[SI-LIST] Re: Effective Radius of Decoupling Capacitor
- From: Jack Si <sij99@xxxxxxxxx>
- To: "istvan.novak@xxxxxxxxxxx" <istvan.novak@xxxxxxxxxxx>
- Date: Tue, 27 Aug 2013 20:40:04 -0700 (PDT)
Hi Istvan,
I put my question in two different views,
1. From the reference, I found that the service area or effective radius is
calculated for the PCB by considering Dk and the rise time of the excitation.
So I infer that the decoupling capacitor should be placed within the service
area so the we can maintain the charge hold by the path. Isn't right?
2. The method ,I brought out taken the effective radius of the capacitor (may
be decoupling or bulk), In case of bulk, how it should be distributed
throughout the plane and for decoupling, how near it should be placed to the
load for its effectiveness. Correct me if my inference is wrong.
Thanks
Jack
________________________________
From: Istvan Novak <istvan.novak@xxxxxxxxxxx>
To: sij99@xxxxxxxxx
Cc: "si-list@xxxxxxxxxxxxx" <si-list@xxxxxxxxxxxxx>
Sent: Tuesday, August 27, 2013 6:28 PM
Subject: [SI-LIST] Re: Effective Radius of Decoupling Capacitor
Jack,
The often mis-used concept of service radius of capacitors dates back to
the days when there were hardly any planes on the PDN, most chips were
fed with wide traces and capacitors were placed as close to the device
pins as possible. To answer the question "How close the capacitor
should be?" under these circumstances, the service area concept might
give some useful insight. With planes and multiple capacitors I find
this concept often very misleading.
It might be more useful to think in terms of extreme-terminated
transmission lines. Imagine that you have a dead short at the location
of the capacitor and you connect this short to the observation point
(pin of chip) with a transmission line, formed by the trace or the plane
shape making the connection. The pin of the chip will see the input
impedance of the shorted transmission line, which, if the line is
electrically shorter than quarter wave (or odd-integer multiples, but we
dont want to go there unless our PDN has to service known comb-line
noise spectrum), will be inductive. The inductive input impedance is
proportional to the electrical distance through the tangent function, so
the farther we place the capacitor, the higher the inductance becomes.
Dependent on where you may loose effectiveness of bypassing, you can
then draw a conclusion of the 'service area' or how far you can place
the capacitor to still do something useful. Instead of using ideal
short, you can also modify the transmission line expressions to include
the realistic impedance of the capacitor.
Though it discusses laminates, the same reasoning can be found in more
detail in
http://www.electrical-integrity.com/Quietpower_files/Quietpower-16.pdf
Regards,
Istvan Novak
Oracle
On 8/27/2013 3:52 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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