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# RE: PD power

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Dave and all,

I agree with your item (3).  Please keep in mind that 12.95W is meant to be
steady state power limit.  The inrush current to charge up the cap is a
transient condition and should be treated separately.

Thong

> -----Original Message-----
> From:	Dave Dwelley [SMTP:ddwelley@xxxxxxxxxx]
> Sent:	Tuesday, March 27, 2001 11:59 AM
> To:	Dieter Knollman; Yair Darshan
> Cc:	'Dieter Knollman'; stds-802-3-pwrviamdi@xxxxxxxx
> Subject:	Re: PD power
>
>
> 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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