[SI-LIST] Re: Drivers on a 50Ohm line
- From: "Bhagwath, Nitin" <nitin.bhagwath@xxxxxx>
- To: <tom@xxxxxxxxxxxxx>, <si-list@xxxxxxxxxxxxx>
- Date: Wed, 28 Jul 2004 20:32:51 -0700
Hi All,
So the three significant point-to-point termination cases are Receiver
termination to ground, Receiver termination to a non-Vcc source (usually
Vcc/2) and finally Source Series termination. I'm again assuming
Vcc=3D2.5 and Zo=3D50Ohms.
End termination to ground
------------------------------
Here, the driver will see the load more or less as a characteristic
impedance from DC to somewhere where skin effect kicks in (I'd rather
not go there for this discussion). To drive 2.5V into a 50Ohm line
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
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
drive 2.5/(2*50) =3D 25mA
Source Series termination
-------------------------
Assuming a series termination, which when added to the internal output
impedance of the driver equals the characteristic impedance of the line,
the driver will need to drive an effective load of 2*Char Impedance =3D
100Ohms. Now, I agree that "over time"/at DC, the effective impedance
of the transmission line changes, but I'm mainly concerned about the
situation over a frequency range where the transmission line behaves as
a resistive impedance (where Zo =3D (L/C)^(1/2)).
Here, if the combined output impedance =3D characteristic impedance of =
the
line, then the input to the line =3D 1/2Vcc, which reflects at 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 =3D 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 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
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
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
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
required current is itself 5/100 =3D 50mA? Wouldn't this seriously =
affect
the voltages at the load?
Thanks
-Nitin
-----Original Message-----
From: Tom Dagostino [mailto:tom@xxxxxxxxxxxxx]=20
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
of. If it is a 1 nsec 50 Ohm trace unterminated then an open
transmission line effect comes into play. The signal that is launched
into the line will have an amplitude of
Vlaunch =3D (Vopen * 50 Ohms)/(Zout + 50 Ohms) or the classic voltage
divider where 50 Ohms is the line impedance and Zout is the output
impedance of the driver.
If the line is unterminated when the signal reaches the line after 1
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 above
equation applies at all times.
See the good explanation of this in Howard Johnson's book "High Speed
Digital Design, a Handbook of Black Magic" chapter 4 or any book that
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
[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 =3D3D 50mA =
=3D
of current on this signal to allow for a full swing of the 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
through a 50ohm line without the voltage being attenuated? This would
get accentuated at 3.3V, or 5V where the current requirement would
increase.
Any thoughts appreciated
Thanks
-Nitin
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