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EDC-UNITVIII Question&answer GRIET-ECE G.Surekha Page 1 UNIT VIII-SPECIAL PURPOSE ELECTRONIC DEVICES 1. Explain tunnel Diode operation with the help of energy band diagrams. TUNNEL DIODE: A tunnel diode or Esaki diode is a type of semiconductor diode which is capable of very fast operation, well into the microwave frequency region, by using quantum mechanical effects. Impurity concentration in normal diode is 1 part in 8 10 in Tunnel diode 1 part in 3 10 .Normally a electron or hole must have energy greater than or equal to potential energy barrier, to move to other side of the barrier. For very thin barrier there is a large probability of electron penetrating through the barrier which is called as “Tunneling:. Under normal Forward bias operation As voltage begins to increase, electrons at first tunnel through the very narrow pn junction barrier because filled electron states in the conduction band on the n-side become aligned with empty valence band hole states on the p-side of the p-n junction. As voltage increases further these states become more misaligned and the current drops this is called negative resistance because current decreases with increasing voltage. As voltage increases yet further, the diode begins to operate as a normal diode, where electrons travel by conduction across the pn junction, and no longer by tunneling through the pn junction barrier. Thus the most important operating region for a tunnel diode is the negative resistance region. Reverse bias operation When used in the reverse direction they are called back diodes and can act as fast rectifiers with zero offset voltage and extreme linearity for power signals.Under reverse bias filled states on the p-side become increasingly aligned with empty states on the n-side and electrons now tunnel through the pn junction barrier in reverse direction. ENERGY BAND STRUCTURE OF HIGHLY DOPED PN DIODE: www.jntuworld.com www.jntuworld.com

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 1

    UNIT VIII-SPECIAL PURPOSE ELECTRONIC DEVICES

    1. Explain tunnel Diode operation with the help of energy band diagrams.

    TUNNEL DIODE:

    A tunnel diode or Esaki diode is a type of semiconductor diode which is capable of very

    fast operation, well into the microwave frequency region, by using quantum mechanical

    effects. Impurity concentration in normal diode is 1 part in 810 in Tunnel diode 1 part in 310 .Normally a electron or hole must have energy greater than or equal to potential

    energy barrier, to move to other side of the barrier. For very thin barrier there is a large

    probability of electron penetrating through the barrier which is called as Tunneling:.

    Under normal Forward bias operation

    As voltage begins to increase, electrons at first tunnel through the very narrow pn junction barrier because filled electron states in the conduction band on the n-side become aligned with

    empty valence band hole states on the p-side of the p-n junction. As voltage increases further

    these states become more misaligned and the current drops this is called negative resistance because current decreases with increasing voltage. As voltage increases yet further, the diode

    begins to operate as a normal diode, where electrons travel by conduction across the pn junction, and no longer by tunneling through the pn junction barrier. Thus the most important operating region for a tunnel diode is the negative resistance region.

    Reverse bias operation

    When used in the reverse direction they are called back diodes and can act as fast rectifiers with

    zero offset voltage and extreme linearity for power signals.Under reverse bias filled states on the

    p-side become increasingly aligned with empty states on the n-side and electrons now tunnel

    through the pn junction barrier in reverse direction.

    ENERGY BAND STRUCTURE OF HIGHLY DOPED PN DIODE:

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 2

    ln CF CD

    NE E KT

    N for N type material If NC < ND, EF > EC

    Similarly for P type EF = EV+KT C

    n

    A

    Nl

    N If NA > NC; EF < EV

    Fermi level lies in conduction band in N type as shown in Fig.5.23 (a).

    Fermi level lies in valence band in P type material. Fermi level is at same energy level

    on both sides. By reverse bailing barrier height increases as shown in fig.5.23 (b).Fermi

    level on N side is lowered. Tunneling of electron from P to N side is the result. (From

    filled sates to empty states). If we increase the reverse bias, reverse current increases as

    shown in characteristics of (Fig.5.25) a.

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 3

    Similarly for forward bias tunneling occurs from N to P type material as shown in

    Fig.5.24. Further increase in forward, the condition shown in 5.24(b)reached and

    maximum current follows (characteristics of Fig.5.25).

    Further increase will reduce the current as shown in fig.5.24(1) till a minimum current

    flows due to the condition shown in fig.5.24(d) section 3 of 5.25(a)

    Besides the above current due to tunneling normal diode current flows as shown in

    dotted lives in fig.5.25(a). Resultant is the graph shown in 5.25(b).

    The symbol and equivalent circuit is shown in the above fig5.26.

    Application: -

    1. Used as oscillator at VHF / UHF.

    2. Ultra high speed switch.

    3. Used as logic memory (storage) device.

    4. Used as amplifier.

    Advantages Disadvantages

    1) Low noise 1) Less voltage range of separation

    2) Ease of operation 2) No isolation of input and output,

    3) High speed as it is a two terminal device.

    4) Low power.

    2. Differences between Tunnel Diode and PN Diode.

    Impurity concentration in normal diode 1 part in 810 Impurity concentration in Tunnel diode 1 part in 310 Width of the junction barrio varies inversely as square root of impurity concentration

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 4

    22B

    A

    VW

    e

    Where VB is barrier potential, is the permitivity of material NA acceptor concentration.

    Width of junction for normal diode 5 microns 65 10 m

    Width of junction Tunnel Diode < 100 A0 10

    -8m

    Normally a electron or hole must have energy greater than or equal to potential energy barrier, to move to other side of the barrier.

    For very thin barrier there is a large probability of electron pert rating through the barrier which is called as Tunneling:.

    3. Explain the principle and operation of varactor Diode.

    Varactor or varicap diodes are used mainly in radio frequency (RF) circuits to be able to

    provide a capacitance that can be varied by changing a voltage in an electronics circuit.

    This can be used for tuning circuits including radio frequency oscillators and filters.

    Although both names: varactor and varicap diode are used, they are both the same form

    of diode. The name varactor meaning variable reactor, or reactance, and varicap meaning

    variable capacitance (vari-cap).

    Operation of a variable capacitor

    They key to understanding how a varactor or varicap diode works is to look at what a

    capacitor is and what can change the capacitance. As can be seen from the diagram

    below, a capacitor consists of two plates with an insulating dielectric between them.

    ... the capacitance and the

    amount of charge that can be

    stored depends on the area of the

    plates and the distance between

    them....

    The capacitance of the capacitor is dependent upon the area of the plates - the larger the

    area the greater the capacitance, and also the distance between them - the greater the

    distance the smaller the level of capacitance.

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 5

    A reverse biased diode has no current flowing between the P-type area and the N-type

    area. The N-type region and the P-type regions can conduct electricity, and can be

    considered to be the two plates, and the region between them - the depletion region is the

    insulating dielectric. This is exactly the same as the capacitor above.

    As with any diode, if the reverse bias is changed so does the size of the depletion region.

    If the reverse voltage on the varactor or varicap diode is increased, the depletion region of

    the diode increases and if the reverse voltage on varactor diode is decreased the depletion

    region narrows. Therefore by changing the reverse bias on the diode it is possible to

    change the capacitance.

    Change of varactor diode capacitance with reverse bias

    Varactor or varicap circuit symbol

    The varactor diode or varicap diode is shown in circuit diagrams or schematics using a

    symbol that combines the diode and capacitor symbols. In this way it is obvious that it is

    being used as a varacor or varicap capacitor rather than a rectifying diode.

    Circuit symbol for a varactor diode / varicap diode

    When operated in a circuit, it is necessary to ensure the varactor diode remains reverse

    biased. This means that the cathode will be positive with respect to the anode, i.e. the

    cathode of the varactor will be more positive than the anode. Varactor Diode or

    Varactors are operated reverse biased so no current flows, but since the thickness of the

    depletion region varies with the applied bias voltage, the capacitance of the diode can be

    made to vary. Generally, the depletion region thickness is proportional to the square root

    of the applied voltage; and capacitance is inversely proportional to the depletion region

    thickness.

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 6

    Applications:

    They are used in PLL, voltage controlled oscillators, harmonic generation, electronic

    tuning devices in tuners for television, mobiles, parametric amplification, AM radios,

    voltage-variable tuning, frequency multipliers, etc.

    4. Describe the working principle of an SCR with V-I characteristics and also explain two

    transistor analogy of an SCR.

    SILICON CONTROLLED RECTIFIER:

    Construction of SCR

    An SCR consists of four layers of alternating P and N type semiconductor materials. Silicon is

    used as the intrinsic semiconductor, to which the proper dopants are added. The junctions are

    either diffused or alloyed. The planar construction is used for low power SCRs (and all the

    junctions are diffused). The mesa type construction is used for high power SCRs. In this case,

    junction J2 is obtained by the diffusion method and then the outer two layers are alloyed to it,

    since the PNPN pellet is required to handle large currents.

    Construction Symbol

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 7

    CHARACTERISTICS OF SCR.

    1) SCR is a three terminal four layer semiconductor device.

    2) Leakage current is very small for SCR compared with germanium.

    3) SCR acts as a switch when it is forward biased.

    4) When gate is open i.e., IG = 0, and anode voltage is applied junctions P1 N1 and P2 N2 are

    forward biased where N1 P2 is reverse biased. Only small reverse current flows.

    5) If we increase anode voltage further, at one stage anode current increases suddenly and

    voltage across the SCR falls to holding voltage VH.

    6) Once SCR fires (conducts), it will remain in conduction till the current through the device is

    reduced less than IH, adding current by reducing applied voltage (to less than holding

    voltage) close to zero.

    7) The firing angle can be varied by varying the Gate voltage. With very large positive (gate

    current break over may occur at very low voltage and SCR works as if it is a normal PN

    diode.

    TWO TRANSISTOR VERSION OF SCR.

    -T1 is PNP and T2 is NPN.

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 8

    Ib1 = IA Ie1 = IA - 1 IA = IA(1 - 1) - (1)

    Ib1 = Ic2 and Ic2 = 2Ik - (2)

    Ib1 = IA(1 - 1) = 2IK - (3)

    We know Ik = IA + Ig. ( IA = IC1 + Ib1) - (4)

    Putting the value of Ik from eqn. (4) in eqn. (3)

    IA(1 - 1) - 2(IA + Ig)

    IA(1 - 1) = 2(IA + Ig)

    IA(1 - 1 - 2) = 2 Ig. Or 2

    1 21

    g

    A

    II -(5)

    Equation 5 indicates that if( 1 + 2) = 1, IA =

    - SCR is also called as Thyrister - Latching current (IL) the min. current required to fire the device - Holding current (IH) min. current to keep the SCR conductivity

    - Voltage safety factor 2 .

    f

    PIVV

    RMS of operating voltage

    Value of Vf is 2 to 2.7.

    Application of SCRs

    1. SCRs are mainly used in devices where the control of high power, possibly coupled with high voltage, is demanded.

    2. Their operation makes them suitable for use in medium to high-voltage AC power control applications, such as regulators and motor control.

    3. SCRs and similar devices are used for rectification of high power AC in high-voltage direct current power transmission.

    5. Explain Rectifier circuits using SCRs.

    SCR Half wave Rectifier:- Below fig shows the circuit of an SCR half wave rectifier.

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 9

    SCR does not conduct during negative half cycle (like normal PN diode)

    Firing angle depends on gate voltage

    Conduction angle is ( - )

    During the positive half cycle of ac voltage appearing across secondary, the SCR will conduct

    provided proper gate current, the lesser the supply voltage at which the SCR is triggered ON.

    Reffering to above fig the gate current is adjusted to such a value that SCR is turned ON at a

    positive voltage V1of ac secondary voltage which is less than the peak voltage Vm. Beyond this

    The SCR will be conducting till the applied voltage becomes zero. The angle at which the SCR

    starts conducting during the positive half cycle is called firing angle . There fore the conduction

    angle is (1800- ).

    Average DC output

    0

    1sin .

    2av mV V wt dwt

    0

    1cos

    2

    1cos cos

    2

    11 cos

    2

    1 cos2

    m

    m

    m

    m

    V wt

    V

    V

    V

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 10

    RMS VOLTAGE: VRMS is given by

    1/21

    sin 22

    m

    RMS

    VV

    SCR FULL WAVE RECTIFIER

    The SCR Full wave Rectifier is shown in below fig. During the Positive half cycle of the input

    signal, anode of the SCR1 becomes positive and the at the same time the anode of SCR2

    becomes negative. When the input voltage reaches V1 as shown in below fig (b), SCR1 starts

    conducting and therefore only the shaded portion of positive half cycle will pass through the

    load. During the negative half cycle of the input, the anode of SCR1 becomes negative and the

    anode of SCR2 becomes positive. Hence SCR1 does not conduct and SCR2 conducts when the

    input voltage becomes V1.

    1 cosDCVm

    V

    6. Explain the principle and working of Photo Diode.

    PHOTO DIODES: The diagrams shown below are construction, biasing and symbol of Photo

    diode.

    Construction Biasing Symbol

    A P N K P N

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 11

    - When light falls on reverse biased PN photo junction, holes and electron pairs are liberated which leads to current flow through the external load.

    - Current will be zero only for a positive voltage VT. Current luminous flux - LEDs are used for displays, including seven-segment display. - A photodiode is a type of photo detector capable of converting light into either

    current or voltage, depending upon the mode of operation. A photodiode is a p-n

    junction or PIN structure. It is designed to operate in reverse bias. When a photon of

    sufficient energy strikes the diode, it excites an electron, thereby creating a free

    electron (and a positively charged electron hole). This mechanism is also known as

    the inner photoelectric effect. If the absorption occurs in the junction's depletion

    region, or one diffusion length away from it, these carriers are swept from the

    junction by the built-in field of the depletion region. Thus holes move toward the

    anode, and electrons toward the cathode, and a photocurrent is produced. This

    photocurrent is the sum of both the dark current (without light) and the light current,

    so the dark current must be minimised to enhance the sensitivity of the device.

    Applications

    P-N photodiodes are used in similar applications to other photo detectors, such as

    photoconductors, charge-coupled devices, and photomultiplier tubes. They may be used to

    generate an output which is dependent upon the illumination .

    Photosensors of all types may be used to respond to incident light, or to a source of light which is

    part of the the same circuit or system. A photodiode is often combined into a single component

    with an emitter of light, usually a light-emitting diode (LED), either to detect the presence of a

    mechanical obstruction to the beam (slotted optical switch), or to couple two digital or analog

    circuits while maintaining extremely high electrical isolation between them, often for safety

    (optocoupler).

    PIN diodes are much faster and more sensitive than p-n junction diodes, and hence are often used

    for optical communications and in lighting regulation.

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 12

    7. Describe the operation of Schottky diode.

    Schottky Diode

    Junction of lightly doped n-type

    semiconductor with a metal electrode.

    The junction of a doped semiconductor (usually n-type)

    with a metal electrode can produce a very fast-switching

    diode which is mainly used in high frequency circuits or

    high speed digital circuits. Under forward bias, the

    electrons move from the n-type material to the metal

    and give up their energy quickly. There are no holes

    (minority carriers), so the conduction quickly stops

    upon change to reverse bias. Schottky diodes find

    application as rectifiers for high frequency signals.

    Construction

    A metal-semiconductor junction is formed between a metal and a semiconductor, creating a

    Schottky barrier (instead of a semiconductorsemiconductor junction as in conventional diodes). Typical metals used are molybdenum, platinum, chromium or tungsten; and the semiconductor

    would typically be N-type silicon.The metal sides acts as the anode and N-type semiconductor

    acts as the cathode of the diode. This Schottky barrier results in both very fast switching and low

    forward voltage drop.

    Reverse recovery time

    The most important difference between p-n and Schottky diode is reverse recovery time, when

    the diode switches from non-conducting to conducting state and vice versa. Where in a p-n diode

    the reverse recovery time can be in the order of hundreds of nanoseconds and less than 100 ns for

    fast diodes, Schottky diodes do not have a recovery time, as there is nothing to recover from (i.e.

    no charge carrier depletion region at the junction). The switching time is ~100 ps for the small

    signal diodes, and up to tens of nanoseconds for special high-capacity power diodes. It is often

    said that the Schottky diode is a "majority carrier" semiconductor device. This means that if the

    semiconductor body is doped n-type, only the n-type carriers (mobile electrons) play a

    significant role in normal operation of the device. The majority carriers are quickly injected into

    the conduction band of the metal contact on the other side of the diode to become free moving

    electrons. Therefore no slow, random recombination of n- and p- type carriers is involved, so that

    this diode can cease conduction faster than an ordinary p-n rectifier diode. This property in turn

    allows a smaller device area, which also makes for a faster transition. This is another reason why

    Schottky diodes are useful in switch-mode power converters; the high speed of the diode means

    that the circuit can operate at frequencies in the range 200 kHz to 2 MHz, allowing the use of

    small inductors and capacitors with greater efficiency than would be possible with other diode

    types. Small-area Schottky diodes are the heart of RF detectors and mixers, which often operate

    up to 50 GHz.

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 13

    Limitations

    The most evident limitations of Schottky diodes are the relatively low reverse voltage rating for

    silicon-metal Schottky diodes, 50 V and below, and a relatively high reverse leakage current.

    Diode designs have been improving over time. Voltage ratings now can reach 200 V. Reverse

    leakage current, because it increases with temperature, leads to a thermal instability.

    Problems

    Q)1. An SCR FWR is connected to 250V. 50 Hz mains to supply ac voltage to resistive load of

    10 for firing angle of 90 . Find DC output voltage and load current.

    Solution: -

    Given VRMS = 230V, RL = 10 , = 90

    VDC = ? IL = ?

    max

    2

    1

    RMS

    mDC

    VV

    VV Cos

    Or Vmax = VRMS 2 = 250 2 = 353.6 volts

    353.61 cos90 112.6

    112.611.26

    10

    DC

    L

    L

    volts

    VI Amps

    R

    Q 2) A sinusoidal voltage V = 200 sin 314 t is applied to an SCR whose forward break down

    voltage is 150V. Determine the time during which SCR remain off.

    Solution: -

    Given V1 = 150V, Vm = 200V W = 314 = ? t = ?

    1 sinmV V or 1 150 3sin

    200 4m

    V

    V

    1sin 3 / 4 48.6

    T = 1/f f = ? w = 2 f = 314 or f = 314/2 = 50Hz.

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  • EDC-UNITVIII Question&answer

    GRIET-ECE G.Surekha Page 14

    T= 1/50 = 0.02sec = 20 m. sec.

    t = 48.6360 360

    20 2.7 secT m

    Q 3) A half wave rectifier employing SCR is adjusted to have a gate current of 1mA and its

    forward breakdown voltage is 150V. If a sinusoidal voltage of 400V peak is applied, determine.

    i) Firing angle (ii) Average output voltage

    iii) Average current for a load resistance of 200 iv) Power output.

    Given

    V1 = 150V, Vm = 400V, = ? VDC = ? IDC = ? PDC = ? RL = 200

    Solution: -

    V1 = Vm sin , or Sin = V1 / Vm = 150/400 = 3/8 = 0.375.

    = Sin-1

    0.375 = 22 .

    400 400

    1 cos (1 cos 22 ) (1.927) 122.62 2 2

    m

    DC

    VV volts

    122.60.613 .

    200

    . 122.6 0.613 75.15 .

    DCDC

    L

    DC DC DC

    VI Amps

    R

    P V I Watts

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