What is Semiconductor Diode – Electronics Tutorial

What is Semiconductor Diode – Electronics Tutorial
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What is Semiconductor Diode – Electronics Tutorial

  • Page 1

    76 Principles of Electronics 6.1 Semiconductor Diode 6.3 Resistance of Crystal Diode 6.5 Crystal Diode Equivalent Circuits 6.7 Crystal Diode Rectifiers 6.9 Output Frequency of Half-WaveRectifier6.11 Full-Wave Rectifier6.13 Full-Wave Bridge Rectifier6.15 Efficiency of Full-Wave Rectifier6.1...

  • Page 2

    Semiconductor Diode 776.1 Semiconductor DiodeA pn junction is known as a semi-conductor or *crystal diode.The outstanding property of a crystal diode to conduct current in onedirection only permits it to be used as a rectifier. A crystal diode is usuallyrepresented by the schematic symbol sh...

  • Page 3

    78 Principles of ElectronicsIt is interesting to see that behaviour of diode is like a switch. When the diode is forward biased,it behaves like a closed switch and connects the a.c. supply to the load RL. However, when the diodeis reverse biased, it behaves like an open switch and disconn...

  • Page 4

    Semiconductor Diode 79(iv) Refer to Fig. 6.4 (iv). During the positive half-cycle of input a.c. voltage, both the diodes arereverse biased. However, during the negative half-cycle of input a.c. voltage, both the diodes areforward biased.6.3 Resistance of Crystal DiodeIt has already been dis...

  • Page 5

    80 Principles of Electronics2.Reverse resistance. The resistance offered by the diode to the reverse bias is known asreverse resistance. It can be d.c. reverse resistance or a.c. reverse resistance depending upon whetherthe reverse bias is direct or changing voltage. Ideally, the reverse...

  • Page 6

    Semiconductor Diode 81(iii) Ideal diode model. An ideal diode is one which behaves as a perfect conductor whenforward biased and as a perfect insulator when reverse biased. Obviously, in such a hypotheticalsituation, forward resistance rf = 0 and potential barrier V0 is considered negligibl...

  • Page 7

    82 Principles of Electronics(i) The peak current through the diode will occur at the instant when the input voltage reachespositive peak i.e. Vin = VF = 20 V.∴VF = V0 + (If)peak [rf + RL]...(i)or(If)peak =0200.719.3 A10500510−−==++FfLVVrR = 37.8 mA(ii)Peak output voltage = (If)peak ...

  • Page 8

    Semiconductor Diode 83Fig. 6.11Net circuit voltage = 10 − 0.7 − 0.7 = 8.6 VTotal circuit resistance = 1 + 48 + 1 = 50 Ω∴Circuit current = 8.6÷50 = 0.172 A = 172 mAExample 6.5. Determine the current I in the circuit shown in Fig. 6.12 (i). Assume the diodes tobe of silicon a...

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    84 Principles of ElectronicsSolution. It appears that when the applied voltage is switched on, both the diodes will turn “on”. Butthat is not so. When voltage is applied, germanium diode (V0 = 0.3 V) will turn on first and a level of 0.3V is maintained across the parallel circuit. T...

  • Page 10

    Semiconductor Diode 8520 mA and a single diode is used in this circuit, a current of 28.6 mA would flow through the diode,thus damaging the device. By placing them in parallel, the current is limited to a safe value of 14.3 mAfor the same terminal voltage.Example 6.9. Determine the currents...

  • Page 11

    86 Principles of ElectronicsFig. 6.19Solution. Let us assume that diode in Fig. 6.17 (i) is OFF i.e. it is reverse biased. The circuitthen becomes as shown in Fig. 6.17 (ii). Referring to Fig. 6.17 (ii), we have,V1 = 10 V8kΩ2k8 kΩ ×Ω+ = 8VV2 = 10 V6kΩ4k6k×Ω+Ω = 6V∴ Voltage acr...

  • Page 12

    Semiconductor Diode 87(ii) Peak inverse voltage. It is the maximum reverse voltage that a diode can withstand with-out destroying the junction.If the reverse voltage across a diode exceeds this value, the reverse current increases sharply andbreaks down the junction due to excessive heat. P...

  • Page 13

    88 Principles of ElectronicsOperation. The a.c. voltage across the secondary winding AB changes polarities after everyhalf-cycle. During the positive half-cycle of input a.c. voltage, end A becomes positive w.r.t. end B.This makes the diode forward biased and hence it conducts current. D...

  • Page 14

    Semiconductor Diode 89d.c. power. The output current is pulsating direct current. Therefore, in order to find d.c. power,average current has to be found out.*Iav = Idc = 00sin1122ππθθ=θππ+∫∫ mfLViddrR=00sin[ cos ]2)2)ππθθ =−θπ( +π( +∫mmfLfLVVdrRrR=122))×=×π( +( +π...

  • Page 15

    90 Principles of Electronicsa.c. power is contained as 50 watts in positive half-cycles and 50 watts in negative half-cycles. The 50watts in the negative half-cycles are not supplied at all. Only 50 watts in the positive half-cycles areconverted into 40 watts.∴Power efficiency =40 10050...

  • Page 16

    Semiconductor Diode 91d.c. power output = I2dc × RL = ()219.41000 × 800 = 0.301 watt(iii)d.c. output voltage = Idc RL = 19.4 mA × 800 Ω = 15.52 volts(iv) Efficiency of rectification = 0.3010.763 × 100 = 39.5%Example 6.15. A half-wave rectifier is used to supply 50V d.c. to a ...

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    92 Principles of ElectronicsFig. 6.24Peak inverse voltage. Suppose Vm is the maxi-mum voltage across the half secondary winding. Fig.6.25 shows the circuit at the instant secondary voltagereaches its maximum value in the positive direction.At this instant, diode D1 is conducting while dio...

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    Semiconductor Diode 93Operation. During the positive half-cycle of secondary voltage, the end P of the secondarywinding becomes positive and end Q negative. This makes diodes D1 and D3 forward biased whilediodes D2 and D4 are reverse biased. Therefore, only diodes D1 and D3 conduct. These...

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    94 Principles of Electronics(ii) As during each half-cycle of a.c. input two diodes that conduct are in series, therefore,voltage drop in the internal resistance of the rectifying unit will be twice as great as in the centre tapcircuit. This is objectionable when secondary voltage is small...

  • Page 20

    Semiconductor Diode 95Idc =2πmI∴d.c. power output, Pdc = I2dc × RL = 22 ×πmLIR...(i)a.c. input power. The a.c. input power is given by :Pac = I2rms (rf + RL)For a full-wave rectified wave, we have,Irms = Im/2∴Pac =22mI(rf + RL) ... (ii)∴ Full-...

  • Page 21

    96 Principles of ElectronicsR.M.S. primary voltage = 230 V∴ R.M.S. secondary voltage= 230 × (1/5) = 46 VMaximum voltage across secondary= 46 × 2 = 65VMaximum voltage across halfsecondary winding isVm = 65/2 = 32.5 V(i)Average current, Idc =2232.5100×=ππ ×mLVR = 0.207 A∴d.c. ou...

  • Page 22

    Semiconductor Diode 97(ii) The peak inverse voltage is equal to the maximum secondary voltage i.e.PIV = 81.3 V(iii) In full-wave rectification, there are two output pulses for each complete cycle of the inputa.c. voltage. Therefore, the output frequency is twice that of the a.c. supply freque...

  • Page 23

    98 Principles of ElectronicsCentre-tap circuitR.M.S. secondary voltage = 230 × 1/5 = 46 VMax. voltage across secondary = 46 × 2 = 65 VMax. voltage across half secondary winding isVm = 65/2 = 32.5 VFig. 6.34∴PIV =2 Vm = 2 × 32.5 = 65 VBridge type circuitR.M.S. secondary voltag...

  • Page 24

    Semiconductor Diode 99voltage (ii) d.c. output current. Use simplified model for the diodes.Fig. 6.35Solution. The conditions of the problem suggest that the a.c voltage across transformer second-ary is 12V r.m.s.∴ Peak secondary voltage isVs (pk) = 12 × 2 = 16.97 V(i) At any instant in ...

  • Page 25

    100 Principles of Electronicsto blow. If the fuse does not blow, the d.c. output from the rectifier will be extremely low and thetransformer itself will be very hot.(ii) When the primary or secondary winding of the transformer opens, the output from therectifier will drop to zero. In this c...

  • Page 26

    Semiconductor Diode 101The fact that a pulsating d.c. contains both d.c. and a.c. compo-nents can be beautifully illustrated by referring to Fig. 6.38. Fig. 6.38(i) shows a pure d.c. component, whereas Fig. 6.38 (ii) shows the *a.c.component. If these two waves are added together, the resul...

  • Page 27

    102 Principles of Electronics∴Ripple factor =2/21/−πmmII = 1.21It is clear that a.c. component exceeds the d.c. component in the output of a half-wave rectifier.This results in greater pulsations in the output. Therefore, half-wave rectifier is ineffective for con-v...

  • Page 28

    Semiconductor Diode 103 *If such a d.c. is applied in an electronic circuit, it will produce a hum.○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○A comparison among the three rectifier circu...

  • Page 29

    104 Principles of Electronics*The shorthand name of inductor coil is choke.** The shape of the circuit diagram of this filter circuit appears like Greek letter π (pi) and hence the nameπ-filter.○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○○...

  • Page 30

    Semiconductor Diode 105another filter capacitor C2 connectedacross the load. Only one filter sectionis shown but several identical sectionsare often used to improve the smooth-ing action.The pulsating output from therectifier is applied across the inputterminals (i.e. terminals 1 and 2) of t...

  • Page 31

    106 Principles of Electronics∴Vdc = Vp(in) 112−LfR C = 14.9 (1 – 0.038) = 14.3VExample 6.24. The choke of Fig. 6.45 has a d.c. resistance of 25 Ω. What is the d.c. voltage if thefull-wave signal into the choke has a peak value of 25.7 V ?Fig. 6.45Solution. The ou...

  • Page 32

    Semiconductor Diode 107greater than the input power. In fact, there are losses in the circuit (e.g. in diodes, capacitors etc.) sothat the output power will actually be less than the input power.6.23 Half-Wave Voltage DoublerA half-wave voltage doubler consists of two diodes and two capacitor...

  • Page 33

    108 Principles of ElectronicsReferring to Fig. 6.50 (ii), it is easy to see that C1 (charged to VS (pk) )and the source voltage (VS)now act as series-aiding voltage sources. Thus C2 will be charged to the sum of the series peakvoltages i.e. 2 VS(pk).(iii) When VS returns to its original po...

  • Page 34

    Semiconductor Diode 109a zener diode.A properly doped crystal diode which has asharp breakdown voltage is known as a zenerdiode.Fig. 6.53 shows the symbol of a zener diode.It may be seen that it is just like an ordinarydiode except that the bar is turned into z-shape.The following points may ...

  • Page 35

    110 Principles of ElectronicsFig. 6.556.27 Zener Diode as Voltage StabiliserA zener diode can be used as a voltage regulator to provide a constant voltage from a source whosevoltage may vary over sufficient range. The circuit arrangement is shown in Fig. 6.56 (i). The zenerdiode of zener ...

  • Page 36

    Semiconductor Diode 111Applying Ohm’s law, we have,R =0iZLEEII−+6.28 Solving Zener Diode CircuitsThe analysis of zener diode circuits is quite similar to that applied to the analysis of semiconductordiodes. The first step is to determine the state of zener diode i.e., whether the zener i...

  • Page 37

    112 Principles of ElectronicsIZ =I − ILwhere IL = 0LER and I = 0−iEERPower dissipated in zener, PZ = VZ IZ(ii) Off state. Referring to the circuit shown in Fig. 6.58 (ii),I = ILand IZ = 0VR = Ei − E0and V = E0(V < VZ)∴PZ = V IZ = V(0) = 02. Fixed Ei and Variable RL. ...

  • Page 38

    Semiconductor Diode 113ILmin = I − IZMandRLmax =0=ZLminLminEVIIIf the load resistance exceeds this limiting value, the current through zener will exceed IZM andthe device may burn out.3. Fixed RL and Variable Ei. This case isshown in Fig. 6.60. Here the load resistance RL isfixed while th...

  • Page 39

    114 Principles of ElectronicsSolution. If you remove the zener diode in Fig. 6.61 (i), the voltage V across the open-circuit isgiven by :V =10 120510×=++LiLRERR = 80 VSince voltage across zener diode is greater than VZ (= 50 V), the zener is in the “on” state. It can,therefore, be ...

  • Page 40

    Semiconductor Diode 115Minimum Zener current. The zener will conduct minimum current when the input voltage isminimum i.e. 80 V. Under such conditions, we have,Voltage across 5 kΩ = 80 − 50 = 30 VCurrent through 5 kΩ, I =30 V5kΩ = 6 mALoad current, IL =5 mA∴Zener current, IZ ...

  • Page 41

    116 Principles of ElectronicsExample 6.29. A 10-V zener diode is used to regulate the voltage across a variable load resistor[See fig. 6.65]. The input voltage varies between 13 V and 16 V and the load current varies between 10mA and 85 mA. The minimum zener current is 15 mA. Calculate th...

  • Page 42

    Semiconductor Diode 117Voltage rating of each zener, VZ = 10 VCurrent rating of each zener, IZ = 1000 mA Input unregulated voltage, Ei = 45 V Regulated output voltage, E0 = 10 + 10 + 10 = 30 VLet R ohms be the required series resistance.Voltage across R = Ei − E0 = 45 − 30 = 15 V...

  • Page 43

    118 Principles of ElectronicsMaximum power rating of zener isPZM= VZ IZM = (12 V) (200 mA) = 2.4 WExample. 6.34. Fig. 6.70 shows the basic zener diode circuits. What will be the circuit behaviourif the zener is (i) working properly (ii) shorted (iii) open-circuited?Fig. 6.70Solution. ...

  • Page 44

    Semiconductor Diode 119If you trace from the high side of the bridge to the low side, you will see that the only resistance acrossthe secondary of the transformer is the forward resistance of the two ON diodes. This effectivelyshorts out the transformer secondary. The result is that excessive...

  • Page 45

    120 Principles of Electronics(i) 2.5 V(ii)3 V(iii) 10 V(iv) 0.7 V6. A crystal diode is used as ........(i) an amplifier(ii) a rectifier(iii) an oscillator(iv) a voltage regulator7. The d.c. resistance of a crystal diode is ........its a.c. resistance.(i) the same as(ii) more than(iii) less ...

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    Semiconductor Diode 121(i) the same as(ii) less than(iii) more than(iv) none of the above24. A zener diode is always .......... connected.(i) reverse(ii) forward(iii) either reverse or forward(iv) none of the above25. A zener diode utilises ........ characteristic forits operation.(i) forward...

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    122 Principles of Electronics41. A 1000 V power supply would use ........ asa filter capacitor.(i) paper capacitor(ii) air capacitor(iii) mica capacitor(iv) electrolytic capacitor42. The ...... filter circuit results in the best volt-age regulation.(i) choke input(ii) capacitor input(iii) r...

  • Page 48

    Semiconductor Diode 123Problems1. What is the current in the circuit in Fig. 6.72 ? Assume the diode to be ideal.[10 mA]Fig. 6.72Fig. 6.732. Using equivalent circuit, determine the current in the circuit shown in Fig. 6.73. Assume the forwardresistance of the diode to be 2 Ω.[358 mA]3. Find ...

  • Page 49

    124 Principles of ElectronicsFig. 6.7812. Calculate the peak voltage across each half of a centre-tapped transformer used in a full-wave rectifierthat has an average output voltage of 110V.[173V]13. What PIV rating is required for the diodes in a bridge rectifier that produces an average ou...