RE: PD power
The numbers that you saw in the draft where the power supply ad-hoc draft
proposal that I have sent month ago and was presented by Karl in March
The 19.5W is an error and it should be 17W. Which is (44V-0.5A*20R)*0.5A=
17W during startup. (0.5A/100mSec for startup only, supplied by PSE)
And you right, it is supporting option 3 which is the natural way to handle
this situation and is being used for many many years in the power supply
> -----Original Message-----
> From: Lynch, Brian [SMTP:brian_lynch@xxxxxx]
> Sent: ג, מרץ 27, 2001 11:55 PM
> To: 'Dave Dwelley'; Dieter Knollman; Yair Darshan
> Cc: stds-802-3-pwrviamdi@xxxxxxxx
> Subject: RE: PD power
> 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
> is already in the spec.
> >-----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...
> >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.
> >At 09:10 AM 3/27/2001 -0700, Dieter Knollman wrote:
> >>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
> >>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
> >>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
> >>Hence, there is no need for 500 ma.