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Design a buck-boost converter with the following specifications:

- Output voltage: -28V
- Input voltage: 100V
- Load: 10 Ohms
- Output voltage ripple: No more than 0.5%

Specify the frequency, duty cycle ratio, and the values of each component.

Answer :

Here's a comprehensive approach incorporating step-by-step calculations and derivations:

1. Converter Selection and Mode Analysis:

Since the desired output voltage (-28V) is negative and has a larger magnitude than the input voltage (100V), a non-inverting buck-boost converter is the most suitable topology.

2. Key Design Parameters:

  • Input voltage (Vin) = 100 V
  • Output voltage (Vo) = -28 V (negative due to non-inverting buck-boost)
  • Load resistance (R) = 10 Ω
  • Output voltage ripple (ΔVo) ≤ 0.5% of Vo (0.005 * -28 V) = 0.14 V

3. Duty Cycle (D) Calculation:

In a non-inverting buck-boost converter, the output voltage relates to the input voltage and duty cycle (D) through the following equation:

Vo = - Vin * D / (1 - D)

We need to solve for D:

D = (Vo + Vin) / (Vo - Vin) = (-28 V + 100 V) / (-28 V - 100 V) ≈ 0.778

4. Inductance (L) Calculation:

To minimize output voltage ripple, we can use the following equation for the minimum inductance:

L ≥ (Vin * D * (1 - D)) / (ΔVo * f)

where f is the switching frequency. We'll choose a common switching frequency (f) of 50 kHz for this design.

L ≥ (100 V * 0.778 * (1 - 0.778)) / (0.14 V * 50 kHz) ≈ 1.1 mH

5. Capacitor (C) Selection:

The output capacitor (C) value is selected based on the allowable ripple current (ΔIL). A larger capacitance reduces ripple. We can estimate ΔIL using the following equation:

ΔIL ≈ (Vin * D) / (f * L) ≈ (100 V * 0.778) / (50 kHz * 1.1 mH) ≈ 14.1 A

Assuming a desired ripple current ratio (ΔIL / Iout) of 10% (typical value), where Iout is the output current:

ΔIL / Iout = 0.1

Iout = Vin / R = 100 V / 10 Ω = 10 A

Therefore, ΔIL ≈ 1 A

Using the formula for capacitor ripple voltage (ΔVc) and assuming a desired ΔVc of 10 mV (a small value relative to ΔVo):

ΔVc = (ΔIL * L) / (f * C) = 10 mV

C ≥ (ΔIL * L) / (f * ΔVc) ≈ (1 A * 1.1 mH) / (50 kHz * 10 mV) ≈ 22 µF

We can choose a standard value slightly higher, such as C = 33 µF.

6. Component Selection:

  • Inductor (L): Use a 1.2 mH or higher inductor with a current rating exceeding the maximum load current (Iout = 10 A). Consider core material, size, and temperature limitations.
  • Capacitor (C): Use a 33 µF or higher electrolytic capacitor with a voltage rating exceeding the output voltage magnitude (|Vo| = 28 V). Consider voltage ripple and temperature specifications.
  • Switch (MOSFET): Select an N-channel MOSFET with a voltage rating exceeding the input voltage (Vin = 100 V) and a current rating exceeding the maximum load current (Iout = 10 A). Choose a MOSFET with low on-resistance (RDS(on)) for better efficiency. Consider switching speed and gate drive requirements.
  • Diode (Schottky): Use a Schottky diode with a voltage rating exceeding the output voltage magnitude (|Vo| = 28 V) and a current rating exceeding the maximum load current (Iout = 10 A). Consider forward voltage drop and switching speed.

7. Additional Considerations:

  • Feedback Loop: Implement a feedback loop using a controller IC (e.g., TL431) and optocoupler to regulate the output voltage and maintain stability. This ensures the output voltage stays at -28V even with input voltage variations.
  • Heat Dissipation: Calculate power dissipation in each component (MOSFET, diode) and choose appropriate heat sinks to prevent overheating.
  • Layout:

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