RE: PD power
Yair:
You have suggested two options at the end of this e-mail:
"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."
I would vote for option C: Constrain the PD to operate within a 350mA
average AND peak current. If you want more power than the specification
originally appeared to intend, you may have to fall back on a wall wart. If
a WAP requires periodic bursts of power, you may have to design a boost/buck
converter to take advantage of 1/2CV^2 effects.
I do not expect my 240V, 40A electric clothes drier to operate from a 100W
lamp socket. It's a special case. I would hope that we can cover as many
cases as possible within the reasonable constraints of 802.3af - but I don't
think it appropriate to keep attempting to stretch the spec to save a dollar
or two in the PD supply,if the PD has special requirements. Standards are a
good thing for interoperability. The downside of standards id that you
can't always do what you want within the confines of a standard.
Respectfully,
Peter Schwartz
Applications Engineer
Micrel Semiconductor
Phone: 408.435.2460
FAX: 408.456.0490
peter.schwartz@xxxxxxxxxx
-----Original Message-----
From: Yair Darshan [SMTP:YairD@xxxxxxxxxxxxxx]
Sent: Wednesday, March 28, 2001 8:21
To: Dieter Knollman
Cc: 'Dieter Knollman'; dave.richkas@xxxxxxxxx;
male@xxxxxxxxxxxxxxxxxxxxxx; Thong_Huynh@xxxxxxxxxxx;
Peter.Schwartz@xxxxxxxxxx; rkaram@xxxxxxxxx; ribrooks@xxxxxxxxxxxxxxxxxx;
stds-802-3-pwrviamdi@xxxxxxxx
Subject: 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 (0.35A x 20Ohm)
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, then 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.), then the PSE voltage is changing due to the load regulation
limitation of PSE supply. This variation 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]
<mailto:[SMTP:djhk@xxxxxxxxxx]>
> Sent: ?, ??? 27, 2001 6:10 PM
> To: Yair Darshan
> Cc: 'Dieter Knollman'; stds-802-3-pwrviamdi@xxxxxxxx
<mailto: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
>
>
>