|Thread Links||Date Links|
|Thread Prev||Thread Next||Thread Index||Date Prev||Date Next||Date Index|
yes the PSE needs a current limit in any case.
I'm saying to set the PSE limit higher than 350 ma, and let the high power PD's make sure that they do not draw more than 350 ma, nor more than 13.9 watts, for example.
However, if we instead make the PSE be the dominant current limit,
then it will have both voltage and current regulation.
Therefore we would have to set the PSE current limit at 350 ma maximum.
Then the nominal current limit will be less than the max, say 1%, 2%, ... x% less than 350 ma,
not to mention the minimum current limit (350 ma - 4%,...). What tolerances are possible?
Again the maximum power output goes down accordingly, do we really want to do this?
What do others think?
The open question is: what accuracy of current limit is possible, and at what cost?
From: Avinom Levy [SMTP:Avinoaml@xxxxxxxxxxxxxxxxx]
Sent: Wednesday, July 19, 2000 10:33 AM
To: 'Sterling Vaden'; Tullius, Nick [CRK:0M90-I:EXCH]
Cc: Brooks, Rick [SC5:321:EXCH]; stds-802-3-pwrviamdi@xxxxxxxx
Subject: RE: power delivery question from Liaison report
Ringing voltage is a special case for the safety agencies, and they even address it with different terminology - "TNV", Telephony Network Voltage. I think we can not apply the same telephony network requirements to the Ethernet network.
A lot had been said about keeping the PD side "simple and stupid". In light of that, and in light of the fact that the PSE side will any way have to have hardware to control the current (inrush current, minimum current, real time power management etc.), I think the burden of this task you mentioned, keeping the maximum current below 350mA, should be on the PSE side. The maximum power capability is a derivative of the voltage and current limit, and will be a parameter to be taken into consideration during the design of any PD.
Phone (631) 756 4680 Ext. 305
Fax (631) 756 4691
From: Sterling Vaden [SMTP:savaden@xxxxxxx]
Sent: Wed, July 19, 2000 10:24 AM
To: Nick Tullius
Cc: Rick Brooks; stds-802-3-pwrviamdi@xxxxxxxx
Subject: 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
> Nick Tullius wrote:
> 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
> -----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.