Diode

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Chapter 1- Diode Circuits

Objective

Upon completion of this chapter you will be able to:

 Understand the basics of Semiconductor Devices

 Understand the operation of PN Junction Diode in Forward and Reverse Bias  Perform DC and Small Signal Analysis of PN Diode.

 Determine the output of various diode circuits like Clipper, Clamper and Voltage Regulator.

Introduction

Diode is dispositive made of a semiconductor material, which has two terminals or electrodes (di-ode) that’s act likes on-off switch. When the switch is on it acts as short circuit and passes all current. When it is off it behaves like an open circuit and passes no current.

Diode is simply PN junction. PN junction is a homo-junction between a p-type and n-type semiconductor. It acts as a diode, which can serve in electronics as a rectifier. Logic gate, voltage regulator (Zener diode), switching or tuner (varactor diode) and in optoelectronics as a light-emitting diode (LED), laser diode, photo detector or solar cell.

Depending up on conductivity the materials are classified into 3 parts (1) Conductor  very high, ρ=Very low (2) Insulator’s  Very low ρ =Very high

(3) Semi converters _insulator _sem _conductor _conductor  _sem _insulator Conductivity   1

 The ability of material to oppose the flow of charged particles

S ,_m m

    

L=length of the conductor A=Cross sectional area

N=No of charged particles present to conductor Q=charge of the charged particle

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d

V Drift (or) Average velocity of charged particle Current densityJ I A  Q NQ I T T   Transient time d l V   d NqV I l  d d NqV NqV I J A lA V    d J nqV n=concentrationn N N lA V   And V_dE d V  E where  mobility J nq E  and J E nq    For conductor l R A  

 As temperature increases mobility decreases ρ increases i.e. R increases.  Conductor’s are called +ve temp coefficient element

For semiconductor

 As temperature increases mobility increases ρ decreases i.e. R decreases.  Semi conductor are called –Ve temp coefficient element

Semiconductors

IV group element, 4 valencies Ex- C,Si, Ge, Sn, Pb & Fm

 Si & Ge selected as semiconductor because (1) These two are most available material on.

(2) These two provides moderated values of voltage & current

 Si is preferable used because

(1) Thermal stability of Silicon is more than that of Ge (2) The cut voltage of Si is more than Ge

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Ex- Si, Ge- intrinsic semiconductors

 In conductors the conduction band is almost full & valence band is almost empty. Valence band & conduction band over lapped in conductor.

g

E 0 For conductors

 Insulators conduction band is almost empty & valence band is almost filled

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There is no free e_{in its crystal structure.}

 At low temperature 00_{K semiconductors acts as insulator when applied energy increases,}

temperature in the semiconductor increases, breakage of covalent band’s increase, e_hole pair generation increases, conductivity increases, Eg decreases.

 At room temperature, semiconductor acts as conductors.

g E eV 0K 300K Ge Si 0.785 1.21 0.72 1.12

 

4 g 4 1.21 3.6 10 k eV Si E 0.785 2.23 10 k eV Ge             

 As Voltage increases, temperature increases, breakage of covalent bonds decrease, e hole pair generation increases, conductivity decreases and it will damage the device at some particular voltage, temperature generated in the semiconductor become large & semiconductor damage permanently.

 In intrinsic semiconductor

Number of free e __{Number of holes}

Number of free e_{/volume = Number of holes/volume} n p For intrinsic semiconductor

Rate of combination=Rate of regeneration

Mass action law

For both intrinsic and extrinsic materials, at equilibrium:

2 i

n np => n n p_i   for intrinsic semiconductor

i

n Intrinsic semiconductor concentration

Ex. In intrinsic semiconductor as intrinsic concentration 6 3 i

n 1.5 10 / cm  and its hole concentration is _{p 2.25 10 / cm}_ _ 6 3_, So, e_{concentration is} 2 i 2 12 6 3 i 6 n n p n 2.25 10 n 10 / cm p 2.25 10       

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semi-conductor. Extrinsic=Intrinsic + Impurities Extrinsic N type P type    _  Impurities

V group III group

Pentavalent Trivalent Phosphorous Boron Arsenic Aluminum Antimony Gallium Bismuth Indium N-type semiconductor

 N type=Intrinsic +V group Impurities

 4 valance e_{of phosphorous forms 4 covalent bonds with the 4 present silicon atoms.} There is no place for the 5th_e_{& it is considered to be free e}_.

 Each atom of phosphorous provides 1 freee_{. If we add small amount of impurity to the} intrinsic semiconductor, then it will provide large no of free e_.

 Majority carriers are e_{and minority carriers are holes}

 Conductivity is n type semiconductor     _N _e _hole nq _e pq_p For N type n>>p and q_e q_p

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 In N type semiconductor n N _D Donor atom concentration i.e.  _N N q_D _e

P type semiconductor

 P type- Intrinsic + III group

 The 3 valence e_{of Al forms 3 covalent bonds with the nearest 3 ‘Si’ atoms. There is no} e_{in the 4}th_{covalent bond and the missing e}_{in 4}th_{covalent bond is considered as hole.}

 Each atom of 3rd_{group provided one hole. If we add a small amount of III group into the}

intrinsic semiconductor then we get large no of holes.  Majority carriers are hole and minority carriers are e_.

 Conductivity is p type semiconductor     _P _e _hole nq _e pq_p For p type semiconductor p>>n and and q_e q_p

 _P pq_p

 The conductivity in P type is mainly due to holes.

 In P type semiconductor P N _A Acceptor atom concentration i.e.  _p N q_A _p

Note:

(1) The conductivity in intrinsic semiconductor

i e hole

    

i nq e pq p

     For intrinsic semi n=p=n _i





i nqi n p

    

(2) The  in N type semiconductor  _N N q_D _e The  in P type semiconductor  _p N q_A _p

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i n np (a) Intrinsic n=p=n_i (b) N type 2i N 0 n P N  (c)P type 2i P A n n N  Solved Examples

Problem: For intrinsic silicon semiconductor at room temperature the charge concentration

16 3

1.5 10 / cm , the mobility of electron & holes are _{0.13m / V sec}2 _ _&_{0.05m / V sec}2 _

respectively its conductivity will be

Solution:





16 19

 

4

i nqi n p 1.5 10 1.6 10 .18 4.31 10 S / m

          

Problem: A heavily doped N type semiconductor has the following data e p

2.5 



 , doping concentration_{4.2 10 / m}_ 8 3_{, intrinsic concentration} _{1.5 10 / m}_ 4 3_{. Calculate the ratio}

conductivity of N type/ conductivity of intrinsic semiconductor.

Solution: i i



e p



i p N e e nq _n 1 nq n          _  _   _  _ 4 4 8 8 1.5 10 ₁ 10 1.5 10 ₁ 2 25 5 4.2 10 4.2 10        _  _ _  _  _ _  _ _

 

4 4 i N 4.2 5 ₁₀ _{2 10} 1.5 7       

Problem: Silicon sample ‘X’ is doped with _{10 atoms/m}8 3_{of indium another silicon sample of}

identical dimensions doped with 1020_{atoms per/m}3_{of Antimony the ratio of hole/e}_mobility

=0.25. Then the ratio of conducting of sample X/Y

Solution: X p 18 p 20 Y e e p q 10 p q 10          2 .25 10  

Problem: In N type silicon bar of 2cm long & has gross sectional area 2mm x 2mm, when 1V battery is connected across it. A current of 8mA flows in it first the doping concentration in that semiconductor. Assume 2

e 0.13m / V s

   and 2

n 0.05m / V s

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© Kreatryx. All Rights Reserved. 8 www.kreatryx.com Solution: R V 1 l I A    D e l 8m _N _q A      3 D 6 19 2 8 10 N 100 4 10 1.6 10 0.13            22 D N .192 10 21 3 D N 1.92 10 / m Fermi level

The Fermi level is the level at which the probability of finding the free e-_{is 50%.}

For Intrinsic semiconductor

i n n p  c F c E E KT n N e            F V v E E KT p N e           

For intrinsic n=p & N_c N_v c F F V c v E E E E KT KT N e N e                      c F F V E E E E c v F E E E 2  

For N type semiconductor  C F c E E KT n N e    D n N

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

C F



c D E E N ln N KT           c F C D N E E KTln N     __ __  

For P type semiconductor F V V E E KT P N e    A p N



F V



V A E E N ln N KT           V F V A N E E KTln N     __ __   PN Junction Diode

Semiconductor pn junction is a two-terminal device. It is the most fundamental device element that forms the basis of many electronic devices such as pn diodes, optoelectronic devices like light-emitting diode and photodetector, field effect transistors and bipolar transistor.

PN junction conducts high current in one direction and conduct very small amount of current in the reversed direction. Thus, pn junction has the property of rectification.

PN junction is formed in a single crystal of semiconductor by making one end of the crystal p-type by doping it with acceptor atom and making the other end n-type by doping with donor atoms. The region where p-type and n-type meet is the junction.

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© Kreatryx. All Rights Reserved. 10 www.kreatryx.com  By combining P type & N type a small force acting on the junction, this force cause reaction in the charged particles. These charged particle always moves from higher concentration to lower concentration, due to this recombination takes place in both the regions and depletion region formed. As P type & N type are equally doped the recombination rate is same in both the region. The region in which there is no availability of charge particles or the region in what there is no further penetration of charged particles is called depletion region or space charge region or Barrier.

 To make the conduction of current in the PN junction diode must be properly bias

Forward bias

A pn junction can be forward biased to lower the voltage across the junction. If a positive forward voltage V is applied to the p-side of the pn junction relative to the n-side, the effective voltage across the junction is (VB–V) not VB. Thus, the energy required by the

majority carrier to overcome the potential barrier is less than earlier zero voltage bias case. As the result more majority carrier will be able to diffuse across the junction.

 All the minority carriers moves away from the junction they can’t from closed loop

minority

I 0

 All the majority carriers move towards the junction.

 As forward bias voltage V increases, the repelling forces to the majority carriers increases and depletion region width (W) decreases. Junction resistance decreases.

 At some particular voltage the depletion region width reduced to zero, the charge particles starts conducting the junction.

 The voltage at which the charged particles starts crossing the junction is called cut in voltage or Threshold voltage or barrier voltage or depletion region voltage or space ohmic region voltage.

0.7 0 S .3 i Ge V_ _{ } 

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When V V _

There exists barrier; I_D0

Case-2: When V V _

This is the minimum voltage to reduce the barrier width to zero. After that charged particle starts crossing the junction.

D

I 0, The current through the diode at this voltage is called significant current (At least 1% of maximum diode current)

Case-3: When V V _y

The number of charged particles crossing the junction increase, the current through the diode increases non-linearly or exponentially & is given by the equation

D T V V D 0 I _I e_  _1_    

Where I =Reverse saturation current ₀ V_D Voltage across diode

 Intrinsic factor when values depend on type of material  1 Ge & 2 Si

T

V Temperature dependent voltage= KT q       At room temperature T=300K V_T 26mV D e p

I  I I Flows from P to N type

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current in the direction of overhead is called forward bias of diode.

Reverse bias

Barrier potential increases under reverse bias voltage V. The bias voltage is effectively subtracted from the potential barrier. Consequently, the voltage across the junction is(VB+ V)

As the result, lesser chance for majority carrier to diffuse across the junction and more easily for the minority carrier to drift across the junction would be happened.

 All the majority carriers moved away from the junction, they can’t form loop

Majority

I 0

 The minority carriers move towards the junction as reverse bias voltage increases. The majority carriers more rapidly moves away from the junction

 The depletion region with (w) increases as V increases and so the junction temperature increases due to this breakage of covalent bonds are also increase and e-_{here pair}

generation is also increases.

 The number of mobility carriers crossing the junction increase i.e. currents due to these minority carriers increase.



I_nv,I_po 



0



no po



0 I I I I also increases   

 I Flows from N to P type. It flows reverse to arrow head direction. It is called reverse ₀ current

Note: Effect of Increase in temperature:

2 1 T T 10 02 01 I I 2          

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In reverse biased mode, the current flowed in the diode is extremely small. However, upon further applying the reverse biased voltage until the point VBD where breakdown occurs. The

current would increase rapidly. Normally diode breakdowns at about negative 25.0V and avalanche breakdown occurs beyond this voltage point.

 I_D I₀

Non – Linear Analysis of Diode

The two-terminal element behaves in a most asymmetric manner: its resistance is very low for currents of more than a few milliamps flowing in one direction through the device, but it has an enormously high resistance to current flow in the opposite direction. This element is very useful for constructing absolute value, peak detection, overvoltage protection, and more general nonlinear resistance circuits. The diode’s current-voltage relationship is actually exponential, so it is also useful for building exponential and logarithmic response amplifiers. Its characteristics are strongly temperature-dependent, so a diode also makes an excellent, accurate temperature sensor. Of course, some types of diodes also can emit and detects light (LEDs, laser diodes, and photodiodes), so the variety of applications of the seemingly simple semiconductor diode is nearly endless.

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Forward Bias:

 When V_& Rf are given

 When V_ is given or type of diode is given

 When both are not given or ideal diode

Reverse Bias Diode resistance Static or DC resistance D DC D V R I    _ _   Dynamic or AC resistance D D AC D D V dV R or I dI   

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D 0 V V I _I e_  _1_     0

I Reverse saturation current

D T 0 D 0 V V I _I _I e  D 0 I I D T D 0 V V I _I e  Differentiate w.r.t to V_D D T V V D 0 D T dI _{I e} 1 dv V     T AC D V R I  

O.C. & S.C. tests for diode circuits

Open Circuit test

All the diode are replaced by open circuit

Calculate V_AV_C for each diode If



V_AV_C



V_ D is ON



VAVC



V  D is OFF

 When number of diodes having



V_AV_C



V_Y the diodes which has more V_A V_C becomes ON first, again calculate V_A V_C for remaining diodes to check whether ON or OFF state. If all the diode have same V_AV_C, diode which have less cut in voltage will ON first again check V_A V_C for remaining diodes

Short Circuit test For ideal diodes

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 Calculate the current through the each diode If I_D  0 D-ON

I_D  0 D-OFF

Solved Examples

Problem: For the circuit calculate static & dynamic resistance for Si diode

Solution: I_D 2.7 0.7 1mA 2    D DC 3 D V 0.7 R 700 I 10     T AC D V 2 26mV R 52 I 1mA      

Problem: For the Si diode circuit shown calculate the I & I_D1 _D2

Solution: V_A V_C V_Y For D1 & D2

D1 D2 10 0.7 I 28.18mA .33k 28.18 I I 14.09mA 2      

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Solution: V_A V_C for Si diode = 4V & for Ge diode- 4V So,V of Ge is less than that of Si, then Ge becomes ON first _Y

4 0.3 _{I 1.85mA} 2K    3 0 V 10 I 1.86V

Problem: calculate I ,I (Assume Si diode) ₁ ₂

Solution: V_A V_C 0V D is OFF and ₁ V_A V_C 10V D is ON ₂

1 2 10 0.7 I 0,I 9.3mA 1K     0 V 10 9.3 0.7 

Problem: The reverse current of A PN junction diode is 10µA at 200_{C. Calculate change in}

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  

2 1



2 T T log 3.5 10  

 

2 1 2 T T 10log 3.5  0 T 18.07 C  

Problem: At room temperature voltage across silicon diode is 0.7V when 2mA of current flows through it. If the voltage across the diode is reverse 0.75V then the current through the diode? (assumeV_T 25mV) Solution: D T V V D 0 I _I e_  _1_     0 0.7 K 2 25 2mA I 1.66nA e  1            3 0.75 9 2 25 10 D I 1.66 10_ e_{ }  1 5.43mA         

Problem: Find valuesI & I₁ ₂, Assume V_D 0.7V

Solution: For D ,V₁ _A V_C 3 & For D ,V₂ _A V_C 10

Since voltage across D is higher than₂ D , ₁ D becomes ON first ₂

2 10 0.7 I 18.6mA 0.5K   

 

0 V 5 0.5 18.6 2     6.3V 1 D is OFF 1 I 0 2 I 18.6mA

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Solution: 1 A C 2 A C 3 A C For D , V V 15 For D , V V 10 For D , V V 5       So D1 will ON first

Draw the circuit where only D1 is ON

1 10 5 0.7 15 0.7 14.3 I 1.43mA 10K 10K 10        A V 10. 0.7 2 1.43 9.3 2.86 6.44V      B V 6.44 6 1.43 6.44 8.48     2.14 Now For D , V₂ _A V_C  1.44V 3 A C For D , V V 2.14V

So , D is remain OFF i.e. I₂ 2=0

and D3 is ON

Again redraw the circuit

1 10 0.7 0.7 I 1.25mA 8K     3 1 0.7 5 I I 1.25 2.15 0.9mA 2K         

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Solution: Diode is forward biased i.e. V1 > 7







i



1 V 3 0.6K V 3 500 0.6K     







i



V 3 0.6K 3 7 500 .6K      



V 3 600i



₄ 1100    i V 10.33V

Problem: For circuit device characteristic is given in the figure. Calculate the value of I in the circuit

Solution: R 0.7 0.5 200 1mA



  

Apply KVL in the circuit 5 0.5

I 3.75mA

1200 

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Solution: D T V nV D1 0 I I e 1     D2 D1 0 I I I D T V nV 2 e 1 T V V 0.693 1 V 18.02mV 2 0 V 4.981 100I Diode Applications Clipping Circuits

Clipper is a circuit which transmits a part of the input signal. Transmitted signal either above of the reference signal or below of the reference signal or between the two reference signals. It is also called Limiting circuit.

 Clippers circuit consist of Diode or Diodes, Reference voltage or voltages and resistor or resistors.

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1. Analyze the working condition of diodes

2. Write the transfer characteristic equation (V0 Vs Vi).

3. Plot the transfer characteristic curve (V0 Vs Vi).

4. Draw the o/p wave form for a given i/p wave.

Clippers are classified into two types: Clippers

Single level clipper Two level clipper

Series clipper Shunt clipper

+ve clipper -ve clipper +ve clipper -ve clipper

Series Clipper

Negative Clipper

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V0=VR When VR>Vin Diode is ON (S.C.) V0=Vin R in R 0 in in R V V V V V V V      _  Transfer characteristics Type 2:

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V0=Vin

When -VR>Vin Diode is OFF (O.C.)

V0=-VR R in R 0 in in R V V V V V V V       _{ }  Transfer characteristics Positive Clipper

When VR<Vin Diode is OFF (O.C.)

V0=VR

When VR>Vin Diode is ON (S.C.)

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When VR<Vin Diode is OFF (O.C.)

V0=Vin When VR>Vin Diode is ON (S.C.) V0= VR R in R 0 in in R V V V V V V V      _ 

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Type 2:

When -VR<Vin Diode is OFF (O.C.)

V0=Vin When -VR>Vin Diode is ON (S.C.) V0= -VR R in R 0 in in R V V V V V V V       _{ }  Transfer characteristics

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When VR<Vin Diode is ON (S.C.)

V0=VR

When VR>Vin Diode is OFF (O.C.)

V0= Vin in in R 0 R in R V V V V V V V      _  Transfer characteristics

Note: To draw the o/p wave form of single level clipper 1. Draw the i/p wave form in doted form.

2. Draw the reference voltage level in dotted form.

3. If P type of diode connected to o/p voltage, then there should be no positive peak in the o/p wave form. Draw the o/p wave form below the reference voltage, as per the given signal shape.

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© Kreatryx. All Rights Reserved. 28 www.kreatryx.com 4. If N type of diode connected to o/p voltage, then should be no negative peak in the o/p wave form. Draw the o/p wave form above the reference voltage, as per the given signal shape.

Two level clipper

1 2 V V When V V_i  ₂ 1 D is OFF & D₂ is ON 0 2 V V When V V & V V_i  ₂ _i  ₁ 2

D is OFF & D₁ is OFF

0 i V V When V V_i  ₁ 2 D is OFF & D₁ is ON 0 1 V V Transfer characteristics Solved Examples

(29)

© Kreatryx. All Rights Reserved. 29 www.kreatryx.com Solution: Method 1: V 7 _{7 3} in _I ₀ 500 600    _  _{  } _     V 7 ₄ in I 500 600    If diode has to be FB. I 0_  sub I 0





0 4200 600V 2000 in      6200 31 V 10.34V in  600  3  Method 2: V_in 3 I 1100   And V_x I 600

 

3



V_in 3 600



 

V_x 3 1100    If V_x 7V diode F.B





V_in 10.33V

Problem: The input voltage _Vin varies linearly from 0 to 150V sketch the transfer characteristics.

Solution: Case-1: V 100V

in D1 RB D₂ RB

(30)

Problem: The input voltage _Vin varies linearly from 0 to 150V sketch the transfer characteristics

(31)





V_o 100 I 200k = 50 V Here V_x 50V Case-2: 100V V 150V in   D₁ FB D₂ RB V 100V o  Case 3: 25V V 100V in   D₁ FB D FB 2 V_o V_in Transfer characteristics

(32)

© Kreatryx. All Rights Reserved. 32 www.kreatryx.com Problem: The input voltage _Vin varies linearly from 0 to 150V sketch the transfer characteristics. Solution: Case-1: V V in x D1 R.B D₂ R.B V 25V o 

For both the diodes in RB V 25V in Case-2: 25V V _inV_x D₁ R.B D₂F.B





in in V 25 2 1 V_o 200k 25 V 25 300K 3 3    _ _     

Max value of Vin for D2 will be in RB

x x 2 1 V V 25 100V =>V 137.5V o 3 3   Case-3 137.5 V 150 in   D F.B 1 D F.B 2 Then V 100 0  Transfer characteristics

(33)

Solution:

Case 1: V_in 3VV_o 3V

Here we use ideal resistor to protect the diode from burning

Case 2: V 1V V 1V in  o 

Case 3: 1 V _in3VV_o V_in

(34)

Solution:

Transfer characteristics

Problem: For the ideal diode circuit, draw the output waveform. Calculate minimum and maximum value of output is?

(35)

During +ve cycle input



V 0_i 



2

D is OFF

And when V 4V_i  D₁ is OFF (O.C)

0 i V V When V 4V_i  D₁ is ON (S.C) i V 4 i 4k   and V₀  4 2ki i 0 V V 2 2   When V 10_i  : V_0max 7V During –Ve cycle input



V 0_i 



1 D is OFF And when



V_i  3



D₂ is ON (S.C) 0 V  3V(minimum) When V_i  3 D is OFF ₂ 0 i V V Clampers

These circuit are used to shift the signal either upwards or downwards. If the signal shifted upwards positive D.C. inserted into the applied signal. If the signal shifted downwards –Ve D.C inserted into the applied signal.

Type of clamper circuit

(i) Negative clamping circuit or Positive peak clamp circuit (ii) Positive clamping circuit or Negative peak clamp circuit

Negative clamping

 In clampers

 

_{P P}

 

_{P P}

i/p o/p

V_  V_

 There is no change in the shape of the signal

 Clampers consists of energy storage element like ‘C’

 Input always given to ‘C’ only. The analysis of clamper circuit always start with the conduction of the diode

(36)

WhenV 0_i  , D is ON (S.C) 0 C i V 0 V V   c i

V V=The “C” charges in accordance with the input At t T, V V_i _m

4

 

C m

V V Max voltage across the capacitor At t T ,V V ,V_i _m _c V_m 4    _{ }     0 i c V  V V 0 V 0

 This V₀ 0 voltage makes the diode OFF (O.C)

 As there is no discharge path in the capacitor, it holds its previous value i.e. V_C V_m

0 i m V V V When T t 4  V V_i  _m V₀ 0 T t 2  V 0_i  V₀ V_m 3T t 4  V_i  V_m V₀  2V_m t T V 0_i  V₀  V_m 5T t 4  V V_i  _m V₀ 0

(37)

1. The signal shifted downwards +ve peak shifted to 0V

-ve peak shifted to

 

2V

_m

Output voltage waveform between 0 to



2V

_m

2. The signal shifted downwards +peak shifted to

V

R

-Peak shifted to



2V

m



V

R

(38)

+Peak shifted to



V

_R -Peak shifted to



2V

m



V

R

The output waveform between



V

R to



2V

m



V

R

Positive clamping

Note:

V

i

 

V to V

m



m 1. The signal shifted upwards -Ve peak shifted to 0

+Ve peak shifted to

2V

m

2V

m to 0

2. The signal shifted upwards -Ve peak shifted to

V

_R

+ve peak shifted to

2V

_m



V

_R

2V

_m



V

_R to

V

_R

3. The signal shifted upwards -Ve peak shifted to



V

R +Ve peak shifted to

2V

m



V

R

(39)

Problem: Draw output waveform

Solution:

Problem: Draw output waveform

(40)

It converts AC signal into pulsating D.C.

Types of rectifier Half wave rectifier

 

m

 

m

V t V sin t ,V







Max voltage

y

V 0

 

Neglected

f

R



Taking into consideration &

R

_f



R

_L

During +ve cycle of V(t)

Diode is F.B & replaced by

R

f

 

_

 

_

f L V t i t R R  

 

m

 

f L V i t sin t R R   _ _    

 

m

 

i t



I sin t



 

m L 0 f L V R V sin t R R   __ __     f L

R



R

(41)

 

0 m

V V sin t





 

D f

V i t R



I R sin t

_{m f}

 



During negative cycle of V(t) D is R.B & replaced by O.C

 

0

 

L

i t 0,V i t R



0  

 

0 m

V

 

V t

 

V sin t



 The maximum voltage across diode is



V

_m.  Peak inverse voltage=

V PIV

_m



 Average (or) DC value:

   

2 DC m m m 0 0 1 1 1 I I sin t d t I sin t d t 2I 2 2 2          







 m m DC dc I V I  & V   

 RMS (or) effective value:

2 2

 



  

rms 0 1 1 1 I i t dt 0 0 2 4 2       



 m m rms rms i V I & V 2 2    D.C O / P Power 100 A.C I/ P Power   

P

_dc



V

_dc



I

_dc



I R

_{dc L}2



V R

_{dc L}2 2 m dc 2 L

I

P



R



 

 

2



ac rms f L m f L

P



V

t R R





I R R







2 m L 2 L 2 2 f L m f L I R R 4 100 R R I R R 4       __ __         _      f L 0.4052 ₁₀₀ R 1 _R

 

_max

40.52%

When

R

_L



R

_f

(42)

 

2 2 _rms dc I RMS value of component in O / P FF 1 1 DC component present in O / P I      __ __   

I

2_rms



I

2_dc



I

2_rms/ac Ripple factor= 2 rms rms/ac dc I 1 I 1.21 I            Note:

1. R.F.>1, AC component present in output is more than DC component present in output. 2. Half wave rectifier is in efficient to convert AC signal in to pulsating D.C.

 Form factor

 

F.F. RMS Value Avg Value 

2  

 Peak factor

 

P.F. Peak Value RMS Value 

Advantage

(1) Circuit is very simple

Disadvantage (1)  is low (2) R.F >1

To overcome the disadvantage of half wave rectifier we use full wave rectifier.

Full wave rectifier

(i) Centre tapped full wave rectifier (ii) Bridge full wave rectifier Centre tapped full wave rectifier

(43)

A=+Ve

D

₁



F.B B=-Ve

D

2



R.B

 

m

 

V t



V sin t



When D F.B replaced by

R

f When D R.B replaced by O.C

 

_

m

 

_

f L f L V t V sin t i t R R R R     

 

m

 

i t



I sin t



 

0 L

V i t R



 

m L 0 f L V R V sin t R R   

R

L



R

f

 

0 m

V V sin t





 

D1 f

V



V t R

m

 

f f L V R sin t R R   

 

D1 m f

V



I R sin t



 

D2 0 V   _ V t V _

 

2V sin t

_m

 



D2 Max m

V

 

2V

D2 m

PIV



2V

During –Ve cycle of AC supply

 

m

 

m f L f L V t V sin t i t I sin t R R R R       

 

0 L m

V i t R



V sin t



 

D2 f

V



i t R



I R sin t

_{m f}

 



 Average value (or) DC value:

m m

dc dc

2I 2V

I  & V 

(44)

m m rms rms I V I & V 2 2    Efficiency dc ac P 100 P 

_

_

2 m L 2 L 2 2 L f m L f I 4 R ₈ _R 100 R R I _R _R 2      _ _ _      f L 0.812 _{100 81.2%} R 1 _R            _       2 Ripple Factor 1 0.483 2 2     _ _     

Form Factor

2 2





 Peak factor m m I 2 2 I   

PIV 2V



m Advantage

(1)  is twice that of half wave rectifier (2) R.F<1

Disadvantage (1) PIV=

2V

m

(2) Centre tapping is difficult

To overcome disadvantage of center tap full wave rectifier we use full wave bridge rectifier

(45)

1 4 3 2 A ve D F.B D R.B B ve D F.B D R.B        

 

m

 

m L f L f V t V i t sin t I sin t R 2R R 2R       

 

m L

 

  



0 L m L f L f V R V R i t sin t V sin t R R R 2R       

 

D1 D3 f m f

V



V



i t R I R sin t





 





m L m f

 

D2 f 0 L f L f V R V R V i t R V sin t R 2R R 2R       _ _  _ _   



   



m D2 L f L f L f V V R R sin t R R R 2R    _ _      

 

D2 m D4 m V V sin t V V sin t      

During –Ve cycle of V(t) A –Ve

D

4is ON

D

1 is OFF B +Ve

D

2is ON

D

2 is OFF

 

m

 

f L V t i t I sin t R R    

  



0 L m L f

V iR



V sin t R





R

 

D1 D2 f m f

V



V



i t R I R sin t





 

f L D3 D1 m m f L R R V V V sin t V sin t 2R R       _ _       



R

L



R

f



 m m DC DC 2I 2V I  , V     m m rms dc I V I ,V 2 2    L 2 2 L f R 8 ₁₀₀ R R     _ _  _  _

(46)

 

max

81.2%

 R.F 0.483 

F.F

2 2







Peak Factor



2

Advantages

(1)  is twice that of half wave rectifier (2) R.F 1

(3)

PIV V



m

(4) No necessary of center tapped Disadvantages

(1) More no of diode required

The bridge rectifier is more efficient to convert the AC signal into pulsating DC

Solved Examples

Problem: Sketch

Vo

for sinusoidal I/P.

Solution:

During positive cycle: D₁FB & D₂ RB. Current path: k   x y z k

V in V

(47)

Current path: z   x y k z V_in V₀ 2  Zener diode Purpose

(1) To conduct significant current in reverse bias region (2) To with stand for large generation of temperature

 Once breakdown occur (large generation of temperature) uniform electric field developed across it terminals, so that the voltage across its terminals become constant. It acts as voltage regulator.

 A heavily doped silicon diode which has sharp (low) breakdown voltage is called zener diode. Zener diode is heavily doped as compare to PN junction diode

 Doping ratio in zener

1 : 10 100 : 10

6



8

 When it is heavily doped the depletion with become very sharp and breakdown occurs earlier compare to P-N diode.

 Zener- Si type only

(48)

 

V 0,R

Z Z



0

When

I

Z minimum value is not given, by default consider

I

Z min



0

Voltage across diode

V V



Z



Zener O.C.

Z rms

I



0 V V



_Z



Zener acts as voltage regulator

(49)

Regulator is circuit which maintained constant output voltage even input voltage varies (or) load resistance varies.

Zener must operate in the break down region i.e.

V V

i



Z Then

V V

0



Z

Z L

I I I

 

Case-1:

V &R

i L are fixed

i z V V I R    _ _   No change (Fixed) 0 L L V I  _R (Fixed) Z L

I

  

I I

(Fixed)

Case-2:

V

_i variable &

R

_L fixed

i Z V V I R         But 0 L L V I R     _ _   Fixed min Z min V V I R        

(50)

Case-3:

V

i fixed &

R

L variable

i z V V I R         Fixed Zmin Lmax Zmax Lmin I I I I I I     0 Lmax min V I R  0 Lmin max V I R  Solved Examples

Problem: Zener is ideal, find the minimum value of

R

L for which the output voltage remain constant Solution: I 24 15 27        A

For output voltage remain constant,

Lmin 24 15 15 R 27    _ _

  &

I

Lmax

 

I I

zmin



I

Lmin 15 27 R 45 9    

(51)

satisfactory operation of Zener diode.

Solution: For

I

Zmax Zmax

6 I





0.03

Zmax 1 I 5mA 200   & I_L 6 6mA 1K  



V 6

i



I

0.5K







V 6 2m 5mA 6mA

i









i

2V 12 11





i 23 V  ₂17.5 For

I

zmin



0 

V 6 2m 6mA

i







i

V 9V



Range 9V < Vi < 17.5V

Problem: Zener voltage is 4.5V. Zener resistance=75Ω. Calculate the value of

R

_L

Solution: I 12 5 7mA 1K

  

_ _

(52)

R 15K



Problem: Zener voltage is 7V, Zener resistance

R 50

z

 

. Calculate the required range of input voltage to obtain output voltage range 7.3V to 7.6V

Solution: For V0=7.3 i i i V 7.3 V 7.3 V 250 7 50 0 250 250       _ _   _ _    





i

5 7.3 7 V 7.3

  

i

V 1.5 7.3





i

V 8.8



For V0=7.6 i i i V 7.6 V 7.6 V 250 7 50 0 250 250                 i

V 3 7.6 10.6

 



Problem: Calculate the output voltage

of the circuit shown in the figure? (assume ideal diode )

(53)

Voltage across diode =10V which is < 12V The Zener is O.C

0

V 10V



Problem:If

V 2.2

_z



. The cut in voltage of diode 0.3V. Calculate

V ?

₀



Solution: Diode is ON 3.7 0.3 I 2K         p 3.7 0.3 3.7 0.3 V 3.7 1K 2V 2K 2K      _ _     p Z

V V



Zener is OFF (O.C)

zener has low cut in voltage as compare to diode therefore diode on first

0 2 0.3 V 0.3 1K 2V 1K     _ _   

Problem: For the circuit shown

V

Z



4.3V

. Cut in voltage of the diodes

V 0.7V

Y



. When diode is forward bias plot transfer curve & calculate

V

0max

& V

0min values. Assume input voltage is

10sin t

 



.

(54)

When 0 V 2, _i  D OFF, D OFF : ₁ ₂ V₀V_i i

2 V 5

 

,

D

1 OFF,

D

2OFF,

Z

OFF,

V V

0



i

5 V



D

1 OFF,

D

2ON, Z is in Break Down region:

V 5V

0



During negative cycle (

D

2OFF) i

V

 

2.7 D

1ON,

D

2OFF,

Z

OFF :

V 2.7

0



i

2.7 V





D

1OFF,

D

2OFF,

Z

OFF :

V V

0



i

Problem: For the ideal zener circuit shown calculate the output voltage & value of current I

Solution: Zener Z1 F.B and Z2 is in R.B.

16 5.3 10.7 I 10.7mA 1K 1K    _ _   



0

V 5.3 V

(55)

For a sinusoidal input V 6sin t

in  Positive cycle: 6 V 0 C1    V 6V C1  After _{ }_{t 90}0