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*To*: "'Dave Dwelley'" <ddwelley@xxxxxxxxxx>, Dieter Knollman <djhk@xxxxxxxxxx>, Yair Darshan <YairD@xxxxxxxxxxxxxx>*Subject*: RE: PD power*From*: "Huynh, Thong" <Thong_Huynh@xxxxxxxxxxx>*Date*: Tue, 27 Mar 2001 15:44:36 -0800*Cc*: "'Dieter Knollman'" <djhk@xxxxxxxxxxx>, stds-802-3-pwrviamdi@xxxxxxxx*Sender*: owner-stds-802-3-pwrviamdi@xxxxxxxx

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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