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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