Instructions:
Complete the three parts—Part 1, Part 2, and Part 3 of Lab 1. When you have completed each part, answer the questions and transcribe/transfer the test results recorded in the lab manual’s tables to the tables provided.

-------------------------------------------------
Part 1—Common Emitter (CE) Amplifier

Theory: 1. What type of bias is used in the common-emitter amplifier shown in Figure 3-1(a)? * Voltage Divider

2. Explain the reason(s) why the AC equivalent circuit appears as it does in Figure 3-1(b). * When an AC input signal is applied to the amplifier, the capacitors will act as shorts. This is why we call this Figure the “AC Equivalent”.

3. What values must be calculated to determine the DC operating parameters? * VB * VE * IE * VC * VCE * VCC

4. What information can be obtained by determining the DC load line of an amplifier? * We use the DC load line to graph all possible current and voltage characteristics of a biased transistor. This includes any measurements that are between saturation and cutoff, including the Q-point.

5. What values must be calculated to determine the AC operating parameters? * Vin = Vb * r’ e * Ve * Av * Vout = Vc * Rin(total) * Rac

6. What information can be obtained by determining the AC load line of an amplifier? * We use the AC load line to graph the maximum/minimum current and voltage range of an amplifier when an AC signal is applied.

7. Why is the current higher in the AC equivalent circuit? * Since the capacitors act as shorts to the AC signal, the circuit in return has a lower overall resistance. This increases the current in the circuit. .

8. What does the dynamic resistance represent? * Dynamic resistance represents the resistance in the base-emitter diode during each half-cycle of the AC input signal.

9. What amplifier components does the AC emitter-collector current flow through in the CE amplifier in Figure 8-1? * RC * RL * RE1

10. What amplifier components does the AC base-emitter current flow through in the CE amplifier in Figure 8-1? * R1 * R2

Preparation: 11. Transcribe the component measurements from Table 3-1 to the table below. Component | Listed Value | Measured Value | R1 | 10 k Ohm | 9.85 K | R2 | 4.7 k Ohm | 4.67 K | RE1 | 100 Ohm | 98.9 | RE2 | 330 Ohm | 323 | RC | 1 k Ohm | .992 K | RL | 10 k Ohm | 9.85 K |
Table 3-1

Test Procedure: 12. Transcribe the calculations and measurements from Table 3-2 to the table below. DC Parameter | Computed Value | Measured Value | VB | 4.8 V | 4.6 V | VE | 4.1 V | 3.9 V | IE | 9.78 mA | Actual | 9.20 mA | VC | 5.3 V | 5.5 V | VCE | 1.2 V | 1.7 V | VCC | 15 V | 15 V |
Table 3-2 13. Why is the calculated value emitter current different from the actual emitter current? Why is it important to use the actual emitter current when calculating the AC parameters? * The calculated value is assuming perfect conditions. The actual value is using the actual measurements taken from the circuit. It’s better to use the actual value so you will be better able to predict how the circuit is going operate.

14. Transcribe the calculations and measurements from Table 3-3 to the table below. AC Parameter | Computed Value | Measured Value | Vin = Vb | 300 mVpp | 302 mVpp | r’e | 2.6 Ohm | | Ve | 4.1 V | 3.9 V | Av | 8.9 | 8.6 | Vout=VC | 5.3 V | 5.6 V | Rin(total) | 2.39 KOhm | 2.67 KOhm |
Table 3-3 15. What is the CE amplifier’s phase relationship for its input and output signals? * They are 180 degrees out of phase with each other.

16. Transcribe the DC and AC load lines from Plot 3-1 to the plot below. * Plot is in attachment.

17. Explain why the input resistance is measured indirectly. Why is the AC output voltage adjusted to be 50% of amplifier’s output? * It is an AC resistance that cannot be measured with an Ohmmeter. * With 50% of the voltage dropped across the amplifier, the other 50% will be dropped across the potentiometer. The potentiometer and amplifier are set up as a Voltage divider when in this configuration. This being the case, we know that anytime R1 and R2 are the same then the voltage output will be 50% of the input. Whatever resistance is measured in the potentiometer at the moment the output voltage is half of the input voltage, will be equal to the total resistance in the amplifier.

18. Transcribe your fault observations and findings from Table 3-4 to the table below.

Measurement | Observations/Findings | C2 Open | Gain was reduced from . 8.6 to 2.2. | RL reduced to 1k Ohm | Gain was reduced from 8.6 to 4.95. | RE1 Open | Transistor entered cutoff due to no ground on Emitter. There was no voltage drop across RC. | R2 Open | Transistor in Saturation. A lower voltage across CE was measured. This means that more voltage is being dropped across RC due to the base-emitter junction becoming forward biased. |
Table 3-4

19. Explain the reason the gain is affected when the bypass capacitor, C2, is open. * With C2 open it will no longer appears as a short when the AC input signal is applied. This means that RE2 will now be taken into considerations when calculating the Gain.

20. In the lab you adjusted input resistance while monitoring the change in output voltage. Why is this procedure better than monitoring the base voltage? * If we were to use the base voltage then we would have to assume that all other connections and devices are operating under perfect conditions. Since this is not the case we needed to measure the output voltage to get the most accurate measurement.

21. Assume the amplifier shown in Figure 3-3 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.

Figure 3-3: Common Emitter Amplifier (a) Is this normal? a. NO

(b) If not, what is the most likely cause of the problem? b. Transistor Failure caused by Resistive Junction inside the transistor.

22. For the circuit above, if C2 were shorted, (a) What DC base voltage would you expect? a. Around 2 V.

(b) What DC collector voltage would you expect? b. Around 1.5 V.

23. Explain a simple test to determine if a transistor is in saturation or cutoff mode. * Measure VC. If VC = VCC then Transistor is in Cutoff mode and almost the whole input voltage is being dropped across VCE. If VC = 0. Then transistor is in Saturation because almost the whole input voltage is being dropped across RC.

-------------------------------------------------
Part 2—Common Collector (CC) Amplifier

Theory: 24. Why is the CC amplifier’s voltage gain approximately 1? * Because the AC output voltage almost perfectly duplicates the input voltage waveform.

25. What is the advantage of connecting the output of a CE amplifier to the input to a CC amplifier? * The CE amplifier will increase the voltage while the CC amplifier will increase the current. When used together this technique is often called power amplifiers.

26. What values must be calculated to determine the DC operating parameters? * VB * VE * IE * VCE

27. What values must be calculated to determine the AC operating parameters? * r’e * Av * VE * Rin(total) * Ap * Rac

28. What amplifier components does the AC emitter-collector current flow through in the CC amplifier in Figure 3-7? * RE * C2 * RL

29. What amplifier components does the AC base-emitter current flow through in the CC amplifier in Figure 3-7? * R1 * R2 * C1

30. What formula is used to calculate the output power of a CC amplifier? *

Preparation: 31. Transcribe the component measurements from Table 3-5 to the table below. Component | Listed Value | Measured Value | R1 | 33 k Ohm | 32.5 K | R2 | 10 k Ohm | 9.86 K | RE | 1 k Ohm | 992 | RL | 1 k Ohm | 1.01 K |
Table 3-5

Test Procedure: 32. Transcribe the calculations and measurements from Table 3-6 to the table below. DC Parameter | Computed Value | Measured Value | VB | 3.5 V | 3.7 V | VE | 4.2 V | 4.3 V | IE | 4.16 mA | Actual | 4.26 mA | VCE | 4.2 v | 4.3 V | VEE | 15 V | 15 V |
Table 3-6

33. Transcribe the calculations and measurements from Table 3-7 to the table below. AC Parameter | Computed Value | Measured Value | Vin = Vb | 1 Vpp | 3.2 V | r’e | 6 | | Av | 1 | 1.2 | Vout=Ve | 4.2 V | 4.3 V | Rin(total) | 6.6 K | 7.2 K | Ap | 6.7 | 7.3 |
Table 3-7 34. What is the CC amplifier’s phase relationship for its input and output signals? * The input and output signals are in sync with one another when used in this type of amplifier.

35. Transcribe the DC and AC load lines from Plot 3-2 to the plot below. * Plot located in attachment.

Plot 3-2 36. Transcribe your fault observations and findings from Table 3-8 to the table below. Trouble | DC Predictions | DC Measurements | Effect of Trouble on Vout | | VB | VE | VCE | VB | VE | VCE | | R1Open | .5 | 1.2 | 1.2 | .5 | 1.2 | 1.2 | Vout reduced to 1.2 Volts | R2Open | 7.6 V | 8.3 V | 8.3 V | 7.7 V | 8.4 V | 8.4 V | Same voltage across each leg except for a .7 V increase in VBE. | R1Shorted | 15 V | 15 V | 15 V | 14.9 V | 14.9 V | 14.9 V | Voltage dropped across Parallel circuit of R2 and RE+Rl is equal. Vout is DC voltage in. | REOpen | 3.2 V | 0 V | 0 V | 3.2 V | 11.2 mV | 11.2 mV | Transistor in Cutoff. Vout will be constant DC until Capacitor is drained. | OpenCollector | 12.8 V | 13.6 V | 0 V | 12.2 V | 12.9 V | 0 V | Waveform Clipped and 180 degrees out of phase. | OpenEmitter | 3.2 V | 0 V | 0 V | 3.2 V | 0 V | 0 V | No output waveform created due to open emitter. |
Table 3-8

37. Describe your observations when increasing and decreasing RL to values larger and smaller than 1k Ohm. Why are the output values clipped? * The output wave form began to rise and fall. As soon as the voltage got too a certain point, the diodes that make up the transistor began to conduct and started clipping the signal.

38. In the lab, you observed the CC amplifier’s phase relationship between the input and output waveforms. Would the phase relationship you observed be the same for an NPN circuit? Explain. * No because the output would have been taken across the Collector instead of the Emitter. The Collector in relation to the Base has a higher voltage gain than the Emitter does. The small swing of the Base waveform is causing a large swing in the Collector waveform which throws it out of phase. Since the Emitter is about the same voltage as the Base you do not have a swing in the voltage waveform. This keeps the waveform in phase with the Base waveform.

39. The circuit used in this experiment used voltage-divider bias. (a) Compared to base bias, what is the advantage? a. Base bias is dependent on the Beta value of the Transistor. This value can vary greatly from transistor to transistor even if they are the same model number. By using Voltage-Divider bias you can almost eliminate the value of Beta from your calculations. When calculating the output of your transistor you don’t even have to factor in the Beta rating. This makes the transistor very predictable in the way that it is going to operate.

(b) What disadvantage does it have? b. Using Voltage-Divider bias requires more components than Base biasing. This will increase the cost of the circuit as well as the heat produced by the extra load of those components. More time must also be taken in order to calculate the output of the transistor since there are more components involved.

40. Explain why common-collector amplifiers do not have voltage gain but still provide power gain. * When finding the Voltage Gain, R1 and R2 do not directly affect the input signal so they are not included in the calculations. The only resistance is seen at r’e in series with RE || RL and as long as r’e is very small compared to RE || RL then you can ignore the voltage drop across r’e all together. The Vout is then taken before it reaches RE || RL and since the Voltage experienced no drops, the Vout is equivalent to the Vin. This is also the same reason that you experience a gain in current. You find Ai by taking RL and Rin(tot) and by finding the ratio of these two measurements will give you the Current Gain. We can neglect the voltage gain because it is equal to 1. As long as RL is smaller than Rin(tot) you will notice a gain in the current. As RL continues to decrease, the current gain will continue to increase.

41. The figure below shows a CC amplifier with voltage-divider bias. Assume βac = βdc = 100. Compute the DC and AC parameters record in the table.

| DC Parameters | AC Parameters | | VB | 10.9 V | r’e | 21.6 | | VE | 11.6 V | Av | .996 | | IE | 1.16 mA | Rin | 24,311 | | VCE | 11.6 V | AP | 2.4 | | | | | |

-------------------------------------------------
Part 3—The Basic Power Supply

Theory 42. When transistors are used in switching applications, in what modes are they operated? * Cutoff or Saturation

43. What is the primary advantage to adding hysteresis to a switching circuit? * The switch is less susceptible to noise.

44. Give at least three advantages of transistor switching circuits over mechanical switches. * Can use low voltages or currents * More reliable * Cost less * Provide faster switching * Can provide isolation when a load is in a dangerous or remote location.

Preparation: 45. Transcribe the component measurements from Table 3-9 to the table below. Component | Listed Value | Measured Value | RB | 10 k Ohm | 9.87 K | RC | 1 k Ohm | 989 | RC1 | 10 k Ohm | 9.88 K | RE | 330 Ohm | 326 |
Table 3-9

Test Procedure: 46. Transcribe the calculations and measurements from Table 3-10 to the table below. Parameter | Computed Value | MeasuredValue | VCE(cutoff) | 15 V | 13.4 V | VCE(sat) | 0 V | 52.1 mV | VRC(sat) | 12.9 V | 12.4 V | Isat | 12.9 mA | |
Table 3-10

47. What is the purpose of RB in the circuit for Figure 3-14? * To keep the transistor from exceeding its maximum operating parameters by limiting the amount of current in the base.

48. Assume you wanted to determine the base current in this circuit. What voltage measurement would you make to do this indirectly? * The 12 V measurement that we took when setting the Vin to 12 Volts. We know that to forward bias the transistor it is going to cost .7 V. That means that 11.3 volts has to be dropped across the Base resistor. We then take that 11.3 over 10,000 to get a base current of 1.13 mA.

49. Transcribe the measurements from Table 3-11 to the table below. Parameter | Measured Value | VIN (LED on) | 0 V | VOUT (LED on) | 49.2 mV | VIN (threshold) | .655 V | VOUT (threshold) | 13.4 V |
Table 3-11 50. What is the relationship between cutoff and saturation for the two transistors in the circuit for Figure 3-15? * When the First Transistor goes into cutoff, the other goes into saturation. They are inversely related. When one is on, the other is off.

51. Transcribe the measurements from Table 3-12 to the table below. Parameter | Measured Value | VIN (LED on) | 0V | VOUT (LED on) | 50 mV | VIN (upper threshold) | .680 V | VOUT (upper threshold) | 13.4 V | VIN (lower threshold) | .647 V | VOUT (lower threshold) | .63 V |
Table 3-12 52. Why is the Vout value higher when Vin is set to 0V in the circuit for Figure 3-17? * The common resistor raised the Voltage threshold.

53. Why is the saturation current for the two transistors different? * The threshold levels have changed.

54. Assume you measured the collector current in a saturated transistor as 10 mA and the base current as 0.25 mA. Why can’t these measurements be used to determine the βDC of the transistor? * Because the transistor always operates at cutoff or saturation. The formula for BDC does not apply to switching amplifiers.