Archived from groups: rec.audio.tech,sci.electronics.design (
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"Fred Bartoli" <fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> writes:
> "Randy Yates" <yates@ieee.org> a écrit dans le message news:
> ekpc8qz2.fsf@ieee.org...
> > John Popelish <jpopelish@rica.net> writes:
> >
> > > Steve Hill wrote:
> > >>
> > >> Can anyone tell me how I should be calculating Parasitic Inductance,
> > >> Resistance and Capacitance? Any equations would be helpful. I am
> > >> revising for a telecommunications module and can't seem to find
> > >> anything anywhere and with no answers to work with, I am confused
> > >> whether I am doing it right.
> > >>
> > >> Typical question:
> > >> A 10.1nH inductor at 1GHz is purely resistive. The measured value of
> > >> resistance is 127ohms. Calculate the parasitic capacitance. [2 Marks]
> > >
> > > If an inductance (assumed to be a lumped inductance) looks resistive,
> > > then it is being resonated (canceled) at that frequency by an equal
> > > magnitude capacitive impedance. Magnitude of inductive impedance is
> > > 2*pi*f*L. Magnitude of capacitive impedance is 1/(2*pi*f*C). Set
> > > them equal and solve for C.
> >
> > The resistance would be irrelevent for this problem, then, but
> > represents the series resistance of the device.
>
>
> Nope, it's the equivalent parallel resistace of the tank, i.e. Rs*Q^2 (when
> Q is high enough).
> Here, estimating Q ~ Rp/L*w0 = 127/(10.1*2*pi) = 2 is clearly not enough for
> the approximation to hold so you have to do the exact maths.
Thanks for the correction, Fred.
So you model the circuit as a capacitor in parallel with the inductor and
a series resistor, in which case we have a parallel resonant circuit instead
of a series resonant circuit? Yup, that makes more sense.
The problem can then be solved exactly as follows:
1. First calculate the series resistance R. One equation for the
total impedance of the circuit is [1]
Z_T = (R^2 + X_L^2) / R
You know Z_T and X_L so you can rearrange this equation in the form
of a quadratic equation and solve for R.
2. Now plug this value of R into the relationship
X_C = (R^2 + X_L^2) / X_L
Note that I used my trusty old book [2] from DeVry, which is now
almost 30 years old.
Man, I would've flunked this question myself without doing some
serious review. I've had my head in the digital stuff way too
long.
--Randy
[1] I am using the somewhat arcane TeX typesetting system syntax here
in which a "_" is used for subscript and a "^" is used for a
superscript.
[2] "Introductory Circuit Analysis," Robert L. Boylestad (2nd edition)
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124