Mastering Switched Mode Power Supplies: A Tutorial for EEET2387

Introduction to Switched Mode Power Supplies

Switched mode power supplies (SMPS) are essential in modern electronics, from smartphone chargers to data center power systems. In May 2026, as AI-powered devices and electric vehicles demand efficient power conversion, understanding SMPS design is more relevant than ever. The EEET2387 final assignment challenges you to analyze and design circuits like the buck-boost converter and flyback converter. This tutorial will guide you through core concepts without solving the assignment directly, helping you build a strong foundation.

Buck-Boost Converter Fundamentals

The buck-boost converter is a versatile DC-DC converter that can step up or step down voltage. In the assignment, you'll analyze a converter with an input voltage range of 7.2 V to 13.2 V, a load of 36 Ω, and a switching frequency of 71.460 kHz. Key parameters include inductor current ripple (ΔiL) and output voltage ripple (Δvo). The assignment states that ΔiL should not exceed 35% of the average inductor current (IL), and Δvo should not exceed 1.5% of the average output voltage (Vo).

Calculating Maximum Inductor Current

To find the maximum inductor current (IL,max), consider all operating conditions: minimum input voltage (Vd,min = 7.2 V) and maximum load (Rload = 36 Ω). For a buck-boost converter in continuous conduction mode (CCM), the average inductor current is given by IL = Vo^2 / (Vd * Rload). With Vo = 9 V, at Vd = 7.2 V, IL = 81 / (7.2 * 36) = 0.3125 A. The ripple current ΔiL = 0.35 * IL = 0.1094 A. Thus, IL,max = IL + ΔiL/2 = 0.3125 + 0.0547 = 0.3672 A. Similarly, calculate for maximum input voltage and ensure the worst-case IL,max is used.

Output Capacitance and Voltage Ripple

The output capacitance (Co) is designed to limit voltage ripple. Using the formula Δvo = (Io * D) / (fs * Co), where Io = Vo/Rload = 9/36 = 0.25 A, and D is the duty cycle. For a buck-boost, D = Vo/(Vo + Vd) = 9/(9+7.2) = 0.5556 at Vd=7.2 V. With Δvo = 0.015 * Vo = 0.135 V, solve for Co: Co = (Io * D) / (fs * Δvo) = (0.25 * 0.5556) / (71.460e3 * 0.135) ≈ 14.4 μF. The assignment requires you to plot inductor current and capacitor voltage over two switching periods, showing IL,max, IL,min, and ripple.

Flyback Converter Transformer Design

The flyback converter is popular in low-power AC-DC applications. In the assignment, you'll design a transformer for a 48 W output with 85% efficiency, using a TDK PC47EE50-Z core. The primary turns (Np) are 340, and the duty cycle is 0.5. You need to calculate secondary turns (Ns) to achieve the nominal output voltage (Vo = 48 V). The turns ratio n = Np/Ns = Vd/(Vo * D) * (1-D). With Vd = 127 Vrms AC rectified, approximate Vd ≈ 180 V DC. Then n = 180/(48*0.5)*0.5 = 3.75, so Ns = Np/n = 340/3.75 ≈ 91 turns.

Magnetising Inductance for Boundary Conduction

To achieve boundary conduction mode (BCM) at 40% load, calculate the required magnetising inductance Lm. The critical inductance Lm,crit = (Vd * D * Ts) / (2 * Io), where Io = Po/Vo = 48/48 = 1 A at full load. At 40% load, Io = 0.4 A. With D=0.5, Ts = 1/fs, assume fs = 100 kHz. Then Lm,crit = (180 * 0.5 * 10e-6) / (2 * 0.4) = 1.125 mH. The assignment specifies Lm = 1.5 mH, which ensures CCM at nominal load and BCM at 40% load.

Air Gap and Number of Turns

Given an air gap length ℓg = 0.456 mm, and required inductance Lm = 1.5 mH, calculate the number of primary turns Np. Using reluctance: R = ℓg/(μ0 * Ae) + ℓc/(μr * μ0 * Ae). For PC47EE50, Ae ≈ 226 mm², ℓc ≈ 100 mm, μr ≈ 2000. Then R ≈ (0.456e-3)/(4πe-7 * 226e-6) + (0.1)/(2000 * 4πe-7 * 226e-6) ≈ 1.606e6 + 176,000 = 1.782e6 A/Wb. Inductance L = N^2 / R, so N = sqrt(L * R) = sqrt(1.5e-3 * 1.782e6) ≈ 51.7 turns. The assignment likely expects Np around 52 turns; note that fringing flux may reduce effective reluctance, but we ignore it per the hint.

PWM Control with UC2842

The UC2842 is a popular current-mode PWM controller. In the assignment, you'll design the oscillator circuit to set the switching frequency (e.g., 95.280 kHz). The oscillator uses Rt and Ct: f = 1.72 / (Rt * Ct). Choose Ct = 10 nF, then Rt = 1.72 / (95.28e3 * 10e-9) ≈ 1.805 kΩ. Use standard resistor values.

Startup Circuit and VCC Supply

The startup circuit must charge the VCC capacitor to the UVLO turn-on threshold (16 V for UC2842) using a resistor from the high-voltage DC bus. At maximum AC input (265 Vrms), Vd ≈ 375 V. The startup resistor Rstart must supply 3 mA to the chip during startup. Rstart = (375 - 16) / 0.003 ≈ 119.7 kΩ. Use a standard value like 120 kΩ. The VCC capacitor should be sized to hold up during startup.

Current Sensing and Spike Suppression

Current-mode control requires sensing the primary current. Use a sense resistor Rcs = 0.75 Ω (given). The peak primary current is 1.3 A, so the sense voltage is 0.975 V. Add a small RC filter (e.g., 1 kΩ and 100 pF) to suppress leading-edge spikes. The UC2842's ISENSE pin has a threshold of 1 V, so the design is safe.

Feedback Compensation with TL431

The 12 V output is regulated using a TL431 shunt regulator and optocoupler. The assignment provides system parameters: Lm = 1.5 mH, Ns? Dmax = 0.45, Rcs = 0.75 Ω, ILED = 5 mA, fs = 95.280 kHz. Calculate the equivalent load resistance Rload = Vo^2 / Po = 12^2 / 48 = 3 Ω. The open-loop transfer function of the power stage can be derived from the flyback small-signal model. A Type 2 compensator (using TL431) provides adequate phase margin.

Conclusion

This tutorial covered key aspects of SMPS design for the EEET2387 assignment: buck-boost ripple calculations, flyback transformer design, UC2842 controller setup, and feedback compensation. By understanding these fundamentals, you can tackle the assignment's problems with confidence. Remember to use allowed tools like spreadsheets and Wolfram Alpha for numerical calculations, but avoid copying solutions. Good luck!