Week 3: Diode Application Circuits

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1 ELE 2110A Electronic Circuits Week 3: Diode Application Circuits Lecture 03-1

2 opics to cover oltage Regulation - Zener Diode Rectifiers DC-to-DC converters Wave shaping circuits Photodiode and LED Reading Assignment: Chap , 3.18 of Jaeger & Blalock Lecture 03-2

3 Breakdown Region Diode Model In breakdown, the diode is modeled with a DC voltage source, Z, and a series resistance, R Z. R Z models the slope of the i-v characteristic. Diodes designed to operate in reverse breakdown are called Zener diodes and use the indicated symbol Lecture 03-3

4 Example 1 Graphic Analysis Load-line equation : (KL along the loop) D I D Using graphic analysis: Choose 2 points (0, -4mA) and (-5, -3mA) to draw the load line. It intersects with i-v characteristic at Q-point (-2.9 ma, -5.2 ). Lecture 03-4

5 Example 2 Piecewise Linear Model Using piecewise linear model: I Z I >0 D Assume in breakdown region I Z (20 5) I 2.94mA Z 5100Ω Since I Z >0 (I D <0), solution is consistent with Zener breakdown assumption. Lecture 03-5

6 oltage Regulator Function: to maintain a constant voltage across the load when the source voltage or the load current vary. Lecture 03-6

7 oltage Regulator Use constant voltage drop model: I S I L I Z S Z R L I Z R (20 5) 3mA 5kΩ 5 1mA 5kΩ S I L 2mA Zener diode keeps voltage across load resistor constant I Z For proper regulation, I Z >0. If I Z < 0, Zener diode no longer controls voltage across load resistor and the regulator is said to have dropped out of regulation. S R Z > 0 R R L R > R R L min 1 S Z Lecture 03-7

8 Example 3 Zener diode model Analysis: KCL at output node: L L L Problem: Find L and I Z using the piece-wise linear model. Given data: S 20, R5 kω, R Z 0.1 kω, Z 5 I Z 5.19 L 5 L mA > Diode is actually operating in breakdown region as assumed. Lecture 03-8

9 Line and Load Regulation Load is modeled as a current drain By superposition: Line regulation L S R z L Z + S ( Rz // R) R + Rz R + Rz R R + z R z R (characterizes how sensitive output voltage is to input voltage changes) I L Load regulation I L L ( R // R) [ Ω] z (characterizes how sensitive output voltage is to load current changes) Lecture 03-9

10 opics to cover oltage Regulation - Zener Diode Rectifiers DC-to-DC converters Wave shaping circuits Photodiode and LED Lecture 03-10

11 AC-DC Converter 220, 50Hz in HK ransformer: step down AC voltage amplitude Isolate equipment from power-line Rectifier: converts an ac input to a unipolar output Filter: convert the pulsating input to a nearly constant dc output Regulator: Reduce the ripple of the dc voltage Lecture 03-11

12 Half-Wave Rectifier Using ideal Diode model: Lecture 03-12

13 Half-Wave Rectifier Using Constant oltage Drop model: When diode is ON: v o ( P sinωt)- on. When diode is OFF: v o 0. ime-varying components in circuit output can be suppressed using a filter capacitor. Lecture 03-13

14 Peak Detector Circuit Replace R with C As input voltage rises, diode is on and capacitor (initially discharged) charges up to s - on At the peak of input, diode current tries to reverse, diode cuts off, capacitor has no discharge path and retains constant, thereby providing a dc output voltage: dc P - on Lecture 03-14

15 Half-Wave Rectifier with RC Load As input voltage rises during first quarter cycle, diode is on capacitor charges up to peak value of input voltage After the peak of input, Input voltage < P Diode cuts off Capacitor discharges exponentially through R Discharge continues till input voltage exceeds output voltage which occurs near peak of next cycle. his process then repeats once every cycle Lecture 03-15

16 Half-Wave Rectifier with RC Load Output voltage is not constant as in ideal peak detector, but has ripple voltage r. Diode conducts for a short time D called conduction interval during each cycle and its angular equivalent is called conduction angle θ c. During discharge period, v o ( t) ( ( P P on on )exp( t / RC) )(1 t / RC) for t << RC r ( on ) ( on ) for << P RC P RC Lecture 03-16

17 Half-Wave Rectifier with RC Load t 0 o find the conduction period, ( P on)exp( ) P cosω( ) RC on Assume <<, It can be solved that, 1 ω 2 RC ( P on ) 1 ω P 2 r P θc ω 2 r P where ω 2π Lecture 03-17

18 Example 4 Problem: Find dc output voltage, output current, ripple voltage, conduction interval, conduction angle of RC loaded half-wave rectifier. Given data: secondary voltage rms 12.6 (60 Hz), R 15 Ω, C25,000µF, on 1 Analysis: dc I dc P on P R on ( ) A 15Ω Using discharge interval 1/60 s, r ( on ) P RC θ c 2 r P θ c θ ω c 2πf rad or 16.6 o 0.769ms π Check of result: he hidden assumptions are justified: << 1 / ms << RC 375ms Lecture 03-18

19 Peak Diode Current Nonzero current exists in diode for only a very small fraction of period An almost constant dc current flows out of filter capacitor to load otal charge lost from capacitor in each cycle is re-supplied by diode during the conduction interval Lecture 03-19

20 Peak Diode Current Let s model the repetitive current pulse is modeled as triangle of height I P and width, Charge supplied during diode conduction: Q I P 2 Charge transferred to the load in each cycle: Q I dc I I 2 P dc dq i Q i( t) dt dt Lecture 03-20

21 Surge Current During first quarter cycle, i d ( t) i t C d c ( ) dt ωc cosωt P P sinωt Peak values of this initial surge current occurs at t 0 + : I SC ωc P Lecture 03-21

22 Peak Inverse oltage (PI) When diode is off, reverse-bias across diode is dc - v s. When v s reaches negative peak, the diode sees the peak inverse voltage (PI) PI vmin on ( ) 2 dc s P P P he diode should have a breakdown voltage greater than the PI in order to avoid entering breakdown region. Lecture 03-22

23 Full-Wave Rectifier Capacitor discharge time halved Require half the filter capacitance to achieve a given ripple voltage Output DC P - on PI 2 P -on 2 P Lecture 03-23

24 Capacitor Size A 0.175F capacitor Lecture 03-24

25 Full-Wave Bridge Rectifier v S P sinωt Current always flows in one direction through the load regardless of whether v S is positive or negative. Which direction? Lecture 03-25

26 v S positive: Full-Wave Bridge Rectifier v S negative: Lecture 03-26

27 Full-Wave Bridge Rectifier v S P sinωt Full wave rectification is achieved without center-tapped transformer by using two extra diodes Filter capacitance is halved compared to that of the halfwave rectifier wo diodes in either conduction path Output DC P -2 on Lecture 03-27

28 (v S positive) Reduced Peak Inverse oltage - on +v S v S P sinωt -2 on +v S - on PI Max(- on +v S ) p on P So diodes with lower breakdown voltage can be used Cheaper Lecture 03-28

29 Rectifier opology Comparison Output DC oltage PI Filter capacitance for a same ripple Half-Wave P ON 2 P ON ( on ) P C r R Center-apped Full-Wave P ON 2 P ON C/2 Bridge Rectifier Full-Wave P 2 ON P ON C/2 Filter capacitor is a major factor in determining cost, size and weight in design of rectifiers. Low PI is important in high-voltage circuits. Lecture 03-29

30 Lecture Example 5 Problem: Design rectifier with given specifications. Specifications: dc 15, r < 0.15, I dc 2 A, Assumptions: on1, input freq50hz Analysis: Use full-wave bridge rectifier 94.6A 0.423ms s 1/50 2A ms 17 2(0.15) F s A /2 2 ) ( dc I P I P r r dc I R r on P C on dc P π ω PI 16 ON P ( P of the transformer output)

31 opics to cover oltage Regulation - Zener Diode Rectifiers DC-to-DC converters Wave shaping circuits Photodiode and LED Lecture 03-31

32 Boost Converter Switching When switch is closed, diode is off. At the end of on : i L ( on ) i L (0 + ) + 1 L on + Sdt il (0 ) 0 S + L on When switch is open, diode is on. Assuming constant v O, at the end of off: S o L S i ( ) i (0+) + L L on + L off In steady state: i ( ) i (0+) L L s o 1 δ where δ on / is duty cycle. Since 0<δ<1, o > S Lecture 03-32

33 Boost Converter Ripple current is given by I S r on or L S S on L on I r I r S I r δ f I r o S L off In ideal converter, power delivered to input of converter is same as power delivered to load resistor. I o I S S I o or I I o I o S o o 1 δ S off Ripple voltage is given as I o o on o on o r on δ C RC RC RC Lecture 03-33

34 Buck Converter When switch is closed, diode is off: on o i ( S dt i S o on ) i (0+) L L (0 ) L L on 0 L i ( ) i (0+) L L When switch is open, diode is on: o s δ on s S o i ( ) i (0+) + L o on L L L off where δ is switch duty cycle. Since 0<δ<1, o < S. Lecture 03-34

35 Buck Converter Ripple current is given by I o r or L off L o o off o ( 1 δ ) I off I r r I f r I S o r on L In ideal converter, power delivered to input of converter is same as power delivered to load resistor. I I o on o I o I o δ S Ripple voltage is given as S r Q 1 on off C on + / 2 / 2 i r dt 1 I on + r off Ir Q C I 2 C r o (1 δ ) 8 r r 8L Lecture 03-35

36 1 2 A second analysis approach In steady state, the average (dc) voltage across the L must be zero, otherwise the inductor current will approach + or infinity as i L takes integration of v L over time. s v 1 o v 2 on off v > S t off < 1 < v 2 > O Q< v v > O off S t Lecture 03-36

37 DC-DC Buck Converter 1 2 s v 1 on off o v 2 v > < 1 on S t O on S v > O < 2 t Lecture 03-37

38 opics to cover oltage Regulation - Zener Diode Rectifiers DC-to-DC converters Wave shaping circuits Photodiode and LED Lecture 03-38

39 Clamping or DC-Restoring Circuit No DC path between i/p and o/p After the initial transient lasting less than one cycle in both circuits, output waveform is an undistorted replica of input Both waveforms are clamped to zero heir dc levels are said to be restored Clamping level can also be shifted away from zero by adding a voltage source in series with diode Lecture 03-39

40 Clipping or Limiting Circuits Have DC path between i/p and o/p O/p voltage cannot exceed the clipping level Dual clipping levels Lecture 03-40

41 Photodiodes If depletion region of pn junction diode is illuminated with light with sufficiently high frequency, photons can provide enough energy to cause electrons to jump the semiconductor bandgap to generate electronhole pairs. Photodectector: convert light signal into electrical form. Usually reverse biased to increase the depletion layer width. Solar cell: convert solar energy into electrical from. Lecture 03-41

42 Light Emitting Diode (LED) When electrons and holes recombine, they release energy his energy is often released as heat into the lattice, but in some materials, known as direct bandgap materials, they release light Engineering LEDs can be difficult, but has been done over a wide range of wavelengths Lecture 03-42

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