Procedure

First, the resistors are labeled with their respective resistances as follows:

Resistor Listed Value Measured ValueR1 10 kΩ 10 kΩR2 4.7 kΩ 4.67 kΩRE1 100 Ω 100 ΩRE2 330 Ω 328.7 ΩRC 1.0 kΩ 996 ΩRL 10 kΩ 10 kΩ

From the table above, it can be seen that the listed value and the measured value of the resistor are very close. Since every resistor has some uncertainty ranging from 1% - 10%, the measured value may differ a little from the listed value.

Next, the dc parameters are computed, measured and simulated as follows:

First, we assume the value of VBE to be 0.7 V.

β = 100

From KCL in transistor node,

IE = IC + IB …………………….. (1)

For a BJT in active mode,

IC = β IB

IC = 100 IB……………………… (2)

From (1) and (2),

IE = 101 IB ……………………. (3)

From KVL in the BJT loop,

VBC = VBE - VCE

VBC = 0.7 - VCE ………………… (4)

VTH = R2

R1+R2 Vcc

VTH = 4.7

10+4.7 (12)

VTH = 3.837 V

RTH = R1 // R2

RTH = 3197 Ω

From KVL along VTH through the BJT to the ground,

VTH - IB RTH - VBE – IE RE = 0

3.837 - 3197 IB – 0.7 – (101 IB) (100 + 330) = 0

IB = 67.273 µA

From KVL along VCC through the BJT to the ground,

VCC – IC RC – VCE - IE RE = 0

12 – (100 IB) (1000) – VCE – (101 IB) (100 + 330) = 0

VCE = 2.3511 V

IC = 6.7273 mA

IE = 6.7945 mA

VBC = -1.6511 V

DC Parameter Computed Value Measured Value Simulated ValueVB 3.622 V 3.72 V 3.70 VVE 2.922V 3.023 V 2.983 VIE 6.795 mA Not Available 6.936 mAVC 5.273 V 4.98 V 5.107 VVCE 2.3511 V 1.957 V 2.124 V

From the above table, the computed value, measured value and simulated value agree with each other quite nicely. The little variation between these 3 values may be due to human error while conducting the experiment or internal heating of the circuit elements. Besides, the β value of the BJT also results in the difference between these 3 values. For the theoretical calculation, we assume that the BJT has a β value of 100 to compute the other DC parameters. However, the BJT that we use in the lab may have a different β value from what we assume. PSPICE may also use other β values for the simulation. Therefore, there will be a slight difference between the computed, measured and simulated values of the DC parameters.

Step 3:

The AC parameters are computed and listed as follows. Then, the signal generator is turned on and the input signal, VIn is set for 300 mVpp at 1000 Hz. The AC signal voltage is also measured at the transistor’s emitter and at the collector. Next, all the AC parameters are also measured and simulated.

re = 26mVIEQ

= 26mV

6.7945mA

= 3.826 Ω

R’L = RL // RC

= 909.1 Ω

AV = - R ’ Lr e+ℜ1

= - 909.1

3.826+100

= -8.756

VOut = AV VIn

= -8.756 (300 mVPP)

= 2.627 VPP

Ve = VC - VCE

= VOut - VCE

= 2.627 VPP - 2.3511 V

= 275 mVPP

RB = R1 // R2

= 3197 Ω

rπ =(β + 1) re

= (100 + 1) 3.826

= 386.426 Ω

RIn = RB // [rπ + (β + 1) RE1]

= 3197 // [386.426 + (100 + 1) 100]

= 2970 Ω

AC Parameter Computed Values Measured Values Simulated ValuesVIn = VB 300 mVPP 302 mVPP 300 mVPP

Ve 275 mVPP 280 mVPP 284 mVPP

r’e 3.826 Ω Not Available Not AvailableAV -8.756 -7.95 mVPP -8.467VOut = VC 2.627 VPP 2.4 VPP 2.54 VPP

RIn (tot) 2450 Ω 2970 Ω Not Available

From the table above, it can be seen that the computed values, measured values and the simulated values do not differ much from each other. During the experiment, there may be some heating of the elements which causes power lose and affect our experimental result. For example, the power loss from heating may decrease the voltage gain of the BJT and reduce its efficiency. Besides, the uncertainty in the equipments may result in some systematic error in our results. Apart from that, the β value for the BJT may also be different for all 3 calculated, measured and simulated values. For our theoretical calculation, we assume the β value to be 100. However, the BJT that we use in the experiment may not have the same β value. So, this will result in a slight change of results compared to our calculation. The simulation from PSPICE may also assume different β values. So, the result may also deviate slightly.

The measurement of RIn (tot) is done indirectly since it is an ac resistance that cannot be measured with an ohmmeter. The output signal (VOut) is measured with an oscilloscope and is recorded with the amplifier operating normally (no clipping or distortion). A rheostat (RTest) is then inserted in series with the source as shown. The rheostat is varied until VOut drops to one-half the value prior to inserting RTest .With this condition, VIn = VTest, and RIn (tot) = RTest. RTest is then removed and measured with an ohmmeter. RIn (tot) is measured and recorded.

Discussion and observation

(1)

What is the phase relationship between VIn and VOut?

The input and output waveforms are compared. We discover that VIn and VOut are 180° out of phase.

(2)Remove the bypass capacitor, C2, from the circuit. Measure the AC signal voltage at the transistor's base, emitter, and collector. Measure the voltage gain of the amplifier. What conclusion can you make about the amplifier's performance with C2 open?

From our measurement, VB = 300 mVPP, VE = 300 mVPP and VC = 600 mVPP.

Voltage gain = VcVB

= -2

Therefore, we can conclude that the voltage gain of the amplifier decreases when C2 is open.

(3)Replace C2, and reduce RL to 1.0 kΩ. Observe the AC signal voltage at the transistor's base, emitter, and collector and measure the voltage gain of the amplifier. What conclusion can you make about the amplifier's performance with RL reduced to 1.0 kΩ?

From our measurement, VB = 300 mVPP, VE = 300 mVPP and VC = 1.24 VPP.

Voltage gain = VcVB

= -4.13

Therefore, we can conclude that the voltage gain of the amplifier decreases when RL is reduced to 1.0 kΩ.

(4)Replace RL with the original 10 kΩ resistor and open RE1. Measure the dc voltages at the base, emitter and collector. Is the transistor in cut-off or in saturation? Explain.

From our measurement, VB = 3.83 V, VE = 3.395 V and VC = 12.02 V.

The transistor is in cut-off mode.

VBE = VB - VE = 3.83 V – 3.395 V = 0.435 V

Since VBE is 0.435 V which is less than 0.7 V, the BE junction is reversed biased and IB cannot flow through the transistor. So, the transistor is in cut-off mode.

(5)Replace RE1 and open R2. Measure the dc voltages at the base, emitter and collector. Is the transistor in cut-off or saturation? Explain.

From our measurement, VB = 0.455 V, VE = 3.81 V and VC = 3.87 V.

The transistor is in saturation mode.

VCE = VC - VE

= 3.87 V – 3.81 V = 0.06 V

Since VCE is 0.06 V which is less than 0.2 V, the BC junction is forward-biased. The equation IC = β IB is no longer valid. So, the transistor is in saturation mode.

Evaluation and review question

(1)When the bypass capacitor, C2 is open, you found that the gain is affected. Explain.

From the appendix given, the voltage gain with the load, AV is given by

AV = VoutputVinput

= R ' L

ℜ+ℜ1

This expression is applicable when the bypass capacitor, C2 is connected as shown in the original circuit. In this case, the value of RE1 is 100 Ω.

When the bypass capacitor, C2 is open, the value of RE1 increases to 430 Ω. So, R’L will be divided by a larger denominator and the quotient will decrease. Therefore, the gain will also decrease.

(2)In step 6, you were instructed to measure the input resistance while monitoring the output voltage. Why is this procedure better than monitoring the base voltage?

By monitoring the output voltage, we will be able to measure a more accurate value of the output voltage after it has dropped by one-half of the value. It is more accurate than by monitoring the base voltage. This is because the base voltage is only of the order of millivolts while the output voltage is of the order of volts. A drop of half the value of the output voltage will be more significant and obvious compared to the drop of half the value of the base voltage.

Besides, the base voltage may be affected by the external noise and this will result in an inaccurate measurement of the drop in the voltage.

Another advantage of monitoring the output voltage instead of the base voltage is that we can check and confirm whether the BJT is functioning in the active mode.

(3)Assume the amplifier shown in Figure 8.2 has +1.8 V DC measured on the base, 1.1 V DC measured on the emitter, and +1.1 V DC measured on the collector.

(a) Is this normal?

No, it is not normal.

(b) If not, what is the most likely cause of the problem?

The BJT is in the saturation mode. If VCE = 0V, then this means the BJT is saturated.

VCE = VC - VE

= 1.1 V – 1.1 V = 0 V

Therefore, the BJT is saturated.

(4) If C2 were shorted,(a) what dc base voltage would you expect?

When C2 is shorted, RE = RE1

= 100 Ω

So, we try to analyse the circuit as before.

From KCL in transistor node,

IE = IC + IB

For a BJT in active mode,

IC = β IB

From KVL along VTH through the BJT to the ground,

VTH - IB RTH - VBE – IE RE = 0

3.837 - 3197 IB – 0.7 – (101 IB) (100) = 0

IB = 235.893 µA

From KVL along VCC through the BJT to the ground,

VCC – IC RC – VCE - IE RE = 0

12 – (100 IB) (1000) – VCE – (101 IB) (100) = 0

VCE = -13.972 V

The value calculated is not plausible since VCE must be positive. This means that our assumption that the BJT is in active mode is wrong. To verify this, we need to use PSPICE to prove that The BJT is indeed not in active mode.

From the simulation, we can see that VB = 1.877, VC = 1.240 V and VE = 1.137 V.

VBE = VB – VE = 1.877 - 1.137 = 0.74 V

VCE = VC - VE = 1.240 – 1.137 = 0.103 V

Since VBE = 0.74 V which is greater than 0.7 V and VCE = 0.103 V which is less than 0.2 V, the BJT is in the saturation mode and not the active mode. This shows that our previous calculation for the DC parameters when C2 is shorted fails to work. The equation IC = β IB does not work when the BJT is in saturation mode.

So, in order to predict the DC base voltage in this case, we have to obtain the result from the simulation.

Expected DC base voltage = 1.877 V

(b) what dc collector voltage would you expect

From the simulation from PSIPCE, the expected DC collector voltage is 1.240 V

(5)Explain a simple test to determine if a transistor is in saturation or in cut-off.

To determine whether a transistor is in cut-off mode, first we need to measure the DC voltage of the base, emitter and collector. The base voltage, emitter voltage and collector voltage are given as VB, VE and VC respectively.

Next, we compute the value of VBE.

VBE = VB - VE

If the value of VBE is less than 0.7 V, then the base-emitter junction will be reversed-biased and no current can flow through. Under these conditions, the transistor is in cut-off mode. Similarly, we can measure the current flowing into the base-emitter junction, IB. If IB = 0 A, then the transistor is in cut-off mode.

To determine whether a transistor is in saturation mode, first we need to measure the DC voltage of the base, emitter and collector. The base voltage, emitter voltage and collector voltage are given as VB, VE and VC respectively.

Next, we compute the value of VBE and VCE.

VBE = VB - VE

VCE = VC – VE

If the value of VBE is greater than 0.7 V and VBE is less than 0.2 V, then both the base-emitter junction and base-collector junction is forward-biased and on. Under these conditions, the transistor is in saturation mode.