I read in sci.electronics.design that Paul Burridge
[email protected]>) 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].
To me, Z(s) means source resistance. I already know that. It's 50
ohms.
'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.
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?
Yes, provided you do it correctly. You have to take the phases into
account.
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?
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)
