Coupled inductor SEPIC converter using UC3843.

  • Vin nominal = 14 V
  • Vout max = 30 V
  • Iout max = 2 A

According calculations based on many resources the MOSFET's current will be

  • Peak = 7.62 A
  • RMS = 5.3 A

To achieve pulse by pulse peak current control, a shunt resistor is put on the switching MOSFET's source, and due to UC3843 Isense voltage bias of 1 V, the shunt resistor will dissipate about 7.62 W which is too much.

Of course an offset voltage can be injected over the Isense pin which will reduce the shunt resistance but also affect the stability somehow and still the power dissipation around 1.5 W.

What if, the shunt resistor put between the secondary inductor & the ground which has a peak current about 2.4 A instead of shunting the MOSFET's source which has that high current peak..

As far as I understand, the two inductors' current is proportional & synchronized.

Anybody has a deeper knowledge about such issue?

  • 1
    \$\begingroup\$ Can you use a smaller resistor and then an amplifier? \$\endgroup\$ – Aaron Nov 10 '20 at 18:28
  • \$\begingroup\$ I agree with Aaron. In my designs, I use a lower value resistor and then use an op amp to amplify the signal. \$\endgroup\$ – bunker89320 Nov 10 '20 at 19:07
  • \$\begingroup\$ @Aaron .. This is the last solution I may settle with .. due to op-amp bandwidth, phase shift, cost & PCB space .. I wonder what will be if the shunt resistor just shifted to the secondary side ? \$\endgroup\$ – Deron Felix Nov 10 '20 at 21:13

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