[SI-LIST] Re: Antwort: Re: Questions about interplane capacitance
- From: steve weir <weirsi@xxxxxxxxxx>
- To: Doug Brooks <doug@xxxxxxxxxx>
- Date: Wed, 12 Mar 2008 23:26:10 -0700
Doug, Istvan's representations are analytically exact. When the
characteristic impedance of the transmission structure is high, and/or
the rise times are slow then capacitors can be placed in close enough
proximity that they load the transmission structure so as to make it
appear a lower impedance with some little bumps. For example if one had
a 10" x 10" 4 mil er4.0 board = 22.5nF and loaded it with bypass every
square inch of 150pH, then for slow enough signals, the distributed
impedance would look like 80mOhms. If the network were constructed from
200 capacitors, an ESR value of 1.6Ohms / cap would make that impedance
uniform down to the RC knee of the parts. Assume that were matched by
an AVP regulator of 80mOhms and the entire thing looks like 80mOhms from
DC to over 1GHz. 80mOhms would support a 32 bit transition w/ about 5%
droop. The impedance scales with dielectric height and the inverse
square root of eR. Scale the dielectric down to 0.1mils and now 320
lines can switch simultaneously in one direction with 5% droop, and at
arbitrary edge rates.
Istvan has shown using analysis with the reverse pulse technique an
inductive PDN shunt impedance acts like a noise high pass filter ( See
DC papers from at least as far back as DC East 2005 ). Put in a square
wave noise pulse ( load current ) and the leading edge changes by Vdelta
= -L*di/dt below the baseline. Allow the pulse to persist long enough
and the system recovers back to the baseline which would be -I*Rpdn.
Return the load current to zero, and now the energy stored in the
inductance kicks back -L*di/dt. The p-p noise is then 2*Ldi/dt -
(Imax-Imin)*Rpdn. If Rpdn is very small then it approximates 2*Ldi/dt.
This behavior is apparent in the transient response plots of virtually
any non-AVP VRM.
Now, suppose that the VRM and PDN can be made to appear resistive right
through Fknee. Then the response to a current pulse of I is simply
Vdelta = -I*Rpdn. There is no component of di/dt, and so the total p-p
noise is just (Imax - Imin)*Rpdn. AVP schemes position the DC operating
point intentionally high so that at Imin they are at their margined high
limits and at Imax they are at their low limits. This allows increasing
Rpdn while still meeting the same current and voltage specifications.
Best Regards,
Steve.
Doug Brooks wrote:
> Istvan,
>
> With all due respect, I would modify your argument a little bit. In very
> simplistic terms, suppose we need x amount of charge to transition from a
> zero to a one in one ns. That amount of charge (I suggest) must be within
> 6 inches of the need (what I think we are referring to as the service
> radius). If not, it takes a little longer to reach the logical one state.
> I look at it, not from the standpoint of a dip in the rail, as much as the
> ability to satisfy the rise time requirement (unless you are referring to
> a dip in the rail that occurs during the rise time itself.) In the
> slightly longer term, the charge will replenish fairly quickly, but not,
> perhaps fast enough to meet the rise time requirement.
>
> Doug Brooks
>
>
>
>
>> Andreas,
>>
>> Yes and no. It is true that charge moves with finite speed, so for any
>> given time
>> duration the charge has to come from locations closer than the ratio of
>> distance
>> over speed. BUT the whole notion of service radius is based on the
>> assumption
>> that as you deplete the charge available in the immediate vicinity of
>> the active
>> device, you have to wait for replenishment, otherwise you get a big dip
>> on the
>> supply rail.
>>
>> Having a matched
>> transmission medium to deliver power to the active device, the charge
>> moves
>> without interruption, and as you deplete the planes close to the device,
>> it gets
>> replenished on the fly from areas further away, so the service area
>> concept is
>> pretty much meaningless in this scenario. Current flows without
>> interruption.
>> The bucket brigade of infinitesimally small inductive and capacitive
>> elements
>> of the transmission line transmits the power continuously. If the load
>> current
>> changes, for any I(t) time function of load current, the transient noise
>> at the load
>> point will be I(t)*Zo, where we assume that Zo is the resistive and
>> frequency
>> independent characteristic impedance of the transmission medium. This is
>> a
>> very simplistic one-dimensional model, but it gives a good insight of
>> why the
>> service radius matters only on PDNs where the network is not matched.
>>
>> Regards,
>>
>> Istvan Novak
>> SUN Microsystems
>>
>>
>>
>>
>>
>> Andreas.Lenkisch@xxxxxxxxxx wrote:
>>
>>> Istvan,
>>> I'm wodering a little about your comments to the service radius.
>>> Independant if the impedance is resistive, we have still a propagation
>>> time which would limit the service radius from my understanding.
>>> Do I'm wrong?
>>>
>>> regards
>>> Andreas
>>>
>>>
>>>
>>> Istvan Novak <istvan.novak@xxxxxxxxxxx>
>>> Gesendet von: si-list-bounce@xxxxxxxxxxxxx
>>> 11.03.2008 13:14
>>>
>>> An
>>> Joel Brown <joel@xxxxxxxxxx>
>>> Kopie
>>> si-list@xxxxxxxxxxxxx
>>> Thema
>>> [SI-LIST] Re: Questions about interplane capacitance
>>>
>>>
>>>
>>>
>>>
>>>
>>> Joel,
>>>
>>> Just one quick comments to the good summary from Steve:
>>>
>>> While considering planes and bypass capacitors in terms of effective
>>> capacitances and inductances is a
>>> valid approach, we need to keep in mind that focusing on the capacitive
>>> or inductive nature of parts
>>> without looking at the wider picture misses a very important and useful
>>> class of solutions, namely that
>>> of matched transmission lines. As it was pointed out earlier several
>>> times on the SI list, the best
>>> (self) impedance for a power distribution network is a resistive one,
>>> neither capacitive, nor inductive.
>>> We can get resistive impedance from a matched transmission line,
>>> regardless of its capacitance and
>>> inductance, and in such cases the notion of 'service area' of parts
>>> become meaningless: you can put
>>> bypass components further away from the active devices without
>>> sacrificing performance.
>>>
>>> Regards,
>>>
>>> Istvan Novak
>>> SUN Microsystems
>>>
>>> Joel Brown wrote:
>>>
>>>
>>>> Interplane capacitance is frequently cited as the only effective bypass
>>>> capacitance on a PCB at frequencies above 200 MHz.
>>>> I am currently working on a design which brings up some questions
>>>>
>>>>
>>> regarding
>>>
>>>
>>>> interplane capacitance.
>>>>
>>>> 1. Power planes normally carry "standard" voltage rails that are used
>>>> throughout a board such as +5V and +3.3V.
>>>> High speed ICs usually have core voltages that are local to the IC and
>>>>
>>>>
>>> are
>>>
>>>
>>>> provided by a local regulator which converts the standard rail to the
>>>>
>>>>
>>> core
>>>
>>>
>>>> voltage (example 3.3 to 1.8V).
>>>> The local core voltage is distributed on a plane area that is local to
>>>>
>>>>
>>> the
>>>
>>>
>>>> IC and therefore is small in area (0.25 sq in or less) which results in
>>>>
>>>>
>>> a
>>>
>>>
>>>> very small amount of interplane capacitance.
>>>> Is this very small amount of capicitance effective for bypassing the
>>>> IC?
>>>>
>>>>
>>> I
>>>
>>>
>>>> am sure it depends somewhat on the current waveform being drawn by the
>>>>
>>>>
>>> IC
>>>
>>>
>>>> but this can only be estimated because semiconductor manufacturers do
>>>>
>>>>
>>> not
>>>
>>>
>>>> provide current consumption profile as a function of frequency. To make
>>>> matters worse, some ICs have several different VCC pins which the
>>>> manufacturer recommends connecting to separate networks of bypass caps
>>>>
>>>>
>>> and
>>>
>>>
>>>> ferrite beads. This cuts the power distributuion up even more resulting
>>>>
>>>>
>>> in
>>>
>>>
>>>> less (practically zero) interplane capacitance. It is somewhat ironic
>>>>
>>>>
>>> that
>>>
>>>
>>>> the the voltages such as +5V and +3.3V which are required at points
>>>>
>>>>
>>> across
>>>
>>>
>>>> the whole board and therefore have the most interplane capacitance are
>>>>
>>>>
>>> also
>>>
>>>
>>>> the voltages which have least requirement for interplane capacitance
>>>>
>>>>
>>> because
>>>
>>>
>>>> they do not directly supply high speed rails.
>>>>
>>>> 2. There has been a lot of emphasis on reducing the mounted inductance
>>>>
>>>>
>>> of
>>>
>>>
>>>> bypass capacitors. Even with this reduced inductance they are still
>>>> only
>>>> effective up to several hundereds of MHz at which point the interplane
>>>> capacitance becomes the only bypass capacitance mechanism. However
>>>> there
>>>>
>>>>
>>> is
>>>
>>>
>>>> inductance between the connection of the IC to the planes. This
>>>>
>>>>
>>> inductance
>>>
>>>
>>>> consists of vias and package inductance. I did look for some numbers
>>>> for
>>>> package inductance and did not find much, it seems to be a closely held
>>>> secret. Also it is unknown how much bypass capacitnace is internal to
>>>>
>>>>
>>> the IC
>>>
>>>
>>>> package. Just for example if we assume 250pH for the vias and 500 pH
>>>> for
>>>>
>>>>
>>> the
>>>
>>>
>>>> package, then the impedance at 500 MHz would be 2.36 Ohms. This seems
>>>>
>>>>
>>> rather
>>>
>>>
>>>> high for the interplane capacitance to be of much benefit.
>>>>
>>>> In summary how much interplane capacitance is needed to be beneficial,
>>>>
>>>>
>>> and
>>>
>>>
>>>> why is it beneficial given the inductance in the vias and package?
>>>>
>>>> Thanks - Joel
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
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Steve Weir
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