RE: PD power
Dave et all,
Option number 3 is the practical option that is being used for many many
years in the power supply field successfully.
Most of power supply topologies work that way, and the source has to supply
In our case the source is the PSE. In other cases the source is the mains,
ac adapters, Battery or DC/DC converter.
> -----Original Message-----
> From: Dave Dwelley [SMTP:ddwelley@xxxxxxxxxx]
> Sent: ג, מרץ 27, 2001 9:59 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
> 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.
> >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
> >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
> >If the PD draws more than 350 ma, it is violating the maximum power
> >Hence, there is no need for 500 ma.