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1. ECE 333 Linear Electronics Chapter 4 Diodes Ideal diode  real Si diode  diode circuits  device modeling  serve as a foundation for modeling transistors and circuits in future chapters 1

2. §4.1 The Ideal Diode 4.1.1 Current-Voltage Characteristics Figure 4.1 The ideal diode: (a) diode circuit symbol; (b) i – v characteristic; (c) equivalent circuit in the reverse direction; (d) equivalent circuit in the forward direction. 2

3. 4.1.1 Current-Voltage Characteristics Figure 4.2 The two modes of operation of ideal diodes and the use of an external circuit to limit (a) the forward current and (b) the reverse voltage. 3

4. 4.1.2 A Simple Application: The Rectifier 4

5. Ex. 4.1 For the circuit in Fig. 4(a), sketch the transfer characteristics. Ex. 4.2 For the circuit in Fig4b, sketch the waveform of V D . 4.1.2 A Simple Application: The Rectifier 5

6. Ex. 4.3 In Fig. 4.3a, let V I have a peak value of 10V and R=1 k Ω . Find the peak value of i D and dc component of v . (Hint: the average value of half-sine waves is V p / π .) Ans. 10 mA; 3.18 V T         6

7. Example 4.1 v s is a sinusoid with 24-V peak amplitude, find the fraction of each cycle during which the diode conducts. Also, find the peak value of the diode current and the max reverse-bias voltage across the diode. Ans. The diode conducts when v s exceeds 12 V, conduction angle is 2 θ : The peak value is:       7

8. 4.1.3 Another Application: Diode Logic Gates Figure 4.5 Diode logic gates: (a) OR gate; (b) AND gate (in a positive-logic system). 8

9. Example 4.2 Assuming the diodes to be ideal, find I and V in the circuits of Fig. 4.6. Ans. Two situations: D1 is on or off? 9

10. The characteristic curve consists of three distinct regions: 1. The forward-bias region, determined by v > 0 2. The reverse-bias region, determined by v < 3. The breakdown region, determined by v < -V ZK 4.2 Terminal Characteristics of Junction Diode 10

11. 4.2.1 The Forward-Bias Region   What is k, T and q? I S is a constant for a given diode as a given temperature For high carrier injection: where n is from 1 to 2.     V T is 25 mV at 20 o C A good approximation:     11

12. Because: 4.2.1 The Forward-Bias Region       Example 4.3 A Si diode said to be a 1-mA device displays a forward voltage of 0.7 V at current of 1 mA. What is I S . How about at a 1-A diode?   12

13. Effect of temperature 4.2.1 The Forward-Bias Region     13

14. Real diodes exhibit reverse current much larger than I S . A small-signal diode whose I S is on the order of 10 -14 A to 10 -15 A could show a reverse current on the order of 1 nA. I S doubles for every 5 o C rise in temperature. Reverse current is mostly due to leakage effects. It doubles for every 10 o C rise in temperature. 4.2.2 The Reverse-Bias Region   14

15. Ex. 4.9 The diode in Fig. E4.9 is a large high-current device whose reverse leakage is reasonably independent of voltage. If V=1 V at 20 o C, find the value V at 40 C and 0 o C. 4.2.2 The Reverse-Bias Region Solve:           15

16. Beyond the breakdown voltage V ZR ~ zener breakdown at the “knee” of i-v curve 4.2.3 The Breakdown Region Normally not destructive Used in voltage regulator 16

17. 4.3 Modeling the Diode Forward Characteristic Why modeling? -- We need to analyze the current and voltage in the circuit. Ideal-diode model and exponential model This part will serve as a foundation for future transistor modeling. 17

18. 4.3.1 The Exponential Model     We can solve the two equations to find V D and I D 18

19. Graphical Analysis using the Exponential Model 4.3.2 Graphical Analysis 19

20. Ex. 4.4 Determine the current I D and V D with V DD =5 V and R=1 k Ω . Assume I D =1 mA at V D =0.7V 4.3.3 Iterative Analysis Using the Exponential Model Solve: First, we assume V D =V 1 = 0.7 V         Use this I D =4.262 and V D =0.738   20

21. 4.3.4 The Need for rapid analysis 4.3.5 The Constant-Voltage-Drop Model   21

22. 4.3.6 The Ideal-Diode Model     22

23. 4.3.7 The Small-Signal Model 23

24. 4.3.7 The Small-Signal Model   Voltage across the diode: Induce a current:       Or simplified as:     24

25. Because: v d /V T <<1 (that mean v d should be much less than 26 mV at room temperature) Use Taylor’s expansion: or 4.3.7 The Small-Signal Model       25

26. 4.3.7 The Small-Signal Model           r d : diode small-signal resistance or incremental resistance 26

27. 4.3.7 The Small-Signal Model         27

28. Example 4.5: power supply has a dc value of 10 V and 60-Hz sinusoid of 1-V peak amplitude. What is the dc voltage and amplitude of sine wave of the diode? 4.3.7 The Small-Signal Model 28

29. Example 4.5: power supply has a dc value of 10 V and 60-Hz sinusoid of 1-V peak amplitude. What is the dc voltage and amplitude of sine wave of the diode (V D =0.7 V at I D =1 mA)? Solve : Assume: V D ≈ 0.7 V, the diode dc current is   This is very close to 1 mA, so the diode voltage V D is indeed very close to 0.7 V. So, at this operation point, the diode incremental resistance r d is:     So, peak amplitude of v d is: 29

30. 4.3.8 Use of the Diode Forward Drop in Voltage Regulation A voltage regulator is a circuit to provide a constant dc voltage between its output terminals. To avoid: (a) changes in the load current drawn from the regulator output terminal and (b) changes in the dc power-supply voltage 30

31. Example 4.6 Consider the circuit in Fig. 4.17. A string of three diodes is used to provide a constant voltage of about 2.1 V. We want to calculate the % change in this regulated voltage caused by (a) a ± 10% change in power supply voltage, and (b) connection of a 1-k Ω load resistance. Solve : (a) With no load (R L ), the current in the diode string is:     So, resistance of 3 diodes is: r = 9.6 Ω   Is this solution precise? 31

32.   Is the constant voltage model precise?         Yes, it is quite precise! This is very close to 7.9 mA. 32

33. Example 4.6 Consider the circuit in Fig. 4.17. A string of three diodes is used to provide a constant voltage of about 2.1 V. We want to calculate the % change in this regulated voltage caused by (a) a ± 10% change in power supply voltage, and (b) connection of a 1-k Ω load resistance. Solve : (b) When a load R L is connected across the diode string, it draws a current of approximately 2.1 mA   So, the current through diodes decreases by 2.1 mA, resulting in a decrease in voltage:   So, this implies that the voltage across each diode decreases by 6.7 mV: Is this solution precise? 33

34. Let’s use exponential model to solve it.   Let’s use exponential model to solve it.             So current in diode decreases by 2.29 mA.   Not much difference 34

35. 4.4 Operation in the Reverse Breakdown Region-Zener Diodes Figure 4.18 Circuit symbol for a zener diode. 35

36. 4.4.1 Specifying and Modeling the Zener Diode   r z is the incremental resistance or dynamic resistance of Zener diode 36

37. V Z = V Z0 + r Z I Z For I Z > I ZK and V Z > V Z0 4.4.1 Specifying and Modeling the Zener Diode 37

38. Example 4.7 The 6.8-V zener diode in the circuit of Fig. 4.21(a) is specified to have V Z = 6.8 V at I Z =5mA, r Z =20 Ω , and I ZK =0.2mA. The supply voltage V + is nominally 10 V but can vary by ±1 V. (a)Find V o with no load and with V + as its nominal value. Solve : V Z = V Z0 + r Z I Z V Z0 = V Z - r Z I Z = 6.8 -0.1 = 6.7 V With no load:   V o = V Z = V Z0 + r Z I Z =6.7 +0.00635×20=6.83 V 38

39. Example 4.7 The 6.8-V zener diode in the circuit of Fig. 4.21(a) is specified to have V Z = 6.8 V at I Z =5mA, r Z =20 Ω , and I ZK =0.2mA. The supply voltage V + is nominally 10 V but can vary by ±1 V. (b) Find the change in V o resulting from the ±1-V change in V + . Solve : So, the line regulation is     39

40. Example 4.7 The 6.8-V zener diode in the circuit of Fig. 4.21(a) is specified to have V Z = 6.8 V at I Z =5mA, r Z =20 Ω , and I ZK =0.2mA. The supply voltage V + is nominally 10 V but can vary by ±1 V. (c) Find the change in V o resulting from connectiong a load resistance R L that draws a current I L =1 mA.. Solve : When R L draws 1 mA, the zener diode current will decrease by 1 mA, the corresponding change in zener diode voltage is:   So, the load regulation is   40

41. (d) Find the change in V when R L = 2 k Ω . (e) Find the value of V when R L = 0.5 k Ω . (f) What is the minimum value of R L for which the diode still operates in the breakdown region? Solve: The load current is about 6.8 V/ 2k Ω = 3.4 mA. So the change in V o is r z × (- 3.4 mA) = -68 mV Solve: Assume zener diode is operated in breakdown region, the load current is 6.8 V/ 0.5 k Ω = 13.6 mA. This is not possible because it is large than supplied current. So zener is cut off. V o = V + (R L /(R L +R))=5V Solve: for the zener diode to be operated at the edge of breakdown region, I z = I zk =0.2 mA and V z = V zk =6.7 V. so the current through R is (9-6.7)/0.5=4.6. so current in R L is 4.6-0.2=4.4. so R L =6.7/4.4 = 1.5 k Ω 41

42. 4.5 Rectifier Circuits Ripple  dc output   Power transformer dc Power supply 42

43. 4.5.1 The Half-Wave Rectifier     1. Current-handling capability required of the diode; 2. Peak inverse voltage (PIV) that the diode must be able to withstand without breakdown PIV = V S (Considering reverse bias)   43

44. Ex. 4.19 For Fig. 4.23(a), show and find: (a) the conduction angle ( π -2 θ ), where conduction begins at an angle θ =sin -1 (V D /V S ) and terminates at ( π - θ ). (b) the average value of v o is . (c) the peak diode current is (V s -V D )/R and PIV. (given: V s is 12-V ( rms ) sinusoidal input, V D =0.7 V, and R=100 Ω . Solve :   π - θ θ   44

45. Ex. 4.19 For Fig. 4.23(a), show and find: (b) the average value of v o is . Solve :   π - θ θ 45

46. Ex. 4.19 For Fig. 4.23(a), show and find : (c) the peak diode current is (V s -V D )/R and PIV. (given: V s is 12-V ( rms ) sinusoidal input, V D =0.7 V, and R=100 Ω . Solve : π - θ θ * Rms : root-mean-squared. 12-V root-mean-squared sinusoid wave is . Here, V s =   46

47. 4.5.2 The Full-Wave Rectifier transformer   PIV = 2V S - V D   47

48. 4.5.2 The Full-Wave Rectifier PIV = V S + V S – V D = 2V S - V D At this point, D 1 feel the PIV V S V S - V D 48

49. 4.5.3 The Bridge Rectifier 49

50. 4.5.3 The Bridge Rectifier What is the PIV? v D3 (reverse) = v o + v D2 (forward) So, PIV = V S - 2V D + V D = V S - V D   + -   D 3 D 2 50

51. 4.5.4 The Rectifier with a Filter Capacitor – The peak Rectifier The filter capacitor serves to reduce substantially the variations in output voltage Once charge, no way to discharge the capacitor 51

52. Consider real application-- with load resistance: R 52 Assume: CR >> T    

53. Consider real application-- with load resistance: R 53 1. Diode conducts for a brief interval Δ t; 2. Assume an ideal diode: from t 1 to t 2 t 1 : v I = v o t 2 : i D = 0, shortly after the peak of v I 3 . During the diode-off interval, C discharges through R. v o decays exponentially with time constant CR ; at the end of discharge interval (≈ T), v = V p - V r 4. When V r is small, v o is almost constant and equal to V p . So, i L is almost constant: Its dc component I L is   * V r is peak-to-peak ripple voltage

54. 54 Average of v o is   Now, we will derive V r During the diode-off interval   At the end of discharge interval         (use Taylor expansion)

55. 55 We can now determine conduction Interval Δ t   (Q: why not using sin(w Δ t )?) use Taylor expansion because Δ t is small    

56. From the above results, we can also determine average diode current during conduction ; this is much larger than load current during diode conduction. And the peak value of diode current The above is for half-wave peak rectifier.   56

57. For a full-wave rectifier circuit with a capacitor, discharge period T is replaced by T/2 : 57      

58. 4.5.5 Precision Half-Wave Rectifier –The superdiode 58 Negative-feedback path Of an op amp Load  

59. 4.6 Limiting and Clamping Circuits (self-reading) 59 Figure 4.30 General transfer characteristic for a limiter circuit. Figure 4.31 Applying a sine wave to a limiter can result in clipping off its two peaks.

60. 60 4.6 Limiting and Clamping Circuits (self-reading)

61. 4.7 Special Diode Types (self-reading) The Schottky-Barrier Diode (SBD) Varactors Photodiodes Light-Emitting Diodes (LEDs) 61

62. Chapter 4 Homework 4.4, 4.9, 4.10, 4.18, 4.23, 4.28, 4.36, 4.54, 4.58, 4.61, 4.76 It is due March 1 st ( Tuesday ). 62