4 ELE2503 – Electronic Systems

Home Experiment 1

Transistor Amplifier 1

Procedure

Steps Notes

1. Construct the circuit shown.

Section A: Biasing and Transistor Gain

1.1

2. Adjust R1 until the voltage across the transistor is about half of the supply voltage.

2.1 If the 1MΩ potentiometer is incapable of achieving this add a 100 kΩ resistor in series and try again.

3. Measure the supply voltage VCC, the voltage across the transistor, VCE, the voltage at point X, Vx, and the voltage at the base, VBE.

3.1 The transistor should be biased by the current in R1 and R2 to a point mid-way on the load line.

4. Disconnect RC, measure its resistance and calculate the collector current, IC.

4.1

5. Disconnect R1 and R2, measure their combined resistance and calculate the base current, IB.

5.1

6. Finally, calculate the transistor gain,hFE (=β).

6.1 Remember, hFE =

R1

R2

1 MΩ

68 kΩ

RCX

560 Ω

LED

BC549

+9v.

IC

VCC VX –

RC-------------------------=

IB

VCC VBE–

R1 R2+--------------------------=

IC

IB-----

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Steps Notes

Section B: Operating Point

7. Determine the two extreme points on the load line i.e. IC max and VCE max.

7.1 Make two assumptions in these calculations:

(i) the LED has no influence on circuit conditions;

(ii) the transistor is short circuited for IC max.

7.2 IC max = and

VCE max = VCC

8. Draw the theoretical load line on axes of IC vs VCE. On this load line indicate the operating point as determined in section A.

8.1 You should note that the actual operating point does not lie on the theoretical line. This is caused by the LED which causes a reduction in the maximum values of IC and VCE.

Section C: use as an Amplifier

9. Vary R1, so that the transistor varies above and below the operating point, and observe the LED.

Note the intensity of the LED for a reduced value of R1, (i.e. IB increased) and for an increased value of R1 (i.e. IB decreased).

9.1 The LED should vary in intensity.

9.2 You should observe a fairly linear response, as long as you do not saturate or cut-off the transistor.

9.3 If you had a signal source, it would be coupled into the above circuit as shown:

Since you may not have a source, we can make the transistor respond just as it responds to an A.C. signal, by varying the base current up and down by simply varying R1 by hand.

VCC

RC----------

LargeCapacitor

SignalSource

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Steps Notes

10. Connect the voltmeter across VCE, vary R1 again, and note the variation in VCE.

10.1

11. Vary R1 over the same range as in steps 9 and 10 and measure maximum and minimum values of VBE and VCE.

11.1 Remember VCE minimum occurs at VBE maximum and vice-versa.

11.2 Do not allow the transistor to saturate or cut-off. A VCE value of 0.3V is too close to saturation to be of use.

11.3 VBE varies only slightly, but you should be able to measure it.

12. Use the measurements in step 11 to calculate the gain of the amplifier circuit.

12.1 Gain =

=

12.2 The variations in VCE and VBE are just like the peak-peak variations of an a.c. signal.

1 MΩ

68 kΩ

560 ΩRc

X

LED

BC549V

+9v.

Variation in VCE

Variation in VBE--------------------------------------------

VCEmax VCEmin–

VBEmax VBEmin–------------------------------------------------

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Home Experiment 1

Report

Section A: Biasing and Transistor Gain

1. Measured voltages VCC = volts

VCE = volts

VX = volts

VBE = volts

2. Measured resistances RC = ohms

R1 + R2 = kohms

3. Calculated currents IC = mA

IB = mA

4. Transistor current gain hFE =

Section B: Operating Point

5. Load line IC max = mA

VCE max = volts

6. Operating point IC mA

VCE volts

hFE

IC

IB-----=

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7. Graph of IC vs VCE

Section C: Use as an Amplifier

8. Variations in LED intensity and VCE(Enter: 1 = increase, 2 = decrease)

R1 LED VCE

increaseddecreased

9. Variations in VBE and VCE.

(i) Extreme point 1 VBE max = volts

VCE min = volts

(ii) Extreme point 2 VBE min = volts

VCE max = volts

10. Calculated voltage gain Gain =

gainΔVCE

ΔVBE--------------=

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Analysis and Questions

Question 1For an amplifier, the operating point is approximately in the centre of the load line. Which is the best explanation of this?

1. So the current can increase or decrease in response to an input.

2. This is just a convenient location.

3. This allows for variations in values of VCC.

Question 2If the hFE of the transistor in the circuit of this experiment is halved, the effect on the voltage gain would be:

1. to double.

2. to suffer no change.

3. to halve.

Question 3With respect to the amplifier circuit used in section C, what would be the effect on the gain of VCC was 18 V and not 9 V (IB remaining the same)?

1. The voltage gain would increase.

2. The voltage gain would decrease.

3. The voltage gain would be unchanged.

Question 4Refer to the circuit used in section A. What would be the approximate maximum size of input signal which could be used?

1. 0.7V p/p

2. 0.07V p/p

3. 0.01V p/p

Question 5Refer to the circuit used in section A. The effect of decreasing R1 is to:

1. increase IB and increase VCE

2. increase IB and decrease VCE

3. decrease IB and increase VCE

4. decrease IB and decrease VCE

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Question 6Refer to the circuit used in section A. The effect of increasing R1 is to:

1. increase VBE and decrease IC

2. increase VBE and increase IC

3. decrease VBE and decrease IC

4. decrease VBE and increase IC

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(V)

(V)

(V)

(mA)

(mA)

(mA)

( )Ω

12 ELE2503 – Electronic Systems

(V)

ELE2503 – Electronic Systems 33

Home Experiment 4

Constant Current Source

Procedure

Home Experiment 4

Report

2.

Steps Notes

1. Construct the circuit shown. 1.1

2. Vary the value of the variable resistor and at a number of points (say five) calculate the current in the transistor by measuring the voltage across the emitter resistor.

2.1 Note that the current remains substantially constant for a large part of this variation.

Spindle Position (approx.)

V(2.2 K)(measured)

volts

IE(calculated)

mA

Min resistance¼ turn½ turn¾ turn

Max. resistance

47 K

10 K 2.2 K

BC547

10 K Potentiometer

+9 V

34 ELE2503 – Electronic Systems

Analysis and Questions

Question 1What percentage of variation was there in the value of current if large variations at either extreme are neglected? (ΔI/I AVERAGE)

Question 2If the 2.2 kOhm resistor was changed to 2.7 kOhm, what would be the new value of current?

Question 3If the power supply voltage connected to the 10 kOhm potentiometer were changed to +20 volts, as shown, what difference would it make to the value of current?

1. No change2. It would be larger3. It would be smaller

%

mA

+20

2.2 K

10K

47K10K

V

10K

47 K

+20 V

10 K

10 K2.2 K

+9 V

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36 ELE2503 – Electronic Systems