1. EE 334 Midterm Review

2. Diode: Why we need to understand diode? The base emitter junction of the BJT behaves as a forward bias diode in amplifying applications. The behavior of the diode when reverse bias is the key to the fabrication of the integrated circuits. The diode is used in many important nonamplifer applications.

3. Departure from ideal behavior The four major reason why the actual diode do not correspond exactly to the ideal. 1.Ohmic resistance and contact resistance in series with the diode cause the VI curve to become linear at high forward current. 2.Avalanche or Zener breakdown take place at high reverse voltage, causing an abrupt increase in reverse current. 3.Surface contaminants cause an ohmic layer to form across the junction, which is Increasing the reverse current as reverse voltage is increased. 4.Recombination of current carrier in the depletion region take place due to traps.

4. The purpose of modeling Nonlinear problems are much more difficult than linear ones. These problems could be impossible to solve manually and could require huge amount of time if solved on a computer. One possible solution of the above mentioned problem is to approximate the nonlinear relationship with a model that has a linear relationship. The trust of nonlinear modeling is direct towards this end. The modeling not only simplifies the solution, it also allows the designer to understand how the circuit behaves. Modeling often increases the conceptual understanding of the circuit operation.

6. Schottky Barrier Diode One semiconductor region of the pn junction diode can be replaced by a non-ohmic rectifying metal contact.A Schottky contact is easily formed on n-type silicon. The metal region becomes the anode. An n + region is added to ensure that the cathode contact is ohmic. Schottky diodes turn on at a lower voltage than pn junction diodes and have significantly reduced internal charge storage under forward bias.

8. Reverse Breakdown Increased reverse bias eventually results in the diode entering the breakdown region, resulting in a sharp increase in the diode current. The voltage at which this occurs is the breakdown voltage, V Z. 2 V < V Z < 2000 V

10. Half Wave Rectification

11. Figure 2.7 A full-wave bridge rectifier: (a) circuit showing the current direction for a positive input cycle, (b) current direction for a negative input cycle, and (c) input and output voltage waveforms

13. Half-wave rectifier with filter

19. % regulation is used to measure how well the regulator is Performing its function. Voltage regulation is the measure of circuit’s ability to maintained a constant output even when input voltage or load current varies.

20. Demonstration of Zener diode as a voltage regulator

25. The large value of V CE decreases the effective base width W. Since I S is inversely propositional to W, which cause increase in I C.

27. Bipolar NOR logic gate Example 3.11 Determine current and voltage in the circuit 3.43(b) R c =1K  R B =20K  V BE (on)=0.7V VCE(sat)=0.2V β=50 Lecture #3

28. The process by which the quiescent output voltage is caused to fall somewhere the cutoff and saturated values is referred to as biasing.

30. Example 3.13

31. Q-point has shifted Substantially. Q-point is not stabilized Against the variation .

40. Chapter 4 Small-Signal Modeling and Linear Amplification

41. DC and AC Analysis DC analysis: –Find dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits. –Find Q-point from dc equivalent circuit by using appropriate large-signal transistor model. AC analysis: –Find ac equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits. –Replace transistor by small-signal model –Use small-signal ac equivalent to analyze ac characteristics of amplifier. –Combine end results of dc and ac analysis to yield total voltages and currents in the network.

42. DC Equivalent for BJT Amplifier All capacitors in original amplifier circuits are replaced by open circuits, disconnecting v I, R I, and R 3 from circuit.

43. AC Equivalent for BJT Amplifier Find ac equivalent circuit by replacing all capacitors by short circuits,

45. Hybrid parameter I: diffusion resistance/input impedance The diffusion resistance r  is define as the reciprocal of the i B -v BE curve, which can be find as,

46. Output terminal characteristics of the bipolar transistor: transconductance If we assume constant collector-emitter voltage the, As we know

47. By using the two hybrid parameters (r , g m ), we can develop a simplified small signal hybrid-  - equivalent circuit for the npn transistor. Voltage -controlled current source g m v be can be transformed into current- controlled current source,

48. From equivalent circuit we can write as Voltage gain

49. If we include the early effect then collector current in terms of early voltage as, then

50. Summary of hybrid-  -model parameters Diffusion resistance transconductance Current gain Output resistance

51. Characteristics of a CE amplifier It has moderately low input impedance (1K to 2K) Its output impedance is moderately large(50K or so) Its current gain is high It has very high voltage gain of the order of 1500 or so It produce very high power gain of the order of 10,000 times or 40dB It produce phase reversal of input signal Uses: many applications because of Large gain in voltage, current and power

52. (by voltage divider Rule)

53. Small-Signal Analysis of Complete C-E Amplifier: AC Equivalent Ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage source is ac ground. Assume that Q- point is already known.

54. Small-Signal Analysis of Complete C-E Amplifier: Small- Signal Equivalent

55. If we include an emitter resistance in the circuit, the Q-point of the circuit will be less dependant on the transistor current gain .

56. In order to determine the input impedance R ib, which is the resistance looking into the base of the transistor. We can write the following loop equation The overall input impedance to the amplifier is now Voltage gain is less dependant on 

57. The voltage gain is Substantially reduced When an emitter resistor is included!!

58. How can we Improve the voltage gain ?

59. A common collector amplifier has following chracteristics: 1High input impedance (20-500K) 2Low output impedance (50-2000 Ohms) 3High current gain (50-300) 4Voltage gain of less than 1 5Power gain of 20 to 20dB 6No phase reversal between input and output signals

60. Apply KVL around the base emitter loop Using above equations we can write voltage gain as Input impedance

61. Clearly the voltage gain is less than 1 and no phase reversal

62. Input and output impedance for EF

63. Very large compared to CE

64. Common base amplifier has Very low input impedance (30-150  ) Very high output resistance (500K) Current gain <1 Large voltage gain of about 1500 Power gain of upto 30dB No phase reversal between input and output Uses: for matching low impedance circuit to high impedance circuit

66. Chapter 5 Field-Effect Transistors

68. The MOS Transistor Polysilicon Aluminum

69. It is to be noted that the V DS measured relative to the source increases from 0 to V DS as we travel along the channel from source to drain. This is because the voltage between the gate and points along the channel decreases from V GS at the source end to V GS -V DS. When V DS is increased to the value that reduces the voltage between the gate and channel at the drain end to V t that is, V GS -V DS =V t or V DS = V GS -V t or V DS (sat) ≥ V GS -V t Concept of Asymmetric Channel

70. NMOS Transistor: Saturation Region is called the saturation or pinch-off voltage

76. Common source circuit with coupling capacitance Cc, which act an an open circuit to the dc DC equivalent circuit. Gate

77. PMOS common source circuit If the device is biased in saturation region

78. If I D =0  V DS =5V If V DS =0  I D =V DD /R D =0.25mA

82. Pinchoff at the drain terminal

87. 

88.   We see 

89. VDS=VDD-ID(RS+RD) 2=3-(0.040)(10+RD RD=15K  ID=K(VGS-VTN) 2 40=250(VGS-0.20) 2 VGS=0.6V VG=VGS+VS=0.60+(0.040)(10)=1.0V VG=(R2)/R1+R2)VDD 1=(R2)/150(3) R2=50K  R1=100k 