Bipolar Junction Transistor


     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 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.