[SI-LIST] Re: Drivers on a 50Ohm line

Steve, I'm afraid I don't see what part you're referring to when you say
"No that is not it" since I'm not sure I found much difference from your
reply and what I wrote.

Again assuming Vcc=3D2.5, and Zo=3D50Ohms,

1) Parallel ground (What I called End termination to ground):- I didn't
take the driving low case, because I'm assuming worst case, i.e. driving
high, in which case as we both stated, Vcc/Zo=3D50mA is required.  - I
agree this is a bad choice.

2) Parallel Thevenin (What I called Thevenin end termination):-
Regardless of termination resistor power requirements, the current drive
requirements of the driver stay the same...Vcc/(2*Zo)=3D25mA in my =
example
(assuming equal pullup/pulldown output impedances, and Vth=3DVcc/2).=20

3) I left out Parallel AC, because my understanding is that the driver
current requirement is similar to regular parallel end termination
requirements at high enough frequencies.

4) Source Series:- I follow you, and agree with you for the round trip
waveform you describe.  Now, for the average power, the equation you
stated holds only for 2*Tflight_one_way > Tperiod_average. So:
  a) If 2*Tflight_one_way < Tperiod_average, then it'd be difficult to
make assumptions
     regarding average current requirements, and the safest route would
be to revert to
     the worst case of Vcc/(2*Zo)=3D25mA.

  b) With 2*Tflight_one_way > Tperiod_average being met, there is still
a portion of
     each clock cycle equal to 2*Tflight_one_way where the driver would
need to drive
     Vcc/(2*Zo)=3D25mA (and ideally drive 0mA on the remaining =
portions).


So in any case, my original question still stands.  If we take the 50MHz
5V (twice the Vcc voltage of the above assumption) CMOS oscillator part
I've mentioned below as an example, how can it drive a regular 50Ohm
trace with a max drive spec of 25mA? =20

With Vcc=3D5V, Zo=3D50Ohms,

1) In the end termination to ground schemes, Vcc/Zo =3D 5/50 =3D 100mA =
is
required
2) In the end Thevenin scheme, Vcc/(2*Zo) =3D 5/100 =3D 50mA is required
3) In the source series scheme
   a) 2*Tflight_one_way < Tperiod_average, worst case assumption 50mA is
required
   b) 2*Tflight_one_way > Tperiod_average, 50mA is required for portions
of a cycle,
      which the driver is unable to meet.  So the waveform must become
distorted
      for the time period 2*Tflight_one_way after each transition.

-Nitin

-----Original Message-----
From: steve weir [mailto:weirsp@xxxxxxxxxx]=20
Sent: Wednesday, July 28, 2004 10:27 PM
To: Bhagwath, Nitin; tom@xxxxxxxxxxxxx; si-list@xxxxxxxxxxxxx
Cc: Cherniski, Mike
Subject: Re: [SI-LIST] Re: Drivers on a 50Ohm line


No that is not it.  The classifications commonly found in texts are:

Source series
Parallel ground, usually a poor choice.
Parallel Thevenin, consumes more power than parallel Vtt Parallel Vtt,
parallel termination of choice for performance. Parallel AC, has
multiple caveats but can be used to reduce power consumption.

Skin resistance is only a factor in long or very high frequency traces.

The characteristic impedance does not appear until a low frequency
cut-off=20
that is a function of the trace length.

Your case of resistor to ground, the driver sees 50 ohms and 2.5V / 50mA

when high and 50 ohms and 0V / 0mA ( more or less ) when low.

Source series, you are closer, the driver sees Vdelta through Rterm +=20
Zline.  For the simple case where everything is linear and Rinternal +=20
Rterm =3D Zline ( the way it is supposed to be ) the driver sees +/-25mA
for=20
2*Tflight_one_way.  Average current =3D 50mA * Tflight_one_way /
Tperiod_average.

Most drivers are quite non-linear and drive high currents at a high
Vdelta,=20
and have weak current capability at a low Vdelta.  The current rating is

the continuous capability.

Steve,

At 08:32 PM 7/28/2004 -0700, Bhagwath, Nitin wrote:
>Hi All,
>
>So the three significant point-to-point termination cases are Receiver=20
>termination to ground, Receiver termination to a non-Vcc source=20
>(usually
>Vcc/2) and finally Source Series termination.  I'm again assuming
>Vcc=3D3D2.5 and Zo=3D3D50Ohms.
>
>End termination to ground
>------------------------------
>Here, the driver will see the load more or less as a characteristic=20
>impedance from DC to somewhere where skin effect kicks in (I'd rather=20
>not go there for this discussion).  To drive 2.5V into a 50Ohm line=20
>would require 50mA of driver strength.
>
>
>Thevenin end termination (say to Vcc/2)
>---------------------------------------
>Whether implemented by placing directly to a second Vcc/2 source, or by

>placing a voltage divider from Vcc to ground, this setup still shows a=20
>characteristic impedance of 50Ohms to the driver.  But since the driver

>only needs to drive to half the voltage, it now needs to be able to=20
>drive 2.5/(2*50) =3D3D 25mA
>
>
>Source Series termination
>-------------------------
>Assuming a series termination, which when added to the internal output=20
>impedance of the driver equals the characteristic impedance of the=20
>line, the driver will need to drive an effective load of 2*Char=20
>Impedance =3D3D 100Ohms.  Now, I agree that "over time"/at DC, the=20
>effective impedance of the transmission line changes, but I'm mainly=20
>concerned about the situation over a frequency range where the=20
>transmission line behaves as a resistive impedance (where Zo =3D3D=20
>(L/C)^(1/2)).
>
>Here, if the combined output impedance =3D3D characteristic impedance =
of=20
>=3D the line, then the input to the line =3D3D 1/2Vcc, which reflects =
at=20
>the unterminated end of the line to yield a full Vcc swing.
>
>This situation requires the driver to be able to drive 2.5/100 =3D3D=20
>25mA, similar to the end termination to Vcc/2.
>
>
>
>Where my confusion comes in is that this is that 25mA is a rather large

>requirement for a driver.  I know there are drivers which can drive=20
>more than this, but there are devices which find this requirement to be

>at or above their drive limits.  An example is a KSS oscillator
>(http://202.226.94.5/pdf/e/fxo31fe.pdf) which has 24mA as its drive=20
>capability (it's shown for a 5V CMOS type part, with the 5V just making

>the current requirement more accute!).  I would assume that these=20
>devices are not driven to/past their limits under normal situations (or

>are they?).  If we take a series terminated 5V driver oscillator (I'm=20
>assuming say a 50MHz KSS part from the above datasheet) which can drive

>24mA, then how can it drive a normal 50Ohm transmission line, if the=20
>required current is itself 5/100 =3D3D 50mA?  Wouldn't this seriously =
=3D=20
>affect the voltages at the load?
>
>Thanks
>-Nitin
>
>
>
>-----Original Message-----
>From: Tom Dagostino [mailto:tom@xxxxxxxxxxxxx]=3D20
>Sent: Wednesday, July 28, 2004 12:52 PM
>To: Bhagwath, Nitin; si-list@xxxxxxxxxxxxx
>Cc: Cherniski, Mike
>Subject: RE: [SI-LIST] Drivers on a 50Ohm line
>
>
>Nitin
>
>The answer depends on the full net that the example you site consists=20
>of. If it is a 1 nsec 50 Ohm trace unterminated then an open=20
>transmission line effect comes into play.  The signal that is launched=20
>into the line will have an amplitude of
>
>Vlaunch =3D3D (Vopen * 50 Ohms)/(Zout + 50 Ohms) or the classic voltage =

>divider where 50 Ohms is the line impedance and Zout is the output=20
>impedance of the driver.
>
>If the line is unterminated when the signal reaches the line after 1=20
>nsec it reflects and doubles in amplitude.  The reflection gets back to

>the driver after another 1 nsec and the full open Voltage Vopen will be

>seen at both ends of the line.
>
>If the line is terminated in the characteristic impedance then the=20
>above equation applies at all times.
>
>See the good explanation of this in Howard Johnson's book "High Speed=20
>Digital Design, a Handbook of Black Magic" chapter 4 or any book that=20
>covers transmission line theory.
>
>
>Tom Dagostino
>Teraspeed Consulting Group LLC
>503-430-1065
>tom@xxxxxxxxxxxxx
>www.teraspeed.com
>
>-----Original Message-----
>From: si-list-bounce@xxxxxxxxxxxxx=20
>[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Bhagwath, Nitin
>Sent: Wednesday, July 28, 2004 11:23 AM
>To: si-list@xxxxxxxxxxxxx
>Cc: Cherniski, Mike
>Subject: [SI-LIST] Drivers on a 50Ohm line
>
>
>Hi All,
>
>I have a simple question, for which I'm yet to find a complete answer.
>
>A standard PCB trace has an impedance of 50Ohms.  Does this imply that,

>say, for a 2.5V signal, a driver must be able to drive 2.5/50 =3D3D3D=20
>50mA =3D =3D3D of current on this signal to allow for a full swing of =
the=20
>signal?
>
>If this is the case, then how can a driver, such as a clock oscillator,

>which has a drive strength in the order of 20-30mA drive a buffer=20
>through a 50ohm line without the voltage being attenuated?  This would=20
>get accentuated at 3.3V, or 5V where the current requirement would=20
>increase.
>
>Any thoughts appreciated
>
>Thanks
>
>-Nitin
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