[SI-LIST] Re: impedance and Characteristic impedanece
- From: agathon <hreidmarkailen@xxxxxxxxx>
- To: wy_k@xxxxxxxxx
- Date: Tue, 10 Apr 2007 23:50:35 -0700
The Zin = 0 for that case. Makes sense; it's the reciprocal.
(Same positive current defined as into CCVS.)
The derivation is left as an exercise for .... Arpad. :-)
The Thevenin source is solved with KVL and you get Zin=0.
If you transform to Norton equivalent you get the original infinite Zin.
These are just the native source impedances of the two types: ideal V
source's Zin = 0, ideal I source's Zin = infinity.
It seems that what either version of the problem is out to prove, by
isolating the source from its load through the -1 control factor.
-Agathon
On 4/10/07, Kevin Ko <wy_k@xxxxxxxxx> wrote:
>
>
> 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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