Chapter 3 Diode Circuits. 3.1 Ideal Diode 3.2 PN Junction as a Diode 3.3 Applications of Diodes

Transcription

1 Chapter 3 Diode Circuits 3.1 deal Diode 3.2 PN Junction as a Diode 3.3 Applications of Diodes 1

2 Diode s Application: Cell Phone Charger An important application of diode is chargers. 充 電 器 Diode acts as the black box (after transformer) that passes only the positive half of the stepped-down d sinusoid. id 變 壓 器 : 隔 離 耦 合 升 壓 降 壓 2

3 Diode s Action in The Black Box (deal Diode) The diode behaves as a short circuit during the positive half cycle (voltage across it tends to exceed zero), and an open circuit during the negative half cycle (voltage across it is less than zero). short circuit open circuit 3

4 deal Diode n an ideal diode, if the voltage across it tends to exceed zero, current flows. t is analogous to a water pipe that allows water to flow in only one direction. The corresponding terminals are called the anode and the cathode, respectively. 陽 極 陰 極 4

5 Diodes in Series The diode must turn on if anode > cathode and off if anode < cathode. Defining anode - cathode = D, we say the diode is forward-biased if D tends to exceed zero and reverse-biased if D < 0. Diodes cannot be connected in series randomly. For the circuits above, only (a) can conduct current from A to C. 5

6 Characteristics of an deal Diode short: R 0 open: R 0 R R f the voltage across anode and cathode is greater than zero, the resistance of an ideal diode d is zero and current becomes infinite. i However, if the voltage is less than zero, the resistance becomes infinite and current is zero. 6

7 Anti-Parallel deal Diodes f two diodes are connected in anti-parallel, it acts as a short for all voltages. f A > 0, D 1 is on and D 2 is off, yielding A =. f A < 0, D 1 is off, but D 2 is on, A 1 2 y g A A 1 2 again leading to A =. 7

8 Diode-Resistor Combination Example 3.4: Plot the / characteristic for the diode-resistor combination. The characteristic of this diode-resistor combination is zero for negative voltages and Ohm s law for positive voltages. 8

9 Diode mplementation of OR Gate Example 3.6: A and B can assume a value of either zero or 3. Determine the response observed at the output, out. A B out f A = 3, and B = 0th 0, then we surmise that thtdd 1 is forward-biased d and D 2 reverse-biased. Thus, out = A = 3. f uncertain, we can assume both D 1 and D 2 are forward-biased, immediately facing a conflict: enforces a voltage of 3 at the output whereas D 2 shorts out to B = 0. This assumption is therefore incorrect. The symmetry of the circuit with respect to A and B suggests that out = B = 3 if A = 0 and B = 3. The circuit operates as a logical OR gate and was in fact used in early digital computers. We have introduced NOT and OR gates. How about AND gates? 9

10 nput/output Characteristics The input/output characteristics of a circuit is constructed by varying the input across an allowable range and plotting the resulting output. When in is less than zero, the diode d opens, so out = in. When in is greater than zero, the diode shorts, so out = 0. 10

11 Diode s Application: Rectifier 整 流 器 A rectifier is a device that passes positive-half cycle of a sinusoid and blocks the negative half-cycle or vice versa. 前 一 頁 即 是 相 反 動 作 的 電 路 When in > 0, diode shorts, so out = in ; however, when in < 0, diode opens, no current flows thru R 1, out = R1 R 1 = 0. 11

12 Signal Strength ndicator 信 號 強 度 指 示 器 Suppose in = p sinωt, where ω = 2π/T denotes the frequency in radians per second and T the period. Then, in the first cycle after t = 0, we have T out p sin t for 0 t 2 T 0 for t T 2 To compute the average, we obtain the area under out and normalize the result to the period: T T / out, avg out t dt p ( ) sin T T T p cos t T 0 / 2 p tdt out,avg is proportional to p, the input signal s amplitude, and can be used as a signal strength indicator for the input. 12

13 Diode s application: Limiter 限 壓 器 The purpose of a limiter is to force the output to remain below certain value. n (a), the addition of a 1 battery forces the diode to turn on after 1 > 1. 13

14 Limiter: When Battery aries (Example 3.10) Sketch the time average of out for a sinusoidal input as the battery voltage, B, varies from - to. Rectification fails if B is greater than the input amplitude. Passes only negative cycles to the output, producing a negative average. 14

15 Different Models for Diode So far we have studied the ideal model of diode. However, there are still (b) the exponential and (c) constant voltage models. 15

16 Example 3.12 Plot the input/output characteristic of the circuit in Figure (a) using (a) the ideal model and (b) the constant-voltage model. 16

17 (a) We begin with in = -, recognizing that D 1 is reverse biased. n fact, for in < 0, the diode remains off and no current flows through the circuit. Thus, the voltage drop across R 1 is zero and out = in. As in exceeds zero, D 1 turns on, operating as a short and reducing the circuit to a voltage divider. That is, R2 out inforin 0 R R 1 2 Figure (b) plots the overall characteristic, revealing a slope equal to unity for in < 0 and R 2 /(R 1 +R 2 ) for in > 0. n other words, the circuit operates as a voltage divider once the diode turns on and loads the output node with R 2. (b) n this case, D 1 is reverse biased for in < D,on, yielding out = in. As in exceeds D,on, D 1 turns on, operating as a constant voltage source with a value D,on. Reducing the circuit to that in Fig. (c), we apply Kirchoff s current law to the output node: in out out Don R R 1 2 t follows that out R2 in R1 2 1 R R 1 Don As expected, out = Don D,on if in = Don D,on. Figure (d) plots the resulting characteristic, displaying the same shape as that in Fig. (b) but with a shift in the break point. 17

18 Example 3.13 n the circuit, D 1 and D 2 have different cross section areas but are otherwise identical. Determine the current flowing through each diode. Solution n this case, we must resort to the exponential equation because the ideal and constantvoltage models do not include the device area. We have in = D1 + D2. We also equate the voltages across D 1 and D 2 : that is, T ln ln D1 D2 T S1 S2 D1 D2 S1 S2 Solving the above equations together yields As expected, D1 = D2 = in /2 if S1 = S2. D1 1 in in D2 S 2 S1 1 S1 S 2 18

19 Example 3.14 Using the constant-voltage model, plot the input/output characteristics of the circuit in Fig. (a). Note that a diode about to turn on carries zero current but sustains D,on. if in v out R2 R R 1 2 in At the point D 1 turns on R2 R R 1 2 and hence in in Don R 1 1 Don R2 19

20 Example 3.15 f in = -, D 1 turns on out in D on R2 R Don 2 R1 0 X R 1 ( in Don ) R 1 D 1 turns off when R1 = R2 n this example, since in is connected to the cathode, the diode conducts when in is very negative. The break point where the slope changes is when the current across R 1 is equal to the current across R 2. R R D on in Don in out 2 1 R1 (1 ) Don R R R 1 R in 20

21 Example 3.16 Plot the input/output characteristic of the circuit shown in Fig. (a) using the constantvoltage diode model. 21

22 Cell Phone Adapter (Example 3.17) out = 3 D,on is used to charge cell phones. However, if x changes, iterative method is often needed to obtain a solution, thus motivating a simpler technique out D 3 ln T X s 22

23 Small-Signal Analysis 小 訊 號 分 析 Small-signal analysis is performed around a bias point by perturbing the voltage by a small amount and observing the resulting linear current perturbation. 偏 壓 點 2 exp D1 D1 D S S exp T T f << T, then exp( / T ) 1 + / T D 1 D 1 D S S T T T e T 1 exp T 2 exp exp D1 D1 T T D T D 1 23

24 Small-Signal Analysis in Detail f two points on the curve of a diode are close enough, the trajectory connecting the first to the second point is like a line, with the slope being the proportionality factor between change in voltage and change in current. Point A is called the bias point or the operating point. Also called the quiescent point. 工 作 點 靜 態 點 D D s T D1 T exp d d D D D1 T D D1 24

25 Example 3.18 A diode is biased at a current of 1 ma. (a) Determine the current change if D changes by 1 m. (b) Determine the voltage change if D changes by 10%. Solution (a) We have D D 384A D (b) Using the same equation yields T T D D 26m D (0 1mA) D 1mA 26m 25

26 Small-Signal ncremental Resistance Since there s a linear relationship between the small signal current and voltage of a diode, the diode can be viewed as a linear resistor when only small changes are of interest. T We define the small-signal resistance of the diode as: rd Also called the incremental resistance to emphasize its validity for small changes. n the above example, D =1mA -> r d =26Ω Ω. D 26

27 Small Sinusoidal Analysis (Example 3.19) ( t ) 0 p cos t D ( t ) 0 S exp T 0 T p r p prd p d 0 T 0 p cos t s exp p cos t T 0 T 0 f a sinusoidal voltage with small amplitude is applied, the resulting current is also a small sinusoid around a DC value. 0 27

28 Cause and Effect (Example 3.20) n previous derivation, we assumed a small change in D and obtained the resulting change in D. Beginning with D = T ln( D / S ), investigate the reverse case, i.e., D changes by a small amount and we wish to compute the change in D. Solution Denoting the change in D by Δ D, we have 1 1 ln D D D1 D D1 D D D ln 1 T T T ln T ln 1 S S D 1 S D 1 For small-signal operation, we assume Δ D << D1 and note that ln(1+ϵ) ϵ if ϵ «1. Thus, D D T D1 Figure below illustrates the two cases, distinguishing between the cause and the effect. (a) voltage is the cause and current is the effect; (b) the other way around. 28

29 Example 3.17 Revisited (Example 3.21) With our understanding of small-signal analysis, we can revisit our cell phone charger example and easily solve it with just algebra instead of iterations. Since each diode carries D1 = 6 ma with an adaptor voltage of 3 and D1 = 800 m, we can construct the small-signal model shown below, where v ad = 100 m and r d = 26m/6mA = 4.33Ω. We can thus write: v out 3rd R1 3 rd 11.5m v ad That is, a 100-m change in ad yields an m change in out. n Example 3.17, solution of nonlinear diode equations predicted an 11-m change in out. The small-signal analysis therefore offers reasonable accuracy while requiring much less computational effort. 29

30 Simple is Beautiful (Example 3.22) n Examples 3.17 and 3.21, the current drawn by the cellphone is neglected. Now suppose, the load pulls a current of 0.5 ma and determine the change in out. Solution Since the current flowing through the diodes decreases by 0.5 ma and since this change is much less than the bias current (6 ma), we write the change in the output voltage as: (3r ) out D ( 0.5mA)(3 6.5m6 5 d 4.33) 30

31 Half-Wave Rectifier 半 波 整 流 器 A very common application of diodes is half-wave rectification, where either the positive or negative half of the input is blocked. But, how do we generate a constant t output? no longer assume diode is ideal, but use a constant-voltage model. for in > D,on, D 1 turns on and out = in - D,on for in < D,on, D 1 turns off and out = 0 31

32 Diode-Capacitor Circuit: Constant oltage Model f the resistor in half-wave rectifier is replaced by a capacitor, a fixed voltage output of ( p - D,on ) is obtained since the capacitor (assumed ideal) has no path to discharge. At t 3, in = - p, applying a maximum reverse bias of (2 p - D,on ) across the diode. Hence, diodes used in rectifiers must withstand a reverse voltage of approximately 2 p with no breakdown. 32

33 Diode-Capacitor Circuit: deal Model (Example 3.24) Note that in Fig. (b) the voltage across the diode, D1, is just like in, only shifted down. + D1 - D1 33

34 Example 3.25 A laptop computer consumes an average power of 25 W with a supply voltage of 3.3. Determine the average current drawn from the batteries or the adapter. Solution Since P = x, we have = 25W/ A. f the laptop is modeled by a resistor, R L,thenR R L = / = 0.436Ω. 34

35 Diode-Capacitor With Load Resistor As suggested by the Example 3.25, the load can be represented by a simple resistor in some cases. 負 載 A path is available for capacitor to discharge. Therefore, out will not be constant and a ripple exists. 漣 波 out behaves as before until t 1 ;as in begins to fall after t 1, so does out because R L provides a discharge path for C 1. C 1 must be large that the current drawn by R L does not reduce out significantly. 35

36 Behavior for Different Capacitor alues The resulting variation in out is called the ripple. Also, C 1 is called the smoothing or filter capacitor. For large C 1, out has small ripple. 濾 波 Example 3.26 Sketch the output waveform as C 1 varies from very large values to very small values. Solution f C 1 is very large, the current drawn by R L when D 1 is off creates only a small change in out. Conversely, if C 1 is very small, the circuit approaches the previous figure, exhibiting large variations in out. Figure below illustrates several cases. 36

37 Peak to Peak Amplitude of Ripple The peak-to-peak (p2p) ripple amplitude, R, is the decaying part of the exponential. 峰 到 峰 值 P2p ripple voltage becomes a problem if it goes above 5 to 10% of the output voltage. f the maximum current drawn by the load is known, the value of C 1 is chosen large enough to yield an acceptable ripple. Since out = p - D,on at t 1, the discharge of C 1 through R L is expressed as: t out () t ( p Don)exp 0t t3 RC L 1 To ensure a small ripple, R L C 1 must be much greater than t 3 -t 1 ; thus, noting that exp(-ϵ) 1 - ϵ for ϵ «1, t p Don t () ( ) 1 out t p Don ( p Don) RLC1 RL C1 The first term on the right hand side represents the initial condition across C 1 and the second term, a falling ramp as as if a constant current equal to ( p - D,on )/R L discharges C 1. 37

38 The p2p amplitude of the ripple is equal to the amount of discharge at t 3. Since t 4 - t 1 is equal to the input period, T in, we write t 4 - t 1 = T in - T, where T (= t 4 - t 3 ) denotes the time during which D 1 is on. Thus, R p Don Tin T R C L 1 Recognizing that if C 1 discharges by a small amount, then the diode turns on for only a brief period, we can assume T << T in and dhence R where f in = T in -1 in p Don T p in R C RCf L 1 in L 1 Don f the load current, L, is known, we can write in R L Cf 1 in 38

39 Example 3.27 A transformer converts the 110-, 60-Hz line voltage to a peak-to-peak swing of 9. A half-wave rectifier follows the transformer to supply the power to the laptop computer of Example Determine the minimum value of the filter capacitor that maintains the ripple below 0.1. Assume D,on = 0.8. Solution We have p = 4.5, R L = 0.436Ω, and T in = 16.7 ms. Thus, C T R p Don 1 in 1 417F R L This is a very large value. The designer must trade the ripple amplitude with the size, weight, and cost of the capacitor. n fact, limitations on size, weight, and cost of the adaptor may dictate a much greater ripple, e.g., 0.5, thereby demanding that the circuit following the rectifier tolerate such a large, periodic variation. 39

40 Maximum Diode Current The diode has its maximum current at t 1, since that s when the slope of out is the greatest. This current has to be carefully controlled so it does not damage the device. The current in forward bias consists of two components: (1) the transient current drawn by C 1, C 1 d out /dt, and (2) the current supplied to R L, approximately equal to ( p - D,on )/R L. The peak diode current occurs at the point D 1 turns on because the slope of the output waveform is maximum. 2 p R p 2 R Check textbook for the derivation p C1 in p ( RLC1 in 1) R R p L L p 40

41 Full-Wave Rectifier 全 波 整 流 器 A full-wave rectifier passes both the negative and positive half cycles of the input, while inverting the negative half of the input. As proved later, a full-wave rectifier reduces the ripple by a factor of two. 41

42 The Evolution of Full-Wave Rectifier (a) two half-wave rectifiers for positive and negative cycles; (b) error trial, no rectification; (c) and (d) inverted rectification (e) and (f) rectification for the positive and negative half cycles of the input. 42

43 Full-Wave Rectifier: Bridge Rectifier 橋 式 整 流 器 Combine (e) and (f) of previous slides to a full-wave rectifier, where D 1 and D 2 pass/invert the negative half cycle and D 3 and D 4 pass the positive half cycle. 43

44 nput/output Characteristics of a Full-Wave Rectifier (Constant-oltage Model) The dead-zone around in arises because in must exceed 2 D,ON to turn on the bridge. A drop of 2 D,on on out. 44

45 Complete Full-Wave Rectifier Since C 1 only gets half of the period to discharge, ripple voltage is decreased by a factor of 2. Also (b) shows that each diode d is subjected to approximately one p reverse bias drop (versus 2 p in half-wave rectifier). R 1 p 2Don 2 RCf L 1 in 45

46 Diode Current in the Full-Wave Rectifier Example 3.30 Plot the currents carried by each diode in a bridge rectifier as a function of time for a sinusoidal input. Assume no smoothing capacitor is connected to the output. Solution From Figs. 3.38(c) 38(c) and (d), we have out = - in + 2 D,on for in < -2 D,on and out = in -2 Don D,on for in > +2 D,on. n each half cycle, two of the diodes carry a current equal to out /R L and the other two remain off. Thus, the diode currents appear as shown in the Figure. 46

47 Summary of Half and Full-Wave Rectifiers While using two more diodes, full-wave rectifiers exhibit a lower ripple and require only half the diode breakdown voltage, more suited to adapter and charger applications. 47

48 Example 3.31 Design a full-wave rectifier to deliver an average power of 2 W to a cellphone with a voltage of 3.6 and a ripple of 0.2. Solution We begin with the required input swing. Since the output voltage is approximately equal to p - 2 Don D,on,wehave in p Don Thus, the transformer preceding the rectifier must step the line voltage (110 or 220 rms ) down to a peak value of 5.2. Next, we determine the minimum value of the smoothing capacitor that ensures R 0.2. For a full-wave rectifier, R 2C f 2W 1 L 1 f in 36 2C 1 f in For R = 0.2 and f in = 60 Hz, C 1 = μf. The diodes must withstand a reverse bias voltage of

49 oltage Regulator 穩 壓 器 The ripple created by the rectifier can be unacceptable to sensitive load; therefore, a regulator is required to obtain a very stable output. Three diodes operate as a primitive regulator. The regulator needs to regulate the variation of the line voltage, line regulation defined das out / in ; and regulate the variation of the load current, load regulation defined as out / L. 49

50 oltage Regulation With Zener Diode oltage regulation can be accomplished with Zener diode operating in reverse breakdown. Since D 1 exhibits a small-signal resistance, r d (1-10Ω), large change in the input will not be reflected at the output. The Zener regulator has poor stability if the load current varies significantly. ifi out r r D R D 1 in 50

51 Example 3.33 n the circuit of Fig. (a), in has a nominal value of 5, R 1 = 100 Ω, and D 2 has a reverse breakdown of 2.7 and a small-signal resistance of 5 Ω. Assuming D,on 0.8 for D 1, determine the line and load regulation of the circuit. Solution We first tdetermine the bias current of fdd 1 and hence its small-signal lresistance: Thus, D1 in Don D2 R 1 15mA T rd D1 51

52 Example 3.33 (cnt d) From the small-signal model of Fig. (b), we compute the line regulation as v out r D 1 r D 2 v r r R in D1 D For load regulation, we assume the input is constant and study the effect of fload current variations. Using the small-signal circuit shown in Fig. (c) (where v in = 0 to represent a constant input), we have That is, vout il ( r r ) R D1 D2 1 v out i L ( r r ) R 631 D 1 D 2 1 This value indicates that a 1-mA change in the load current results in a 6.31-m change in the output voltage. 52

53 Limiting Circuits The motivation of having limiting circuits is to keep the signal below a threshold so it will not saturate the entire circuitry. 飽 和 When a receiver is close to a base station, signals are large and limiting circuits may be required. 53

54 nput/output Characteristics For small input levels, the circuit must simply pass the input to the output. As the input level exceeds a threshold or limit, the output must remain constant. 臨 界 值 This behavior must hold for both positive and negative inputs. The input signal is clipped at ± L. 剪 波 器 54

55 Limiting Circuit Using a Diode: Positive Cycle Clipping As was studied in the past, the combination of resistor-diode creates limiting effect. The battery B1 performs a level l shift. 移 位 : 準 位 移 動 55

56 Limiting Circuit Using a Diode: Negative Cycle Clipping 56

57 Limiting Circuit Using a Diode: Positive and Negative Cycle Clipping 57

58 General oltage Limiting Circuit Two batteries in series with the antiparalle diodes control the limiting voltages. 58

59 Example 3.34 A signal must be limited at ± 100 m. Assuming D,on = 800 m, design the required limiting circuit. Solution Figure (a) illustrates how the voltage sources must shift the break points. Since the positive limiting point must shift to the left, the voltage source in series with D 1 must be negative and equal to 700 m. Similarly, the source in series with D 2 must be positive and equal to 700 m. Figure (b) shows the result. 59

60 Non-idealities in Limiting Circuits The clipping region is not exactly flat since as in increases, the currents through diodes change, and so does the voltage drop. ln( ) D T D S 60

61 Example diode circuits Multiple-diode Circuits Turn-on region and turn-off region. Diode in series with the resistance Diode in series with the resistance and the voltage source. 61

62 Two diode circuit Assume, D 2 ON, D 1 OFF, D 1 just ON,» when v v v O (1) v R i (1) R 2 R 2 v O (1) 2 v R1 R 2 v (1) O v O (1) v 62

63 D 1 ON, D 2 ON (1) (2)» when v v v, D 2 just OFF, D 1 ON, D 2 OFF,» when v O v v v v O v (2) O (3) O v O ) (2 v v v v v (2) O v O (2) (3) O v O (1) v 63

64 Example Assume R 1 = 5k, R 2 = 10k. = 0.7, + = +5, - = -5. Determinate v O, i D1 and i D2 for v =0 and v = 4. Solution: 0 : D 1 off,d 2 on, i D1 0, i D 2 i R 2 R 1 R v O i R 1 R v v O r 64

65 4 :D 1 and D 2 on,then v O = v =4 i D v R O i 2 R v v i R 2 O 2 r v ( 5) 0.83mA R 10 i i i v D 1 R 2 D2 R

66 Problem-solving Technique: Multiple Diode Circuits We must initially guess the state of each device, then analyze the circuit to determine if we have a solution consistent with our initial guess. Assume the state of a diode. f a diode is assume to be on, the voltage across the diode is. f a diode is assume to be off, the current through the diode is zero. Analyze the linear circuit with the assumed diode states. Evaluate the resulting state of each diode. f the initial assumption were that a diode is off and the analysis shows that D = 0 and D, then the assumption is correct. f, however, D >0and/or D >, then the initial assumption is wrong. Similarly, if the initial assumption were that a diode is on and the analysis shows that D 0 and D =, then the assumption is correct. f, however, D < 0 and/or D >, then the initial assumption is wrong. f any assumption is proven incorrect, a new assumption must be made and a new linear circuit must be analyzed. 66

67 Example Demonstrate how inconsistencies develop in a solution with incorrect assumptions. Solution: Assume initially that both D1 and D2 are conducting. Then v = -0.7 and v O = 0. Two currents are i O R1 id 2 R 1 v v ' i R 2 R 2 Summing the current at v, i D t is an inconsistent solution. 1mA 0.43mA 1 ir 2 id ma 67

68 Example Determine the D1, D2 and D3 and A and B. Let = 0.7 for each diode. Solution: D 1, D 2 and D 3 are on, B -0.7 A 0 currents at the node,we find 5 A ( A 0.7) ( 10) D A D2 D2 0.86mA 5 5 which is inconsistent with the D2 on 68

69 D 1 and D 3 are on and D 2 is off. D D mA 5 5 (0 0.7) ma 5 We find the voltages as B 0.7 A 5 (1.43)(5) 2.15 D

70 Capacitive Divider Typically a global supply voltage, e.g., 3, is used in electronic systems. However, some circuits in the system needs a higher supply voltage, e.g., 6. oltage doublers may serve this purpose. 倍 壓 器 n Fig. (a), the voltage at C 1 cannot change even if in changes because the right plate of C 1 cannot receive or release charge (Q = C). Since C1 remains constant, t an input change in directly appears at the output. t n Fig. (b), if in becomes more positive, the left plate of C 1 receives positive charge from in,, thus requiring that the right plate absorb negative charge of the same magnitude from the top plate of C 2. Having lost negative charge, the top plate of C 2 equivalently holds more positive charge, and hence the bottom plate absorbs negative charge from ground. Thus, v out is a capacitive division of v in. out in C1 C C out in

71 Waveform Shifter: Peak at -2 p As in increases, D 1 turns on and out is zero. As in decreases, D 1 turns off, and out drops with in from zero. in 1 out p in The lowest out can go is -2 p, doubling the voltage. + p - out = - p + in 71

72 Waveform Shifter: Peak at 2 p Similarly, when the terminals of the diode are switched, a voltage doubler with peak value at 2 p can be conceived. - p + out = p + in 72

73 oltage Doubler The output increases by p, ½ p, ¼ p, etc in each input cycle, eventually settling to 2 p. (C 1 = C 2 ) p 1 1 out p p 73

74 oltage Shifter Shift the average level of a signal up or down because the subsequent stage (e.g., an amplifier) may not operate properly with the present dc level. Adi diode d can be viewed as a battery and dhence a device capable of shifting hifi the signal level, D,on. 74

75 Example 3.37 Design a circuit that shifts up the dc level of a signal by 2 D,on. Solution To shift the level up, we apply the input to the cathode. Also, to obtain a shift of 2 D,on, we place two diodes in series. Figure below shows the result. 75

76 Diode as Electronic Switch Diode as a switch finds application in logic circuits and data converters. (a) is a simple sample and hold, can be regarded as analog memory Diode is a fast switch, but more generally MOSFET is used as the switch 取 樣 與 保 持 開 關 76

77 Junction Feedthrough * For (e) of the previous slide, a small feedthrough from input to output via the junction capacitors exists even if the diodes are reverse biased. 穿 饋 out C C j / 2 / 2 C j 1 in if C j1 = C j2 = C j Therefore, C 1 has to be large enough to minimize this feedthrough. 77

78 Photodiode (PD) 光 二 極 體 偵 測 器 Light electricity The photodiode circuit PD under reverse-biased. No photon intensity only reverse-saturation current. Photons striking the diode create excess electrons and holes in the space- charge region. The photocurrent #/cm 2 -s (quantum efficiency electronic charge photon flux density ( ph = ea ) junction area) The electric field quickly separates these excess carriers and sweeps them out of the space-chargecharge region. Assumed that linear relationship between photocurrent and photon flux. oltage drop across R must be small. Namely photocurrent and value of R is very small. 78

79 Light Emitting Diode (LED) 發 光 二 極 體 Electricity light Electrons and holes are injected across the space-charge region under forward-biased Became excess minority carriers and diffused into the neutral n- and p- region. t recombined with majority carriers, and then emitted of a photon. Fabricated from compound semiconductor materials (GaAs or GaAsP) These are direct-bandgap semiconductor with higher bandgap than silicon. The forward-bias junction voltage is large than that in silicon-based diode. 79

80 Seven-segment Display 七 段 顯 示 器 Numeric readout of digital instrument. Common-anode display The anodes of all LEDs are connected to a 5 source. The input are controlled by logic gates. dark:high h input, the diode d is turn-off. bright:low input, the diode is turn-on. 80

81 Example Assume that a diode current of 10 ma produces the desired light output, and that the the corresponding forward-bias voltage drop is 1.7. Solution: f 0.2 in the low state,then the diode current is R 5 r The resistance R is then determined as R 5 r

82 Optoisolators 光 隔 離 器 The input and the output are electrical isolation. The input signal applied to the LED generates light, which is subsequently detected by the photodiode. The photodiode then converts the light back to an electrical signal. There is no electrical feedback or interaction between the output and input portions of the circuit. 82