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Experiment 2: AC and DCTransistor Gain
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1. To show how to determine the AC andDC current gain values of a transistorfrom its characteristics curves
2. To give more experience plottingcharacteristic curves
3. To demonstrate the characteristicsamong various transistors of the sametype can vary considerably
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Previously, you learned to plot thecommon-emitter characteristics of atypical NPN transistor.
As part of this experiment, you will plottwo other sets of curves, to show thecharacteristics vary among transistors
of the same type
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You will also learn to determinetransistor gain by merely picking out
appropriate points on a graph and doingsome simple calculations.
The D-C current gain of a transistor is
simply the collector current divided bythe base current.
=
The Greek letter (Beta) represents thecurrent gain which is also represented ashFE
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The current gain can be determined bypicking VCEand IBvalues from the
characteristic curves and then readingthe collector currentExample: Look at the following (A-C and
D-C Current Gain Calculation Graph);when VCE = 5.5 V and IB = 30mA, thecollector current is 6 mA. The D-C beta
is: D-C= 6
3 = 200
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Note: The actual values of willdepend somewhat on the point on thegraph. The curves shown in this caseare for an ideal transistor, so thatalmost any point we pick will give usthe same betaExpect some real world variationswhen completing the experiment, sincepractical transistors are always less thanperfect
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The a-c current gain calculation differsfrom the d-c current gain calculation.The a-c gain is calculated in terms of
specific increments of base current
This technique give a more accuratepicture of how a transistor reacts to a-csignals
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Use the previous graph (A-C and D-CCurrent Gain Calculation Graph) todetermine the A-C current gainDetermine how the collector current
changes when the base current is varied
between 20 A and 40 A with VCEheldconstant at 5 V
Use the following formula to determine
the A-C Current Gain.
=
The (delta) symbol means
change in
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Using the values from our current a-cexample, we have:
=
8 (4 )
4 ( )=
4
= 200
The A-C current is sometimes call hfe&
is the same as the D-C gain in this case,since we are using ideal characteristics.We usually expect a considerable
difference in values with practicaltransistors, but the A-C and D-C valuesmay be the same
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Please note! The current gains we havebeen discussing are for the transistoritself, not the amplifier circuit.Biasing networks, input, output and
loading will usually decrease the currentgain of a practical CE amplifier
You will plot the characteristics of twomore transistors, then compare theresults with the curves you obtainedfrom the first experiments transistor
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Finally, you will calculate the AC andDC gains for all three transistors.
You should see that generally speaking,no two transistors are entirely alike.Matched sets are available for special
purchase
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1 330 W, Watt Resistor (orange, orange, brown)
1 100 kW, Watt Resistor (brown, black, yellow)
1 1 kW potentiometer (on trainer)
1 100 kW potentiometer (on trainer)1 MPSA20 NPN transistor (or similar)
Note: Your results may be considerably
different than those found by CIE. This isbecause the current gain for each transistormay be quite different . Do not be surprised ifyour results are not close to ours.
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The circuit for this experiment (Exp. 2)is the same circuit as you built in Exp. 1.You will be measuring the
characteristics of two more MPSA20transistors and then compare the datawith the characteristics of the MPSA20
transistor used in the first experiment.(1432 Exp. 1)
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1. Locate the MPSA20 transistor usedin 1432 Exp. 1 and set it aside forfuture use. Mark this transistor as Q1
a) Find two more MPSA20 transistorsand mark them Q2 and Q3, so youwill be able to identify them later
2. Construct the circuit on thefollowing slide and use the transistorthat you marked as Q2
a) Connect the transistors properly
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1. Transistors wired incorrectly is acommon mistake
3. Measure the collector current valuesfor Q2 using the given base currentsand VCE setting
a) Mark the measured values in the datatable for transistors Q2 characteristicson the next slide
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4. Complete all the readings for Q2 andthen replace the transistor with Q3
as demonstrated in the followingschematic diagram
a) Repeat the measurements for Q3 and
then turn off the power supply.
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5. Plot the data points for Q2 on the Q2Characteristic Curve Graph.
a) Plot the data points for Q2 on the Q2Characteristic Curve Graph.
b) Connect each set of data points and
draw smooth lines for each set after allthe data has been plotted.
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6. Compare your curves with the oneyou developed from the first
experiment. Are the three family ofcurves identical? This means thethree family of curves (do)(dont)have identical characteristics.
7. You can now see how the DC and AC
gains differ, since you have completedthe sets/families of characteristiccurves for the three transistors.
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a) Now, based on the three transistorcurves, determine the value of IC
when IB= 20 A and VCE= 5 V.b) Record each value of ICin line 1 of the
Data Table for Current Gain
Calculations, under the appropriateheading
8. We can calculate the DC gain, with
IB equaling 20A.
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a) Use the following formula to calculatefor DCfor each transistor
1. DC=
9. Calculate each DC current gain value,
and enter it in line 2 of the CurrentGain Calculations Data Table
10. The AC current gain is calculated by
finding out how much the collectorcurrent varies when the base currentchanges by a certain amount
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a) We will use a base current of 10 for this experiment. Use the graphs
for all three transistors as you didpreviously and record the values inline 3 of the Current Gain Calculations
Data Table11. Now find the value of ICfor all three
transistors at a base current of 30 A
and a VCEof 5 V.a) Record the value for each transistor in
line 4 of the data table
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12. The AC current gain of eachtransistor may be found using the
following formula: =
a) In this case, IBis 20 A (30 10 A),while ICis the difference betweenline 4 and line 3 of the Current Gain
Calculation Data Tableb) Record the results of the calculations
in the Current Gain Calculation Data
Table
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Our results can be seen on thefollowing four slides. Considerablevariation can be expected due to the
wide variations of transistor gain.Do not let the variation worry you as
long as you show the same general
pattern we do.
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6. No: dontOf course it is possible fortwo samples of the same transistortype to have the same characteristics.
However, such a possibility is veryslight
13. DontSee the Current Gain
Calculations Data Table14. DontSee the I Gain Calc. Data Table
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In this experiment, you measured thecharacteristics of two MPSA 20transistors.
Using the family of characteristic curvesobtained from these two transistorsalong with the family of curves from thetransistor in experiment 1, you wereable to make direct comparisons of thecharacteristics of a total of 3 transistors
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You were also able to use the curves tocalculate the AC and DC current gains
and see how the comparedYou probably obtained different
characteristics for each of the three
MPSA20 transistors.Although, it is possible, just barely, that
the characteristics of two transistors
were exactly the same
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However, it is far more likely, thedifferences of the three transistors were
readily apparent.This allowed you to see the extremevariability that can be expected in the
characteristics of such solid statedevices
The current gain calculations for all
three devices should have beensomewhat different as well.
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Again, the probability of identical gaincharacteristics is remote.
The stated tolerance for the currentgain of an MPSA20 transistor is in therange of 40-400.
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Rosenow, K. (2001). Lesson 1432:
Common-Emitter AmplifierCharacteristics. Cleveland: ClevelandInstitute of Electronics.
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Developed and Produced by the
Instructors in the CIE InstructionDepartment.
Copyright 12/2012
All Rights Reserved / Dec. 2012
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