# RE: PD power

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I was just checking the draft spec. on the web, It says we are
limited to 12.95 watts continuous power except during startup.
During startup, the PD is limited to drawing up to 19.5 watts for
up to 100ms. (some paraphrasing done here.....)

Having a 20 ohm line impedance complicates analysis somewhat, but
remember that the PD does not see the full 57 volts at startup.
It sees something less. At 400ma, the PD sees 57-.4*20=49 volts
and draws 49*.4= 19.6 watts. The remaining power is lost in the
distribution system.

...just another way of saying (3) is the reasonable choice. It

Brian

>-----Original Message-----
>From: Dave Dwelley [mailto:ddwelley@xxxxxxxxxx]
>Sent: Tuesday, March 27, 2001 2:59 PM
>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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