FreeLists / si-list / [SI-LIST] Re: impedance and Characteristic impedanece

[Kevin Ko <wy_k@xxxxxxxxx>]

Will the answer be same/different with CCVS, instead of VCCS, parallel to 1
ohm?

regards,
Kevin

--- "Muranyi, Arpad" <arpad.muranyi@xxxxxxxxx> wrote:

> Thanks for all the great replies!  I enjoyed
> reading them.  There are many ways to skin the
> cat, right?  Another analysis would be to make
> a Thevenin equivalent of the circuit in which
> the Thevenin source generates the same voltage
> on one side of the resistor that exists on
> the other side, i.e. the current through the
> resistor is always zero.
> 
> Now, regarding how this relates to the original
> topic of T-line impedance:
> 
> The resistor in this "trick question" corresponds
> to the characteristic impedance of the T-line.
> 
> The source (current for Norton, or voltage for
> Thevenin) corresponds to the reflected wave in
> the T-line (with zero delay).  So the apparent
> impedance (which I also like to call "electrical
> impedance") can be calculated by looking at what
> the circuit does, i.e. taking all voltage and
> current relationships into account.  I find this
> example a good way to illustrate what goes on
> in a T-line without having to go into Maxwell's
> wave equations.  To extend this example for
> T-lines, all you have to do is add in some
> delays (or phase if you do it in the frequency
> domain), and you got it...
> 
> By the way, you can turn a capacitor into an
> inductor with an op-amp, configured as a NIC,
> often used for making inductors on the die.
> Same thing as in the above discussion, right?
> You have a physical device, a capacitor, which
> looks completely different in the circuit.
> 
> We could call all of this "electrical illusions"...
> 
> The lesson from this, which is a big pet peeve
> of mine (and I have commented on this before
> in this list), is that when we talk about
> impedance, we must be clear which one we are
> talking about.  RF engineers tend to talk about
> the electrical impedance as it is seen at a certain
> frequency, taking into account all of the reflection
> and standing wave effects, etc..., not mentioning
> this underlying assumption most of the time.
> Board layout guys or time domain thinkers tend
> to talk about the characteristic impedance (most
> often not mentioning that assumption either).
> 
> Imagine what happens when you put these two types
> of guys into the same conversation...  Or imagine
> what happens when a newbie takes two T-line basics
> classes, one from each of these guys.  Mass confusion,
> total nightmare, which I have had the "pleasure" of
> experiencing way too many times...
> 
> I hope this thread helped to clarify a few basics...
> 
> Arpad
> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
> =3D=3D=3D
> 
> -----Original Message-----
> From: Chris Padilla (cpad) [mailto:cpad@xxxxxxxxx]=20
> Sent: Tuesday, April 10, 2007 9:01 AM
> To: Muranyi, Arpad; si-list@xxxxxxxxxxxxx
> Subject: RE: [SI-LIST] Re: impedance and Characteristic impedanece
> 
> The main thing to notice in this simple circuit is the DEPENDENT SOURCE.
> The amps supplied by the current source depend on the voltage developed
> across the resistor.  I'm sure too many folks glossed over that very
> important detail because it appears to be such a simple, "by inspection"
> circuit.  In reality it is simple if you just note that small but very
> important detail!
> 
> I'll generalize the circuit a tad more:
> 
> Let the current source be Alpha*v.  Note that Alpha is in dimensions of
> amps per volt.
> Let the voltage across the current source and the resistor be v.
> Let the resistor be R.
> 
> Now, look into this circuit to figure out the input impedance.  In
> sophomore circuits class, we did this by "hooking on" an arbitray V
> source with a labeled driving i.  Figure out V/i and you have Zin.
> 
> As my circuit theory professor would say:  Now thrash around a bit!
> 
> I did a Kirchoff's Current Law at the top node of the resistor to get: i
> =3D Alpha*v + v/R.  Note that v =3D V.
> 
> I find that V/i =3D R/(1 + R*Alpha).  Plug in R =3D 1, and Alpha =3D -1 =
> and
> sure enough, V/i =3D 1/0. =20
> 
> Note that Polarity is VERY important here!!
> 
> Chris
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