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LM3478: Current Sense resistor

Dwight Pray
Prodigy 80 points
Part Number: LM3478


I am somewhat confused about selecting a current sense resistor.  Here is my question:

 Is Vsl the difference between Vsense and the line intersecting the duty-cycle? (see Fig. 30)

 The power supply that I am developing is a flyback with a maximum DC of 0.83 and a peak primary current of 190mA.  The Fs is 150kHz.

Any help in understanding this will be appreciated.

  • 0
    TI__Guru* 78675 points
    Yes your understanding is correct. The slope compensation is subtracted from the error voltage, which is equivalent to superimpose the slope comp signal onto the real current sense signal and compare with a fixed voltage.
    Hope this helps.
  • 0 in reply to Youhao Xi
    Prodigy 80 points

    Thank you but that doesn't make it any clearer.  An example would be more clear. Let's say that Dmax = 0.81, ipk = 175mA.  From this, can you tell me Vsl, Rsense?

  • 0 in reply to Youhao Xi
    Prodigy 80 points
    Thank you but that doesn't make it any clearer. An example would be more clear. Let's say that Dmax = 0.81, ipk = 175mA. From this, can you tell me Vsl, Rsense?
  • 0 in reply to Dwight Pray
    TI__Guru* 78675 points
    The slope compensation signal is a triangle between 0 to VSL. At the end of each cycle the ramp reaches VSL and reset to zero by the clock and start ramp up again. Assume Dmax=0.81, then the amount of slope compensation is Dmax x VSL = 0.8 x 92 mV = ~ 75 mV.

    Ipk=175mA. Assuming you are using 0.2 Ohm current sense resistor, your peak Vsense =175mA x 0.2 Ohm= ~35mV.

    To be at steady state operation, your error voltage should be 35mV+57mV at 81% of the switching cycle, namely the intersection at 81%, in order to produce 81% duty cycle.

    Hope this clarifies.