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I'm designing a couple of step-down voltage regulators based on a LM22677 chip (datasheet) and following the example:

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I want to get 5V and 10V from a 12V supply, but I have concerns regarding the power rating I should have on the resistors: considering a maximum output current of 5A, how much W would I dissipate at max?

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    \$\begingroup\$ Which resistor are you concerned about? \$\endgroup\$ – The Photon Sep 30 '14 at 17:07
  • \$\begingroup\$ Wattage of resistor can be calculated using the formula IIR.where I is the current flowing through the resistor. \$\endgroup\$ – Sanjeev Kumar Sep 30 '14 at 17:16
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The feedback voltage on this part is about 1.2V, so nominal power dissipation in the 976 ohm resistor is about 1.5mW. (V^2/R) There's 3.3V-1.2V or 2.1V across the 1.54K resistor which will dissipate just under 3mW. There's no value given for R3, but it will also dissipate negligable power if chosen correctly. You can use the same technique to calculate the power at other output voltages with other resistor values. Generally you can pick a feedback divider so that the power dissipation in the resistors is negligable.

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The current out will depend on the load connected between VOUT and ground. Assuming VOUT is regulated to 10V, you can use ohms law and see that Rfbt and Rfbb will pull ~4mA (assuming FB pin is high impedance)

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