Lab 8 bipolar junction transistor characterstics

1. Bipolar Junction Transistor Characterstics
Experiment - #8
Kehali B. Haileselassie and Kou Vue
11/14/2013
ELC ENG 330 – Electronics I
Fall 201

2. Purpose:
The purpose of this laboratory is to become familiar with the D.C. operation of the
bipolar junction transistor (BJT), and a basic D.C. circuit using the BJT, namely, the current
mirror.
Equipment
External Multi-meter (for current measurements)
Regulated Power Supply
2N3904
Resistors
Introduction
The objective of this lab is to analyze the characteristics of Transistor. A transistor is a three-
terminal device which allows a small current to control a much larger current, thus achieving a
“current gain.” this lab, we focus on BJTs. The three terminals of a transistor are the base (usually
used as the control terminal), the emitter and the collector. There are three parts of the lab and
voltage across Rc is measured and current Rc is computed. Once the measurement is taken results are
plotted on the graphs and DC current Gain, Incremental resistance etc. are found.
A bipolar junction transistor is formed by joining three sections of semiconductors
withalternatively differentdoping. The middle section (base) is narrow and one of the other
tworegions (emitter) is heavily doped. Two variants of BJT are possible: NPN and PNP.

3. In NPN transistors,electron flow is dominant while PNP transistors rely mostly on the flow of
holes. Therefore,to zeroth order, NPN and PNP transistors behave similarly except the sign of
current andvoltages are reversed.! In practice, NPN transistors are much morepopular than PNP
transistors because electrons move faster in a semiconductor. As a result,a NPN transistor has a
faster response time compared to a PNP transistor.
The BJT is a three-terminal semiconductor device containing two pn junctions. If
checked with an ohmmeter it appears to be two diodes of opposite polarity connected in series.
However, unlike two series diodes, the BJT can be used to amplify
Procedure
Components
Transistor (2N3904)
Value
-
Rc
Rb
Vbb DC Voltage Supply
Vcc Voltage Supply
100 Ohms
33KOhms
Vary
Vary
Figure_1
1. Connect the 2N3904 transistor up into the circuit as shown above; for your base and
Collectorresistors measure and record the exact values. (Set VBB and VCC voltages

4. set to zero).
2. Measure the voltage across RB while increasing the VBB supply voltage. Continue to
increase VBB until the voltage across RB reaches 1.65 volts in order to set up a base current
of 50 μA, which you can confirm by Ohm’s law.
3. Without changing VBB, increase VCC until the voltage from collector to emitter is +2.0
volts. Then, record the value of voltage across the resistor RC, and compute the collector
current (which equals the current through RC) by using Ohm’s Law applied to RC.
4. Without changing VBB, repeat step 3 by increasing VCC until the voltage from collector
to emitter is +4.0 volts. Record the value of voltage across the resistor RC, compute the
collector current as before, and enter the values into the table.
5. Repeat for VCC values of 6, 8 and 10 volts, and complete the table.
6. Next, reset VCC to a value of zero volts, and increase VBB until the voltage across RB is
3.3 volts, giving a base current of 100 μA.
7. Repeat steps 3, 4 and 5, completing the following table:
8. Next, reset VCC to a value of zero volts, and increase VBB until the voltage across RB is
4.95 volts, giving a base current of 150 μA.
9. Repeat steps 3, 4 and 5, completing the following table:
10. Plot the transistor characteristics on the following graph, using the values from the

5. three tables above. You should have three curves, labeled with IB = 50μA, 100μA and
150μA.
11. From your plotted curves, estimate the DC current gain (defined as βDC ≡ IC/IB) at a
value of VCE = 3 volts, for each value of base current (50μA, 100μA and 150μA).
12. Repeat step 11 at a value of VCE = 7.5 volts, for each value of base current (50μA,
100μA and 150μA).
13. Finally, for the case where IB = 150μA, compute the “incremental resistance”
ΔVCE/ΔIC = (VCE1 – VCE2)/(IC1 – IC2). Use the points at VCE = 10 volts and VCE = 2 volts,
on the IB = 150μA curve.

6. Results
Ib 50uA
Vce(Volts)Vrc (Volts) Irc = Ic = Vrc/Rc
2
1.12
0.0112
4
1.17
0.0117
6
1.23
0.0123
8
1.31
0.0131
10
1.41
0.0141
Ib 100uA
Vce(Volts)Vrc (Volts) Irc = Ic = Vrc/Rc
2
2.25
0.0225
4
2.5
0.025
6
2.69
0.0269
8
2.88
0.0288
10
3.1
0.031
Ib 150uA
Vce(Volts)Vrc (Volts) Irc = Ic = Vrc/Rc
2
3.09
0.0309
4
3.64
0.0364
6
3.96
0.0396
8
4.28
0.0428
10
4.5
0.045
Figure_2Voltage Vs Current Rb
Table_2

7. Voltage
IB = 150uA Ic A
3
7.5
(Ic/Ib) DC Current Gain
0.0309
206
0.0418
278.6666667
Table_3 DC Current Gain
Yes, the DC current gain is depending on IB.
Yes, the DC current gain is significantly depending on IB.
Change V
Ib = 150uA
Stimulation Result
Change I
8
Table_4 Incremental Resistance
Incremental Resistance
0.0141
567.3758865

8. Discussion &Analysis:
One of the Task of the transistor is to provide DC current gain. The measurement shows
exactly that. In the circuit input current Ib and output current is Ic and DC current gain can be
found using Ic/Ib and its shown in the figure 5 for voltage 3 and voltage 7.5. Also the
incremental resistance is found by dividing change in voltage by change in current. I think
current gain does not depend on the Ib, because if Ib increase then Icincreases with similar ratio.
As seen in figure 5, as voltage increases current gain increases as well. Resistance value is
medium and that is 568 ohms.Since BJT are useful in discrete circuit design, we analyzedthestaticcharacteristicof

9. BJT.We measured and plotted the I-V curve for collector current and collector-emitter voltage, while keeping baseemitter voltage constant.
Conclusion
A single npn BJT was used to drive this differential amplifier. The collector current entering
the npn BJT will be the current source driving the differential amplifier. The transistor is good
component to get amplified current using a very small current. Dc current gain increases as Voltage
is increases. But Dc current gain for the same voltage with the different base current is same. Each
part has different base current but if DC current gain is found for same voltage on different curves it
will come out to be about the same.