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In the following post 230V AC to 5V DC converter, lossless , the 1.44 W Universal Input Buck Converter was mentioned. When R1=2k the output voltage is about 3.3V.

In the LNK304 datasheet (applications section) , http://www.mouser.com/ds/2/328/lnk302_304-306-179954.pdf, they mentioned that the voltage divider that consist of R1 and R3 will determine the value of the output voltage.

My question is what R1,R3 values will result in 5V output voltage and what will the supplied current be for this output voltage?

Thanks

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From the datasheet I read, The feedback input circuit at the FEEDBACK pin consists of a low impedance source follower output set at 1.65 V.

My guess is this FB is used to provide a gain ratio to multiply Vref to the output , (Rf+Rs)/Rs * Vref = (13k+2.05k)/2.05k * 1.65 = 12.375V But the Si diode drop must be added. But the App Note ought to give better info.

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  • \$\begingroup\$ I want to implement this circuit on a PCB that will be integrated in a consumer product to convert from 220V to 5V & 3.3V. I know that this circuit is non-isolated, but what makes it unsafe. If I covered this circuit in a box within the product enclosure and dragged the output wires out of the box to power the main pcb. Will it be safe? \$\endgroup\$ – ousama kanawati Jan 14 '17 at 2:58
  • \$\begingroup\$ not unless it is designed properly \$\endgroup\$ – Sunnyskyguy EE75 Jan 14 '17 at 3:01
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Check the design guide not just the datasheet. Many companies have information in app notes not just the datasheets and each company is different.

High side buck converters need not just the feedback path to be optimized but the inductor and capacitor as well. There is a an equation on page 10:

$$ R_{fb} = \frac{V_o - V_fb}{\frac{Vfb}{R_{bias}}+I_{fb}} $$

\$ R_{fb} \$ is R1 in the diagram and \$ R_{bias} \$ is R3, \$ I_{fb} \$ is 49uA

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