FREE ESSAY ON BIPOLAR JUNCTION TRANSISTOR

BIPOLAR JUNCTION TRANSISTOR

Purpose
The objective of this lab is to introduce the Bipolar Junction Transistor (BJT). A BJT is
a three terminal device composed of an emitter, base, and collector terminals. In this
lab we will introduce two major types of BJT's : npn and pnp. The first, npn, has an
n-type emitter, a p-type base and a n-type collector. On the other hand the pnp has a
p-type emitter, a n-type base, and a p-type collector. Also the transistor consists of
two major pn junctions, the emitter-base junction (EBJ) and the collector-base junction
(CBJ). Depending on the bias condition of each of these junctions, there are different
modes of operation. We will show that the basic principle of a BJT is the use of the
voltage between two terminals on order to control the current in the third terminal. 
Activity One 
Diagram of a npn BJT 2N3904
Theory
In this part of the lab we will use the curve tracer to display the common-emitter BJT
family of curves. We will see the i-v characteristics of Ic vs. Vce for steps of IB. The
i-v characteristic showing Ic vs. Vce for different values of VBE are not linear. Thus we
will see that the output resistance of the BJT change slightly with current. Then using
the values of Ic and Ro, we can calculate the early voltage, Va. The important feature of
this device is that the i-v characteristics are not perfect linear.
Data Section
Outline of procedures:
1) Use the curve tracer to display the common-emitter BJT family of curves (ic vs vCE for
steps of iB).
2) Determine IB needed to set the Q-Point for Ic=0.5mA and VCE=5 Volts.
3) Determine DC.
4) Determine AC = IC/IB.
5) Determine the output resistance, Ro, by measuring the slope of the i-v curve and
taking the inverse of that.
6) Does the output resistance change with voltage on the same curve?
7) Does the output resistance change with current on different curves?
8) Determine the early voltage.
Data Table / Calculations / Analysis
1) Completed in lab.
2) IC = 560 A
VCE = 5 Volts
IB is found to be 5A according to curve tracer.
3) DC = IB / IC
560 A / 5 A = 112
4) AC = IC/IB.
IC1 = 420A , IB1 = 4A, VCE = 5 Volts
IC2 = 680A , IB2 = 6A, VCE = 5 Volts
AC = IC2 - IC1 /IB2 - IB1
AC = (680-420)/(6-4) = 130
A diagram is attached explaining the origin of the values clearly.
5) Q point is 5 Volts
Point A Point B
VCE = 1 Volt VCE = 9 Volts
IC = 660A IC = 700A
IB = 6A IB = 6A
Ro = [(700A - 660A)/(9-1)]-1 = 200000
A diagram is attached explaining the origin of the values clearly.
6) Q1 is 3 Volts
Point A Point B
VCE = 1 Volt VCE = 5 Volts
IC = 660A IC = 680A
IB = 6A IB = 6A
Ro = [(680A - 660A)/(5-1)]-1 = 200000
Q2 is 7 Volts
Point A Point B
VCE = 5 Volt VCE = 9 Volts
IC = 680A IC = 700A
IB = 6A IB = 6A
Ro = [(700A - 680A)/(9-5)]-1 = 200000
The output resistance values determined with 2 different Q points along the same IB value
(curve) shows that voltage does not change resistance.
A diagram is attached explaining the origin of the values clearly.
7) Q1 is 5 Volts
Point A Point B
VCE = 1 Volt VCE = 9 Volts
IC = 1.16mA IC = 1.22mA
IB = 10A IB = 10A
Ro = [(1.22mA - 1.16mA)/(9-1)]-1 = 133333
Q2 point is 5 Volts
Point A Point B
VCE = 1 Volt VCE = 9 Volts
IC = 660A IC = 700A
IB = 6A IB = 6A
Ro = [(700A - 660A)/(9-1)]-1 = 200000
The output resistance values determined with the same Q point on two different IB values
(different curves) shows as current increases, IC, resistance decreases.
A diagram is attached explaining the origin of the values clearly.
9) Early Voltage (VA)
Ro = VA / IC -* IC * Ro = VA
560A * 200000 = 112
This value matches up with the value determined at the beginning of Activity 1 (3).
Conclusion
In conclusion the BJT characteristics were as expected. As current increased the output
resistance decreased, and as voltage changes the output resistance did not change. Hence
current change and not voltage change affect the output resistance.
Activity Two
Diagram of a npn BJT
Theory
In this part of the lab we will set the dc voltages to the terminals of the BJT and
measure the corresponding voltages at the nodes. Then we will calculate the currents
through the emitter, base and the collector terminals. Next, we will calculate 
and  from these currents. We will see that even though the resistor values are
not completely matched we will have some discrepancies in the currents. But for the most
important part, we will show that when we will calculate  from
and  from there will be a big change. In
the second part, when we change the dc voltages we will show that the transistor current
is more dependent on the emitter potential than the collector potential for both npn and
pnp BJT's.
Data Section
Outline of procedures:
ESTABLISHING DEVICE CURRENTS:
1) Choose RC and RE to be well matched.
2) Adjust dc supplies to +10 Volts and -10 Volts.
3) Measure the dc voltages with the DVM at points E, B, C.
4) Calculate VBE, IE, IB, IC
5) Calculate  and  from currents in part (4)
6) Calculate  from 
7) Calculate  from 
IDENTIFYING THE CONTROLLING JUNCTION:
8) Set V+ = +10 Volts and V- = -5 Volts
9) Measure VBE, VE, VB, VC
10) Calculate all terminal currents,  and 
11) Set V+ = +5 Volts and V- = -5 Volts
12) Measure VBE, VE, VB, VC
13) Calculate all terminal currents,  and 
14) Compare this data with the data found at + 10 Volts.
15) Do the transistor currents depend more on the conditions in the emitter of the
collector?
16) Set up two-equations-in-two-unknowns and solve simultaneously for n and IS. 
17) Are these values reasonable? Why or why not?
MEASURING EFFECTS OF CIRCUIT RESISTANCE
18) Set V+ = +10 Volts and V- = -10 Volts
19) Verify VE, VB, VC.
20) Shunt Rb by another 10000 resistor, and measure VE, VB, VC.
21) Calculate all terminal currents,  and 
22) Remove the resistor, and shunt Rc by another 10000 resistor, and measure VE,
VB, VC.
23) Calculate all terminal currents,  and 
24) Remove the resistor, and shunt RE by another 10000 resistor, and measure VE,
VB, VC.
25) Calculate all terminal currents,  and 
26) Change V- to -5 Volts. Measure and calculate again.
27) Compile a neat table of all data.
Data table
1) Completed in lab
2) Completed in lab
3) Attached
4) Attached
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8) Completed in lab
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11) Completed in lab
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18) Completed in lab
19) Completed in lab
20) Attached
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24) Attached
25) Attached
26) Attached
Calculations *are attached
Conclusion
In conclusion, the circuit worked as expected. VC changes according to the difference
between V+ and V-. Since VB is grounded very little voltage is lost through the base
collector so the voltage between the emitter and collector terminals remain almost the
same while V+ and V- are equal but of opposite sign values. Also when VC is less than VE
saturation occurs in the circuit, hence it is forward biased as opposed to being reversed
biased in active mode.  and  remain nearly the same no matter what the
conditions of the V+ and V- while in active mode. When saturation occurs  and
 are affected greatly.
Activity Three
Diagram of a pnp BJT
Theory
Refer to the theory statement listed in Activity Two.
Data Section
Outline of procedures:
ESTABLISHING DEVICE CURRENTS:
1) Choose RC and RE to be well matched.
2) Adjust dc supplies to +10 Volts and -10 Volts.
3) Measure the dc voltages with the DVM at points E, B, C.
4) Calculate VBE, IE, IB, IC
5) Calculate  and  from currents in part (4)
6) Calculate  from 
7) Calculate  from 
Data Table
1) Completed in lab.
2) Completed in lab.
3) Attached
4) Attached
5) Attached
6) Attached
7) Attached
Calculations *are attached
Conclusion
In conclusion from the results obtained in Activities Two and Three, the branch currents
and node voltages of npn and pnp transistors, it can be said that the transistor currents
depend more on the emitter potential than the collector potential. Also the error that is
seen in calculating  from and  from  is caused by
the fact that the resistor do not all have the same values, they are not completely
matched.