UNIVERSITY OF NORTH CAROLINA AT CHARLOTTE
Department of Electrical and Computer Engineering
Experiment No. 5 - BJT Basics
Overview: The purpose of this lab is to familiarize the student with the three operation regions
of a Bipolar Junction Transistor: cutoff, active, and saturation.
The transistor is a three-terminal semiconductor device, in specific the BJT essentially consists
of a pair of PN junction diodes that are connected in series, back to back either PN-NP to form
PNP transistor or NP-PN to form NPN transistor. Both electrons and holes conduct current, and
this is why it is called bipolar. The circuit symbols are
In both symbols the emitter is distinguished by an arrowhead which indicates the polarity of the
device (PNP or NPN), and that is indicated by the direction of the arrowhead on the emitter
where the arrowhead points in the direction of normal current flow in the emitter.
There are three regions of operation for the transistor:
Figure 1. BJT transistor biasing
(1) Active Region:
To bias a BJT transistor for operation in the active region, the base-emitter junction must be
forward biased and the collector-base junction reverse biased. The collector-base current is very
close in magnitude to the emitter-base current, and therefore, the collector-base voltage for a
particular collector current will be much greater than the emitter-base voltage for that same
current. This multiplication of the emitter-base voltage seen across the collector-base junction is
voltage amplification. The active region is the desired region of operation for a linear amplifier.
In Figure 1a it is clear that the NPN transistor is biased for operate in the active region. If the
collector voltage falls below the base voltage by an amount that exceeds the threshold voltage of
that junction, the collector-base junction will become forward biased and the transistor will enter
the saturation region. In a similar manner, the PNP transistor in Figure 1b operates in the active
region so long as the collector potential is no more than the junction threshold voltage above the
base potential. In addition to the active and saturation regions, the transistor will enter the cutoff
region if both the base-emitter and base-collector junctions become reverse biased.
(2) Saturation Region:
As stated above the transistor will enter the saturation region when the collector-base junction
becomes forward biased. As base-emitter current is increased there is a point where forward bias
of the collector-base junction occurs, and beyond that point only small increases in collector
current and corresponding small decrease in the collector voltage will result from further
increases in the base-emitter current. Transistor switching circuits are generally driven into
saturation for the ON state in order to minimize the voltage drop across the collector-emitter
junction. However, for application requiring rapid switching the distance driven into saturation
must be limited to prevent long charge storage time and the resulting reduced switching time.
(3) Cutoff Region:
The cutoff region is entered when both junctions are in reverse biased. In this case, all terminal
currents are extremely small and the transistor is said to be off. Switching circuits are driven into
cutoff when the desired state of the switch is OPEN.
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Pre-Lab – BJT Basics
1. Assume the transistor you will be using has a nominal β of 200 and VBEon = 0.7V. The
transistor will enter the active region when Vin reaches the turn-on voltage Von . For the
circuit of Figure 2, calculate the value of Vin that will cause the transistor to enter saturation.
If you were to measure the base-emitter voltage and find it to be 0.2V, what would you
expect to measure for the values of Vin and Vc ?
Ic
+
Rc
Vout
1k
Vcc
10V
Rb
Vcc
b
Q1
c
Rpot
+
10V
10k
100k
Vin
Q2N3904
Ib
e
-
Figure 2. BJT transistor circuit configuration
2. Bring a compact flash card to lab session.
(INSTRUCTOR’S SIGNATURE_____________________________DATE
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)
Lab Session – BJT Basics
I Measurement of Gain and Transfer Characteristics
1) Adjust the power supply VCC so that it produces +10 V. Turn off the power supply.
2) Measure and record the actual value of the 1kΩ and 100kΩ resistors that will be used in
your circuit.
3) On a protoboard, build the circuit of Figure 2. Turn on the power supply and adjust the
potentiometer so that Vout is 5 V.
4) Measure the values of VCC and Vout to write a KVL loop equation around the collector
emitter side of the circuit. Solve for VCE. The voltage across the collector-emitter junction VCE
indicates the region of operation of the transistor; in the cutoff region VCE  VCC, and in the
saturation region VCE  0; in the active region, 0 < VCE < VCC. Use these three equations to
determine the operating region of the transistor.
5) With Vout still at 5V, measure the voltage across RB. Use Ohm’s Law and the measured
resistance values for RB and RC to compute IB and IC. Compute the DC forward current gain,
β, from the dependent source equation for the transistor, IC = β IB, and record this value in
your lab report.
6) The value of β is subject to drift with time and temperature. The following procedure will
demonstrate the dependence of β on the temperature of the transistor. Maintain Vin so that
Vout is still 5V. While observing Vout, warm the transistor by pinching the package between
your thumb and forefinger. You will notice a change in Vout. How is β related to the
temperature of the transistor?
7) Prepare a four-column table in your lab report to record about thirty data points, labeling
the columns Vin, Vout, VBE and VCE. For Vin from 0 to 1.5 volts, take data at intervals of 0.25
V (0 V, 0.25 V, 0.5 V, etc.). For Vin from 1.5 V to 10V, take data at 0.5 V intervals.
8) Plot the transfer characteristics.
a. Graph the data you collected on a full page. Place Vin on the X-axis, and plot Vout, VBE, and
VCE as functions of Vin . The graph should be similar to the one in Figure 3, but should also
show VBE and VCE. With enough data points collected, the plot will show that the transition
between the cutoff and active regions, and between the active and saturated regions are
somewhat rounded. Use different color ink to fit straight lines to each of the three operating
regions for the Vout data. The two intersections of these three straight lines represent the best
approximations for the transitions from the active to the cutoff region and from the active to
the saturation region. Draw a vertical line extending from the top to the bottom of your graph,
at each of these intersections. Label the mode of operation for each of the three regions.
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Figure 3. Output voltage vs. input voltage
b. Now, use the Tektronix 576 or equivalent Curve Tracer to display the transistor common
emitter characteristic curves for base currents from 0µA though 60µA in 10µA steps. Also,
record the value of β given by the curve tracer, and compare this value with that computed in
procedure 5; assuming the value of β displayed by the curve tracer is the actual value,
calculate the percent error.
Figure 4. BJT curve tracer circuit
*Note: The source VC needs to be sinusoidal with a DC offset such that it varies
between 0V and 10V. On the AFG 310, an offset setting of 2.5V is needed.
II. An Oscilloscope Curve Tracer
1) Construct the curve tracer circuit of Figure 4. Put your oscilloscope in the X-Y Mode.
Channel 1 becomes the X-axis input, and Channel 2 becomes the Y-axis input. Channel 1 (the
X-axis) will display the transistor’s collector-emitter voltage, VCE, and Channel 2 (the Y-axis)
will display the emitter current, IE.
2) Begin with VBB adjusted so that IB = 40µA. Using 1X Oscilloscope probes connect Channel
2 of the oscilloscope to the emitter of the BJT, and Channel 1 to the collector of the BJT.
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3) Adjust the oscilloscope (volts/division and position controls) for both Channel 1 and
Channel 2 to obtain a display like the one shown below. If the display doesn’t look as it
should, check the time varying signal applied to the collector to be sure it is sinusoidal and
varies between 0V and 10V when displayed with DC coupling.
4) Notice in the display above that a single horizontal line (IC vs. VCE) is present: it starts 1.4
divisions above the origin, and runs horizontally for ten divisions (0 volts to +10 volts) on the
VCE axis. Since IC is caused by IB, the DC current gain can be calculated. One division
vertically = 5mA, and the current gain is: βDC = IC/IB = 7mA/40µA = 175A/A.
5) Save your display to the flash card. Do this by pressing the Save/Recall key to display the
Save/Recall menu.
6) Determine βDC for your transistor. Show all work.
7) In order to display a family of curves as shown by a curve tracer, “Infinite Persist” must be
used to retain the display as the base current is changed to add another characteristic curve.
The Tektronix TDS 2000 oscilloscope does not allow signal averaging or single sweep when
used in the X-Y Mode. Thus, with the vertical gain needed to display the characteristics, these
curves will be broadened and blurred by the random noise floor. Furthermore, since the base
current must be change without creating a continuous smear on the display, it will be
necessary to open the base circuit while adjusting the base current. Of course, this prevents
watching the base current meter reading as the current is adjustment, and it will be necessary
to determine, in advance, a VBB for each base current curve to be display. Before creating the
display of curves, make a table with two columns (IB and VBB) for the IB values: 10 µA, 20µA,
30µA, 40µA, 50µA, and 60µA.
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8) With IB adjusted to 0µA (i.e., VBB = 0) press display and apply “Infinite Persist”
9) Now, open the base circuit and adjust VBB to the value corresponding to10µA, and
momentarily reconnect the base circuit to add the 10µA characteristic.
10) Repeat step 9 until the full set of curves up through IB = 60µA is obtained.
11) When completed, the display should look as follows:
*Note: The height of the curves as they appear on the oscilloscope will vary from
one transistor to another.
12) Save your family of curves to the flash card. Do this by pressing the Save/Recall key to
display the Save/Recall menu.
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Lab Session – BJT Basics (Data Sheet)
INSTRUCTOR'S INITIALS
DATE:
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Post Lab – BJT Basics
1. The straight line drawn in the active region on the transistor input/output characteristic can be
represented by an equation in the point-slope form, y = mx + b (where x equals Vin and y equals
Vout). Provide a mathematical proof showing that β is related to the slope of the line in the active
region by:
β = mRB/ RC
Hint: use IC = β IB, and Ohm’s law.
Measure the slope of the line in the active region of the transistor input/output
characteristic for your transistor. Using the equation above and the measured slope,
calculate the average β. How well does this value agree with that calculated in part 5 and
with the curve tracer value?
2. What region of operation is best suited for a linear, class A, amplifier?
3. What uses can be made of a BJT operating in the cutoff and saturation regions?
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