Calculating total impedence....

[Paul Hovnanian P.E.]

I read in sci.electronics.design that Bill Sloman <[email protected]>


Well, it's lucky that he wasn't a Windows kind of person, otherwise
gravity would keep crashing. (;-)

I think he preferred an Apple.

 

[Jim Thompson]

s = a + jw = a + 2*PI*j*f

This is particularly important if you are trying to solve for a
transient response rather than a steady state solution. If not, the jw
approach results in simpler algebra which may be important in the wee
hours of the morning.

It is quite easy to derive transient behavior from the "S"
representation... just do a partial-fraction expansion and the
exponentials and sinusoids are readily evident.

...Jim Thompson

 

[Baphomet]

I read in sci.electronics.design that Bill Sloman <[email protected]>


Well, it's lucky that he wasn't a Windows kind of person, otherwise
gravity would keep crashing. (;-)

Now, that's funny! ROTFLMAO

 

[Fred Abse]

Hi,

It's midnight here and I'm suffering from brain failure. Can anyone come
to my rescue and show me how to work out the impudence of this circuit
fragment? (I know, but everyone's brain's entitled to not function once
in a while and I'm getting my -js and +js all mixed up). :-(

Sine wave input to left of 50R resistor (representing Rgen) at frequency
of 10Mhz.
Need to know total impedence from point A through to ground. Thanks.

<knackered>



50R 33R 10uH
___ A ___ ___
Sig ---------|___|---------|___|----------UUU--+ input
|
|
|
|
+---+-----+
| |
| .-.
--- | |
--- | |
25p | 50k'-'
| |
| |
| |
+----+----+
=+=
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

The reactances almost cancel.

At 10 MHz:

Taking the parallel arm, we have 50k in parallel with -j636.62, which
equates to 8.1 -j636.52 series.

10 uH at 10 MHz is +j628.32, so the impedance looking right from point A
is:

33 + 8.1 + j(628.32-636.52) = 41.1 - j8.2

I assume you wanted the impedance looking right from point A, not the
impedance including the parallel path through the generator.

I did a spreadsheet for doing series - parallel impedance conversions.
I'll post it to A.B.S.E if anyone wants it.

I did it longhand, too (same result), I don't really want to type all that
out, but if you actually wanted an expression rather than numerical
values.................

E&TE

 

[Baphomet]

I think he preferred an Apple.

 

[Michael]

Great stuff! So the impedance of the network above at 10Mhz is.....?
(shouldn't take you more than 2 minutes, I would imagine.)

91.472R @ -5.142deg
or 91.104-8.198j

 

[Michael]

Great stuff! So the impedance of the network above at 10Mhz is.....?
(shouldn't take you more than 2 minutes, I would imagine.)

Sorrry I didn't pay enough attention, my last post was the impedance
from the point "signal input" not from A

50R//(33R+10uH+25pF//50k) is 22.911R @ -6.137deg (point A)

The impeadance seen at point A without your signal source connected
would be 41.914R @ -11.279deg

 

[Mike]

On Tue, 28 Oct 2003 15:24:33 GMT, Fred Bloggs wrote:

....

This is not how it's done in practice. First off, change that 50K to
60K, as I told him that the DC bias is to be set at 2.0VDC with a
100K+150K. Then you consider the parallel combination of 25p||60K- and
its equivalent Qc of Y/G=60KWC at frequency W. This is an impedance of
60K/(1+jWC60K)=60k/(1+(60KWC)^2)- j*(WC60K^2)/(1+(60KWC)^2) which is
recognized as a series combination of equivalent Rc and Cc, where
Rc=60k/(1+(60KWC)^2) and Cc=C*(1+(60KWC)^2)/(WC60k)^2. Substituting Qc,
these are now: Rc=60K/Qc^2 and Cc=C*(1+Qc^2)/Qc^2. Then at 10MHz, Qc=95,
so that Rc=7 ohms, and Cc=C*1.0001. Taking Cc=C for this high-Qc results
in less than 0.0005 fractional error in Wo or 0.05%- well go ahead and
use 25.0025p if you want. You will have the same effect with L and its
Q. The main thing is that these components Q's be an order of magnitude
in square ratio larger than the circuit Q at resonance.

Well, I dunno Fred, I think you practice somewhere different than I do. My
s-domain transfer function makes sense to me, and I understand it very
well. It works for me, so I'll continue to practice on a different stage.

-- Mike --

 

[John Woodgate]

I read in sci.electronics.design that Michael <[email protected]>

The impeadance seen at point A without your signal source connected
would be 41.914R @ -11.279deg

Yes, but that doesn't help the OP knowing how to calculate it.

 

[Bill Sloman]

I think he preferred an Apple.

But he used it as an analog device, rather than for digital
computation, and the crash was an integral part of his procedure.

 

[Paul Burridge]

I made a number of approximations and at least one big assumption (that you
didn't want to know the precise value of impedance at 10MHz, instead you
really wanted to know it at resonance, which 10 MHz is nearly at).

Others with different experience than mine will tell me it's a mistake
to assume that you meant "at resonance" when you clearly asked about 10MHz.
But *all the other numbers you gave us were only given to 2 significant
digits*...

This problem actually originated from Fred Bloggs' solution to someone
else's problem in the which the OP was trying to achieve a greater
input signal swing to his schmitt trigger. I basically just simplified
Fred's design in order to attempt to analyse how it worked in detail.
I thought it might prove instructive. It certainly has!

Being able to plug numbers into the calculator and get a number out to
8 decimal places does have some value... but in my experience, getting
some understanding out (instead of 8 decimal places) has vastly more
value. Even though my answer probably wasn't even good to two decimal
places

Your approach was of the intuitive kind that a trouble-shooter would
use, I guess. In fault-finding, one doen't need to know the answer to
umpteen decimal places. But what you came up with jogged my memory as
to some long-forgotten stuff, in addition to giving me a valuable,
different way of seeing the problem.
But John Jardine probably sums up problems like this quite nicely with
this little snippet from several months back:
"Calculating real and imaginary circuit values, magnitudes and
phase-shifts for more than a few RCL components is an odious way to
waste a couple of hours and points up the real benefit of using Spice
programs."
Way to go, JJ.

 

[Paul Burridge]

The reactances almost cancel.

At 10 MHz:

Taking the parallel arm, we have 50k in parallel with -j636.62, which
equates to 8.1 -j636.52 series.

10 uH at 10 MHz is +j628.32, so the impedance looking right from point A
is:

33 + 8.1 + j(628.32-636.52) = 41.1 - j8.2

I assume you wanted the impedance looking right from point A, not the
impedance including the parallel path through the generator.

I did a spreadsheet for doing series - parallel impedance conversions.
I'll post it to A.B.S.E if anyone wants it.

I did it longhand, too (same result), I don't really want to type all that
out, but if you actually wanted an expression rather than numerical
values.................

Okay, chaps. I have enough info now from the contributions to this
thread to be able to tacke this sort of problem in future (I
believe!).
Many thanks indeed to all who responded.

 

[Paul Hovnanian P.E.]

It is quite easy to derive transient behavior from the "S"
representation... just do a partial-fraction expansion and the
exponentials and sinusoids are readily evident.

Yes. That's what I was trying to say (sort of). The jw approach is only
useful for steady-state solutions. However, if this is all one wants, it
is somewhat easier to calculate. When someone asks for an impedance,
they usually want the steady-state part.

 

[Paul Hovnanian P.E.]

But he used it as an analog device, rather than for digital
computation, and the crash was an integral part of his procedure.

Thank goodness Galileo didn't use three PCs at the Tower of Pizza for
his gravity experiments.
The Windows machine went down much faster than the Apple. The Linux
machine is still up.

 

[Fred Abse]

Actually I believe it was Heaviside who came up with the notational
trickiness using "S"

It was, but he called it "p"

 

[Jim Thompson]

It was, but he called it "p"

Isn't "P" just a *frequency-normalized* "S" ?

I.E., the center frequency is unity.

...Jim Thompson

 

[gwhite]

Actually I believe it was Heaviside who came up with the notational
trickiness using "S".

I think he came up with "operational calculus," where he used the Laplace
transform. The mathematicians scoffed at him and he in turn called them wooden
headed. Of course, after he died (they couldn't do it before, since they would
have to admit being wooden headed) the math folks figured out "how it worked"
(and it *did* work) and put operational calculus on a rigorous foundation.

 

[gwhite]

Well I've had a peek at your link and one of my reference books and it
appears this Laplace stuff involves calculus. Algebra's one thing, but
I never got around to studying Leibnitz's little contribution to
science and don't plan to start now. So I guess I'm stuck with the
messy way!

The whole purpose of this 's' domain stuff is actually to *avoid* calculus
(specifically solving differential equations). By solving in the 's' domain,
the problem is reduced to an algebraic one. Note that if you just want
frequency response, or circuit impedance, 's' is simply replaced by 'jw'. No
calculus there -- just a substitution. Solving with 's' is a more complete
solution, in the sense that it includes the transient part. Not only that, it
is notationally simpler too.

If one wants the _complete_ (time domain) response, meaning including the
transients, then 's' is not replaced by 'jw' before performing the
transformation to the time domain. Even though the Laplace Transform is by
definition calculus (an integral to be specific), there exists a big
table/library of canned transforms for the purpose of avoiding calculus. So use
*algebra* to put the 's' domain solution into a recognized canned form. Use the
canned form to convert to the time domain. Voilà, no calculus. The Laplace
Transform is your best friend.

 

[Baphomet]

Okay, chaps. I have enough info now from the contributions to this
thread to be able to tacke this sort of problem in future (I
believe!).
Many thanks indeed to all who responded.
--

And if I may go slightly off topic (I've gotten really good at that in my
declining years), I have started to re-read "The World of Mathematics" by
James R. Newman which I haven't opened since I was sixteen. It's more than a
tool...it's a foolosophy....er.....philosophy.

 

[Paul Burridge]

Sorrry I didn't pay enough attention, my last post was the impedance
from the point "signal input" not from A

50R//(33R+10uH+25pF//50k) is 22.911R @ -6.137deg (point A)

The impeadance seen at point A without your signal source connected
would be 41.914R @ -11.279deg

Thanks. I've managed to dig up a Smith Chart from my other computer's
hard disk. It'll be instructive to see if the graphical approach I'm
about to take here will agree with the predictions of your
arithmetical one.