# Input current and output current in boost regulator

How can the input current of a boost regulator be more then the output current of the same regulator when the load has a set current draw.

For example going from 5V to 12V with a load of 2A. The power would be 24W and the input would be [email protected] and the output would be 12V@2A.

If the input and output are in series through the inductor, how can they have different currents?

On the right the switch is off, and the inductor is providing 2A to the load per load requirement. This means the input power supply is providing Vin = 5V * 2A = 10W and the load is taking Vout = 12V * 2A = 24W. Obviously this isn't true. The answer is that the current changes, but how can the current change if the input is in series with the output through the inductor?

Can someone explain?

• You'll need to read more carefully about the mode of operation of a boost converter. In just about any "tutorial"-like explanations you'll see how each element contributes. Mar 11, 2022 at 21:32

The input and output have different duty cycles.

Current at the input is flowing 100% of the time but the output current through the diode is only non-zero for a small percentage of the time depending upon the boost ratio. For example with a 2:1 voltage ratio there will be about 50% duty cycle so the average output current will only be about half of the average input current. The additional current is flowing through the switch.

With a step-down converter the ratio is reversed with the output current higher than the input current.

If the input and output are in series through the inductor, how can they have different currents?

They are not therefore they can have different currents.

When the switch is closed, there is clearly current in the inductor that goes nowhere near the load. That's called the charge cycle and energy is acquired by the inductor. And, while the charge cycle is happening, the load is taking whatever current it needs from the output capacitor i.e. the diode is reverse biased and that current cannot come from the inductor.

When the switch opens, the inductor current is partially supplying the load and also topping up the stored energy in the output capacitor. Again load current is a fraction of inductor current with the remainder feeding the output capacitor.