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California University of Pennsylvania
Department of Applied Engineering & Technology
Electrical / Computer Engineering Technology
EET 215: Introduction to Instrumentations
Lab No.3
Bipolar Junction Transistor (BJT)
Multisim Simulation
Due Date: 9/21/2015
Names:
Signature:
Date
1. Korbin Barker
Korbin Barker
9/21/15
2. Brian Swanson
Brian Swanson
9/21/15
3. Aaron Morgan
Aaron Morgan
9/21/15
Bipolar Junction Transistor
Objective of the Experiment
The main objective of this experiment is to understand the basic functions of the bipolar junction
transistor
Learning Outcomes
Students will demonstrate:
-
the ability to understand the BJT current gain
the ability to calculate transistor voltages and currents
understanding of the operation of BJT in saturation and cut-off (as a switch)
Understanding and analyzing a DC sweep example
Instructions
- Read all steps
- Answer all questions. Type your answers and make sure they are clear and to the pint.
Introduction
A BJT transistor is a 3-terminal semiconductor device used in amplification or switching applications. A
transistor is referred to by the letter Q.
Figure -1:
BJT terminals
And the electronic symbol is shown along with the current directions.
Figure -2:
BJT Electronic symbol with current directions.
Background Information
- With choosing the base as the input side, the collector current, IC, is calculated as a function of the
base current, IB, and the transistor’s current gain, β (Beta).
IC = β IB
It is noticed from the schematic of Figure -2 that IE = IB + IC
Substituting the relationship IC = β IB in to the equation IE = IB + IC , we get:
IE = IB + IC = IB + β IB = (β+1)IB
A-
DC Sweep Analysis
A well-known method of understanding the operation of a transistor circuit is to investigate the output
and how it responds to a specified input.
For the NPN BJT transistor, one typically wishes to examine how the output (the collector current)
behaves when the collector to emitter voltage (VCE) sweeps between two specified values while the base
current is fixed to a pre-specified value.
A -1 DC Sweep Analysis: Single Source Sweep.
In this case, the base current is fixed at 1uA while the voltage VCE sweeps between 0 to 1 V in increments
of 0.01 Volts.
Steps:
1- Start Multisim and place an NPN transistor (2N2222) and a current and voltage sources as shown.
Q1
2N2222A
I1
1uA
2- Set-up the DC sweep as follows:
* on the menu bar, click on Simulate>>Analysis>>DC Sweep
* Set up the analysis parameters as shown
V1
12V
* click on the output tab and click onI(QA1(IC)) on the left and then click Add to select it as an output
variable
* Click on Simulate. You should see a plot of Ic vs. VCE as shown
Place your results here as follows:
The circuit:
The DC Sweep
Questions:
1- From the DC sweep, approximately at what VCE value does the collector current start to rise?
VCE =
50 mV
2- The maximum value of Ic is ?
200µA
3- This maximum collector current is called the Saturation current or the cut-off current (highlioght the
correct answer)
A -2 DC Sweep Analysis: Nested Source Sweep.
The analysis of A-1 were for a fixed value of base current (IB) One may be interested in investigating a
family of curves of Ic vs. VCE for different values of IB
Thus, we will have to add to the DC sweep analysis another source that will sweep between 1uA to 8uA
in steps of 1uA. This will give us a family of 8 curves.
Steps:
* in Multisim for the same circuit:
Simulate>>Analysis>>DC sweep
* Set up the parameters as shown:
* Click Simulate. you should obtain a curve as shown:
Results:
Place your family of curves below
Questions:
1- What are the minimum and maximum saturation current (IC sat) and what are the corresponding
base currents: Use proper units and prefixes.
IC (sat)
IB
.2 mA
1 µA
1.65 mA
1 µA
2- In all cases, the value of VCE at which the collector currents begin to rise is: 50 mV
B- Transistor as a switch
When the transistor is operating as a switch, then we have two modes:
-
Saturation:
o Ic = Ic(sat) -- Saturation (maximum) current allowed by the circuit
o VCE = VCE(sat.) = 0.2V (almost 0 V)
o Thus, one can assume that the switch between the collector and the emitter is closed
(0V across the switch.)
C
E
-
Cut-off:
o IC = IB = IE = 0 A No current is flowing.
o The switch acts like and open circuit between the collector and the emitter.
C
E
1Using Multisim, construct the circuit shown. Here, the transistor is used as a switch. The 330Ω
resistor is used for current limiting. Also, notice the LED across the C-E of the transistor.
Note, although the circuit shows 2N3904 transistor, you may use the 2N3904 or the 2N2222A that you
used in part A.
5V
VCC
330Ω
5V
2
VCC
3
1
2N3904
10kΩ
0
Input
0
0
Fig -3:
Transistor switch demonstration
Place your circuit here:
2-
Test your circuit and complete the table below.
Vin (Volt)
Input Side
0V (LOW)
+5V (High)
Q (Saturation or Cut-Off)
Sat. or OFF ?
OFF
Sat
LED (ON or OFF)?
ON
OFF
3Clearly explain your results of the table above. Why is the LED ON and why is it OFF whenever
the input is high or low ?
If the switch is off, current from the right side Vcc will flow through the LED because the transistor is
deactivated. When the switch is on, the transistor collects current from that Vcc and drains it through the
ground because it is an easier path for the current to travel.
C- Q operating in active region.
When the transistor is operating in between the saturation and cut-off, it is said that the transistor is in
ACTIVE region of operation. Here, the collector current is proportional to the base current according to
the equation:
Ic = β IB
`
Also,
IE = IB + IC = IB + β IB = (β+1)IB
Then, Ohm’s law apply to circuit calculations.
1-
Determine the current gain, β, for the transistor as follows:
aUsing Multisim, construct the circuit shown in Figure -4 Use the Multisim DMM
measure and record the values listed in the table below. (make sure to use popper units)
Note: Use the random figure (Figure –x) to help with how the measurements are taken.
Figure -4:
Transistor in active region of operation
Figure –x:
Random circuit as an aide for measurements.
Parameter
Value
VRB
4.355 V
VRC
2.491 V
VBE
644.847 mV
VCE
9.509 V
b- Calculate the base, collector, and emitter currents.
Using Ohm’s law, calculate IB, IC
IB = VRB/ Rb = 9.266 µA
IC = VRC/RC = 2.076 mA
and IE = IC + IB = 2.085 mA
List your results accurately in the table below (use the prefix mA or uA appropriately)
Parameter
Value
IB
9.266 µA
IC
2.076 mA
IE
2.085 mA
c- Determine the current gain, β.
β = IC/ IB
β = 224.0
d- For reasons that will be apparent later, from the table above, write down the measured value of VBE
VBE = 644.847 mV
e- Repeat and complete the measurements for the table below for RC = 1KΩ and RC = 2.2 KΩ.
Table 1:
Beta calculations
VRB (measured)
4.517 V
4.355 V
IB = VRB/RB
9.611 µA
9.266 µA
VCE(measured)
9.917 V
7.516 V
RC
1KΩ
2.2KΩ
VRC(measured)
2.083 V
4.484 V
IC = VRC/RC
2.083 mA
2.038 mA
5- Calculate the average value of β as the actual value to be used in calculations
Βaverage = 218.35
6- Did VRB or IB change as RC was changed ? Why or Why not ?
β= IC/IB
216.7
220.0
Yes they changed. The changing of resistance caused there to be a change in both the voltage drops
across resistors as well as a change in current.
7- Using IB of Table -2 above and the average value of β, theoretically predict the value of VRC and VCE
for different values of RC Place your results in table -3 below.
Helpful Equations:
VRC = Ic Rc
VCE = Vcc – VRC = 12V - VRC
Table 2: Theoretical Results
RC
500Ω
1.6KΩ
10KΩ (!!)
VRC (Volts)
1.011 V
3.237 V
20.232 V
VCE (Volts)
10.989 V
8.763 V
-8.232 V
Circuit simulations
In Multisim, change RC to the values in the Table and record your measurements in Table 3-B.
Table 3-B:
Simulation Results
RC
500Ω
1.6KΩ
10KΩ (!!)
VRC (Volts)
1.051 V
3.297 V
11.816 V
9- For which values of Rc did the theoretical and simulated results almost aagree?
500Ω and 1.6kΩ
10- For which case of Rc did the simulated and theoretical results did not agree?
Why? 10kΩ, due to it reaching saturation before this value was needed.
VCE (Volts)
10.949 V
8.703 V
184.107 mV
9-
Explain the result for the measurement of VRC and VCE in Table -3.
The measurements excluding the 10kΩ values were close because they did not cause the transistor to
reach full saturation.
Challenges
AFor the circuit of Figure 4 above, what is the maximum value of Rc for the transistor to just reach
saturation (VCE = 0.2 V)
RC(max.) =
5.665kΩ ?
B- Suppose Rc = 5KΩ, what is the minimum value of RB so that the transistor is operating in linear
region (active region)?
RB(min) =
447.4Ω ?