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------------------------------------------------ Problem 6

A buck-boost converter has an input voltage of 10 V. The switching frequency of its transistor is 6.5 kHz, and its duty ratio is 45%. Compute the following:

a) The transistor on time, [tex]t_{on}[/tex]

b) The average voltage across the load, [tex]V_t[/tex]

Answer :

To solve the given problem, we'll compute the transistor on time [tex]t_{on}[/tex] and the average voltage across the load [tex]V_t[/tex] for a buck-boost converter with specified parameters.

a) Transistor On Time, [tex]t_{on}[/tex]

  1. Definition: The "on time" of the transistor in a switching converter is the duration for which the transistor is conducting during one cycle of the switching frequency.

  2. Given Data:

    • Switching frequency [tex]f = 6.5 \text{ kHz}[/tex]
    • Duty ratio [tex]D = 45\% = 0.45[/tex]

  3. Formula: The switching period [tex]T[/tex] is given by:
    [tex]T = \frac{1}{f}[/tex]
    Subsequently, the on time [tex]t_{on}[/tex] can be expressed by:
    [tex]t_{on} = D \times T[/tex]

  4. Calculations:

    • First, compute the switching period:
      [tex]T = \frac{1}{6.5 \times 10^3} \approx 1.538 \times 10^{-4} \text{ seconds}[/tex]
    • Then, compute the on time:
      [tex]t_{on} = 0.45 \times 1.538 \times 10^{-4} \approx 6.921 \times 10^{-5} \text{ seconds}[/tex]
    • Therefore, the transistor on time [tex]t_{on} \approx 69.21 \text{ microseconds}[/tex].

b) Average Voltage Across the Load, [tex]V_t[/tex]

  1. Definition: The average voltage across the load in a buck-boost converter is influenced by the duty cycle and the input voltage.

  2. Given Data:

    • Input voltage [tex]V_{in} = 10 \text{ V}[/tex]
    • Duty ratio [tex]D = 0.45[/tex]

  3. Formula: The average output voltage [tex]V_t[/tex] for a buck-boost converter is:
    [tex]V_t = \frac{-D}{1-D} \times V_{in}[/tex]
    (Note: The negative sign indicates a reversal of polarity typical in buck-boost configurations.)

  4. Calculations:

    • Plug the values into the formula:
      [tex]V_t = \frac{-0.45}{1-0.45} \times 10 = \frac{-0.45}{0.55} \times 10 \approx -8.182 \text{ V}[/tex]
    • Thus, the average voltage across the load [tex]V_t \approx -8.18 \text{ V}[/tex].

Summary:

  • The transistor on time [tex]t_{on}[/tex] is approximately 69.21 microseconds.
  • The average voltage across the load [tex]V_t[/tex] is approximately -8.18 V, indicating an inversion of polarity from the input voltage.

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