Re: power delivery question from Liaison report
What about ringing voltage, typically 105 to 110 V P-P?
Obviously, the telephone network sees higher voltages than 52V, albeit
current limited.
> Nick Tullius wrote:
>
> Team,
>
> just a reminder that 48 Vdc is a NOMINAL voltage (number of battery
> cells x 2) and has little to do with the limits of the operating
> range. The actual float voltage of the battery is 52.08 V (2.17 V/cell
> x 24 cells) for flooded batteries, and typically 54.00 V (2.27 V/cell
> x 24 cells). This is obviously the highest voltage ever appearing at
> the power-to-telecom equipment interface.
>
> These voltages are all within the maximum SELV value of 60 Vdc (see
> IEC 60950).
>
> For an example of the parameters needed to define a generic 48 V bus,
> see ANSI T1.315 Voltage Levels for DC-Powered Equipment Used in the
> Telecommunications Environment.
>
> Best of luck,
>
> Nick Tullius
> Astec Advanced Power Systems
> Tel 613 763-2359
> Fax 613 763-7155
> ntullius@xxxxxxxxxxxxxxxxxx
>
> -----Original Message-----
> From: Brooks, Rick [SC5:321:EXCH]
> Sent: Tuesday, July 18, 2000 7:00 PM
> To: stds-802-3-pwrviamdi@xxxxxxxx
> Subject: power delivery question from Liaison report
>
> I was reading in the Draft Liaison report from ISO/IEC JTC 1/SC
> 25/WG 3 to IEEE802.3 on power feeding that was
> handed out at the July Plenary.
>
> IEEE802.3af had question 4: Info on parameter limits (voltage,
> current, power, source impedance, ...) for world wide standards.
>
> i.e. what are the restriction beyond SELV.
>
> The response back was 48 VDC max, 175 ma max per pin.
>
> My question is:
> Is the 48VDC output from the port really 48VDC max as the
> response to the question indicates?
>
> If so, my thoughts are the following:
>
> We would have to spec our power output at the PSE as 48 VDC + 0%,
> - 8%, or something like that,
> so that it never exceeds 48 VDC continuously.
>
> This will further limit the available load power;
> it would be less than the load power that was discussed at the
> last meeting namely 14.6 watts.
>
> So, in that case the PD must be designed to draw at most 350 ma,
> as we discussed.
> And the power delivered at 100 meter cable would then be:
> Pwr = [44.2 - (12.5 x 0.35)] 0.35 = 13.9 Watts. (where 44.2 VDC
> is the lowest output voltage to still be in spec)
> For long cable lengths, the current per pin will be balanced, and
> we don't exceed the 175 ma per pin.
>
> For short cable lengths, we probably need an additional power
> spec, so that neither RJ-45 pin exceeds 175 ma.
> Say that due to connector imbalance, one pin is 175 ma, and the
> other is 20% below that, or 140 ma, which is a total of 315 ma.
>
> Then the power for a short cable would be (at least) 13.9 watts
> (44.2 * 0.315).
>
> This would say that the PD device should be designed not to draw
> more than 350 ma,
> and at the same time not to draw more than 13.9 watts.
> That way we never exceed 48 VDC nor 175 ma per pin on a
> continuous basis.
>
> This puts the burden on the PD end to meet these current and
> power requirements.
> The PSE end would have a max voltage of 48 VDC, but it's current
> limit would be set slightly higher than 350 ma
> by some appropriate margin.
>
> If, on the other hand, we put the burden at the PSE end, then the
> available power goes down even more, but that may be OK also.
>
> comments?