# Re: PD power

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Dieter -

Good point. This could complicate the PD inrush limiter, if the PD has to
vary its inrush limit to follow the input voltage...

Possibilities:
1) PD controls its inrush current to keep <12.95W at startup. This is a
bear for an analog circuit - but it's really easy for a switching regulator.
2) PD sets inrush current <225mA (12.95W/57v) for all input voltages - but
now it's a 6.8W PD at 30V...
3) We write the spec to waive the 12.95W limit for the first 100ms, and
allow up to 350mA (or maybe 500mA) for that time. 12.95W after that - easy
for the switching regulator, once the input cap is charged.

3 looks like the only practical answer to me.

Dave

At 09:10 AM 3/27/2001 -0700, Dieter Knollman wrote:

>Yair,
>
>I have a problem with your numbers.
>If the maximum PD power is 12.95 watts, and the minimum PD input voltage is 37
>volts, then the maximum PD current is 350 ma.
>
>The PD input voltage varies from 37 to 50 volts with a 7 volt line drop.  This
>is an average input of 43.5 volts.  At this input voltage the PD is only
>allowed
>to draw 300 ma.  On a short loop, the PD input could be as high as 57
>volts.  If
>the PD draws 350 ma at this input, then the PD is consuming 19.95 watts.
>
>The point that I am trying to make is that for a 12.95 watt PD, the maximum
>current is 350 ma.  This value occurs only at the longest loop with the
>minimum
>voltage.  For other cases, the PD current must be less than 350 ma.
>
>Having a 350 ma PSE current limit allows the PD to draw at least 12.95 watts.
>If the PD draws more than 350 ma, it is violating the maximum power
>specification.
>Hence, there is no need for 500 ma.
>
>Dieter

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