[SI-LIST] Re: Effective Radius of Decoupling Capacitor
- From: steve weir <weirsi@xxxxxxxxxx>
- To: si-list@xxxxxxxxxxxxx
- Date: Tue, 27 Aug 2013 06:57:46 -0700
Jonathan:
For what it is worth (not much) the equation is not quite complete:
radius = 1/(200 * 2pi * Fsrf)
should be:
radius = propagation velocity / (200 * 2 pi * Fsrf)
or
radius = c/(eR)^0.5 / (200 * 2 pi * Fsrf )
Fsrf is in Hz
200 and 2 pi are dimensionless.
Honestly, I don't know why anyone would want to use this formula to make any
decisions.
Steve.
On 8/27/2013 2:39 AM, Jonathan Riley wrote:
> Hi Steve
> I would be grateful if you could clarify the units being used in the
> equation given in the second scenario. For example, if I try this with the
> familiar 100nF decouplers (self resonant frequency around the 10MHz mark),
> the value for the radius becomes incredibly small.
>
> Like you say, rules of thumb should be used with caution. Could you suggest
> things to check for that would warn when the limits for this rule are
> exceeded?
>
> Could you also recommend any papers/books that look into this in detail?
>
> Regards
> Jonathan
>
>
> On Tue, Aug 27, 2013 at 10:21 AM, steve weir <weirsi@xxxxxxxxxx> wrote:
>
>> There are a couple of different time/frequency dependent mechanisms: 1)
>> That arises due to the transmission delay between a power disturbance
>> and the capacitor, and 2) that arises from the frequency dependent
>> impedance of the capacitor.
>>
>> 1) If you consider a situation where a bypass capacitor is some distance
>> from a load, the capacitor itself cannot deliver charge to that load
>> faster than the round-trip transit time through the interconnect between
>> the capacitor and the load. For a capacitor to be effective at all its
>> impedance needs to be low compared to the load. A quarter wave
>> resonator translates a low impedance at one end: the capacitor to a
>> high impedance at the other end: the load. Therefore a given capacitor
>> placed 1/4 wavelength away from a load actually raises the impedance
>> seen by that load. For things with very fast edges this distance is an
>> issue.
>>
>> 2) Above the capacitor's SRF, a capacitor's impedance quickly becomes
>> inductive, and eventually the impedance rises linearly with frequency.
>> But just as important is the effective impedance and delay of the
>> interconnect, which is why we physically distribute bypass capacitors in
>> the first place. A fixed relationship such as radius = 1/(200 * 2pi
>> Fsrf) is an attempt to state a coarse distance away from the capacitor
>> where it will still appear capacitive. It is based on a presumption of
>> the interconnect characteristics and like any rule of thumb should be
>> used with caution.
>>
>> Steve.
>> On 8/27/2013 12:52 AM, Jack Si wrote:
>>> Hi experts,
>>> I read from an application note that the effective radius of the
>> capacitance if 0.005*lamda. lamda is the actual wavelength of the
>> capacitor's resonance frequency.i.e, 2pi*vp*sqrt(LC). where vp is the
>> propagation velocity. From this equation, i infer that the inductance(ESL)
>> and capacitance are directly proportional in square root. i.e, radius
>> increase with the increase of ESL or C.
>>>
>>> But in paper "Effective Decoupling Radius of Decoupling Capacitor" i
>> found it is inversely proportional. Please suggest me where i miss the way.
>>>
>>> Thanks and Regards,
>>>
>>> Jack
>>>
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>>
>> --
>> Steve Weir
>> IPBLOX, LLC
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>
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--
Steve Weir
IPBLOX, LLC
1580 Grand Point Way
MS 34689
Reno, NV 89523-9998
www.ipblox.com
(775) 299-4236 Business
(866) 675-4630 Toll-free
(707) 780-1951 Fax
All contents Copyright (c)2013 IPBLOX, LLC. All Rights Reserved.
This e-mail may contain confidential material.
If you are not the intended recipient, please destroy all records
and notify the sender.
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