RE: PD power
Dieter,
I explain my numbers.
1. We have the following requirements:
PSE output voltage 44V to 57V.
PSE output current 350mA max. continuous.
2. We don't have a requirement limiting the power from the PSE. This
limitation is function of the above requirement (1).
It means that PSE can supply 44Vx0.35A=15.4W to 57Vx0.35A=19.95W..
3. In the PD side, the PD designer should design for the worst case i.e.
with max. link length, which is 7V drop (0.35 Ax20Ohm)
3.1 Thus the worst case is 37V input at PD with 0.35A continuous current
max.
4. In your calculation you are assuming that the PD power supply is a switch
mode power supply. In this case when the PSE output is 57V
or in case that the loop is short than you right that the current will
be lower than 350mA average.
5. Since we are not limiting the implementation of the power supply type, We
could have "crazy implementation" that will consume 350mA regardless
of the input voltage. You can argue what are the chances that it may
happen but you and I do not know how future application will evolve.
6. Since we don't have specification for max. power, we are not violating
anything. We have spec for max. current and voltage range.
Statements 1 to 6 are not relevant to your questions, it meant only to
express my view on the facts as I understand them.
Now to your last paragraph: You have said that:
Dieter:
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
specification.
Hence, there is no need for 500 ma.
Yair:
and my comment to that is
7. If you set the current limit at PSE to max. 350mA, it could be O.K under
the following conditions:
You set the current limit to max. average current of 350mA.
The reasons for it are:
7.1 In any case, you will sense the current through some low pass filter in
order to reduce your sensitivity to noise. Now we can argue about
the filter time constant.
7.2 In the worst case condition, PSE output is 44V and PD gets 12.95
average max. and suppose we have an application that needs the 11-12 Watts
most of the time.
It means that we are working around 350mA continuously.(Many
applications needs this power level and many will be in the future.). If you
specify 350mA peak, than
we will be sensitive to small transients in the current due too load
changes (wireless access point for example, or other types.) and the PD
will be shut of randomly without good reason, hence system
performance will degrade.
7.3 If you have PSE output voltage transients due to dynamic load in port n
(assuming the current is changing within the range you want 350mA peak.)
Than the PSE voltage is changing due to the load regulation limitation
of PSE supply.
This variations will change the current through port (n+1) . Now, if
you sensitive to peaks above 350mA for short duration, than this port will
be disconnected.
(See example for this situation in my presentation).
Now, if you filter this phenomena with your current sensing circuit,
than you are O.K. hence again we are need to spec. the numbers and to
recognize that we should talk average current and not peak.
7.4 If you have PSE changing is voltage between 44V TO 57 due to line load
regulation, you will have the same effect as presented on 7.3
7.5 If you combine all the effects above you can not escape from the
following conclusions:
1. During normal operation, the PSE will limit the current to 350mA
average.
2. Current peak above 350mA up to TBD mA is aloud as long as their are
limited in pulse width to TBD mSec , Limited to TBD duty cycle etc.
I suggest the following numbers:
Max. average 350mA (We already decide max. continuous of 350mA
which the same as average if the current ripple is zero.)
Max peak current 500mA. for 10mSec max. at duty max. of 10%
3. I also planning to make some lab tests to verify the above numbers and
may modify them a little bit in order to allow this average and peaks
without adding
too much burden on the PSE power supply.
4. Dieter, Even if you are using power supply for general purpose
application you have max. average with some margins for peaks and valley
limited by the integration factor of the power supply consisting of
the power supply output cap and control loop dynamics. It should be the same
approach for PSE output. After all PSE output is a power supply
output.
The question is, how I handle this and support as many as possible
application types.
a- by killing the application (and many others with the same behavior)
?
b- by supporting this application or others with similar
characteristics by specifying the relevant parameters?
I choose (b) since this is our job in the IEEE802.3af.
Thanks
Yair.
> -----Original Message-----
> From: Dieter Knollman [SMTP:djhk@xxxxxxxxxx]
> Sent: ?, ??? 27, 2001 6:10 PM
> To: Yair Darshan
> Cc: 'Dieter Knollman'; stds-802-3-pwrviamdi@xxxxxxxx
> Subject: PD power
>
>
> 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 only
> 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 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
> 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
> watts.
> If the PD draws more than 350 ma, it is violating the maximum power
> specification.
> Hence, there is no need for 500 ma.
>
> Dieter
>
>
>