Rec101 unit 1 (part iii) diode applications
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The document provides a comprehensive overview of diode applications in electronics, focusing on fundamental concepts of PN junction diodes and their characteristics. It discusses various configurations for rectification, including half-wave and full-wave rectifiers, along with specialized applications such as clippers and voltage regulators. The document also includes detailed mathematical analysis and practical examples related to diode operations.
Rec101 unit 1 (part iii) diode applications
- 1. Diode Applications REC 101: Basic Electronics Unit 1 PN junction diode: Introduction of Semiconductor Materials Semiconductor Diode: Depletion layer, V-I characteristics, ideal and practical, diode resistance, capacitance, Diode Equivalent Circuits, Transition and Diffusion Capacitance, Zener Diodes breakdown mechanism (Zener and avalanche) Diode Application: Series , Parallel and Series, Parallel Diode Configuration, Half and Full Wave rectification, Clippers, Clampers, Zener diode as shunt regulator, Voltage- Multiplier Circuits Special Purpose two terminal Devices :Light-Emitting Diodes, Varactor (Varicap) Diodes, Tunnel Diodes, Liquid-Crystal Displays. 9/11/2017 1 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad
- 2. Diode Application: Series, Parallel and Series, Parallel Configuration • The forward resistance of diode is so small compared to the other series elements of network that it can be ignored (if not specified) • In general, a diode is in the ON state if the current established by the sources is such that, its direction matches to the arrow of diode symbol, and VD VK otherwise it is in OFF state 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 2 Diode VK (V) Ideal 0 Ge 0.3 Si 0.7 GaAs 1.2 Steps to find the state of diode 1. Replace the diode with equivalent circuit. 2. mentally replace diode with a resistance and check direction of current through it. 3. If current direction matches to diode arrow then diode is ON otherwise it is OFF Practical VK ideal ideal VK Rf OR Diode Equivalent circuit VK knee voltage Rf diode forward resistance
- 3. Diode Application: Series , Parallel and Series, Parallel Diode Configuration, 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 3 ideal 0.7 V 10 V 100 Si 10 V 100 1.Replace diode with equivalent circuit ideal 0.7 V 10 V 100 2.Replace ideal diode with fictitious resistance and find the current direction 3. As current matches with diode direction, it is ON. Short-circuit the diode and solve the circuit mA93A093.0 100 7.010 I 0.7 V 10 V 100 I
- 4. Diode Application: rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 4 • One of the common application of diode is ‘rectifier’ which converts AC signal into DC. • There are many rectifier circuits – Half Wave rectifier – Full wave rectifier; Bridge type & Centre tapped • Practical rectifier consists of; Transformer, Rectifier, Filter & Regulator stage • Rectifier output for one cycle (period) of input is analyzed AC Input + DC output - Transformer Stage Rectifier Stage Regulator stage Filter Stage
- 5. Diode Application: Half Wave rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 5 t Vi(t) T Vm -Vm + Vo(t)=Vi(t) - + Vi(t) - Rt Vi(t) T Vm -Vm Vo T Vm t Positive Half cycle (0 t< T/2) Vi(t) >0 and diode is ON So Vo(t) = Vi(t) + - + V0(t) - + Vi(t) - R Half wave rectifier circuit T tSinVtV mi 2 where,)( The process of removing one- half the input ac signal to establish a dc level is called halfwave rectification V0(t) T Vm t + Vi(t) - Rt Vi(t) T Vm -Vm Negative Half cycle (T/2 t< T) Vi(t) <0 and diode is OFF So Vo(t) = 0 + Vo(t)=0 - TtTfor TtfortV tV i 2/0 2/0)( )(periodofoutputcombinedTherefore 0
- 6. Diode Application: Half Wave rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 6 t Vi(t) T Vm -Vm V0(t) T Vm t 2T 2T Vdc= 0.318Vm mm T m dc T m TT i T dc V Cos T Cos V tCos T V V dttSinV T dtdttV T dttV T V )0( 22 1 .0)( 1 )( 1 2 0 2 0 2 0 2 00 0 VDC: is periodic average of output wave mdc VV 318.0 TtTfor TtfortV tV i 2/0 2/0)( )(0 2 0 4 2sin 42 2cos1 2 .0)( 1 )( 1 2 2 0 2 2 0 22 0 2 2 0 2 22 0 2 0 2 0 2 mm T m T m T m dc T m TT i T orms VV t T V t T V dtt T V V dttSin T V dtdttV T dttV T V Vrms: square root of periodic average of squared output mrms VV 5.0
- 7. Diode Application: Half Wave rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 7 Peak inverse voltage (PIV or PRV) : Applying KVL, in loop of reverse bias diode. PIV Vm - Vm + R - PIV + V0(t)=0 V Ripple factor: ratio of Vac to Vdc 21.1157.11 2 12 factorformiskwhere 11factorRipple 2 2 2 f 2 222 222 m m f dc rms dc dcrms dc ac dcacrms V V k V V V VV V V VVV
- 8. Diode Application: Bridge type Full Wave rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 8 t Vi(t) T Vm -Vm t Vi(t) T Vm -Vm Vo(t) T Vm t + Vi(t) - R - Vo(t) + Positive Half cycle (0 t< T/2) Vi(t)>0 so D2 and D3 is ON D1 and D4 is ON So Vo(t) = Vi(t) + Vi(t) - R - Vo(t) + D1 D2 D3 D4 The process of utilizing full period of ac signal to establish a dc level is called fullwave rectification A bridge with 4 diodes as shown in figure is most familiar full wave rectifier.
- 9. Diode Application: Bridge type Full Wave rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 9 t Vi(t) T Vm -Vm V0(t) T Vm t t Vi(t) T Vm -Vm Vo(t) T Vm t + Vi(t) - R - Vo(t) + Negative Half cycle (T/2 t< T) Vi(t)<0 and D2 and D3 is ON D1 and D4 is ON So Vo(t) = - Vi(t) m T m T i T dc V dttSinV T dttV T dttV T V 22 )(2 1 )( 1 2 0 2 00 0 VDC: As the area above the axis for one full cycle is now twice of that for a half-wave system, dc level has also been doubled mdc VV 636.0 TtTfortV TtfortV tV i i 2/)( 2/0)( )(periodofoutputcombinedTherefore 0 Vdc= 0.636Vm
- 10. Diode Application: Bridge type Full Wave rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 10 22 2 )( 2 )( 1 22 0 2 22 0 2 0 2 mm T m T i T orms VV dttSin T V dttV T dttV T V Vrms: mrms VV 707.0 Peak inverse voltage (PIV or PRV) Applying KVL, in loop containing input and reverse bias diode. PIV Vm + Vm - R - Vo(t) + - Vm + R - Vo(t) + Ripple factor: ratio of Vac to Vdc 48.01109.11 22 1 2 21 2 2 2 2 m m dc rms V V V V
- 11. Diode Application: Centre tapped Full Wave rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 11 t Vi(t) T Vm -Vm t Vi(t) T Vm -Vm Vo(t) T Vm t R 1:2 + Vi(t) - - Vo(t) + + Vi(t) - + Vi(t) - Positive Half cycle (0 t< T/2) Vi(t) >0 so D1 is ON & D2 is OFF So V0(t) = Vi(t) + Vi(t) - R - Vo(t) + + Vi(t) - + Vi(t) - 1:2 D1 D2 Another type of fullwave rectifier is Centre tapped It uses 2 diodes with a centre tapped transformer. t Vi(t) T Vm -Vm Vo(t) T Vm t R 1:2 + Vi(t) - - Vo(t) + + Vi(t) - + Vi(t) - Negative Half cycle (T/2 t< T) Vi(t) <0 so D1 is OFF & D2 is ON So V0 (t) = -Vi(t)
- 12. Diode Application: Centre tapped Full Wave rectifier 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 12 VDC: equal to that of bridge type full wave rectifier mdc VV 636.0 TtTfortV TtfortV tV i i 2/)( 2/0)( )(periodofoutputcombinedTherefore 0 t Vi(t) T Vm -Vm Vo(t) T Vm t Vdc= 0.636VmVrms: equal to that of bridge type full wave rectifier mrms VV 707.0 Peak inverse voltage (PIV or PRV) Applying KVL, in loop containing input and reverse bias diode. PIV 2Vm R 1:2 + Vm - - Vm + + Vm - + Vm -- PIV + Ripple factor: ratio of Vac to Vdc 48.0
- 13. Diode Application: Clippers 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 13 Clippers are networks that employ diodes to “clip” away a portion of an input signal without distorting the remaining part of the applied waveform. t Vi(t) T Vm -Vm + V0(t) - + Vi(t) - R t Vo(t) T Vm -Vm If Vi(t) 0, Diode ON, Vo(t)= Vi(t) If Vi(t) <0, Diode OFF, Vo(t) =0 t Vi(t) T Vm -Vm + V0(t) - + Vi(t) - R t Vo(t) T -Vm If Vi(t) >0, Diode OFF, Vo(t)= 0 If Vi(t) 0, Diode ON, Vo(t) =Vi(t) Negative Series Clipper Positive Series Clipper
- 14. Diode Application: Clippers 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 14 t Vi(t) T Vm -Vm + V0(t) - + Vi(t) - R t Vo(t) T Vm-VVV V -V t Vi(t) T Vm -Vm + V0(t) - + Vi(t) - R t Vo(t) T -(Vm+V) If Vi(t) -V0, Diode ON, Vo(t)= Vi(t) -V If Vi(t) -V<0, Diode OFF, Vo(t) =0 If Vi(t) -V>0, Diode OFF, Vo(t)= 0 If Vi(t) -V 0, Diode ON, Vo(t) =Vi(t) -V t Vi(t) T Vm -Vm + V0(t) - + Vi(t) - R t Vo(t) Vm+V V V V -V t Vi(t) T Vm -Vm + V0(t) - + Vi(t) - R t Vo(t) T -(Vm-V) If Vi(t) +V0, Diode ON, Vo(t)= Vi(t) +V If Vi(t) +V<0, Diode OFF, Vo(t) =0 If Vi(t)+V>0, Diode OFF, Vo(t)= 0 If Vi(t) +V 0, Diode ON, Vo(t) =Vi(t) +V T Negative biased Series Clipper Positive biased Series Clipper
- 15. Diode Application: Clippers 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 15 + V0(t) - + Vi(t) - R t Vi(t) T Vm -Vm t Vo(t) T Vm If Vi(t) >0, Diode OFF, Vo(t)= Vi(t) If Vi(t) 0, Diode ON, Vo(t) =0 + V0(t) - + Vi(t) - R t Vi(t) T Vm -Vm t Vo(t) T -Vm If Vi(t) 0, Diode ON, Vo(t)= 0 If Vi(t) <0, Diode OFF, Vo(t) =Vi(t) + V0(t) - + Vi(t) - R t Vi(t) T Vm -Vm t Vo(t) T -Vm If Vi(t)-V 0, Diode ON, Vo(t)= V If Vi(t)-V <0, Diode OFF, Vo(t) = Vi(t) V V V + V0(t) - + Vi(t) - R t Vi(t) T Vm -Vm t Vo(t) T Vm If Vi(t)-V >0, Diode OFF, Vo(t)= Vi(t) If Vi(t)-V 0, Diode ON, Vo(t) = V V V V Negative parallel Clipper Positive parallel Clipper Negative biased parallel Clipper Positive biased parallel Clipper
- 16. Diode Application: Clippers 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 16 + V0(t) - + Vi(t) - R t Vi(t) T Vm -Vm t Vo(t) T -Vm If Vi(t)+V 0, Diode ON, Vo(t)= -V If Vi(t)+V <0, Diode OFF, Vo(t) = Vi(t) VV -V + V0(t) - + Vi(t) - R t Vi(t) T Vm -Vm t Vo(t) T Vm If Vi(t)+V >0, Diode OFF, Vo(t)= Vi(t) If Vi(t)+V 0, Diode ON, Vo(t) = -V V-V Negative biased parallel Clipper Positive biased parallel Clipper -V + V0(t) - + Vi(t) - R t Vi(t) T Vm -Vm If Vi(t)- V1 0, Diode D1 ON, & Vi(t)+ V2 >0, Diode D2 OFF, Vo(t)= V1 If Vi(t)- V1 <0, Diode D1 OFF, & Vi(t)+ V2 >0, Diode D2 OFF, Vo(t)= Vi(t) If Vi(t)- V1 <0, Diode D1 OFF, & Vi(t)+ V2 0, Diode D2 ON, Vo(t)= -V2 V1 V1 t Vo(t) T V1 V2 -V2 -V2 D1 D2
- 17. Diode Application: Clampers Clamping networks have capacitor connected in series with a diode and resistance in parallel. Steps to solve clamper circuit 1. Start analysis by examining the response of the portion of the input signal that will forward bias the diode 2. determine maximum voltage with polarity on capacitor (assumed that capacitor is charged instantaneously) 3. Assume that during the period when the diode is in OFF state the capacitor holds on to its established voltage level 4. Now determine the output 5. Check that total swing of the output matches to the input 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 17
- 18. Diode Application: Clampers 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 18 Step 1. In +ve half cycle, Diode is ON and maximum voltage on charged capacitor voltage Vm with polarities shown V1 + V0(t) - + Vi(t) - R C + V0(t) - + Vi(t) - R C + Vm - t Vi(t) T Vm -Vm t Vi(t) T Vm -Vm
- 19. Diode Application: Clampers 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 19 2. Assuming capacitor has charged upto to Vm and holds in when diode is in OFF state t Vi(t) T Vm -Vm t Vo(t) T Vm -Vm Now as Vi(t)-Vm0, diode is ON and Vo(t)=0 + V0(t) - + Vi(t) - R C + Vm - + V0(t) - + Vi(t) - R C + Vm - t Vi(t) T Vm -Vm t Vo(t) T -2Vm Now as Vi(t)-Vm<0, diode is OFF and Vo(t)= Vi(t)-Vm =-2Vm Positive part of cycle (0 t< T/2) Vi(t) =Vm so Diode is ON So V0(t) = 0 Negative part of cycle (T/2 t< 0) Vi(t) =-Vm so Diode is OFF So V0(t) = Vi(t)-Vm= -2Vm
- 20. Diode Application: Clampers 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 20 The output is drawn t Vi(t) T Vm -Vm t Vo(t) T -2Vm -2Vm -2Vm
- 21. Diode Application: Clampers 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 21 t Vi(t) T Vm -Vm t Vo(t) T -2Vm + V0(t) - + Vi(t) - R C Vm + - Vm + V0(t) - + Vi(t) - R C t Vi(t) T Vm -Vm t Vo(t) T 2Vm - m + t Vi(t) T Vm -Vm Vm+V + V0(t) - + Vi(t) - R C V - + t Vo(t) T 2Vm+V V t Vi(t) T Vm -Vm V + V0(t) - + Vi(t) - R C (Vm+V) + - t Vo(t) T -V -2Vm-V t Vi(t) T Vm -Vm Vm-V + V0(t) - + Vi(t) - R C V - + t Vo(t) T 2Vm-V -V t Vi(t) T Vm -Vm V + V0(t) - + Vi(t) - R C (Vm-V) + - t Vo(t) T V -2Vm+V
- 22. Diode Application: Zener diode as shunt regulator 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 22 The use of the Zener diode as a regulator is very common. Steps to analyse Zener diode networks 1. Determine the state of the Zener diode by removing it from the network and calculating the voltage across the resulting open circuit 2. Substitute the appropriate equivalent circuit and solve for the desired unknowns + VD - Forward bias VD 0 + VK - + VD - + VD - - VZ + Reverse bias VZ <VD < 0 Breakdown VD = VZ
- 23. 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 23 + V0 - R + VZ - IZ RL Vi + V0 - R + V - RL Vi L iL RR VR V If V VZ Then Zener is ON (breakdown region) If V < VZ Then Zener is OFF If diode is ON, then circuit shall be + V0 - R RL Vi VZ IR IZ IL ZZZ L Z i L ZZi Z LRZ ZO IVP RR V R V R V R VV I III VVV 11 Diode Application: Zener diode as shunt regulator
- 24. 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 24 L iL RR VR V If V VZ Then Zener is ON (breakdown region) If V < VZ Then Zener is OFF If diode is ON, then circuit shall be sheetdataperascurrentdiodemaxiswhere soconstant,isas breakdowninONbetodiodeforresistanceLoadMin thus min minmin max min min ZMZMRL R L Z L L L L Zi Z L L iL ZO IIII I R V R V I R VV RV R RR VR VVV + V0 - R + VZ - IZ RL Vi + V0 - R + V - RL Vi + V0 - R RL Vi VZ IR IZ IL Diode Application: Zener diode as shunt regulator
- 25. Diode Application: Voltage-Multiplier Circuits 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 25 • Voltage multipliers use clamping action to increase peak rectified voltages without increasing the transformer’s voltage rating. • Multiplication factors of two, three, and four are common. • Voltage multipliers are used in high-voltage, low-current • applications such as cathode-ray tubes (CRTs) and particle accelerators
- 26. Diode Application: Voltage-Multiplier Circuits 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 26 Half wave Voltage doubler t Vi(t) T Vm -Vm D1 C1 + Vi(t) - - 2Vm + D2 C2 t Vi(t) T Vm -Vm C1 + Vi(t) - + Vo(t) - C2 + - Vm+ Vm - - Vm + + - C1 + Vi(t) - - 2Vm + C2 Vm Half wave Voltage doubler operation: Positive half cycle Half wave Voltage doubler operation: Negative half cycle t Vi(t) T Vm -Vm
- 27. Diode Application: Voltage-Multiplier Circuits 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 27 Full wave Voltage doubler Full wave Voltage doubler operation: Positive half cycle Full wave Voltage doubler operation: Negative half cycle t Vi(t) T Vm -Vm C1 D1 + 2Vm - C2 + Vi(t) - D2 t Vi(t) T Vm -Vm C1 D1 + - C2 + Vi(t) - D2 Vm + - t Vi(t) T Vm -Vm C1 D1 + - C2 + Vi(t) - D2 Vm - + + - Vm + - Vm + - Vm + - Vm
- 28. Diode Application: Voltage-Multiplier Circuits 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 28 Voltage Tripler and Quadrupler t Vi(t) T Vm -Vm Vm + Vi(t) - D1 C2 D2 C3 D3 C4 D4 C1 + - + - 2Vm + - 2Vm + - 2Vm doubler Quadrupler Tripler
- 29. Thanks 9/11/2017 REC 101 Unit I by Dr Naim R Kidwai, Professor & Dean, JIT Jahangirabad 29 Contact Dr Naim R Kidwai Professor & Dean JETGI, Faculty of Engineering Barabanki, (UP) Email: naim.kidwai@gmail.com Naim.Kidwai@jit.edu.in