[SI-LIST] Re: Why Termination at Both End ?
- From: Rohit MISHRA <rohit.mishra@xxxxxx>
- To: Rajan Hansa <all.si.list@xxxxxxxxx>, "si-list@xxxxxxxxxxxxx" <si-list@xxxxxxxxxxxxx>
- Date: Wed, 10 Aug 2011 16:40:13 +0800
Rajan,
Benjamin Disraeli once said, "Only the fool wonders but the wise man asks" so
your question may look silly But you are a wise chap :-)
I would suggest you that when you start a new question, always start it with
new thread and with new subject in mail so that it'd be easier for group
members to track it down.
Now comes to your question, why distributed model has high bandwidth ? I am not
sure if it's possible to explain that in few words but let me try, To
simplifying things, I assume that it's lossless transmission line i.e. only L &
C and it has a time delay of 1 ns.
Now If you drive a sinusoidal signal of frequency 1 GHZ ( i.e. time period of 1
ns) through this line, you will see that line sees two voltage peaks & two
valleys for this signal. If you try to model this transmission line with LC
sections, you will need minimum 8 LC sections to get that voltage/impedance
profile and with one LC section(lumped model) you will get 1/8 of Bandwith i.e.
1/8 X 1 GHz = 125 MHz so you can see that with increasing LC sections you can
model more peak & valley and that's why more bandwidth.
Regarding your second question, Why minimum 10 LC sections are needed ? As
already explained that minimum 8 LC sections are needed when time delay of
transmission line & BW of signal are same. 10 sections is basically a thumb
rule as it's easy to remember and can be used only when TD & BW are same But if
you increase the bandwidth of the signal, you will need more LC sections i.e.
for 2 GHZ signal, line will see 4 peaks & 4 valleys and in this case you will
need 16 LC sections.
Hope that helps.
Rgds,
Rohit
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of Rajan Hansa
Sent: Wednesday, August 10, 2011 12:44 PM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Why Termination at Both End ?
Thanks everyone for clearing my doubt !!
I have one more question, I know it may look silly and is very basic but I
am sure experts here can enlighten me. Can anyone explain that why it's
said that transmission line can be modeled as distributed RLGC model ? Ok I
know distributed model has high bandwidth but why distributed ckt has more
bandwidth than lumped ckt ? Why minimum 10 LC sections are needed for
transmission line ?
Rajan
On Tue, Aug 9, 2011 at 3:55 PM, Rohit MISHRA <rohit.mishra@xxxxxx> wrote:
> Rajan,
>
> Here's my 2 cents :
>
> I believe you can use a receiver without matching it's input impedance with
> transmission line impedance But Only If the system satisfies these
> conditions :
>
> 1) Transmitter is connected to single receiver.
> 2) Transmission line is a controlled impedance line and has no
> unintentional impedance discontinuities(like via, stub, package lead, etc)
>
> In real system even though board is designed with controlled-impedance
> traces, there is still the opportunity for a signal to see an unintentional
> impedance discontinuity and in this case just source termination won't stop
> the ringing that's why termination at both end.
>
> Rohit
>
>
>
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
> On Behalf Of Rajan Hansa
> Sent: Tuesday, August 09, 2011 1:57 PM
> To: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Why Termination at Both End ?
>
> Experts,
>
> Can anyone explain that why in some designs we see source as well as load
> terminations, I mean termination at both side of traces ? If ringing is a
> issue then only source termination should be sufficient to control it and
> we
> can use a receiver with very high input impedance i.e. no need to match
> input impedance of receiver with transmission line characteristics
> impedance.
>
> Rajan
>
>
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