703V 0V 0V
EXPERIMENT No.7
Transistor as a switch
Transistors amplify current.It can be used as a switch (either fully on with maximum current or fully off with no current).It can be used as an amplifier (always partly on)
Components:1 x Small Signal NPN transistor,2 x resistors
Exercise:Connect the circuit as shown in Fig 12 and switch on the power.
Connect the multimeter between base and emitter.
Note the voltage reading and explain what this reading is indicating.
The reading with the multimeter was 0.8v
The current flows from base to emitter.
Connect the multimeter between collector and emitter.
Note the voltage reading and explain what is the reading is indicating.
The reading with the multimeter was 0.6v.
The gate is opened between collector and emitter.
So a higher current flow from the collector to the emitter.
In the diagram below what are the regions indicated by the arrows A and B?
How does a transistor work in these regions?Explain in detail:
When the transistor is between saturation and cut-off.it operates in the active region.
Cut -off region is when there is no collector current.It is completely off.
Base current is too low or zero.
Saturation region is when the transistor is turned on at the maximum level.
What is the power dissipated by the transistor at Vce of 3 volts?
P = Ic x Vce Ic = 13 mA
= 0.013 x 3 = 0.013 A
= 0.039 Watts
What is the Beta of this transistor at Vce 2,3,& 4 volts?
Beta=Ic/Ib Vce=2v Vce=3v Vce=4v
Ic=20mA Ic=13mA Ic=5mA
Ib=0.8mA Ib=0.5mA Ib=0.2mA
B2=20/0.8 B3=13/0.5 B4=5/0.2
=25 =26 =25
EXPERIMENT No.8
Summary:Vary the base resistor and measure changes in voltage and current for Vce,Vbe,Ic, and lb.Then plot a load line.
Set up the following circuit on a bread board.Use a 470R fot Rc and a BC547 NPN transistor.
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 across Vce and Vbe.Also measure and record the current for lc and lb.Then change the Rb resistor and do all the measurements and record the new readings.Do this for each of the resistor values above.
Rb:47K Vbe:0.7V Vce:0.09V lb:0.11mA lc:6.17mA
Rb:220K Vbe:0.68V Vce:0.16V lb:0.03mA lc:6.02mA
Rb:270K Vbe:0.68V Vce:0.20V lb:15.9uA lc:5.71mA
Rb:330K Vbe:0.68V Vce:0.32V lb:13.1uA lc:5.71mA
Rb:1M Vbe:0.65V Vce:2.16V lb:4.2uA lc:2.02mA
Your voltage drop measurements across Vce should vary from below 0.3v (showing the transistor is in the saturated switch region) to above 2.0v (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?
During the experiment when the Rb value was increased,the Vce value also went up.
Discuss what happened for Vbe during this experiment.What change took place if any,and what caused the change?
During the experiment when the resistance was increased,the Vbe value dropped.
The Vbe has a constant 0.6V operating voltage.
Discuss what happened for lb during this experiment.What change took place,and what caused the change?
During the experiment ,when the Rb values were increased,the lb values decreased.
Discuss what happened for Ic during this experiment.What change took place,and what caused the change?
During the experiment when the Rb value was increased,the Ic value dropped.
The Ic values are higher than the Ib values.
Plot the points for lc and Vce on the graph below to create a load line.Plan the values for you so use up the graph space.Use lc as your vertical value,and Vce as your horizontal value.
Using Vbe on the Vce scale,plot the values of lb so the finished graph looks similar to fig 13.
Calculate the Beta (Hfe) of this transistor using the above graph.
B = Ic/Ib
47K B = 6.17/0.11 220 K B=6.02/0.03
= 56.09 = 200.7
270K B=5.95mA/0.0159 330 K B=5.71/0.0131
= 374 = 435.9
1 M B=2.02/0.0042
= 480.9
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 transistor is cut-off when the Vce is very large on the load line graph,and Ib will be nearly zero,and no collector current will flow.
On the other end of the load line,the value of lb increases with the maximum current passing through is 0.11 mA.Then Vce will be nearly zero,and the transistor is saturated.
In both cut-off and saturation.the minmum power is consumed by the transistor.
the transistor can be operated correctly in the active area with different values of lb.
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