by S K Rai Electronics Engineering Department BKBIET, CEERI Road Pilani (Raj.)-333 031 Q.
Download ReportTranscript by S K Rai Electronics Engineering Department BKBIET, CEERI Road Pilani (Raj.)-333 031 Q.
by S K Rai Electronics Engineering Department BKBIET, CEERI Road Pilani (Raj.)-333 031 Q. Why DC Power Supply? Q. Source of DC……….. Q. What is regulation? It is the process to maintain the terminal voltage as constant even if input voltage varies or load current is varying. Q. When regulation is required? EE2603-01 BKBIET, Pilani Electronic Circuit Analysis DC Power Supply DC Power supply is a circuit that provides a steady dc voltage obtained by rectifying the ac voltage. The following is the block diagram of a power supply stage. Step-down transformer High ac voltage Low ac voltage High ripple dc voltage Vm Ripple Filter Rectifier Voltage regulator Low ripple dc voltage Pure constant dc voltage Vm Vm Load using dc voltage Vdc n:1+1 Vac fS C Ri VZ EE2603-01 BKBIET, Pilani Electronic Circuit Analysis IB 0.7 RL IL Vdc Ripple and dc voltage n:1+1 Vac=220V f=50Hz Vmax Idc Vmax C Vmax VO V= - m2t2 V= m1t1 RL Vmin Vdc V is higher for high Idc dc voltage Vmax V 2 ( from slope ) V dc voltage Vdc Vmax I dc 2 fC I dc t2 of ac voltage t1 T Ripple voltage ( rms ) ( from slope ) V 4 fC V where I dc dc , f frequency RL Vac 2 and Vmax n Vdc V V 2 3 ( triangular I dc 2 fC Ripple voltage Vr ( rms ) I dc 4 fC 3 Vdc where I and f frequency dc R L EE2603-01 BKBIET, Pilani Electronic Circuit Analysis wave ) of ac voltage Ripple Factor We have Vr ( rms ) Ripple factor r Idc Vr ( rms ) Vdc Vdc and I dc 4 fC 3 Vmax Vdc 4 fC 3 1 1 Vdc 1 4 fR L C Vdc 31 . 1 1 . 05 (c) r I dc 4 fC 220 2 Vdc 10 220 10 2 Vdc t2 T 10:1+1 4 fR L C 3 Idc Vmax Vac=220V f=50Hz C RL 100 1k Vdc F Vmax 4 fR L C 1 1 4 50 1k 100 31 . 1V ( b ) Ripple voltage Vr ( rms ) 1 t1 Vdc 29 . 6 V Vr ( rms ) Vmin 4 fR L C 3 Example: Given the Power supply circuit shown, find (a) output dc voltage Vdc (b) output ripple voltage Vr (rms) (c) ripple factor “r” ( a ) dc voltage Vdc Vmax Vdc V RL Idc VO V= - m2t2 V= m1t1 1 4 50 1k 100 F 29 . 6 1k 4 50 100 0 . 05 5 % EE2603-01 BKBIET, Pilani Electronic Circuit Analysis 3 I dc 4 fC 3 0 . 85 V Vdc R L 4 fC 3 RC Filter in Power Supply VC1 n:1+1 t Idc Idc Vmax Vac=220V f=50Hz VC2 C1 Vmax R Vdc1 t If C 2 is in F and X C 2 is in k CR2LVdc2 RL XC2 XC2 Vdc 1 can be found from Vdc 1 Vmax ripple r1 V1 ( rms ) VC1 Vdc 1 V1 ( rms ) can be found I dc 4 fC 1 VC2 C2 t from V1 ( rms ) I dc 4 fC 1 3 C2 1 .3 C2 k ( for f 60 Hz ) or 60 50 C 2 opens for dc Vdc 2 Vdc 1 V1(rms) R 1 .3 RL C2 k ( for f 50 Hz ) RL R RL V2(rms) ripple r2 t V2 ( rms ) V1 ( rms ) EE2603-01 BKBIET, Pilani Electronic Circuit Analysis 1 . 56 XC2 R V2 ( rms ) Vdc 2 Series Voltage Regulators n:1+1 Vac=220V f=50Hz Idc Vmax C1 Vmax Series voltage regulator Vi Vo Regulated power supply R RL VZ Unregulated power supply Vm Vdc Vdc EE2603-01 BKBIET, Pilani Electronic Circuit Analysis Example: Given the Series regulator circuit shown, find (a) output dc voltage Vo (b) Zener current IZ (3 ) I L IB 50 Vi 20V (2 ) IR 20 V 12 V 0 . 22 k 36 . 4 mA R 0.22k VZ 12V - 0.7V + + - + 11 . 3V 1k 11 . 3 mA 50 11 . 3 mA 0 . 226 mA Vo=? RL (1) Vo VZ 0 . 7 12 0 . 7 11 . 3V 1k - IZ=? ( 4 ) I Z I R IB 36 . 4 0 . 226 36 . 174 mA EE2603-01 BKBIET, Pilani Electronic Circuit Analysis Improved Series Regulator Analysis of the Improved Series Regulator + Vi VB 1 Vo 0 . 7 IR4 Vi VB 1 IR 4 I C 2 IB1 R4 IB1 IL 1 - VCE1 Q1 Q2 2 IR1 - + - & I C 2 2 IB 2 IB 2 I Z 0 .7 R3 R3 VZ R3 0.7V IL IB1 VB1 R4 IC IL 0 .7 R3 I Z 0 .1IR 1 + R1 + - + - R2 - but I B 2 0 . 1 I Z 0 .7 0 .9 I Z EE2603-01 BKBIET, Pilani Electronic Circuit Analysis IR 1 0 .1I L Vo RL + I Z 0 .1I Z 0 .9 I Z & VR 2 Vo R2 R1 Vo VR 2 1 R1 R 2 R2 VR 2 VZ 0 . 7 IC Series Regulator n:1+1 Vac=220V f=50Hz Idc Vmax Vmax IC Series Regulator Vi C1 R3 VZ Because Ii=0, a small power Zener may be used to regulate the whole load current in RL Ii=0 VZ R1 + VZ R2 Vo RL Ii=0 EE2603-01 BKBIET, Pilani Electronic Circuit Analysis VR 2 VZ Vo R2 R1 Vo VZ 1 R1 R 2 R2 Current-limiting IC Series Regulator Analysis of the Circuit Vi RSC Q1 R3 Ii=0 + VZ VZ IB1 + + IL Q IC2 2 IO High IL IL + R1 VZ + Vo + RL R2 Ii=0 1. Vi and R3 will produce regulated voltage VZ 2. Because of the Op. Amp. Property, VZ will present at the junction of R1 and R2 3. VZ=VR2 will result Vo which will be higher than VZ 4. Normal value of IL will produce normal VBE at Q2 whose collector current is small so that Io from IC will go mostly to IB1 5. High value of IL will produce high VBE at Q2 6. Q2 collector current IC2 becomes large so that Io from IC will go mostly to IC2 reducing IB1 and consequently IL down to the limited current value. EE2603-01 BKBIET, Pilani Electronic Circuit Analysis Shunt Voltage Regulators n:1+1 Vac=220V f=50Hz Idc Vmax C1 Vmax Vi + voltage regulator RShunt s Vo Regulated power supply + VZ + RL + Unregulated power supply Vo = VZ+VBE = Regulated voltage VBE Vdc Vm Vdc VRs = Vi – VZ – VBE which is constant if Vi is constant IRs = (VRs/Rs) is also constant IC = IRs will all flow through BJT (when no load RL and BJT is hot) if load RL is connected IRs will mostly drain by the load RL making IC small (when there is load BJT is cold) Therefore Shunt Regulators are suitable to supply circuits where RL is always connected. EE2603-01 BKBIET, Pilani Electronic Circuit Analysis IC Shunt Voltage Regulator n:1+1 Vac=220V f=50Hz Idc Vmax C1 Vmax Unregulated power supply Vm Rs R3 VZ Vdc IC Shunt voltage regulator Vi Ii=0 VZ + Vo + VZ Regulated power supply R1 RL R2 Vdc Ii=0 R R 2 1 Because Ii=0, a small power Zener VR 2 VZ Vo Vo VZ 1 R1 R 2 R2 may be used to regulate the whole load current in RL VRs = Vi – Vo which is constant if Vi is constant IRs = (VRs/Rs) is also constant IC = IRs will all flow through BJT (when no load RL and BJT is hot) if load RL is connected IRs will mostly drain by the load RL making IC small (when there is load BJT is cold) Therefore Shunt Regulators are suitable to supply circuits where RL is always connected. EE2603-01 BKBIET, Pilani Electronic Circuit Analysis Voltage Regulator IC Fixed Positive Voltage Regulator IC n:1+1 Vac=220V f=50Hz Idc Vmax +Vi C1 Vmax Voltage regulator IC 1 7812 C2 is a high frequency filter (C2 < 0.1F) Regulated power supply +12V 2 C2 RL 3 Unregulated power supply C1 is a ripple filter capacitor (C1 > 100F) Vo=+12V Vm Vdc Vdc IC Part Number Output Voltage Min input Voltage 7805 +5V +7.3V 7806 +6V +8.3V 7808 +8V +10.5V 7810 +10V +12.5V 7812 +12V 14.6V 7815 +15V +17.7V 7818 +18V +21V 7824 +24V +27.1V EE2603-01 BKBIET, Pilani Electronic Circuit Analysis Example: Given the IC regulator circuit shown, 7805 requires a minimum input voltage of 7.3V. (a) find the maximum load current IL that can be used. (b) if a maximum load current IL = 500mA is required, what is the size of the filter capacitor? Take ac supply frequency 50Hz. Voltage regulator IC n:1+1 Idc +V =15Vmax i Vmax Vac=220V f=50Hz Vmax 1 C1 C1 = 250F 2 7805 Vm 15V Vdc 7.3V V 15 V 7 . 3V 7 . 7 V Vr ( rms ) I dc 2 . 22 4 fC IL = ? 3 Unregulated power supply ( a ) 2 . 22 V Ripple Vo=+5V voltage Vr ( rms ) I dc 4 fC 2 3 7 .7 2 3 2 . 22 V ( b ) 2 . 22 V Ripple 3 3 2 . 22 4 50 100 V 3 0 . 077 A 77 mA EE2603-01 BKBIET, Pilani Electronic Circuit Analysis C I dc 2 . 22 4 f 3 voltage Vr ( rms ) I dc 4 fC 0 .5 A 2 . 22 4 50 3 3 650 F Fixed Negative Voltage Regulator IC n:1+1 Vac=220V f=50Hz Idc Vmax C1 Vmax -Vi Voltage regulator IC 1 7912 C2 is a high frequency filter (C2 < 0.1F) -Vm Regulated power supply -12V C2 RL 3 Unregulated power supply C1 is a ripple filter capacitor (C1 > 100F) 2 Vo=-12V -Vdc -Vdc IC Part Number Output Voltage Min input Voltage 7905 -5V -7.3V 7906 -6V -8.3V 7908 -8V -10.5V 7910 -10V -12.5V 7912 -12V -14.6V 7915 -15V -17.7V 7918 -18V -21V 7924 -24V -27.1V EE2603-01 BKBIET, Pilani Electronic Circuit Analysis Adjustable Voltage Regulator IC The IC regulator LM317 can be used to produce any regulated output voltage between 1.2V to 37V. Typical IC values are: Vref = 1.25V and Iadj = 100A Adjustable Voltage regulator IC n:1+1 Vac=220V f=50Hz Vmax -Vi Vmax C1 Idc VIN 1.2V < Vo< 37V LM317 VOUT R1 ADJ Iadj = 100A Vref = 1.25V R Vo 1 . 25 V 1 2 100 A R 2 R 1 R2 Vref Vo VR 1 VR 2 Vref I R 1 I adj R 2 Vref R 2 Vref R 1 EE2603-01 BKBIET, Pilani Electronic Circuit Analysis R 1 2 I adj R 2 R 1 Analysis of the LM317 IC Regulator circuit 1. Unregulated power supply enters LM317 with a dc voltage of VIN and a ripple peak-to-peak voltage of V=Vmax-Vmin depending upon the value of C1 and IL used. n:1+1 Vac=220V f=50Hz Vmax IL 1.25V Vmax C1 470 F IL VIN 2. Note that LM317 data are Vref = 1.25V and Iadj = 100A and will produce a regulated dc voltage VOUT determined by R1 and R2 LM317 VOUT R1 ADJ 100A C2 R2 C3 RL 3. Unwanted voltage spikes are suppressed by the two diodes connected from VOUT to ADJ and VIN terminals of the IC. 4. C2 is a ripple filter across R2 . 5. C3 is unwanted high-frequency low capacitance bypass capacitor which is always necessary at the output of all regulator circuits. EE2603-01 BKBIET, Pilani Electronic Circuit Analysis Example: Given the LM317 IC regulator circuit shown, (a) find the maximum load current IL that can be used, if a ripple of peak-to-peak voltage of 2V is present at the input of the regulator. (b) what is the output voltage of the variable IC regulator if R 1 = 0.1k and R2= 1k ? Take ac supply frequency 50Hz. n:1+1 Vac=220V f=50Hz Vmax VIN IL C1 470 F LM317 VOUT R1 ADJ 100A In case if V Vmax Vmin 2V Vr ( rms ) ( a ) 0 . 577 V Ripple voltage I dc 0 . 577 4 fC IL 1.25V Vmax Vr ( rms ) V 2 3 2 2 3 C2 R2 C3 RL 0 . 577 V I dc 4 fC 3 3 0 . 577 4 50 470 3 93 . 942 mA R 1k ( b ) Vo 1 . 25 V 1 2 100 A R 2 1 . 25 V 1 100 A 1k 13 . 75 0 . 1 13 . 85 V R 100 1 EE2603-01 BKBIET, Pilani Electronic Circuit Analysis EE2603-01 BKBIET, Pilani Electronic Circuit Analysis