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by. A Lecture On Voltage Regulators. S K Rai Electronics Engineering Department BKBIET, CEERI Road Pilani (Raj.)-333 031.
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by A LectureOnVoltage Regulators 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? BKBIET, Pilani
Step-down transformer Ripple Filter Voltage regulator Load using dc voltage Rectifier High ac voltage Low ac voltage High ripple dc voltage Low ripple dc voltage Pure constant dc voltage 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. BKBIET, Pilani
VO DV= - m2t2 DV= m1t1 DV is higher for high Idc Ripple and dc voltage BKBIET, Pilani
VO DV= - m2t2 DV= m1t1 Ripple Factor Example: Given the Power supply circuit shown, find (a) output dc voltage Vdc (b) output ripple voltage Vr (rms) (c) ripple factor “r” BKBIET, Pilani
VC1 VC1 Idc t t R C2 RL RL VC2 VC2 t t V1(rms) V2(rms) R C2 RL RC Filter in Power Supply Vdc1 Vdc2 BKBIET, Pilani
Vi R RL VZ Series Voltage Regulators Series voltage regulator Vo Regulated power supply Unregulated power supply BKBIET, Pilani
b=50 Vi Vo=? 20V R RL 0.22kW 1kW VZ IZ=? 12V - + 0.7V + - + - Example: Given the Series regulator circuit shown, find (a) output dc voltage Vo (b) Zener current IZ BKBIET, Pilani
Vi Q1 Vo RL RL R4 R1 VZ - + IB1 + IR4 R3 0.7V - R2 IC2 + + + Q2 IR1 - - - + VCE1 - IL IL Improved Series Regulator Analysis of the Improved Series Regulator VB1 BKBIET, Pilani
+ - Vi Vo R3 R1 Ii=0 VZ RL R2 VZ VZ Ii=0 Because Ii=0, a small power Zener may be used to regulate the whole load current in RL IC Series Regulator IC Series Regulator BKBIET, Pilani
+ - Q1 + + + - IB1 - Ii=0 - Q2 VZ + + VZ - - - Ii=0 RSC High IL IL 1. Vi and R3 will produce regulated voltage VZ Vi 2. Because of the Op. Amp. Property, VZ will present at the junction of R1 and R2 Vo IL R3 R1 3. VZ=VR2 will result Vo which will be higher than VZ IC2 RL 5. High value of IL will produce high VBE at Q2 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 VZ R2 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. IO Current-limiting IC Series Regulator Analysis of the Circuit BKBIET, Pilani
Regulated power supply Unregulated power supply Rs - + 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) Vi Vo + + VZ RL if load RL is connected IRs will mostly drain by the load RL making IC small (when there is load BJT is cold) - - + VBE - Shunt Voltage Regulators Shunt voltage regulator Vo = VZ+VBE = Regulated voltage Therefore Shunt Regulators are suitable to supply circuits where RL is always connected. BKBIET, Pilani
+ Vi - Vo Rs R3 R1 VZ Ii=0 VZ R2 VZ Regulated power supply Unregulated power supply Ii=0 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) IC Shunt voltage regulator Because Ii=0, a small power Zener may be used to regulate the whole load current in RL + if load RL is connected IRs will mostly drain by the load RL making IC small (when there is load BJT is cold) RL - IC Shunt Voltage Regulator Therefore Shunt Regulators are suitable to supply circuits where RL is always connected. BKBIET, Pilani
Voltage regulator IC 1 2 7812 C2 RL +Vi Vo=+12V Regulated power supply +12V 3 Unregulated power supply Voltage Regulator IC Fixed Positive Voltage Regulator IC C1 is a ripple filter capacitor(C1 > 100mF) C2 is a high frequency filter (C2 < 0.1mF) BKBIET, Pilani
Voltage regulator IC 1 2 7805 +Vi=15Vmax Vo=+5V IL = ? 3 15V 7.3V 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. C1 = 250mF Unregulated power supply BKBIET, Pilani
Voltage regulator IC Regulated power supply -12V 1 2 7912 C2 RL -Vi Vo=-12V 3 Unregulated power supply Fixed Negative Voltage Regulator IC C1 is a ripple filter capacitor(C1 > 100mF) C2 is a high frequency filter (C2 < 0.1mF) BKBIET, Pilani
-Vi 1.2V < Vo< 37V VIN VOUT LM317 R1 ADJ Iadj = 100mA R2 Vref = 1.25V 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 = 100mA Adjustable Voltage regulator IC BKBIET, Pilani
IL 1.25V VIN VOUT LM317 100mA C3 RL C2 R1 ADJ 2. Note that LM317 data are Vref = 1.25V and Iadj = 100mA and will produce a regulated dc voltage VOUT determined by R1 and R2 R2 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. 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 DV=Vmax-Vmin depending upon the value of C1 and IL used. BKBIET, Pilani
IL 1.25V VIN VOUT LM317 100mA C3 RL R1 C2 ADJ R2 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 R1 = 0.1kW and R2= 1kW ? Take ac supply frequency 50Hz. BKBIET, Pilani
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