RE: PD power
We can easily reduce the time to 50mSec by specifying that the PD input cap
will be 470uF max. which is more than enough for the typical low cost 15W
(Tested analyzed and verified).
I do not agree to you guess that most devices would not consume max. power.
Today, most devices that uses the POL technology are around the 10W load.
Tomorrow it could be other applications which less power (IP phones)
and day after tomorrow ? who knows. We need to be prepared to wide range of
According to that if the max. power expected from PSE at PD input is 12.95W
than startup energy might be easily above this number.
> -----Original Message-----
> From: Wael William Diab [SMTP:wdiab@xxxxxxxxx]
> Sent: ד, מרץ 28, 2001 9:28 PM
> To: Dave Dwelley
> Cc: Dieter Knollman; Yair Darshan; 'Dieter Knollman';
> Subject: Re: PD power
> Dave / Dieter and all,
> Regarding the 3 options mentioned below:
> Option 2 would limit the device to a 6.8W device only at startup. I am
> not sure that that is a bad thing since we are not limiting the maximum
> power of the device during operation. My guess is that most devices
> would not consume maximum power at startup (the first xxx msec of
> My concern with Option 3 is the actual time that we have to deal with
> the inrush current on the PSE side. During startup when the FET is
> partially on it will be dissipating alot of power. 100 msec seems to be
> an excessive time to hold 0.5A. If we go down this path I would suggest
> that we look at reducing the inrush current time.
> that's my 2c
> Dave Dwelley wrote:
> > Dieter -
> > Good point. This could complicate the PD inrush limiter, if the PD has
> > 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
> > 2) PD sets inrush current <225mA (12.95W/57v) for all input voltages -
> > 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 -
> > 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
> > >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
> > >
> > >The point that I am trying to make is that for a 12.95 watt PD, the
> > >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
> > >If the PD draws more than 350 ma, it is violating the maximum power
> > >specification.
> > >Hence, there is no need for 500 ma.
> > >
> > >Dieter