Sedra/Smith Microelectronic Circuits 6/E. Chapter 3: Diodes. S. C. Lin, EE National Chin-Yi University of Technology

Transcription

1 Sedra/Smith Microelectronic Circuits 6/E Chapter 3: iodes 1

2 Outline 3.1 The Ideal iode 3.2 Terminal Characteristics of Junction iodes 3.3 Modeling the iode Forward Characteristic 3.4 Operation in the Reverse Breakdown Region Zener iodes 3.5 Rectifier Circuits 3.6 Limiting and Clamping Circuits 3.7 Special iode Types 2

3 3.1 The Ideal iode i + v (a) diode circuit symbol (b) i v characteristic i i v + v < 0 i = 0 i + v > 0 v= 0 (c) equivalent circuit in the reverse direction (d) equivalent circuit in the forward direction. 3

4 The two modes of operation of ideal diodes and the use of an external circuit. +10V +10V 10mA 1kΩ 0mA 1kΩ + + 0V 10V (a) the forward current (b) the reverse voltage 4

5 v i + v V p 0 t v i i R v + o (a) Input waveform. (b) Rectifier circuit. + v = 0 + v = vi i + v i vo = v v i = 0 R v + i i R o = 0 vi 0 vi 0 (c) Equivalent circuit when v 0. (d) Equivalent circuit when vi 0. i v o V p 0 t (e) Output waveform. 5

6 Example 3.1 Fig.3.4(a), shows a circuit for charging a 12-V battery. If v s is sinusoid with 24-V peak amplitude, (a) Find the fraction of each cycle during which the diode conducts. (b) Also find the peak value of the diode current and (c) the maximum reverse-bias voltage that appears across the diode. 100Ω i Method 1: v s 12V 24V 12V α θ V ( a) 24sinα = 12 α = 30 o 2θ = 180 2α = 120 o o 6

7 Method 2: 24V 12V V θ θ (a)24 cosθ = 12 θ = 60 (b) Conduction angle = 2θ = 120 I (max) (max) = = 0.12A 100 o (c)the maximum reverse voltage across the diode: V = = 36V o 7

8 3.1.3 Another Application: iode Logic Gates + 5V v A v B R v C v Y (a) OR gate v A (b) AN gate R v B v Y v C va vb v C vy va vb v C vy 8

9 Example 3.2 Assuming the diode to be ideal, Find the values of I and V in the circuit of Fig. (a) and (b) I I V 5kΩ +10V 10kΩ V + Sol: Asssume that both diodes are conducting VB = 0 and V = 0 (10 0)V I = = 2 1mA 10kΩ Writting a node equation at B (K.C.L) 0 ( 10) I + 1= I = 1mA 5 Results in I = 1mA, Thus 1 is conducting as originally assumed and the find resuit is I = 1mA and V = 0V 9

10 . I +10V 5kΩ I V + Sol: We asssume that both diodes are conducting then V = 0 and V = 0 I B 2 (10 0) V = = 2mA 5kΩ The node equation at B is 10kΩ 0 ( 10) I + 2 = I = 1mA 10 (This is not possible) 10V Our assumption is not correct, we start again assuming that is OFF and is ON. 10 ( 10) The current I is given by I 1.33mA 2 = = 2 15 V = = 3.3V 1 B Thus is reverse biased as assumed I = 0A and V = 3.3V

11 3.2 Terminal Characteristics junction diode i 1 (1) The forward-bias region determined by v > 0 (2) The reverse-bias region determined by V < v< 0 ZK V ZK 0 0.5V 0.7V v (3) The breakdown-bias region 3 2 Expanded scale i + v determi ned by v < V ZK Figure 3.8 The diode i v relationship with some scales expanded and others compressed in order to reveal details. 11

12 3.2.1 The Forward-bias region ( v/ nv T ) i = I e 1 (3.1) s ( a) I is saturation current, another name for I is the scale current. s ( b) I s is a constant for a given diode at a given temperature. ( c) I s is directly propotional to the across-sectional area of the diode. s (d) I For "small-signal" diode intened for low-power application, I s s is on the order of A ( e) I a very strong function of temperature. As a rule of thumb, I s s doubles in value for every 5 C rise in temperature. o 12

13 V T : Thermal voltage V T = K T q (3.2) -23 where K = Boltzmann's constant = Joules/Kelvin T = The absolute temperature in Kelvin = 273+temperature q = The magnitude of electronic charge = Coulomb 19 o C At room temperature(20 C), At room temperature(27 C), o o V V T T 25.2mV. 26mV. n n :1 2, epending on the material and the physical structure of the diode. 13

14 the forward direction i >> v/ nv i T i Ise v= nvt ln Is If the voltage is v ( v ), the diode current I ( I ) thus i I e and i I e i i v1/ nvt v2/ nv 1 s 2 s 2 1 = e ( )/ v2 v1 nv T i v v = nv = nv 2 1 T I s 2 2 ln 2.3 T log (3.5) i1 i1 i T This equation simply states that for a decade changes in current, the diode voltage drop changes by 2.3nV 60mV,for n = 1 T 14

15 Example: esign the circuit in below Figure to provide an output voltage of 2.4V. Assume that diode available have 0.7V drop at 1mA and that V= 0.1V/decade changes in current. +10V R v o Sol: Refer Fig.a to obtain v o = 2.4V each diode must have a voltage drop of 0.8V thus i2 v2 v1 = 2.3nVT log i1 3 i2 i2 0.1 = log = 0.1log i1 i1 i2 = 10i1 = 10mA The value of R is determined from (10 2.4)V R = = 760Ω 10mA 15

16 T2 T1 i > o 2mV/ C I v Figure 3.9 Illustrating the temperature dependence of the diode forward characteristic. At a constant current, the voltage drop decreases by approximately 2 mv for every 1 C increase in temperature. 16

17 Example 3.3 : A silicon diode said to be a 1-mA device displays a forward voltage of 0.7V at a current of 1mA. Evaluate the junction scaling constant Is in the event that n is either 1or 2, what scaling constants would apply for a 1-A diode of the same manufacture that conducts 1A at 0.7V Si nce for the 1 ma diode. If n 1 i I e I = i e s v/ nvt ( v/ nvt ) s /25 16 = Is = e = A / If n = 2 I = 1 10 e = A s The diode conducting 1A at o.7v corresponds to mA in parallel with a total junction area 1000 times greater. Thus it is also 1000 times greater being 1pA and 1µA, respectively for n = 1and n = 2. 17

18 3.2.2 The Reverse-Bias Region If v is negative and a few times large then V (25mV) in magnitude the diode current becomes i I s (Saturation current) I s o : It doubles for every 10 C rise in temperature I s = I 2 s 2 1 ( T T /10) 2 1 T The Breakdown Region Z: Zener v V K: Knee < ZK In the breakdown region the reverse current increase rapidly with the associated increase in voltage drop very small 18

19 3.3 Modeling the diode forward characteristic The Exponential Model. I R + V V V > 0.5V V / nvt Assuming that I = Ise (3.6) I > Is The other equation that governs circuit operation is obtained by writing a V V loop equation by KVL, resulting in I = (3. 7) R 19

20 3.3.2 Graphical Analysis Using the Exponential Model. i V R Q(Operating point) I Slop = 1 R 0 V V v Figure 3.11 Graphical analysis of the circuit in Fig using the exponential diode model. 20

21 3.3.3 Iterative analysis using the exponential model Example3.4 I determine the current and the diode voltage V for the circuit in Fig.3.10 with V = 5V and R= 1kΩ, assume that the diode has a current of 1mA at voltage of 0.7V and that its voltage drop changes by 0.1V for every decade change in current. Sol : We assume that V = 0.7V, I = 1mA V V (5 0.7)V I = = = 4. 3mA R 1kΩ I2 V2 V1 = 2.3 n VT log I I V = V + n= n V V 1 ( ) log 2, 2.3 T 0.1V I1 4.3mA = log 1mA = 0.763V The first iteration are I = 4.3mA, V = 0.763V 21

22 We assume that V = 0.763V, I = 4.3mA V V ( )V I = = = 4.237mA R 1kΩ log 4.237mA V2 = + = 0.762V 4.3m A The second iteration are I = 4.237mA, V = 0.762V We assume that V = 0.762V, I = 4.237mA I V 2 V V ( )V = = = 4.238mA R 1KΩ 4.238mA = log = 0.762V mA Because did not change by much stop here and the solution is I = mA and V = V I 22

23 3.3.5 The Constant-Voltage-rop Model. (Rapid analysis) i (ma) V = 0.7V i = 0, v 0.7V i = ( v 0.7V) / R v 0.7V v (V) V 23

24 The constant-voltage-drop model of the diode forward characteristics and its equivalent-circuit representation. i (ma) I V = 0.7V v V = 0.7V v (V) 24

25 Example: Repeat the problem in example 3.4, utilizing the constant-voltage-drop model whose parameters are given in Fig Find I and V. I I R + + V V v V = 0.7V Sol: I V V V (5 0.7)V = = = 4.3mA R 1kΩ = V = 0.7V 25

26 3.3.7 The iode Small-Signal Model. i = I e S v / nv T (3.8) v () t = V + v () t (3.9) d 26

27 i() t = Ie = Ie v () t / nvt S S V / nvt = Ie S I = I e v d e () t / nv v ()/ t nv T d [ V + v ( t)] / nv T d T (3.12) If the signal such that v d v d nv () t T ( t) is kept sufficiently small << 1 Thus, the approximate expression v d i () t = I 1 + (3.14) nvt Small-signal approximation 27

28 We have I i () t = I + vd nv T where i r d d I = nv v T d = i d v d nv I T i d (3.15) (3.17) ( 3.18) The r d is diode smal-signal resistance, and called incremental resistance) In Fig.3-13(b), which is equal to the slope of the tangent to the i- v curve at the operating point Q is equal to the small-signal conductance. that is, r d 1 = id v d i = I d (3.19) 28

29 Example 3.6: Consider the circuit shown in Fig 3.15 a string of three diode is used to provide a constant voltage of about 2.1V, we want to calculate the percentage change in this regulated voltage cause by : (a) 10% change in the power-supply voltage (b) Connection of a 1k Ω load resistance. Assume n=1 Sol: with no load (10 2.1)V I = = 7.9mA 1kΩ nvt 25mV rd = = = 3.2Ω I 7.9mA The total incremental resistance r = 3r = 9. 6Ω d 29

30 (a) r v =± o mV r R =± = ± + + This implies a change of about ± 3.2mV per diode. (b)when a load resistance of 1k Ω is connected across the diode string. it draws a current of 2.1mA v = 2. 1 r = = 20mV. o this implies that the voltage across each diode decreses by about 6. 7mV. 30

31 3.4 Operation in the reverse breakdown region -Zener diode (1) rz is the inverse of the slop of the almost-linear i - v curve at poin t Q (2) r Z : increase resistance (3) Typically, r Z is in the range of a few ohms to a few ten ohms 31

32 Vz = Vzo + Izrz (3.20) 32

33 3.4.2 Use of the Zener as a Shunt Regulator Example 3.7: The 6.8V Zener diode in the circuit of Fig.3.19(a) is specified to have V = 6.8V at I = 5mA, r = 20 Ω and I = 0.2mA, the V is 10V ± 1V. ZK + Z Z z 33

34 (a) Find Vo with no load and with V + at its nominal value. Sol: first we must determine the value of Iz = I= = 6.35mA Thus V = V = V + I r = 6.83V o z zo z z 3 V = V I r = = 6.7V zo z z z V zo (b) Find change in Vo resulting from the ± 1V change in V +. + rz 20 Sol: Vo = V =± 1 =± 38.5mV R + r z line regulation V / V =± 38.5mV/1V = ± 38.5mV/V o i 34

35 (c) Find change in V resulting from connecting a load resistance R. that draws a current ( V / I ) in mv/ma. o L o I L = 1mA, and hence find the load regulation Sol: V = I r = 20Ω 1mA = 20mV o load regulation z z V / I = 20mV/mA (d)find change in V when R = 2kΩ. Sol: R = 2kΩ, I I L RL z o o 6.8V / 2kΩ = 3.4mA V = I r = = 68mV o z z L L L 35

36 (e)find the value of V when R = 0.5k Ω. Sol: o (f) What is the minmum value of R L for which the diode still operates region in the breakdown. L RL = 0.5kΩ, would draw a IL = 6.8/ 0.5 = 13.6mA.This is impassible. because I supplied through R is only 6.4mA, therefore, + RL the zener must be cut-off. He nc e Vo = V = 5V R + R L Sol: I z = I = 0.2mA V = V V = 6.7V zk z zk zo I(min) = = 4.6mA I(max) = ( )mA = 4.4mA, RL = = 1.5kΩ

37 3.4.3 Temperature effects, T.C or Temco ( 1) V < 5V, dv / dt < 0. Negative T.C Z z (2) V > 5V, dv / dt > 0. Positive T.C Z z (3) V = 5V, dv / dt = 0. Zero T.C Z z 37

38 3.5 Rectifier Circuits Figure 3.20 Block diagram of a dc power supply. (1) The Half-wave rectifier. (2) The full-wave rectifier (a) The center tap rectifier (b) The bridge rectifier 38

39 3.5.1 Half-Wave Rectifier. (a) Half-wave rectifier. (b) Equivalent circuit of the half-wave rectifier with the diode replaced with its battery-plus-resistance model. 39

40 (c) Transfer characteristic of the rectifier circuit. v = 0, v < V R v = ( v V ), v V R+ r O S O O S O S O φ i = sin 1 V V m φ i (d) Input and output waveforms, assuming that r << R. 40

41 We assume v = V sin α thus V C 1 π = v 2 0 Sdα π 1 π = V sin 2 0 m αdα π Vm π = ( cos α ) 0 2π Vm = = 0.318V m π S m 1/2 1 π 2 = v 0 S dα Vrms 2π 2 1 π 2 2 Vrm s = V sin 2 0 m αdα π 2 V π m 1 1 = ( cos 2 α) dα 2π Vm 1 1 = ( α sin 2 α) 2π Vm = V = 0. 5V 4 rms m π 0 PIV = V φ i = sin m 1 (PIV : Peak inverse voltage) V V O m 41

42 3.5.2 The full-wave rectifier (a) circuit (b) transfer characteristic assuming a constant-voltage-drop model for the diodes 42

43 (c) input and output waveforms. 43

44 1/2 2 1 π 2 = v 0 S dα We assume vs = Vmsin α thus Vrms 1 π 2π V C = 2 v 2 0 Sdα π π 2 2 V = rms V 0 m sin αdα 2 1 2π π = V sin 2 0 m αdα 2 π 2 V π = m 1 1 ( cos2 α) dα V 2π m π = ( cos α ) 0 2 π π Vm 1 1 = ( α sin 2 α) 2 Vm π 2 4 = = 0.636V 0 m π 2 Vm = Vrms = Vm 2 PIV = 2 V (PIV : Peak inverse voltage) φ i = sin 1 m V V m 44

45 3.5.3 The Bridge Rectifier Figure 3.27 The bridge rectifier: (a) circuit; (b) input and output waveforms. 45

46 3.5.4 The Rectifier with a Filter Capacitor-The Peak etector A simple circuit used to illustrate the effect of a filter capacitor 46

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49 We assumption that RC >> T in Fig3.29(b) The load current il = vo/ R dvi The diode current i = ic + il = C + il dt (1) erivative the T/ RC r( rms) uring the diode cutoff interval V V = V e V P r P r V T RC V V frc V P = P = r( rms) = V cos( ω t) = V V P P r v = V V V 2 3fRC (2) erivative the conduction interval t o P r 1 The angle ω t is very small, cos( ω t) 1 ( ω t) 2 1 V ω t = V V ω t V V 2 = 2 P 1 ( ) P r 2 r / P (3.30) P 2 We note that when V << V the conduction angle ω t will be small r P 49

50 (3) ertermine the average diode current during conduction, i the charge that the diode supplies to the capacitor Q Supplied = ic( av) t (a) the charge that the capacitor loses Q lost = C Vr (b) CV r The(a) equal to (b) ic( av) = t VP VP Using(3.29 a) Vr = C =, frc V fr 2V and(3.30) ω t = 2 Vr / VP t = ω V V C V f r we obtain: ic( av) = = t 2 r r P P V r R V 2π f V r r av 2π VP VP I π 2 V / V 2V R = i( av) = ic( av) + IL= I L 1 π 2 VP/ V + r (3.31) V P = r L P r 50

51 (4) ertermine the peak value of the diode current, i dv dt I i(max) = C + il (max) dvi d ( t = t1 = t) C = C ( VPcos( ω t) ) = ωcvpsin( ω t) dt dt 3 ( ω t) = ω C VP ( ω t) [sin( ω t) ] = ω t + + 3! 2 r P L P r V V P = 2πf V = I 2π 2 V / V fv R V r i(max) = i + I = I 1+ 2π 2 V / V P C L L P r (3.32) 51

52 the full-wave rectifier circuit. f r = 2 f i T /2 RC VP Vr = VPe T /2 VP Vr = VP = RC 2 f RC VP Vr( rms) = 4 3 frc (3.33) 52

53 2.4V P Vr( rms) = f = RL= = RC when the ripple small then 6. 5% V = V V r( rms) r% = 2.4VC 2.4I = = RC C Vr( rms) 2.4 = % V RC C If r% > 6.5% ( 60Hz, kω, C µf) C i( av) = ic( av) + IL= I L 1 π VP/ 2 V + r (3.34) i(max) = ic + IL = I L 1+ 2 π VP / 2 V r (3. 35) P C 1 VP 4.17 V = V V = V = 1 V 2 4fRC RC C P r P P 53

54 Figure 3.27 The superdiode precision half-wave rectifier and its almost-ideal transfer characteristic. Note that when v I > 0 and the diode conducts, the op amp supplies the load current, and the source is conveniently buffered, an added advantage. Not shown are the op-amp power supplies. 54

55 3.6 limiting and clamping circuits limiter circuit. L v K v = Kv + I O I L K Ingeneral K > 1 In this section have K 1 L+ ( a) If vi > (the upper threshold), K the v is limited or clamping to the upper limiting level O L ( b) If vi < (the lower limited threshold), K the v is limited to the lower limiting level L O. L + 55

56 Hard limiting. Soft limiting. 56

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58 Example.3.26 Assuming the diode to be ideal, describe the transfer characteristic of the circuit shown in Figure E3.26. Figure E3.26 Sol: (1) v 5V, v O I : conducts, cut-off = vi = 0.5v 2. 5 I (2) 5V v 5V, both diode are cut-off and v O = v (3) v 5V, v O I I I : cut-off, conducts = vi = 0.5v I 58

59 . v o v O = 0.5v I v I v O = 0.5v 2.5 I 59

60 Example : Assuming the diode to be ideal, describe the transfer characteristic of the circuit shown in Fig.a. 0 < v i 150V A v i 100kΩ kΩ + Sol: For VA 100V, both diodes are conducts 2 25 vi + = 100 vi = 137.5V 3 3 v o (1)For 0 < Vi 25V, both diodes are cut-off v O = 25V (2)For 25V< Vi < 137.5V, 1 conducts and 2 cut-off 2 25 vo = VA = vi + (V) 3 3 (3)For 137.5V< Vi < 150V, both diodes are conducts v = 100V O 60

61 v o v O 2 25 = vi v i 61

62 3.6.2 The Clamped Capacitor or C Restorer Figure 3.32 The clamped capacitor or dc restorer with a square-wave input and no load. 62

63 Figure 3.33 The clamped capacitor with a load resistance R. 63

64 3.6.3 Voltage oubler Figure 3.34 Voltage doubler: (a) circuit; (b) waveform of the voltage across 1. 64

65 65