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I am using LT3748 for Flyback converter design. In datasheet (https://www.analog.com/media/en/technical-documentation/data-sheets/3748fb.pdf), given formula for Rsns value (pg 21) is Rsns=100mV/Ilim, Ilim being the desired maximum output current. Instructions for Rsns selection are given on pg17.

For 2A max output current, Rsns should be 100mV/2A = 0.05R. My problem is that I can't figure out what is the min power rating I need for resistor (I have 330mW power rated resistors that I would like to use).

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  • \$\begingroup\$ No, \$I_{LIM}\$ is not the output current; it's the primary peak maximum current. \$\endgroup\$
    – Andy aka
    Commented Apr 1, 2020 at 12:32
  • \$\begingroup\$ OK, that makes more sense than my initial logic. \$\endgroup\$
    – DenR
    Commented Apr 1, 2020 at 12:42

3 Answers 3

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Per datasheet, the Rsense current is not output current, but peak switch current. When you have this current selected, then calculate your sense resistor per specification and apply Ohms law to calculate power dissipation in the resistor.

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  • \$\begingroup\$ This makes sense, as @Andy aka pointed out too. I have a schematic that I am analyzing. So my problem is that I don't know what they have selected this peak current to be. However, I understand that would be a different question, so I would accept our answer. \$\endgroup\$
    – DenR
    Commented Apr 1, 2020 at 12:45
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I've never used this device but I suggest that the power in the primary = the power in the secondary + a bit for losses. That means that if you know the input and output voltages and the output current you should be able to figure it out.

You haven't specified any voltages so let's do an example:

  • VOUT = 12 V.
  • IOUT = 2 A.
  • POUT = 12 × 2 = 24 W.
  • PIN = 24 W / Efficiency = 24 / 0.9 = 26.7 W.
  • Let's say VIN = 53.4 V.
  • IIN = PIN / VIN = 26.7 / 53.4 = 0.5 A.
  • Since you've calculated 0.05 Ω we can calculate the power.
    PRsense = I2R = 0.52 × 0.05 = 12.5 mW.

Remember that that's average so you need to check that the resistor can actually handle the peak.

Also remember that I could be talking nonsense.

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  • \$\begingroup\$ I gave you a wrong premise to start with. (I am analyzing an already existing circuit, and seeing LT3748 for the first time). Iout is a primary peak maximum current, not an output current, as others pointed out (which makes much more sense than what I suggested). \$\endgroup\$
    – DenR
    Commented Apr 1, 2020 at 12:47
  • \$\begingroup\$ I know. I calculated for Rsense on the primary. Read my answer again. \$\endgroup\$
    – Transistor
    Commented Apr 1, 2020 at 12:49
  • \$\begingroup\$ I did, and I now see your point. Thanks for fixing my wrong initial logic, and providing an example. \$\endgroup\$
    – DenR
    Commented Apr 1, 2020 at 13:12
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Ilim being the desired maximum output current

No, \$I_{LIM}\$ is not the output current; it's the primary peak maximum current.

To calculate what this is you need to understand what the primary saturating current is for your flyback transformer and choose a value of resistance that prevents the ramping primary current exceeding that value.

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