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I am using this IC in boost converter configuration : boost converter

It steps up the voltage from 16.5V to 23V.

It drives a load current of 270mA.

I need to calculate the input current of the converter.

I understand that output power/input power is efficiency.

Since I know the output power, can I take the efficiency value from the graph 21 of the attached datasheet (86.8% approx - obtained from fig 21 graph) and find the input power?

With the input power, and since I know the input voltage, I can calculate the input current.

Is this the right way?

And I want to check if all the components at the output of the IC are properly rated.

Which parameters and components should I check and how to check?

Boost Converter

Sorry for the poor image quality. The two transistors are taken from the SWE pin.

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I would make this a comment but I can't (yet).

Your procedure is correct. You have to keep in mind that the efficiency value varies among different ICs, Vin, Vout, temperature... so take it as a very good estimation which is probably enough for most purposes.

Edit: The input current you would be calculating is an average value. Also, you can only do this simple calculation if Vin has negiglible ripple.

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  • \$\begingroup\$ Thank you. How to check if all the components at the output of the IC are properly rated. Which parameters and components should I check and how to check? \$\endgroup\$ – Freshman Jan 24 at 8:50
  • \$\begingroup\$ I am not sure what you mean by properly rated. \$\endgroup\$ – Ivan Perez-Rodriguez Jan 24 at 8:52
  • \$\begingroup\$ The component is properly derated at the output section \$\endgroup\$ – Freshman Jan 24 at 9:19
  • \$\begingroup\$ I'll try to answer what I think you mean: L and C should be big enough to get a low enough ripple in output current and outpur voltage, respectively. That is the main factor to choose them. Given that neither current nor voltage are very high, I would say frequency is the main factor to choose both transistor and diode. In other words given this low boost factor, those components should not affect dc levels at the output. \$\endgroup\$ – Ivan Perez-Rodriguez Jan 24 at 10:08
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Using the efficiency rating of the IC is a good way to calculate the input power with respect to the output power but, as @Ivan Perez-Rodriguez has said, the efficiency value will vary so treat it as an average.

In terms of component ratings, a rule of thumb is to pick a component that has a voltage rating at least double the circuit voltage or a current rating 20-30% above what you expect the current flowing through the device to be. This will allow for significant derating of the component before you get into any trouble.

For a capacitor this is called derating a capacitor by 50%. If you don't do this the the capacitance may drop significantly as you approach the rated voltage, causing instability in your circuit or allowing lots of ripple to appear in the outputs.

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