FreeLists / si-list / [SI-LIST] Re: Ideal driver characteristics

["Ingraham, Andrew" <Andrew.Ingraham@xxxxxxxxxx>]

> We were talking about driver characteristics a few weeks back, and Arpad
> questioned why someone would consider a current source to be an ideal
> characteristic driver.
> 
> I'm wondering the same thing.  Since a current source's output is
> independent of the voltage at its output, it seems to me that such a
> driver
> would do a pretty poor job of absorbing reflections.  In fact, it seems to
> me that such a driver would't absorb them at all, it would reflect them
> the
> same as an open-ended line.  Am I correct in this assumption?
 
You are correct.  A current source driver is not intended to terminate a
line.  Ideally it is like an open circuit.  If you wanted your driver to
terminate reflections, then a current source itself is not a good choice.

But there might be other reasons for wanting a current source driver.

For example, if you have a multi-drop line with separate terminations on the
ends, current-source drivers might look pretty good.

Or if you had bipolar transistors.


> Looking at the IBIS curves for any number of models, it seems to me that
> the
> transistors are usually saturated, and therefore acting as current
> sources,
> when they first turn on.  Thus, it follows (I think) that any reflections
> that hit the driver when it is first transitioning are likely to be
> reflected instead of absorbed - providing one very good reason to avoid
> line
> lengths that are 1/2 of the cycle time.
 
I don't know if it is valid to deduce the switching behavior (including the
impedance while switching) based on the static I/V characteristics.  I kind
of doubt it.

If you really want a driver whose impedance is linear and time-invariant,
you may need to start with a non-switching linear circuit or RF amp.


> It seems to me if that I'm looking for an "ideal" driver, I'm looking for
> one whose V/I characteristic is a staight line from the zero-current point
> out to the maximum current (driving + absorbing reflections) at which it
> will ever be operated.  I'm not likely to find such a device, but that
> seems
> to me to be the perfect "linear" driver we'd like to have.
 
That is not sufficient.  What happens to the instantaneous V/I
characteristics as the gates of the output transistors ramp up or down?

Here is a simplistic example.  Let's say you had push-pull output
transistors with your "ideal" V/I curves.  If one switches off a nanosecond
before the other switches on, then you have a nanosecond "dead" period when
neither transistor is on and the driver provides no termination.  If offset
in the other direction, then in addition to the momentary "crowbar" current,
you have a nanosecond where both are on and the impedance is half what you
want it to be.  Getting the impedance to not change, requires careful
timing, in addition to having linear devices.  (Or, making your devices not
switch off at all.)

Andy

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