[SI-LIST] Re: impedance and Characteristic impedanece
- From: Kevin Ko <wy_k@xxxxxxxxx>
- To: arpad.muranyi@xxxxxxxxx, si-list@xxxxxxxxxxxxx
- Date: Tue, 10 Apr 2007 19:26:58 -0700 (PDT)
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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