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
ntype emitter, a ptype base and a ntype
collector. On the other hand the
pnp has a ptype emitter, a ntype base, and a
ptype collector. Also the
transistor consists of two major pn junctions, the
emitterbase junction
(EBJ) and the collectorbase 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 commonemitter BJT family of
curves. We will see
the iv characteristics of Ic vs. Vce for steps of IB. The iv
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 iv
characteristics are not
perfect linear. Data Section Outline of procedures:
1) Use the curve tracer to
display the commonemitter BJT family of curves
(ic vs vCE for steps of iB). 2)
Determine IB needed to set the QPoint
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 iv 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 = (680420)/(64) = 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)/(91)]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)/(51)]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)/(95)]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)/(91)]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)/(91)]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
twoequationsintwounknowns 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 5) Attached 6) Attached 7)
Attached 8)
Completed in lab 9) Attached 10) Attached 11) Completed in lab
12)
Attached 13) Attached 14) Attached 15) Attached 16) Attached 17)
Attached 18)
Completed in lab 19) Completed in lab 20) Attached 21)
Attached 22) Attached 23)
Attached 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.