# Boost Converter Input Current

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#### mengghee

##### Full Member level 3
Hello everybody,

I have build a boost converter from 10 to 20V and my load is 40Ohm resistor. which i assume a load current of 500mA. but when run it. the readings on the input power supplies says that the whole circuit is taking around 1.5A or 2A. can anybody tell me why is that ? and is it normal ? does that mean that the components rating for my boost converter needs to be higher ? thank you.

regards,
mengghee

#### v_c

If you apply power conservation, you get

$V_{in} I_{in} = V_{out} I_{out} + P_{loss}$
where $V_{in} I_{in}$ is the power input to the converter,
$V_{out} I_{out}$ is the power output of the converter (delivered to the load resistor) and $P_{loss}$ is the total of power losses in the converter. Note that these are all average values. If your converter has an efficiency of 80%, then your output power is 80% of the input power and power losses are 20% of input power. If your efficiency was 100%, power loss would be zero and
$I_{in} = \frac{V_{out} I_{out}}{V_{in}} = \frac{(20V)(500mA)}{10V} =1A$

This is the average current, the peak current will be higher. The input current in the boost converter will have a peak and valley, so it will have a peak to peak ripple which will depend on the inductor value, input voltage and operating frequency. But the average should be approximately 1A. It may be a bit higher
due to losses (see the next paragraph).

So let's say you have a 80% efficiency and your output voltage and currents are 20V and 500mA. Your output power would be 10W and your input power would have to be 12.5W (since output power is 80% of power input, Pin=Pout/0.80). Now since Pin = Vin Iin =12.5W, and Vin=10V, Iin would have to be 1.25A. So as you can see, if the efficiency of your converter is low (losses are high), you will draw higher currents from your input power supply.

Best regards,
v_c

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