1. by S K Rai Electronics Engineering Department BKBIET, CEERI Road Pilani (Raj.)-333 031

2. EE2603-01 Electronic Circuit Analysis 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? BKBIET, Pilani

3. EE2603-01 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 Rectifier Low ac voltage Ripple Filter High ripple dc voltage Voltage regulator Low ripple dc voltage Load using dc voltage Pure constant dc voltage BKBIET, Pilani

4. EE2603-01 Electronic Circuit Analysis  V= - m 2 t 2  V= m 1 t 1 VOVO Ripple and dc voltage  V is higher for high I dc BKBIET, Pilani

5. EE2603-01 Electronic Circuit Analysis Ripple Factor  V= - m 2 t 2  V= m 1 t 1 VOVOExample: Given the Power supply circuit shown, find (a) output dc voltage V dc (b) output ripple voltage V r (rms) (c) ripple factor “r” BKBIET, Pilani

6. EE2603-01 Electronic Circuit Analysis RC Filter in Power Supply RLRLRLRL t V C1 R C2C2C2C2 RLRLRLRL I dc t V C2 V dc1 V dc2 R C2C2C2C2 RLRLRLRL t V C1 V 1 (rms) t V C2 V 2 (rms) BKBIET, Pilani

7. EE2603-01 Electronic Circuit Analysis Series Voltage Regulators RLRLRLRLR VZVZVZVZ ViViViVi VoVoVoVo Unregulated power supply Series voltage regulator Regulated power supply BKBIET, Pilani

8. EE2603-01 Electronic Circuit Analysis RLRLRLRLR VZVZVZVZ ViViViVi 1k  0.22k  12V  50 20V V o =? I Z =? Example: Given the Series regulator circuit shown, find (a) output dc voltage V o (b) Zener current I Z 0.7V + - + - + - BKBIET, Pilani

9. EE2603-01 Electronic Circuit Analysis Improved Series Regulator 0.7V+- +- VZVZVZVZ V B1 I R1 V CE1 +- ILILILIL ILILILIL Q2Q2Q2Q2 IC2IC2IC2IC2 I B1 I R4 R1R1R1R1 R2R2R2R2 RLRLRLRL VoVoVoVo R3R3R3R3 Q1Q1Q1Q1 R4R4R4R4 ViViViVi RLRLRLRL +- + - + - Analysis of the Improved Series Regulator BKBIET, Pilani

10. EE2603-01 Electronic Circuit Analysis IC Series Regulator +  R1R1R1R1 ViViViVi R2R2R2R2 R3R3R3R3 RLRLRLRL VZVZVZVZ VoVoVoVo VZVZVZVZ VZVZVZVZ I i =0 Because I i =0, a small power Zener may be used to regulate the whole load current in R L IC Series Regulator BKBIET, Pilani

11. EE2603-01 Electronic Circuit Analysis +  +  + +   +  +  R1R1R1R1 R2R2R2R2 RLRLRLRL VoVoVoVo ViViViVi R3R3R3R3 VZVZVZVZ VZVZVZVZ VZVZVZVZ I i =0 ILILILIL ILILILIL R SC Current-limiting IC Series Regulator IOIOIOIO I B1 I C2 High I L Analysis of the Circuit 1. V i and R 3 will produce regulated voltage V Z 2. Because of the Op. Amp. Property, V Z will present at the junction of R 1 and R 2 3. V Z =V R2 will result V o which will be higher than V Z 4. Normal value of I L will produce normal V BE at Q 2 whose collector current is small so that I o from IC will go mostly to I B1 6. Q 2 collector current I C2 becomes large so that I o from IC will go mostly to I C2 reducing I B1 and consequently I L down to the limited current value. Q2Q2Q2Q2 Q1Q1Q1Q1 5. High value of I L will produce high V BE at Q 2 BKBIET, Pilani

12. EE2603-01 Electronic Circuit Analysis Shunt Voltage Regulators Shunt voltage regulator Unregulated power supply Regulated power supply RsRsRsRs V BE ViViViVi VZVZVZVZ VoVoVoVo+ +  +  RLRLRLRL +  V o = V Z +V BE = Regulated voltage V Rs = V i – V Z – V BE which is constant if V i is constant I Rs = (V Rs /R s ) is also constant I C = I Rs will all flow through BJT (when no load R L and BJT is hot) if load R L is connected I Rs will mostly drain by the load R L making I C small (when there is load BJT is cold) Therefore Shunt Regulators are suitable to supply circuits where R L is always connected. BKBIET, Pilani

13. EE2603-01 Electronic Circuit Analysis Because I i =0, a small power Zener may be used to regulate the whole load current in R L +  IC Shunt voltage regulator Unregulated power supply Regulated power supply RLRLRLRL +  V Rs = V i – V o which is constant if V i is constant I Rs = (V Rs /R s ) is also constant I C = I Rs will all flow through BJT (when no load R L and BJT is hot) IC Shunt Voltage Regulator if load R L is connected I Rs will mostly drain by the load R L making I C small (when there is load BJT is cold) Therefore Shunt Regulators are suitable to supply circuits where R L is always connected. R1R1R1R1 ViViViVi R2R2R2R2 R3R3R3R3 VZVZVZVZ VoVoVoVo VZVZVZVZ VZVZVZVZ I i =0 RsRsRsRs BKBIET, Pilani

14. EE2603-01 Electronic Circuit Analysis Voltage Regulator IC 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+12V14.6V 7815+15V+17.7V 7818+18V+21V 7824+24V+27.1V Fixed Positive Voltage Regulator IC C2C2C2C2 RLRLRLRL 1 Voltage regulator IC 2 3 7812 V o =+12V +V i Unregulated power supply Regulated power supply +12V C 1 is a ripple filter capacitor (C 1 > 100  F) C 2 is a high frequency filter (C 2 < 0.1  F) BKBIET, Pilani

15. EE2603-01 Electronic Circuit Analysis Unregulated power supply 15V 7.3V 1 Voltage regulator IC 2 3 7805 V o =+5V +V i =15Vmax I L = ? Example: Given the IC regulator circuit shown, 7805 requires a minimum input voltage of 7.3V. (a) find the maximum load current I L that can be used. (b) if a maximum load current I L = 500mA is required, what is the size of the filter capacitor? Take ac supply frequency 50Hz. C 1 = 250  F BKBIET, Pilani

16. EE2603-01 Electronic Circuit Analysis Fixed Negative Voltage Regulator IC C2C2C2C2 RLRLRLRL 1 Voltage regulator IC 2 3 7912 V o =-12V -V i Unregulated power supply Regulated power supply -12V 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 C 1 is a ripple filter capacitor (C 1 > 100  F) C 2 is a high frequency filter (C 2 < 0.1  F) BKBIET, Pilani

17. EE2603-01 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: V ref = 1.25V and I adj = 100  A Adjustable Voltage regulator IC 1.2V < V o < 37V -V i V IN V OUT ADJ LM317 R1R1R1R1 R2R2R2R2 I adj = 100  A V ref = 1.25V BKBIET, Pilani

18. EE2603-01 Electronic Circuit Analysis V IN V OUT ADJ LM317 R1R1R1R1 R2R2R2R2 C2C2C2C2 ILILILIL RLRLRLRL C3C3C3C3 100  A 1.25V Analysis of the LM317 IC Regulator circuit 1. Unregulated power supply enters LM317 with a dc voltage of V IN and a ripple peak-to-peak voltage of  V=V max -V min depending upon the value of C 1 and I L used. 5. C 3 is unwanted high-frequency low capacitance bypass capacitor which is always necessary at the output of all regulator circuits. 3. Unwanted voltage spikes are suppressed by the two diodes connected from V OUT to ADJ and V IN terminals of the IC. 4. C 2 is a ripple filter across R 2. 2. Note that LM317 data are V ref = 1.25V and I adj = 100  A and will produce a regulated dc voltage V OUT determined by R 1 and R 2 BKBIET, Pilani

19. EE2603-01 Electronic Circuit Analysis V IN V OUT ADJ LM317 R1R1R1R1 R2R2R2R2 C2C2C2C2 ILILILIL RLRLRLRL C3C3C3C3 100  A 1.25VExample: Given the LM317 IC regulator circuit shown, (a) find the maximum load current I L 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 R 2 = 1k  ? Take ac supply frequency 50Hz. BKBIET, Pilani

20. EE2603-01 Electronic Circuit Analysis Thanks BKBIET, Pilani