# Calculating total impedence....

Discussion in 'Electronic Design' started by Paul Burridge, Oct 28, 2003.

1. ### Paul BurridgeGuest

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

2. ### MikeGuest

I say forget the j's, and stick with s.

Looking in at point A, toward the 33R resistor:

Z(s) = R33 + sL + R50k/(sR50kC + 1)

R33 == 33R
R50k == 50k
C == 25pf
L == 10uH

If you insist on combining everything together:

Z(s) = ((sL + R33)(sR50kC + 1) + R50k)/(sR50kC + 1)
= (s^2(LCR50k) + s(R50kR33C + L) + R33 + R50k)/(sR50kC + 1)
= ((R33 + R50k)(s^2(LCR50k/(R33 + R50k) +
s(R50kR33C + L)/(R33 + R50k) + 1)/(sR50kC + 1)

Now you can substitute s = jw:

Z(jw) = ((R33 + R50k)(1 - w^2(LCR50k/(R33 + R50k) +
jw(R50kR33C + L)/(R33 + R50k))/(1 + jwR50kC)

Is that what you were looking for?

-- Mike --

3. ### Paul BurridgeGuest

"s"?? What's "s"??
To me, Z(s) means source resistance. I already know that. It's 50
ohms.
And "s" is....what???
No.
Listen, Mike, I'm sure you're trying to be helpful, but none of the
above makes the slightest bit of sense to me. Thanks for your efforts
but I'm still none the wiser.
Can anyone else explain it in a simpler way? It's hardly rocket
science after all. For a start, is it okay to work out the impedence
of the series elements seperately from the parallel elements and just
add the two afterwards? I end up with 629 ohms for the series part and
625 ohms for the parallel. But since I expect Z(total) for the two
combined to be around the same as Z(source) this seems way off beam.
Any ideas?

4. ### Kevin AylwardGuest

"S" is the circuit expresd via Laplace transforms.

http://www.anasoft.co.uk/EE/laplace/laplace.html
Once you undertsnd the laplace bit, you plug in s=jw.

You then need to know complex math, i.e. how to handle (a+jb)/(c+jd)
If you cant understand how to handle (a+jb)/(c+jd), we are not going to
get any further with this problem. Its a must. Do you understand this or
not?

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

5. ### John WoodgateGuest

I read in sci.electronics.design that Paul Burridge
>) about 'Calculating total impedence....', on Tue, 28 Oct
2003:
That's the trouble with offering too-sophisticated solutions. If Paul
can't recall how to work out a rather simple impedance, what's the point
of introducing a more advanced technique without any explanation?

's' is the complex frequency variable s = [sigma] + j[omega].
'Z(s)' is 'Z as a function of s'. The source impedance would have the
symbol 'Zs', with s maybe as a suffix.
Yes, it is, but the damage has been done.
Yes, provided you do it correctly. You have to take the phases into
account.
You haven't taken the phases into account. You can do it two ways,
either expressing the impedances as magnitude and phase OR as real and
imaginary parts. Since you mentioned -j and +j, I suppose you are
reasonably happy with real and imaginary parts.

25 pF at 10 MHz is 637 ohms, so you can probably forget the 50 kohms for
practical purposes. That makes it a lot easier. 10 uH at 10 MHz is 628
ohms, so the circuit is nearly series-resonant. You need to be careful
here. The residual reactance is 628 - 637 = -9 ohms, which is a
capacitive reactance, and the capacitance value is 1.77 nF (1770 pF).
Yes, it's a LARGE value, because the residual reactance is SMALL.

We don't know the resistance of the 10 uH coil so we can only assume
that it's low compared with 33 ohms. So the impedance at A is 33-j9
ohms.

If you really wanted to take the 50 kohms into account, you would best
switch to admittances. The conductance of 50 kohms is 20 uS
(microsiemens) and the susceptance of the capacitor is 1/637 S = 1570
uS. So the total impedance is 1/(20 + j1570) Mohms.

This requires a bit of complex number lore to change to (20-
j1570)/(20^2 + 1570^2)**. This then evaluates to 8 + j 637 ohms. OK, 8
ohms isn't all that negligible compared with 33 ohms, but you now know
how to allow for it anyway. The Q of the coil matters as well: it may
well not be satisfactory to assume that its series resistance at 10 MHz
is the same as the d.c. resistance, and there may be losses that are
best expressed as a parallel resistance as well. But we can only use the
data we are given.

There is also a loss resistance associated with the 25 pF, and expressed
as parallel resistance it might not be negligible compared with the 50
kohms (or, expressed as a series resistance, it might not be negligible
with the 8 ohms). Again, we don't have data on that.

** We need to make the denominator real, so we do this:

1/(x + jy) = (x-jy)/{(x - jy)(x + jy)} = (x - jy)/(x^2 + y^2)

6. ### Paul BurridgeGuest

Right, thanks for the link. I'll check it out. I've often heard the
term 'Laplace' bandied about but have no knowledge of it as yet.
I don't have any problem with algebra, Kev. I'm just wondering though,
why this Laplace business seems to be the preferred way of solving the
problem. What advantage does it confer over the '+j/-j' way of
approaching such problems?

7. ### Paul BurridgeGuest

Fair enough. I'd never heard of it.
Yes, I did remember that much but seem to have forgotten *how* to take
phases vectorially; i.e, the correct outcome being given by the square
root of the sum of the squares. Does that not apply in this instance?
Thanks, John. Obviously I'm going to need a while to absorb this lot
and look into the background of this Laplace lark as well. It could
take a while. :-/

impudence? seems like a well behaved network to me
25p at 10MHz => -636.6j
10u at 10MHz => 628.3j

First 25p // 50k

1/(-636.6j) = 1.57 X 10E-3 j
1/(50,000) = 0.02 X 10E-3

add and find reciprocal to get impedence (i assume you have a
scientific calculator if not, multiply numerator and denominator by
complex conjugate
of denominator)

8.11 - 636.8j

8.11 - 8.3j

41.1 - 8.3j

I havent checked this - hopefully the answer is right. But the method
is certainly right.

If you dont have a sci calculator, you might want to use a Smith Chart
for stuff like this to save a lot of calculations. However it is
efficient only if the resistance / impedence values are not too widely
separated from one another (as in this case), otherwise selection of a
normalizing impedence may be difficult.

Thanks
- Kais

9. ### Joe McElvenneyGuest

Hi,

If you haven't got it already, do a search for AppCad 3.0.2 on
the Agilent site. In the engineering tools section there is a RPN
calculator with a parallel-series-parallel conversion function
offering V2.5.1 but if you poke around you will find the later
version.

Cheers - Joe

10. ### Tim ShoppaGuest

The impedance of the 25pF capacitor at 10MHz is 636 ohms (times -i).
This means that the parallel resistance of 50K is almost negligible.

The impedance of the 10 uH inductor at 10 MHz is 628 ohms (times i).

You are very near resonance at 10 MHz... as a result the combination of
the L and the C gives a near-zero impedance. The 33 ohm resistor then
dominates. So effectively at resonance (which is I assume why you're
asking about 10MHz) the stuff to the right of "A" is almost exactly like
a 33 ohm resistor to ground.

Does this help any?

Tim.

11. ### Kevin AylwardGuest

Because the j way is messy on paper. Work out all the stuff with sL,
1/sc and R, then plug in the s=jw at the end.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

12. ### Fred BloggsGuest

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.

13. ### GenomeGuest

Bear in mind that the necro asked for the impedance from point A to ground.
You missed out the R50 from the siggy gen.

Whilst we're on the subject, what's all that knobrot about s?

Taking your sum, plus your nice notation...., ignoring R50, and bending it a
bit...

XA = R33 + XL + XC//R50k

Or, including the 50 ohm sig gen impedance....

XA = (R33 + XL + XC//R50k)//R50

OK, so XC//R50K is,

[R50K/jwC]/[R50k + 1/jwC]

You multiply the top and bottom by jwC. What's that then? some sort of
hidden trick. Whoopsy.... that's multiple negative points for not showing
your method....... and end up with,

R50k/[jwCR50k+1]

Oooh impessive.

XL is jwL and R33 is R33 so, if you wanted to get your original sum

XA = R33 + jwL + R50k/[jwCR50k + 1]

Oh dear, I now have to substitute s for jw

XA = R33 + sL + R50k/[sCR50k + 1]

What the **** changed, should I also convert XA to Z(jw) and then into Z(s)?

That's..... impedance as a function of jw or impedance as a function of s.
Excuse me,

No clothes before..... substituted for...... no clothes afterwards. And then
you still have to go through the shitty algebra with a net benefit of no
extra clothes.

Now, don't get me wrong. Before I switched off as some mathematician went
through the concept of Laplace transforms I managed to hear something about
it converting things from the 'have to divide and multiply in the frequency
domain' to 'only need to add and subtract in the time domain'. Which really
sounds super spiffy wonderful, like log tables, but I never understood the
point of it.........

And, in this case, I still don't. Seriously it's got to be as much use as
farting in an upside down bucket. Unless, of course, you do so with your
left trouser leg rolled up above your kneecap.

DNA

14. ### GenomeGuest

So Mr French bloke was just saving ink?

QED

DNA

15. ### Bill SlomanGuest

Your circuit includes a capacitor and an inductor. To calculate the
impedance oof the whole circuit you have to recognise the fact that
the impedance of a circuit including reactive elements has to be
represented by a complex number

a+ib

where "i" is the square root of minus one.

The impedance of a capacitor at a particular frequency f Z(c)= -i/wC

where w is the frequency in radians per second, which is two pi times
the frequency in Hz, and c is the capacitance in farads.

The impedance of an inductor is similarly Z(L)= iwL

Complex impedances, so defined, can be added in series and in
parallel, just like simple real resistances.

"s" is just electrical engineers shorthand for iw or 2.i.pi.f.

Hope this helps. But I guess that what you really needed was a good
night's sleep.

16. ### Eugene KaplounovskiGuest

Use full impedances for all components (Zr=R; Zc=1/(jwC); Zl=jwL).
Then combine the full impedances in usual manner, i.e.
Zpar=(Z1*Z2)/(Z1 + Z2);
Zseries= Z1 + Z2. Use standard complex number arithmetics.
This is what the previous poster has suggested.
- work out impedance of parallel combination of resistor and
capacitor;
- add impedance of series inductor;
- add impedance of series resistor. This is your impedance from point
A toward the right. If you want to include the generator's impedance
(50 Ohm), it would be parallel to the whole thing as impedance of the
voltage source is zero.

Regards,
Eugene.

17. ### GenomeGuest

My apologies, of course what you meant to say was

Substitute s for jw to begin with.... do the sums and then.....

1/s^n becomes -j/w^n when n is odd.
1/s^n becomes 1/w^n when n is even.
1/s^n is -1 when n is 0
s^n becomes -jw^n for n is odd.
s^n becomes w^n for n is even.
s^n is 1 when n is 0

ARGGGGHHHHHHHHHH!!!!!!!!

It's so bloody simple I'm surprised I ever managed without it. All I have to
do is remember the transformation bits and I will be saved. Thankyou,
thankyou, thankyou!!!!

DNA

18. ### GenomeGuest

to do is remember the transformation bits and I will be saved. >Thankyou,
thankyou, thankyou!!!!
Oh shit, did I miss a minus sign somewhere?

DNA

19. ### John WoodgateGuest

I read in sci.electronics.design that Paul Burridge
>) about 'Calculating total impedence....', on Tue, 28 Oct
2003:

[big snip]

I mean you were already confused.
Sort of, but if you are happy with using 'j', do it that way.
You don't need to bother with Laplace unless you are going to study what
happens to LCR circuits with non-sinusoidal signals, such as steps,
impulses and rectangular waves. While you can use Laplace with sine-
waves, you don't need to, especially not if all you use it for is