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Timestamp: 2019-04-23 18:40:07+00:00

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direction but not the other.
. Diode Equivalent circuit in the reverse direction Equivalent circuit in the forward direction.
Operation Reverse Bias • -ve voltage is applied to Anode • Current through diode = 0 (cut off operation) • Diode act as open circuit Forward Bias +ve voltage applied to Anode • Current flows through diode • voltage Drop is zero (Turned on) • Diode is short circuit .
The two modes of operation of ideal diodes Forward biased Reverse biased Forward Current 10 mA Reverse Voltage 10 V .
5v vD 1.2 1 1 iP   vD 1.5v vD  0 .5  1. Ex 3.5v   iD  1.5 A iD  0 1 v D  1.
Rectifier circuit Input waveform Equivalent circuit when vi  0 Output waveform. Equivalent circuit when vi ≤ 0 Waveform across diode .
18V 2 2  . Exercise 3-3 10  0 iD   10mA 1k 1 t2 vD   vi dt t 2  t1 t1 1   2  vD    10 sin    0dt  2  0   1 10 10 vD    10 cos  0     1  1   3.
Battery Charger 24sin  12 V 1 sin      30 0 2 Conduction Angle    2  120 0 one  third of cycle .
6 Circuits for Example 3.Figure 3. Find the value of I and V . Diodes are ideal .2.
Assumption Both Diodes are conducting .2.Example 3.
VB  0 Node A 10  0 I D2   1mA 10k Node B 0    10 I 5 k  I D1  I D 2   2mA 5k From above equation I D1 should be 1mA It is not possible Not Possible Thus assumption of both diode . Assumption Both Diodes are conducting V  0.
3310k   10  3.Example 3.33 5k   3. I D1  0.2(b).3 V.33mA 15 15 VA  10  1. I D 2  1.33mA . Assumption # 2 Diodes 1 is not conducting Diodes 2 is conducting 10    10  20 I D2    1.3v Assumption is correct VB  VA  3.3v VB  1.
Figure E3. Find the value of I and V .4 Diodes are ideal .
4 Diodes are ideal . Find the value of I and V I= 2mA I= 0A I= 0A I= 2mA V= 0V V= 5V V= -5V V= 0V .Figure E3.
4 Diodes are ideal . Find the value of I and V I= 3mA I= 4mA V= 3V V= 1V .Figure E3.
Find the value of I and V .Figure P3.2 Diodes are ideal .
Find the value of I and V Diode is conducting I = 0.2 Diodes are ideal . Figure P3.6 mA V = -3V Diode is cut-off I = 0 mA Diode is conducting V = 3V I = 0.6 mA V = 3V Diode is cut-off I = 0 mA V = -3V .
V=1 V . Problem 3-3 Diodes are ideal . Find the value of I and V D1 Cut-Off & D2 Conducting I = 3mA D1 Cut-Off & D2 Conducting I = 1mA .
In ideal diodes circuits. 10V peak sine wave. Sketch the waveform of vo .4 v1 is a 1-kHz.Figure P3.
10V Vo = 0V f = 1 K-Hz f = 1 K-Hz .= 0V Vp. Sketch the waveform of vo Vp+ = 10V Vp+ = 0V Vp. In ideal diodes circuits. v1 is a 1-kHz.= . 10V peak sine wave.
Figure P3.= 0V Vp. 10V peak sine wave. Sketch the waveform of vo Vp+ = 10V Vp.= 0V f = 1 K-Hz f = 1 K-Hz .= -10V Vp+ = 10V Vp+ = 10V f = 1 K-Hz Vp. v1 s a 1-kHz.4 In ideal diodes circuits.
4 In ideal diodes circuits. Sketch the waveform of vo . 10V peak sine wave. v1 s a 1-kHz.Figure P3.
In ideal diodes circuits.= -10V V0 = 0V Vp+ = 10V f = 1 K-Hz Vp. Figure P3. Sketch the waveform of vo Vp+ = 0V Vp.4 v1 s a 1-kHz. 10V peak sine wave.= -5V f = 1 K-Hz .
10V peak sine wave. Sketch the waveform of vo Vp+ = 10V Vp.4 In ideal diodes circuits.= -5V f = 1 K-Hz .Figure P3. v1 s a 1-kHz.
Problem 3-4(k) vi  10V peak @ frequency 1000 H z vi  10 sin 2000t For Vi >0 V D1 is cutoff D2 is conducting vo=1V For Vi < 0 V is conducting D2 is cutoff vo=vi+1V -9V .
Figure P3.B X=A+B .6 X=A.
Problem 3-4 (c) vi  10Vpeak @ frequency1000 H z vi  10 sin 2000t vo=zer o .
Problem 3-4(f) Vi is a 1kHz 10-V peak sine wave. +ve Half Cycle with 10 V peak at 1 KHz .
Problem 3-4(h) vi  10Vpeak @ frequency1000 H z vi  10 sin 2000t vo=z .
5 v vi  10Vpeak @ frequency1000 H z vi  10 sin 2000t . B is battery i of 4. Sketch and label the B 100 mA 4. Problem 3.5 V .5 vi is 10 V peak sine wave and I = 100 mA current source.
5V    sin 1 (0. D1 conducts D2 cutoff 4.6 Fraction of cycle that i B of 100mA flows   0. D1 cutoff all current flows thru battery Conduction angle 10 sin   4.35 360 .30 Conduction angle    2  126.5V .45)  26.Solution P3-5 100 mA vi  10Vpeak @ frequency1000 H z vi  10 sin 2000t B  4. iB  0 A vi  4.5 V vi  4.5 v All curren t flows thru D1.153.60 126.7 0 .5V .
Problem 3-5 100 mA 4.5 v .
Problem 3-5 10 4.35T   35mA T T .5 100 mA 1 1 iBaverage   iB dt  100  0.
. the diode does not allow current to flow.. In the event that the battery or power source is connected the wrong way round. . REVERSE POLARITY PROTECTOR • The diode in this circuit protects a radio or a recorder etc.
D2=On I3=0.5 mA I1= I3=0. Problem 3-9 I1 I1 2 2 I3 I3 D1& D2 Conducting I1=1mA D1=off.7 V .66 mA I2=0.5 mA V = -1.
5V V=-2V .225 mA I=0A V=4. Problem 3-10 D conducting D is not conducting I=0.
Problem 3-16 V RED GREEN 3V On Off D1 conducts 0 V Off Off -3 V Off On D2 conducts .
Quiz No 3 DE 28 EE -A Sketch vO if vi is 8 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducting .
Solution Quiz No 3 8  4 I1  2 I 2  2  2 I1  3I 2 2 I 2  2  I 2  1mA 8V Vo  1 1  2  3V I1 I2 8 2 I 2  1mA vi/2 2 I Vo  1 1  2  3V Conduction angle  2  60o 4 sin   2    30   2 1 Fraction of Cycle the diode conducts    33% 10-10-07 2 3 .
Vo=5V D1 Vi>5V D2 is 10conducts 5 Vomax  5   7.5V 2 D2 Conduction angle  2  60o 10 sin   5    30 5   2 1 Fraction of Cycle the diode conducts    33% 2 3 22-10-07 . Sketch vO if vi is 10 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducts D never conducts +12 V 1 Vi<5V D2 is cut-off.
Vo=Vi 5 10conducts Vi>5V D2 is Vomax  5   7.5V 2 Conduction angle  2  60o 10 sin   5    30   2 1 Fraction of Cycle the diode conducts    33% 2 3 . Quiz No 3 DE 27 CE -B Sketch vO if vi is 10 sin  Find out the conduction angle for the diode & fraction of the cycle the diode is conducts D never conducts 1 Vi<5V D2 is cut-off.
sketch vo if the input is 10sin (9) • Find out the conduction angles for Diode D1 & D2 (4) and the fraction of the cycle these diodes conduct. Problem •Assume the diodes are ideal. (2) .
 2  vi  1  vo  vi vi  1 vi  1V  v0  1  1 4 vipeak  10V  vopeak  4.25V vi  2V  vo  2V .
25V vi  1 v0  1  1V vi  2V  vo  2V 4 -2V . 2  vi  1  vo  vi vi  1 vi  1V  v0  1  1 4 vi  2V  vo  1.
Two-dimensional representation of the silicon crystal. 14 Electrons .
At room temperature. some of the covalent bonds are broken by thermal ionization. both of which become available for current conduction. Each broken bond gives rise to a free electron and a hole. .
Intrinsic Semiconductor Electrons and holes .
. The Doping of Semiconductors .
n and p Type Semiconductors .
p-n Junction • P Junction – Concentration of holes is high – Majority charge carrier are hole • N Junction – Concentration of electron is high – Majority charge carrier are electron .
Diffusion Current ID • Hole diffuse across the junction from the p side to the n side & similarly electron • Two current components add together to form the diffusion current with direction from p to n side .
Drift Current Is • Diffusion current due to majority carrier diffusion • A component due to minority carrier drift exists across the junction .
.(a)The pn junction with no applied voltage (open-circuited terminals). (b) The potential distribution along an axis perpendicular to the junction.
.• The polarity of applied voltage which can't produce any current is called Reverse Bias. • The polarity of applied voltage which causes charge to flow through the diode is called Forward Bias.
Terminal Characteristics of a Junction Diode .
The diode i–v relationship The diode i–v relationship with some scales expanded and others compressed in order to reveal details. .
Terminal Characteristics of a Junction Diode • Forward Biased Region v > 0 • Reversed Biased Region v < 0 • Breakdown Region v < -VZK .
Forward Biased Region  v  i  Is  e nVT  1     • Is Saturation current – Scale Current – Is is constant at a given temperature – Is is directly proportional to Cross-Sectional region of the diode. Is doubles if cross-sectional area is double – Is is 10-15 A for small size diode – Doubles in value for every 10OC rise in temperature .
~ 25 mV • n is 1 or 2 depending on the material and the physical structure of the diode – n = 1 for Germanium Diode & n=2 for Silicon .6 X 10-19 Coulombs – VT @ 20oC is 25. Forward Biased Region  v  • Thermal Voltage VT i  Is  e nVT  1   – VT = kT/q   • K = Boltzmann’s constant = 1.38 X 10-23 Joules/Kelvin • T = Absolute Temperature in Kelvin (273 +Temp in Co) • q = Magnitude of charge = 1.2mV.
Forward Biased Region  v  i >> Is v i  Is  e   nVT  1   i  I se nVT  v  ln i  ln I s e nVT   v ln I   nV s   T i v  nVT ln Is b Relationship of the current i to the voltage v holds good over many decades of current (seven decades. a factor of 107 .
Forward Biased Region v1 I1  I s e nVT v2 I2  I se nVT  v v  I2 2 1 e nVT I1 I2 I2  v2  v1   nVT ln  2.3nVT log I1 I1 .
7v .8v  0. Forward Biased Region I2  v2  v1   2.6v 0.3nVT  60mV for n = 2 120mV  v  0.5v  cut  in  voltage v  0.3nVT log I1 I2 • for I1  10 v drop changes by for n = 1 2.
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.
0 V At 0 C I = ¼ μ A V = 0. At 400 C I = 4*1 = 4 μ A V = 4 μ A * 1MΩ = 4.9 If V=1V at 20o C.Figure E3.25 V . Find V at 400C and 00C Is At 20o C Reverse current Is = 1V/1M Ω= 1μ A Since the reverse leakage current doubles for every 100 C increase.
7 V at a current of 1mA.7 3  1 I s  10 e 2510  3  6. Find Is at n=1 & 2 v v  i  I se nVT  I s  ie nVT  0.7 3 10 2 I s  10 e 2 2510  3  8.Forward biased Diode Characteristics Example 3.3  10 A .3 • A silicon diode displays a forward voltage of 0.9  10 16 A  0.
9  10 15 A i 10  4 For i  0.9  10 .7V @ i=1mA. Ex 3.7  1 3 I s  10 e 2510 3  6.76V Is 6.7 Silicon Diode with n=1 has VD=0.1mA & 10mA v v  i  I se nVT  I s  ie nVT  0.9  10 2 i 10 For i  10mA  V1  VT ln  25  10  3 ln 16  0.1mA  V1  VT ln 3  25  10 ln 16  0. Find voltage drop at i=0.64V Is 6.
05  1.7V v 0. Solution P3-18 (a) At what forward voltage does a diode for which n=2 conduct a current equal to 1000Is? (b) In term if Is what current flows in the same diode when its forward voltage is 0. Diode current  i  1000IS v v i  ISe nVT  1000 I S  I s e 22510 3 v  0.7 V (a )   2.7 i  ISe nVT  I se 0.2 106 I S .345V (b) v  0.
Find the value of the current I required to obtain an output voltage Vo=2 V. Problem 3-23 • The circuit shown utilizes three identical diodes having n=1 and Is= 10 -14 A. Assume n=1 . what if the change in the output voltage. Assume n=1 • If a current of 1mA is drawn away from the output terminal by a load.
Solution 3-23 The circuit shown utilizes three identical diodes having n=1 and Is= 10 -14 A.025 I DX ΔvoY  vO 2  v01  22. Find the value of the current I required to obtain an output voltage Vo=2 V.025  3. what if the change in the output voltage. theref ore I DY  2. Info available n  1.8mV .81mA (v DY  v DX ) (v DY  2 / 3 ) I DY e 0 .025 e 0 .81mA If a current of 1mA is drawn away from the output terminal by a load. I S  10 14 A.Vo  2V vo 2 The voltage across e ach diode is  vDX  3 3 v DX 2 3 14 I DX  I S e ηVT  10 e 0 . (b) Load current  1mA.
both diode have n=1. but D1 has 10 times the junction area of D2. Problem 3-25 • In the circuit shown. What value of V results? .
.. both diode have n=1.........025  ln  92..2mV 2 ......2 I D1 I1  I D 2  I D1  I D 2  I1  I D1...... I D 2  10  2  8mA 80 V0  VD 2  VD1  0....3  I D1  2mA...1e .. What value of V results? VD 1 VD 2 VT VT I D1  I S 1e I D 2  I S 2e I S 1  10 I S 2 VD 1 VT I D1  10 I S 2e VD 2 VT VD 2 VD 2 I D2 I e VT  S 2 VD1  0...... In the circuit shown...... but D1 has 10 times the Solution 3-25(a) junction area of D2..1 I D1 10 I S 2 eVT 10 I D 2 V0  VD 2  VD1  VT ln .
1e   0. I D 2 I D 2  0.1e 2 I D1 0. Vo  50mA.75mA .01  I D 2 I D 2  4. what current solution 2-25 (b) I2 id needed.01  I D1 VD 2 VD 2 I D2 VT I D2  0. Find I D1 . To obtain a value of 50 mV.25)  5.25mA I D1  (10  4.
• Find ‘n’ • Find the value of R for which V = 80 m V. conducting 10mA at 0.7 V and 100 mA at 0. both diodes are identical.8 V. .Problem 3-26 • For the circuit shown.
7    0.8V @ I D 2  100mA I D2 VD 2  VD1  VT ln I D1 Find η 100 0. VT are same S For Diode 1  VD1  0. .7V @ I D1  10mA For Diode 2  VD2  0.737  0.025  ln 10   1.739 I D2 Find R if Vo=80mV V  VD 2  VD1  VT ln I D1 0. Solution 3-26 (a) Diodes are identical therefore I .8  0.08  1.01  I D1  0.4mA 80 R  57.1 1.025  ln I D1 I D1  1.4 .
75  0.7  VT ln  I D 2  7.389  10 3 .75  0.36 Assuming identical diodes for which VD =0.389mA 10 3 10  3 R  947 7. Problem 3. Find R if V0 = 3 V 3 VDx   0.7V @ ID=1mA.75V 4 VDX V I DX  I S e T VD 2 V (VD 2 VD 2 ) I D2 e T V  V e T I D1 D1 V e T (V D 2 V D 2 ) ( 0.389mA I D2 .7 ) V I D 2  I D1  e T  1 e 2510  3  7.
Modeling the Diode Forward Characteristics .
VD VT ID  ISe VDD  VD ID  R .A simple circuit used to illustrate the analysis of circuits in which the diode is forward conducting.
.Graphical analysis of the circuit using the exponential diode model.
and that its voltage drop changes by 0. . Iterative Analysis using the Exponential Model Determined the diode current ID and Diode voltage VD with VDD =5V and R =1000 ohms.1 V for every decade change in current.7 V. Diode has a current of 1mA @ a V D of .
237 mA.3 .1 log  0.3 V2  V1  0.3VT log I1 V  2. I2 V2  V1  2.237 mA I D  4.763V 1.763V Solution VD ID  ISe V T  4.3VT  0.3mA I2 V2  V1  2.1V For Every decade change in current 4.3VT ln I1 VD  0. Solution First iteration VD  0.763  0.1 log  0.7V VD VT ID  ISe  4.237 V2  0.762V 4.762V 4.0 Second iteration VD  0.
.Approximating the diode forward characteristic with two straight lines: the piecewise-linear model.
1 mA to 10 mA. The Piecewise-Linear Model • Exponential curve is approx into two straight lines • Line No 1 with zero slope & Line 2 with a slope of 1/rd • The voltage change of less than 50 mV is observed in case the current change from 0. i 0 D v  0V D (v D  V D 0 ) iD  v D  VD 0 rD .
.Piecewise-linear model of the diode forward characteristic and its equivalent circuit representation.
The Constant – Voltage Drop Model .
1 mA to 10 mA. • Model is used when – Detailed information about diode characteristics in not available .Constant – Voltage Drop Model • Forward conducting diode exhibits a constant voltage drop VD • The voltage change of less than 50 mV is observed in case the current change from 0.
The constant-voltage-drop model of the diode forward characteristics and its equivalent-circuit representation. .
• First determined dc Operating Point • Then small signal operation around the operating point – Small portion of the curve is approximated as almost linear segment of the diode characteristics. The Small – Signal Model • A small ac signal is superimposed on the DC components. .
The Small – Signal Model .
.Figure 3. Note that the numerical values shown are for a diode with n = 2.17 Development of the diode small-signal model.
The Small – Signal Model In absence of signal Once signal is applied v D (t )  VD  v d (t ) I D  I s e VD VT i D (t )  I s e v D VT i D (t )  I s e VD  vd (t ) VT VD vd ( t ) V V i D (t )  I s e T e T vd ( t ) V i D (t )  I D e T vd For very small signal  1 VT vd i D (t )  I D (1  ) VT i D (t )  I D  id (t ) .
The Small – Signal Model vd i D (t )  I D (1  ) ηVT i D (t )  I D  id (t ) I D vd id (t )  VT VT rd  ID rd is inversely proportional to I D .
Modeling the Diode Forward Characteristic .
vd (t ) + vd + ID - VD - . Exp 3-6 VDD  10V.7 V. n  2 Find rd . VD .v d  1V peak amplitude @ 60 Hz Diode has a current of 1mA @ a VD of .
per 1mA of diode current.
• Small signal model is used.
1 V. Example 3-7 • A string of three diodes is used to provide a constant voltage of about 2. We want to calculate the percentage change in this regulated voltage caused by • (a) a + 10 % change to the power supply voltage • (b) Connection of a 1 K ohms load resistance . Assume n=2 .
For the diodes closely matched.25 10 .01A 10 What is the corresponding small signal resistance of each diode and nVT rdx   2. ten “20 mA diodes” ( a 20 mA diode is a diode that provides a 0.7 V drop when the current thru it is 20 mA) connected in parallel operate at a total current of 0.5 of the combination? I Dx 2.1 A.5 req   0. P 3-53 • In a particular cct application. with n=1. 0.1 iDx   0. what current flows in each.
2  0.• If each of the 20 mA diode has a series resistance of 0.5  0.27 10 What connection resistance would single diode need in order to be totally equivalent? .2 ohm associated with the wire bonds to the junction. What is the equivalent resistance of the 10 parallel connected diodes? 1 Re q   2.
The diode i–v relationship .
. • Diodes are intended to operate below their breakdown voltage. • This leakage is undesirable. obviously the lower the better. Leakage current: • In the reverse direction there is a small leakage current up until the reverse breakdown voltage is reached.
The Reversed Biased Region  v  i  Is  e nVT  1     v is negative &  VT (25mV ) i  I S Current in reserved biased diode circuit is due to leakage current & increases with increase in reverse voltage Leakage current is proportional to the junction area & temperature but doubles for every 10oC rise in temperature .
Breakdown Region • Once reverse voltage exceeds a threshold value of diode V ZK. this voltage is called breakdown voltage. • Vertical line for current gives property of voltage regulation . K – Knee • At breakdown knee reverse current increases rapidly with associated small increase in voltage drop • Diode breakdown is not destructive if power dissipated by diode is limited by external circuitry. VZK Z – Zener.
The diode i–v characteristic with the breakdown region shown in some detail. .
Zener Diode • Operation in the Reverse Breakdown Region • Very steep i-v curve at breakdown with almost constant voltage drop region • Used the designing voltage regulator • Diode manufactured to operate specifically in the Breakdown region called Zener or Breakdown Diode .
Zener Diode : Symbol IZ .VZ + .
Model: Zener • Manufacturer specify Zener Voltage Vz at a specified Zener test current Iz. power that the device can safely dissipate 0. .5 W @ 6.8 v at max 70mA Vz  I z rz • rz Dynamic resistance of the Zener and is the inverse of the slope of the almost linear i-v curve at operating point Q • Lower rz.3 volts to 75 volts. the more constant Zener Voltage • The most common range of zener voltage is 3. the Max.
Model for the zener diode. .
Model: Zener Vz  Vzo  rz I z I z  I zk Vz  Vzo .
Designing of the Zener shunt regulator + Supply voltage includes a large ripple component Vo - Zener regulator Vo is an output of the zener regulator that is as constant as possible in spite of the ripples in the supply voltage VS and the variations in the load current Voltage regulator performance can be measured Line Regulation & Load Regulation Line Regulation = ΔVo/ΔVs Load Regulation = ΔV /ΔI .
IL R r Vo  Vzo( )  VS ( z ) .(rz ||R) ΔI L . Expression of performance : Zener regulator I + (Vs -Vo ) (Vo -V zo)   IL R rz V o .I L(rz ||R) R  rz R  rz • Only the first term on right hand side is desirable one Second and third terms depend upon Supply Voltage Vs and Load current IL Vo rz • Line Regulation = /  Vs (rz  R) • Load Regulation = ΔVo  .
The circuit with the zener diode replaced with its equivalent circuit model.
8  5  20 10  6.7v .8 V & Iz=5mA. rZ=20 ohm Vz  Vzo  rz I z 3 Vzo  Vz  I z rz  6. a) Find No Load VoRegulation Vo &Line Depending upon the manufacturer provide Data First calculate Vzo if Vz =6.
Now connecting the Zener diode in the Cct as shown Calculate actual Iz and resulting Vo Thus establishing operating Point V  Vzo 10  6.35  20  10 3  6.827V  6.5mv R  rz 520 Vo 38.35mA R  rz 500  20 Vo  Vzo  I z rz  6.7 Iz    6.7  6.83V Now carry out Small Signal Analysis Suppress DC source and calculate resultant change in Vo Use voltage divider rule V  rz  1 20 Vo    38.5 Line Regulation   3.85mv / v V 1 .
b) Find vO if load resistance RL connected & draws 1mA and load regulation .
21mA  210 A .14mA R  RL || rZ 500  19. 1mA drawn by load would decrease by same amount so Vo  rz I z  20  1mA  20mV Iz Vo Load Regulation   20mV / mA I z 6.807V Z o ZO Z Z Vs  VZ 10  6.807 IZ    6.35  20  6.94 I Z  6.7  5.83v RL   6.94 6850 exact CalculationsV  V  V  I r  6.14  .83k 1mA Check RL || R  20  6830  19.35  6.
4mA RL I Z  3.4mA Vo r Z I Z  68mV .c) Vo for RL  2k VZ I RL   3.
• 1) Check  10 500 2000 2000 Vo   10  8v 2500 Zener at Breakdown region .
7v 6. 10v 10 0.5k A 500  6.8 .63v  2k  20 Vo B 19.
8 .A B 6.7  2000 Voc   6.63v 2020 Re q  19.
d ) RL  500 10v 10  500 Vo   5v 500 1000 Zener is not operating 500 @ V  V o zk 5  6.8v .
6  0.7 RL    1.5k I RL 4.7v Iz Iz VDD  9v min 9  6.7 6.2  4.4m .e) Min value of R for which the diode still L operates in the breakdown region • at Breakdown Region  10  1v I z  I zk  0.6mA 500 0.4mA Vzk 6.7v RL I  4.2mA I  I zk  I RL I RL  4.2mA 500 Vz  Vzk  6.
The regulator operates from a 10-V supply and has a 1.5-V zener specified at 12mA. Problem D3. The zener has an incremental resistance rz = 30 Ω and a knee current of 0. (a) What is the value of R you have chosen? (b) What is the regulator output voltage when the supply is 10% high? Is 10% low? (c) What is the output voltage when both the supply is 10% high and the load is removed? (d) What is the smallest possible load resistor that can be used while the zener operates at a current no lower than the knee current while the supply is 10% low? .5-V zener regulator circuit using a 7.68 Design a 7.5mA.2-kΩ load.
5  VZO  12  30  10  VZO  7. Solution 3-68 rz  30 I Zk  0.5 I RL   6.5V I Z  12mA 3 7.25mA 1.2 .5mA VZ  7.14V 7.
The zener has an incremental resistance rz = 30 Ω and a knee current of 0.2 So that I Z  3.5 I RL   6. The regulator operates from a 10-V supply and has a 1.5 R  250 10 .5-V zener regulator circuit using a 7. (a) What is the value of R you have chosen? Select I  10mA 7.5-V zener specified at 12mA.25mA 1.5mA.75mA Which is  I Zk 10  7. Design a 7.2-kΩ load.
03 0.28  V O  7.03)  0.250  (1.03 V O  1 0.55V .2 // 0.2 // .1V Thus V O  7.4V to  7.6V (c) What is the output voltage when both the supply is 10% high and the load is removed? With V   11 V and I L  0 11  V O V O  V ZO  X 0. (b) What is the regulator output voltage when the supply is 10% high? Is 10% low? For V   1V 1.
14  0.38mA VO 7.(c) What is the smallest possible load resistor that can be used while the zener operates at a current no lower than the knee current while the supply is 10% low? IZK=0.14 V 9  7.03 X 0.38  0.25  7.155 RL min  7.5 2 0.5  1. VZO=7.155V 1 7.155 11V 3 250 R 0.04k .5mA.5mA RL min 7.
Rectifier Circuit Power Supply • Power supply must supply dc voltage to be constant in spite of – variation is ac line voltage – Variation in current drawn by load. that is variable load resistance .
Rectifier Circuits • Filter – Smoothes out pulsating dc but still some time-dependent components-(ripple) remain in the output • Voltage Regulation – Reduces ripples – Stabilizes magnitude of dc output against variation in load current – Regulation by Zener Diode or Voltage regulator I.C .
. Half Wave Rectifier Transfer characteristic of the rectifier circuit Input and output waveforms. assuming that rD >> R.
Input and output waveforms. .
Full Wave Rectifier • Diode in Reverse biased state Anode @ .VDO Twice as in case of half wave rectifier .Vs Cathode @ + Vo • PIV = 2Vs .
Bridge Rectifier D1 • Peak Inverse Voltage D4 D2 D3 – PIV => consider loop D3. R & D2 – VD3(res) = Vo + VD2 – Vo = Vs – 2VD – PIV = Vs – 2VD + VD = Vs – VD – Half of PIV for Full wave Rectifier .
waveforms assuming an ideal diode.
Peak detector with Load Vo iL  R iD  iC  i L dVs iD  C  iL dt .
.Figure 3.30 Waveforms in the full-wave peak rectifier.
Peak Rectifier : Output Voltage • When Vr is small – Vo = Vpeak – iL is almost constant VP – DC components of iL iL  R • Accurate value of output 1 dc voltage V V  V Average Value o 2 P r .
provided Vr  V p fC . Charge / Discharge Cycle t  vo  VP e CR T  Vo  VP  Vr  VP e CR T  T e CR  1 CR   T   Vr  VP  1  e CR     VP iL   Vr  VP  1  1  T   R  CR  1 VP T V Vo  VP  Vr Vr  CR  P fCR 2 IL  VP R IL Vr  .
Peak Rectifier : Ripple Voltage • During Discharge cycle t  vo  V P e CR • At the end of discharge cycle T  Vo  VP  Vr  VP e CR   T   Vr  VP  1  e CR     • Since CR >> T T  T e CR  1 CR .
Peak Rectifier : Ripple Voltage   T   T T Vr  VP  1  e CR   e CR  1   CR  T  Vr  VP  1  1    CR  Ripple Voltage VPT VP Vr   CR fCR VP Vr  fCR VP IL  R IL Vr  . provided Vr  V p fC .
the conduction angle will be small ... Peak Rectifier : Conduction Interval VP cos t   VP  Vr Hence t is small  wt  2 Cos  wt   1   . 2!   t   2 VP  1    V P  Vr   2  2Vr  t  VP When Vr<<Vp.
Average Diode Current –During Conduction iD  iC  i L iDav  iCav  I L iCav  i Dav  I L During Charge Qsupplied  icav t iCav  i Dav  I L During Discharge Qlost  CVr Qsupplied  Qsupplied  icav t  CVr .
Average Diode Current –During Conduction Qlost  Qsup plied  CVr  icav t VP T VPT Vr   CVr  CR R VPT   iDav  I L  t R 1 2Vr T 2Vr t   2f VP 2 VP VPT  T 2Vr    i Dav  I L    R  2 VP  2 VP  2VP  i Dav   I L  i L  1    2Vr R  Vr  VP Vr  VP  i Dav  I L .
Deduction • As waveform of is almost right angle r triangle Vr  VP iD max  2i Dav .
Observations • Diode current flows for short interval and must replenish the charge lost by the capacitor. Discharge interval is long & discharge is through high resistance rD  RL • Maximum diode current CdVi iD   iL dt Assuming that i L is almost constant  I L & CR  T  2V p  i D max  i L  1  2   2i Dav   Vr  .
Let the load resistance R =10 k Ohms. Example N0 3-9 Consider a peak rectifier fed by a 60 Hz sinusoidal having a peak value of Vp = 100 V. (a) Find the value of the capacitance C that will result in peak to peak ripple of 2 V (b) Calculate the fraction of the cycle during which the diode is conduction (c) Calculate the average and peak value of the diode current. .
9 100 Sin 2 60t 10k • Find value of C for Vr=2V (peak to peak) VP 100 C   83.2radian VP • => Diode conducts 0.cycle 18% 2 .2  100of  3.3F Vr fR 2  60  10 4 • Find fraction of cycles that diode conducts 2Vr t   0. Example 3.
Solution Exp 3-9 • Find iD max &i Dav  2VP  i Dav  I L  1    Vr  VP 100 IL    10mA R 10000  2  100  i Dav  10 1     324mA  2  imax  2iDav  648mA .
the capacitor discharge for almost T/2 time interval. so V  r V P 2 fCR  VP  i Dav  I L  1     2Vr   VP  imax  2i Dav  I L  1  2   2Vr  . Full wave peak Detector In full wave rectifier. that mean ripple frequency is twice the input.
Applications • Peak Rectifier – Peak detector is used for – Detecting the peak of the an input signal for signal processing systems – Demodulator for amplitude modulated (AM) signals .
Precision Half Wave Rectifier Super Diode • Normal Diodes VD= 0. demodulation or rectification Operational Amplifiers (Op Amp) are used .7v are used for rectifier of input of much larger amplitude then VD • For smaller signals detection.
Wave form Generation Limiting Clamping • Limiter Circuit – Vo is limited between two levels – upper (L+) and lower (L-) thresholds .
33 Applying a sine wave to a limiter can result in clipping off its two peaks. .Figure 3.
Wave form Generation Limiting / Clamping • Double Limiter – Clips off both negative & positive peaks • Single Limiter • Clips off only one side of the input peak • Application – Limits the inputs to operation Amplifier to a limit lower than the breakdown voltage of transistors of input stage of operational Amplifier – Half / Full Rectifier for Battery Charger – Threshold and limiting .
.Figure 3.35 A variety of basic limiting circuits.
Solution Ex 3-27 (a)  5  vi  5  v o  vi (b) VI  5 Vo D2 Conduct.5 2 (c ) vi  5 V D1 Conducts & D2 is off  10  1  vi  5    v i  5  vR     10  10  2 1 1 vo  vi  5  5   vi  2.off  10  vR    vi  5  1  vi  5  10  10  2 1 vo  5  v R  vi  2.5   . D1 cut .
D C Restorer • The output waveform will have its lower peak “Clamped” to O V therefore known as “Clamped Capacitor” • Output waveform will have a finite average value & is entirely different and unrelated to the average value of the input waveform .
Application 6  2v 4 TXR  4v 0v 4v DC Restorers .
.Figure 3.36 The clamped capacitor or dc restorer with a square-wave input and no load.
Figure 3. .37 The clamped capacitor with a load resistance R.
38 Voltage doubler: (a) circuit. (b) waveform of the voltage across D1. .Figure 3.
4 vi 6 VC D off Diode Off V0  Vi  Vc D on vo 0 Diode On Vo  0.7v .
The Voltage Doubler  VP - C1   D1 VD1 2VP VP sin t   C1 D1  a Clamp circuit  VP - DC Restorer VP sin t .
Special Diode Type Schottky-Barrier Diode (SBD) • Shottky-Barrier Diode is formed by bringing metal into contact with a moderately doped ‘n’ type semiconductor material • Resulting in flow of the conducting current in one direction from metal anode to the semiconductor cathode and acts as an open circuit in the other direction .
Schottky-Barrier Diode (SBD) • Gets two important properties – SBD switches on-off faster due to current conducts due to majority carrier b (electrons) – Forward voltage drop is lower then P-n junction diode .
Varactor • Variable Capacitor – Depletion layer acts as junction capacitance – Depletion layer varies Capacitance Depletion Region D Metallic Plate Dielectric – Used for voltage controlled Tuning Circuit .
. • The depletion region increases as reverse voltage across it increases. is essentially devoid of carriers and behaves as the dielectric of a capacitor. the junction capacitance will decrease as the voltage across the p-n junction increases. • By varying the reverse voltage across a p-n junction the junction capacitance can be varied . and since capacitance varies inversely as dielectric thickness. the depletion region. Varactor • When a reverse voltage is applied to a p-n junction .
this property. the voltage drop across them depends on the amount of light that strikes them. • Gunn diodes are negative-resistance diodes that are the basis of some microwave oscillators. the current through the device decreases as the voltage is increased within a certain range. • Light-sensitive or photosensitive diodes can be used to measure illumination. Semiconductor diodes • The tunnel diode. makes it useful as an amplifier. known as negative resistance. .
SCR (Thyristor) • The Silicon Controlled Rectifier (SCR) is simply a conventional rectifier controlled by a gate signal. . • Most SCR applications are in power switching. the gate. and inverter circuits. • The rectifier circuit (anode-cathode) has a low forward resistance and a high reverse resistance. • A gate signal controls the rectifier conduction. phase control. • It is controlled from an off state (high resistance) to the on state (low resistance) by a signal applied to the third terminal. chopper.
Converts Light energy into a electrical signals . Photodiode If reversed biased PN junction is exposed to incident light – the photons impacting the junction cause covalent – bond to break thus give rise to current known as a photocurrent & is proportional to the intensity of incident light.
storage & transmission . Photodiode • Photodiode are manufactured using Gallium Arsenide (GaAs) • Photodiodes are important element of optoelectronics or photonics circuit (Combination of Electronics & optics) used for signal processing.
• Solar Cell – light energy into Electrical energy . Photodiode : Applications • Fiber optics Transmission of telephonic & TV signals • Opto-storage are CD ROM computer disks • Wide bandwidth & low signal attenuation.
Light Emitting Diode (LED) • Inverse of Photodiode • Converts a forward biased current into light • GaAs used for manufacturing LEDs • Used as electronics displays • Coherent light into a narrow bandwidth laser diodes • Fiber Optics & CD ROM .
Optoisolator • LED & Photodiode Electrical to light Light to electrical • Provides complete electrical isolation between electrical circuits • Reduces the effects of electrical interference on signal being fixed within a system • Reduces risk of shock • Can be implemented over long distance fiber optics communication links .
What are the slopes of the characteristic at the extreme + 10 V levels? . Problem 3-103 Sketch and label the transfer Characteristics of the circuit shown over a + 10 V range of the input signal. All diodes are VD =0.7 V @ 1 mA with n=1.
+1 V Vi V0 -2 V -2 V .
Problem 3-103 0  Vi  1 Vo  0 .
10 V peak to peak sine wave. • (b) What are its positive and negative peak values? . input voltage is a 1kHz. The diode is an ideal diode. Ist Sessional • Q No 1 (12 Marks) In the circuit shown. • (a) Sketch the waveform resulting at output terminal vO.
• (b) If a current of 1 mA is drawn away from the output terminal by a load • (i) What is the change in output voltage? • (ii) What is the value of the load? . Ist Sessional • Q No 2 (15 Marks) A circuit utilizes three identical diodes connected in series having n=1 and IS= 10-14 A. • (a) Find the value of current required to obtain an output voltage of 2 V across the three diodes combined.
Ist Sessional • Q No 3 (13 Marks) For the circuit shown. Label the positive and negative peak values assuming that CR >>T. . sketch the output for the sine wave input of 10 volts peak.
. Ist Sessional • Q No 4 (10 Marks) 9. At this current the incremental resistance is specified as 7 ohms. – (a) Find VZO of the zener model.25 V zener diode exhibits its nominal voltage at a test current of 28 mA. – (b) Find the zener voltage at a current of 10 mA.
– Find the diode conduction angle. – What is the average load current? . – Find the load current.8 V and a load resistance of 100 ohms. – Find the value of C that results in a ripple voltage no larger than 1 V peak to peak. Ist Sessional • Q No 5 (20 Marks) Consider a bridge rectifier circuit with a filter capacitor C placed across the load resistor R for the case in which the transformer secondary delivers a sinusoid of 12 V (rms) having the 60 Hz frequency and assuming VD = 0.
– (b) What the value of resistor R. the output voltage is 2. Ist Sessional • Q No 6 (10 Marks) In a circuit shown.7 V drop at 1mA. – (a) Find the current following through the resistor R. Assuming that the diodes are identical and are having 0. .4 V.
an added advantage. Note that when vI > 0 and the diode conducts.Figure 3.31 The “superdiode” precision half-wave rectifier and its almost-ideal transfer characteristic. the op amp supplies the load current. and the source is conveniently buffered. . Not shown are the op-amp power supplies.
–(a) Find VZO of the zener model. Quiz DE28 EE -B (10 Marks) 9. At this current the incremental resistance is specified as 7 ohms. –(b) Find the zener voltage at a current of 10 mA. .25 V zener diode exhibits its nominal voltage at a test current of 28 mA.
(a) What is the value of VZO of the zener model? (b) What voltage do you expect if the diode current is doubled? . Quiz DE 28 EE -A A zener diode whose nominal voltage is 10 V at 10 mA has an incremental resistance of 50 Ω.

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