Power Electronics Circuits
For an ideal buck-boost chopper (inverting buck-boost converter) with an input voltage of 12 V and a duty cycle of 0.6, which of the following is the magnitude of the output voltage in continuous conduction mode?
- A7.2 V
- B18 V
- C30 V
Show answer and explanation
Answer: B. 18 V
In an inverting buck-boost converter, energy from the input is stored in the inductor while the switch is on and released to the output through the diode while it is off. From the volt-second balance of the inductor, Vin × D = Vout × (1 − D), the magnitude of the output voltage is Vout = Vin × D/(1 − D) = 12 × 0.6/0.4 = 18 V. The output polarity is opposite to that of the input (negative with respect to ground). It works as a step-down converter for D < 0.5 and a step-up converter for D > 0.5, and at D = 0.5 the output magnitude equals the input.
- A. This is D × Vin, the buck-converter formula.
- B. Correct. 12 × 0.6 ÷ 0.4 = 18 V, with the polarity opposite to the input.
- C. This is Vin/(1 − D), the boost-converter formula.