# Why boost converter efficiency increases at low D and load resistance?

This slide (page 19) analyzes the efficiency of nonideal boost converter. As you can see from the formula below, the efficiency of boost converter is dependent on duty cycle D and load resistance RL.

The smaller the duty cycle D and load resistance RL, the higher the efficiency.

I would like to understand intuitively how D and RL affect the efficiency. For duty cycle, the smaller duty cycle, the less time that current flows through inductor resistance. So the efficiency will increase.

Is this right?

However, how to explain the similar for load resistance, why smaller load resistance results in higher efficiency?

• R_L isn't the load resistance, its the series resistance of the inductor (inductor = L). As it grows smaller, the inductor wastes less energy through that resistor so that more can be transferred to the actual load, R. Curious that this "non-ideal" model neglects the voltage drop of the switch, as no real solid state switch has zero voltage drop. – Los Frijoles Dec 3 '16 at 6:46
• @LosFrijoles The slide comes from a college class. This is the introduction to losses. Later lessons describe more conduction and switching losses. – Adam Haun Dec 3 '16 at 7:11
• Ah makes sense. – Los Frijoles Dec 3 '16 at 7:12
• Thank you, Los Frijoles. I made mistake in considering RL is load resistance. – anhnha Dec 3 '16 at 7:54