[SI-LIST] Re: Drivers on a 50Ohm line
- From: Clifford van Dyk <clifford@xxxxxxxxxxxxx>
- To: si-list@xxxxxxxxxxxxx
- Date: Thu, 29 Jul 2004 00:19:12 +0200
Hello Nitin
The most basic answer to your question, is that a 50 ohm PCB trace is
not 50 ohm at DC (i.e. a static signal), but rather somewhere close to
0R (copper has a very low DC resistance!). At higher frequencies, when
the signal rise time becomes of the same order as the propagation delay
of the trace (the trace then looks like a classical transmission line),
the trace impedence appears to be 50 ohms (this impedence could be
resistive, reactive etc, depending on the frequency and trace geometry).
It is the edges (that contain high frequency components if one uses
Fourier decomposition) of the signal that see a 50 ohm impedence. To
avoid reflection (energy loss/partial power transfer) at the boundary
between the driver/receiver and the trace, the trace and driver/receiver
ideally need to be matched in impedence. If thsi cannot be achieved,
termination methods (series/parallel termination) are used to get as
close to this ideal as possible.
Kind regards,
Clifford
Bhagwath, Nitin wrote:
>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 =3D 50mA =
>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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