Power Electronics Circuits
In a DC-DC converter, which of the following correctly combines the main reason efficiency drops sharply at very small load currents with a countermeasure?
- AThe lighter the load, the larger the conduction loss, in proportion to the square of the current, so the switch is replaced by one with a higher on-resistance to limit the current
- BAt light load the inductor ripple current becomes zero and the control loop stops, so the inductance is reduced to increase the ripple and keep the control running
- CLosses that do not depend on current, such as gate driving, become a larger share, so the converter switches to PFM or burst mode to reduce the number of switching events
Show answer and explanation
Answer: C. Losses that do not depend on current, such as gate driving, become a larger share, so the converter switches to PFM or burst mode to reduce the number of switching events
Conduction loss is proportional to the square of the current, so it becomes small at light load. The losses for charging and discharging the gates, part of the switching losses and the supply current of the control circuit, however, stay roughly constant regardless of the load current, so they pull efficiency down sharply when the output power is small. At light load the converter therefore switches to PFM (pulse-frequency modulation), which lowers the frequency, or burst mode, which switches for several periods and then rests, to reduce the number of switching events. In exchange, the ripple frequency changes and the ripple tends to grow.
- A. Conduction loss actually becomes smaller at light load. What stands out at light load are the losses that do not depend on current.
- B. Reducing the inductance does not raise efficiency. More ripple actually increases losses.
- C. Correct. The number of switching events is reduced to cut the losses that do not depend on current.