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RE: PD power




Peter, 
I think you miss my point and I would like to clear it. 

1. During normal operation, I do not want more power than we specified
already. I would like to keep the average power or the rms power  (what ever
it is..)
    bellow what was specified ( PD should be design for the
(44V-0.35Aavg*20)*0.35Aavg=12.95Wavg.)

2. It means that load can be changed from 350mA to 500mA for few msec or
less and still the average will be 12.95W. The parameters that we have to
defined here 
    are the peak, the duration, and duty cycle. Nothing is new here, you as
power supply designer do it with out give second thought.

    Example: 150W, 50V power supply can deliver 3A average/continuous/DC
current. You are aware that this power supply is current limit protected 
    to TBD% above this number.

    Now, if you connect a load that changes from 500mA to zero at duty of
350/500 at frequency of let say 500KHZ it will be seen by the above power
supply,
    as 3A average with out any degradation in performance. Why? since the
load frequency change is bellow the power supply output bandwidth, it is 
    averaged by the power supply output cap.

    Now, lets find the frequency (lower than 500K...) , pulse width, duty
etc. that the standard power supply will stop averaging and you will see
that 
    the ratio between the RMS and the average start to increase. This is the
point that I am looking for.

    This point is described by average value, peak value pulse width etc.

    It is true to any power supply, since it is natural behavior of its
physics.

3. Remember that even when you will limit the power supply current limit to
350mA. The current sense circuit is filtered in order to 
    eliminate false shut off of the power supply. It is required to increase
the operational reliability. 

    If you agree to it, we need only to set the bandwidth of this filter.
When you have it, you have also the peak, width, duty etc. 
    required to maintain this average and keeping low RMS/AVG ratio.

   Yair.

  



> -----Original Message-----
> From:	Schwartz, Peter [SMTP:Peter.Schwartz@xxxxxxxxxx]
> Sent:	ד, מרץ 28, 2001 6:55 PM
> To:	Yair Darshan
> Cc:	stds-802-3-pwrviamdi@xxxxxxxx
> Subject:	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
> 		> 
> 		> 
> 		>