7.4 Power Supply Design & Voltage Regulators

PRC Electronics Engineering Licensure Examination Study Guide Exam

Track your progress as you read

Finished sections get a check mark.

1 Chapter 1: Introduction & Board Exam Strategy

1.1 PRC ECE Licensure Exam Overview & Regulation 1.2 Table of Specifications & Exam Strategy

2 Chapter 2: Mathematics Foundations

2.1 College & Advanced Algebra 2.2 Trigonometry & Solid Mensuration 2.3 Analytic Geometry & Differential Calculus 2.4 Integral Calculus & Applications

3 Chapter 3: Advanced Mathematics for ECE

3.1 Differential Equations & Circuit Transients 3.2 Complex Numbers & Laplace Transforms 3.3 Fourier Series, Fourier Transforms & Z-Transforms 3.4 Vector Analysis, Matrices & Probability

4 Chapter 4: GEAS - Engineering Sciences & Economics

4.1 Engineering Physics & Materials Chemistry 4.2 Engineering Mechanics & Strength of Materials 4.3 Engineering Economics & Financial Analysis 4.4 Computer Fundamentals & C/C++ Programming

5 Chapter 5: GEAS - Management, Materials, Environment & CAD

5.1 Engineering Management & Technopreneurship 101 5.2 Materials Science & Engineering 5.3 Environment Science & Engineering 5.4 Computer-Aided Drawing (CAD) & Engineering Graphics

6 Chapter 6: GEAS - ECE Laws, Ethics & Contracts

6.1 Republic Act No. 9292 (Electronics Engineering Law of 2004) 6.2 ECE Code of Ethics & Professional Conduct 6.3 Contracts, Specifications & National Building Code (PD 1096)

7 Chapter 7: Electronics Engineering - Circuits & Semiconductor Diodes

7.1 DC & AC Circuit Analysis & Network Theorems 7.2 Electromagnetics & Passive Components 7.3 Semiconductor Physics & Diode Circuits 7.4 Power Supply Design & Voltage Regulators

8 Chapter 8: Electronics Engineering - Transistors & Analog Circuits

8.1 BJT Operation, Biasing & Small-Signal Amplifiers 8.2 FET Characteristics, Biasing & Amplifier Circuits 8.3 Multistage Amplifiers, Frequency Response & Power Amplifiers 8.4 Operational Amplifiers & Active Filter Design

9 Chapter 9: Electronics Engineering - Digital Electronics & Control Systems

9.1 Feedback Topologies & Oscillator Circuits 9.2 Boolean Algebra, Logic Gates & Combinational Circuits 9.3 Sequential Logic, Flip-Flops & Microprocessor Architecture 9.4 Industrial Power Electronics & Control System Theory

10 Chapter 10: EST - Communications Fundamentals & Transmission Lines

10.1 Signal Fundamentals, Decibel Calculations & Noise Analysis 10.2 Amplitude Modulation (AM) Principles & Receivers 10.3 Frequency Modulation (FM), Phase Modulation & Discriminators 10.4 Transmission Line Parameters, SWR & Smith Chart

11 Chapter 11: EST - Digital Communications, Wireless & Optical Systems

11.1 Pulse & Digital Modulation Systems 11.2 Data Communications, OSI Model & Protocols 11.3 Antenna Principles & Electromagnetic Wave Propagation 11.4 Optical Fiber, Microwave, Satellite & Philippine NTC Regulations

7.4 Power Supply Design & Voltage Regulators

Key Takeaways Explain This Topic AI

Power supply filter circuits (C-filter, LC-choke input, π \pi π -filter) attenuate AC ripple; simple C-filters exhibit ripple voltage V r ( p − p ) = I D C f r i p p l e C V_{r(p-p)} = \frac{I_{DC}}{f_{ripple} C} V r ( p − p ) ​ = f r i ppl e ​ C I D C ​ ​ .

Zener diode shunt regulators maintain constant output voltage V_Z under line and load variations provided Zener current remains within bounds I Z ( m i n ) ≤ I Z ≤ I Z ( m a x ) I_{Z(min)} \le I_Z \le I_{Z(max)} I Z ( min ) ​ ≤ I Z ​ ≤ I Z ( ma x ) ​ .

Fixed linear IC regulators (78xx positive, 79xx negative) provide fixed DC outputs (e.g., 7805 for +5V, 7912 for -12V) with built-in thermal overload and short-circuit protection.

Adjustable 3-terminal regulators like the LM317 regulate output via reference voltage V r e f = 1.25 V_{ref} = 1.25 V re f ​ = 1.25 V using formula V o u t = 1.25 ( 1 + R 2 R 1 ) + I a d j R 2 V_{out} = 1.25 \left(1 + \frac{R_2}{R_1}\right) + I_{adj} R_2 V o u t ​ = 1.25 ( 1 + R 1 ​ R 2 ​ ​ ) + I a d j ​ R 2 ​ .

Switched-Mode Power Supplies (SMPS) utilize high-frequency PWM switching (Buck, Boost, Buck-Boost topologies) to achieve high conversion efficiency (80-95%) compared to linear regulators (30-60%).

Last updated: July 2026

7.4 Power Supply Design & Voltage Regulators

Quick Answer: Power supply filters smooth pulsating DC from rectifiers into steady DC. Capacitor filter ripple is V r ( p − p ) = I D C f r i p p l e C V_{r(p-p)} = \frac{I_{DC}}{f_{ripple} C} V r ( p − p ) ​ = f r i ppl e ​ C I D C ​ ​ . Zener regulators hold voltage V Z V_Z V Z ​ constant across load changes. Fixed 3-terminal regulators (78xx positive, 79xx negative) deliver fixed voltages. Adjustable regulators like LM317 produce V o u t = 1.25 V ( 1 + R 2 R 1 ) + I a d j R 2 V_{out} = 1.25\text{V} \left(1 + \frac{R_2}{R_1}\right) + I_{adj} R_2 V o u t ​ = 1.25 V ( 1 + R 1 ​ R 2 ​ ​ ) + I a d j ​ R 2 ​ . Switched-Mode Power Supplies (SMPS)—Buck (step-down: V o u t = D V i n V_{out} = D V_{in} V o u t ​ = D V in ​ ), Boost (step-up: V o u t = V i n 1 − D V_{out} = \frac{V_{in}}{1-D} V o u t ​ = 1 − D V in ​ ​ ), and Buck-Boost ( V o u t = − D 1 − D V i n V_{out} = -\frac{D}{1-D} V_{in} V o u t ​ = − 1 − D D ​ V in ​ )—achieve 85 − 95 % 85-95\% 85 − 95% efficiency compared to 40 − 60 % 40-60\% 40 − 60% for linear regulators.

Power Supply Filtering & Smoothing Circuits

The output of a rectifier is pulsating DC containing a desired DC component V d c V_{dc} V d c ​ and an undesirable AC ripple voltage component V r V_r V r ​ .

1. Filter Topologies & Ripple Expressions

Capacitor Input Filter (C-Filter): A large electrolytic capacitor placed in parallel across the load charges to peak voltage V m V_m V m ​ during diode conduction and discharges into R L R_L R L ​ during non-conduction.

Peak-to-Peak Ripple Voltage: V r ( p − p ) = I D C f r i p p l e C = V D C f r i p p l e R L C V_{r(p-p)} = \frac{I_{DC}}{f_{ripple} C} = \frac{V_{DC}}{f_{ripple} R_L C} V r ( p − p ) ​ = f r i ppl e ​ C I D C ​ ​ = f r i ppl e ​ R L ​ C V D C ​ ​

DC Output Voltage under Load: V D C = V m − V r ( p − p ) 2 = V m − I D C 2 f r i p p l e C V_{DC} = V_m - \frac{V_{r(p-p)}}{2} = V_m - \frac{I_{DC}}{2 f_{ripple} C} V D C ​ = V m ​ − 2 V r ( p − p ) ​ ​ = V m ​ − 2 f r i ppl e ​ C I D C ​ ​

RMS Ripple Voltage & Ripple Factor: V r ( r m s ) = V r ( p − p ) 2 3 ⟹ γ = V r ( r m s ) V D C = 1 2 3 f r i p p l e R L C V_{r(rms)} = \frac{V_{r(p-p)}}{2\sqrt{3}} \implies \gamma = \frac{V_{r(rms)}}{V_{DC}} = \frac{1}{2\sqrt{3} f_{ripple} R_L C} V r ( r m s ) ​ = 2 3 ​ V r ( p − p ) ​ ​ ⟹ γ = V D C ​ V r ( r m s ) ​ ​ = 2 3 ​ f r i ppl e ​ R L ​ C 1 ​

LC Filter (Choke Input Filter): Series inductor L L L blocks high-frequency AC ripple currents, while shunt capacitor C C C bypasses remaining AC ripple to ground. Ripple factor is independent of load current: γ = 2 3 ω 2 L C ( for full-wave rectifier at 2 ω ) \gamma = \frac{\sqrt{2}}{3 \omega^2 L C} \quad (\text{for full-wave rectifier at } 2\omega) γ = 3 ω 2 L C 2 ​ ​ ( for full-wave rectifier at 2 ω )

π \pi π -Filter (CRC / CLC Filter): Combines an input capacitor, series choke or resistor, and output capacitor, yielding extremely low ripple factor at the expense of higher component cost and size.

Linear Voltage Regulators: Zener & Integrated Circuits

Voltage regulation ensures the output DC voltage remains stable despite changes in AC line voltage (Line Regulation) or load current (Load Regulation).

Line Regulation (%): Δ V o u t Δ V i n × 100 % \frac{\Delta V_{out}}{\Delta V_{in}} \times 100\% Δ V in ​ Δ V o u t ​ ​ × 100%

Load Regulation (%): V N L − V F L V F L × 100 % \frac{V_{NL} - V_{FL}}{V_{FL}} \times 100\% V F L ​ V N L ​ − V F L ​ ​ × 100%

1. Zener Diode Shunt Regulator

A Zener diode connected in reverse bias across the load regulates output voltage at V Z V_Z V Z ​ .

Series current-limiting resistor R S R_S R S ​ : R S = V i n − V Z I S = V i n − V Z I Z + I L R_S = \frac{V_{in} - V_Z}{I_S} = \frac{V_{in} - V_Z}{I_Z + I_L} R S ​ = I S ​ V in ​ − V Z ​ ​ = I Z ​ + I L ​ V in ​ − V Z ​ ​

Operational Bounds:

To maintain regulation at maximum load ( I L ( m a x ) I_{L(max)} I L ( ma x ) ​ ), Zener current must not fall below I Z ( m i n ) I_{Z(min)} I Z ( min ) ​ (typically 5 mA 5\text{ mA} 5 mA ): R S ( m a x ) = V i n ( m i n ) − V Z I Z ( m i n ) + I L ( m a x ) R_{S(max)} = \frac{V_{in(min)} - V_Z}{I_{Z(min)} + I_{L(max)}} R S ( ma x ) ​ = I Z ( min ) ​ + I L ( ma x ) ​ V in ( min ) ​ − V Z ​ ​

At zero load ( I L = 0 I_L = 0 I L ​ = 0 ), Zener current must not exceed maximum power dissipation I Z ( m a x ) = P Z ( m a x ) V Z I_{Z(max)} = \frac{P_{Z(max)}}{V_Z} I Z ( ma x ) ​ = V Z ​ P Z ( ma x ) ​ ​ : R S ( m i n ) = V i n ( m a x ) − V Z I Z ( m a x ) R_{S(min)} = \frac{V_{in(max)} - V_Z}{I_{Z(max)}} R S ( min ) ​ = I Z ( ma x ) ​ V in ( ma x ) ​ − V Z ​ ​

2. Fixed IC Regulators (78xx & 79xx Series)

Standard 3-terminal linear voltage regulators feature internal thermal shutdown, current limiting, and safe-area protection.

78xx Series: Positive fixed voltage regulators (e.g., 7805 = + 5 V +5\text{V} + 5 V , 7812 = + 12 V +12\text{V} + 12 V , 7824 = + 24 V +24\text{V} + 24 V ).

79xx Series: Negative fixed voltage regulators (e.g., 7905 = − 5 V -5\text{V} − 5 V , 7912 = − 12 V -12\text{V} − 12 V ).

Pinouts (TO-220 Package):

78xx: Pin 1 = Input, Pin 2 = Ground, Pin 3 = Output.

79xx: Pin 1 = Ground, Pin 2 = Input, Pin 3 = Output.

3. Adjustable IC Regulator (LM317)

The LM317 is a 3-terminal positive adjustable regulator maintaining a precise reference voltage V r e f = 1.25 V V_{ref} = 1.25\text{ V} V re f ​ = 1.25 V between its Output and Adjust pins.

Output Voltage Formula: V o u t = V r e f ( 1 + R 2 R 1 ) + I a d j R 2 = 1.25 V ( 1 + R 2 R 1 ) + I a d j R 2 V_{out} = V_{ref} \left(1 + \frac{R_2}{R_1}\right) + I_{adj} R_2 = 1.25\text{V} \left(1 + \frac{R_2}{R_1}\right) + I_{adj} R_2 V o u t ​ = V re f ​ ( 1 + R 1 ​ R 2 ​ ​ ) + I a d j ​ R 2 ​ = 1.25 V ( 1 + R 1 ​ R 2 ​ ​ ) + I a d j ​ R 2 ​

Since adjustment pin current I a d j I_{adj} I a d j ​ is small ( ≈ 50 μ A \approx 50\ \mu\text{A} ≈ 50 μ A ), it is often neglected: V o u t ≈ 1.25 V ( 1 + R 2 R 1 ) V_{out} \approx 1.25\text{V} \left(1 + \frac{R_2}{R_1}\right) V o u t ​ ≈ 1.25 V ( 1 + R 1 ​ R 2 ​ ​ )

Recommended setting: Choose R 1 ≈ 240 Ω R_1 \approx 240\ \Omega R 1 ​ ≈ 240 Ω to ensure minimum load current ( ≈ 5 mA \approx 5\text{ mA} ≈ 5 mA ).

Switched-Mode Power Supplies (SMPS) & DC-DC Converters

Switched-Mode Power Supplies switch a power transistor (MOSFET) fully ON and fully OFF at high frequencies ( 50 kHz − 1 MHz 50\text{ kHz} - 1\text{ MHz} 50 kHz − 1 MHz ), controlling output voltage by varying the Pulse-Width Modulation (PWM) duty cycle D = t o n T D = \frac{t_{on}}{T} D = T t o n ​ ​ .

Basic Non-Isolated Converter Topologies

Topology Conversion Function ( V o u t V_{out} V o u t ​ vs V i n V_{in} V in ​ ) Voltage Level Duty Cycle Range ( D D D ) Key Components & Operation

Buck Converter V o u t = D ⋅ V i n V_{out} = D \cdot V_{in} V o u t ​ = D ⋅ V in ​ Step-Down ( V o u t < V i n V_{out} < V_{in} V o u t ​ < V in ​ ) 0 < D < 1 0 < D < 1 0 < D < 1 Transistor in series with input; inductor-capacitor output filter smooths current.

Boost Converter V o u t = V i n 1 − D V_{out} = \frac{V_{in}}{1 - D} V o u t ​ = 1 − D V in ​ ​ Step-Up ( V o u t > V i n V_{out} > V_{in} V o u t ​ > V in ​ ) 0 < D < 1 0 < D < 1 0 < D < 1 Inductor stores energy during t o n t_{on} t o n ​ ; discharges into output capacitor during t o f f t_{off} t o ff ​ .

Buck-Boost Converter V o u t = − D 1 − D V i n V_{out} = -\frac{D}{1 - D} V_{in} V o u t ​ = − 1 − D D ​ V in ​ Inverting Step-Up/Step-Down 0 < D < 1 0 < D < 1 0 < D < 1 Produces negative polarity voltage relative to ground; step-down for D < 0.5 D < 0.5 D < 0.5 , step-up for D > 0.5 D > 0.5 D > 0.5 .

Linear vs. Switched-Mode Power Supply Comparison

Feature Linear Regulator Switched-Mode Power Supply (SMPS)

Efficiency Low ( 30 % − 60 % 30\% - 60\% 30% − 60% ), power lost as heat ( P l o s s = ( V i n − V o u t ) I L P_{loss} = (V_{in}-V_{out})I_L P l oss ​ = ( V in ​ − V o u t ​ ) I L ​ ) High ( 80 % − 95 % 80\% - 95\% 80% − 95% ), minimal power loss

Size & Weight Large & heavy due to low-frequency ( 60 Hz 60\text{Hz} 60 Hz ) transformers & heat sinks Small & lightweight due to high-frequency operation

Output Noise / Ripple Exceptionally low noise, high ripple rejection ( > 60 dB > 60\text{ dB} > 60 dB ) Higher switching noise & EMI requiring filtering

• Design Complexity Very simple (3-terminal ICs) Complex (PWM controller, magnetics, feedback loop)
• Step-by-Step Worked Examples
• Example 1: LM317 Output Voltage Calculation

Problem: An LM317 regulator circuit uses R 1 = 240 Ω R_1 = 240\ \Omega R 1 ​ = 240 Ω and R 2 = 2.4 k Ω R_2 = 2.4\text{ k}\Omega R 2 ​ = 2.4 k Ω ( 2400 Ω 2400\ \Omega 2400 Ω ). The adjust pin current is I a d j = 50 μ A I_{adj} = 50\ \mu\text{A} I a d j ​ = 50 μ A . Calculate the regulated output voltage V o u t V_{out} V o u t ​ .

Solution:

Identify Formula: V o u t = 1.25 ( 1 + R 2 R 1 ) + I a d j R 2 V_{out} = 1.25 \left(1 + \frac{R_2}{R_1}\right) + I_{adj} R_2 V o u t ​ = 1.25 ( 1 + R 1 ​ R 2 ​ ​ ) + I a d j ​ R 2 ​

Calculate Main Term: 1.25 ( 1 + 2400 240 ) = 1.25 ( 1 + 10 ) = 1.25 × 11 = 13.75 V 1.25 \left(1 + \frac{2400}{240}\right) = 1.25 (1 + 10) = 1.25 \times 11 = 13.75\text{ V} 1.25 ( 1 + 240 2400 ​ ) = 1.25 ( 1 + 10 ) = 1.25 × 11 = 13.75 V

Calculate I a d j I_{adj} I a d j ​ Offset Term: I a d j R 2 = ( 50 × 10 − 6 A ) × ( 2400 Ω ) = 0.12 V I_{adj} R_2 = (50 \times 10^{-6}\text{ A}) \times (2400\ \Omega) = 0.12\text{ V} I a d j ​ R 2 ​ = ( 50 × 1 0 − 6 A ) × ( 2400 Ω ) = 0.12 V

Total Output Voltage: V o u t = 13.75 V + 0.12 V = 13.87 V V_{out} = 13.75\text{ V} + 0.12\text{ V} = 13.87\text{ V} V o u t ​ = 13.75 V + 0.12 V = 13.87 V

Example 2: Buck Converter Duty Cycle & Output Voltage

Problem: A Buck DC-DC converter operates with an input voltage V i n = 24 V V_{in} = 24\text{ V} V in ​ = 24 V and a switching frequency of 100 kHz 100\text{ kHz} 100 kHz . If the switch is ON for t o n = 3 μ s t_{on} = 3\ \mu\text{s} t o n ​ = 3 μ s during each period, calculate: (a) Switching period T T T , (b) Duty cycle D D D , and (c) Regulated output voltage V o u t V_{out} V o u t ​ .

Solution:

Calculate Switching Period T T T : T = 1 f s = 1 100 × 10 3 Hz = 10 μ s T = \frac{1}{f_s} = \frac{1}{100 \times 10^3\text{ Hz}} = 10\ \mu\text{s} T = f s ​ 1 ​ = 100 × 1 0 3 Hz 1 ​ = 10 μ s

Calculate Duty Cycle D D D : D = t o n T = 3 μ s 10 μ s = 0.30 ( 30 % ) D = \frac{t_{on}}{T} = \frac{3\ \mu\text{s}}{10\ \mu\text{s}} = 0.30 \quad (30\%) D = T t o n ​ ​ = 10 μ s 3 μ s ​ = 0.30 ( 30% )

Calculate Output Voltage V o u t V_{out} V o u t ​ : V o u t = D × V i n = 0.30 × 24 V = 7.2 V V_{out} = D \times V_{in} = 0.30 \times 24\text{ V} = 7.2\text{ V} V o u t ​ = D × V in ​ = 0.30 × 24 V = 7.2 V