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LM2651: LM2651

Mitch Tilbury
Prodigy 90 points
Part Number: LM2651
Other Parts Discussed in Thread: LM2671,

Can the soft start input to the LM2651 be left floating as stated in the LM2671 datasheet? Couldn't find that in the LM2651 data.

  • 0
    TI__Guru**** 191445 points

    Hi Mitch,

    Even thought the LM2651 and LM2671 part numbers seem similar, the functionality of the two devices are quite different.  LM2671 has a dedicated SS pin with a default SS time when the SS pin is left open.  LM2651 uses a dual function _SD/SS pin with a unique SS operation.  Normally the SS pin voltage is a ramp voltage dV/dt = Css *Iss.  This voltage ramp is presented to a second non-inverting input of the internal error amplifier so that the effective non-inverting input is the lower of Vss and Vref.  The output voltage tracks the SS ramp voltage until the SS ramp reaches Vref at which point the converter regulates to the Vref voltage.  LM2651 charges the _SD/SS pin up above the SD threshold, then the SS voltage controls the current limit of the converter.  The current limit is gradually increased from 0 A as the _SD/SS voltage is ramped up.  So a controlled ramp voltage and the SS capacitor is required for LM2651.  Datasheet section 8.2.2.5 details this operation.

  • 0 in reply to JohnTucker
    Prodigy 90 points

    John:

    Thanks for the additional detail. If the SS pin is left floating, say through a failure of the capacitor (opens), would that prevent the LM2651 from ramping-up this input to the SD threshold of 0.6V and the converter will not start? And that is different from the LM2671, which has default circuitry to complete this part of the startup process if the pin is left floating or its capacitor fails, correct? Thanks again,

    Mitch

  • 0 in reply to Mitch Tilbury
    TI__Guru**** 191445 points
    Mitch,

    I have not tried it personally, but here is what I would expect to happen. With no SS capacitor, the SS pin should be very rapidly pulled up to the internal 2 V rail. When the _SD/SS pin voltage is above .6 V the device is enabled and when the voltage is above 1.3 V, the current limit is proportional to the voltage. The linear voltage ramp WITH a capacitor is what allows the current limit (and the duty cycle along with the output voltage) to increase in a controlled manner. With NO capacitor, the device will be immediately enabled with the current limit set to the maximum. The output voltage should increase rapidly and most likely overshoot the intended regulation point, possibly up to nearly the input voltage. Eventually the control loop would bring the out put back into regulation. Plainly, this type of behavior at start up is not wanted.