Parasitic values

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]



Thanks all.

 

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.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% <> % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

 

| 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.
|
| --
| John Popelish

Is that the same as solving C for XC=127R and F=1GHz.. 1p25?

DNA

 

Not at all. If the inductor looks purely resistive, it is the
inductance and capacitance that have canceled each other, leaving
whatever series resistance the inductor had as the only remaining
visible impedance. The value of that resistance really isn't involved
in the calculation of that capacitance. If no inductance remains,
then XC=XL.

 

| Genome wrote:
| >
| > John Popelish wrote:
|
| > | 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.
| >
| > Is that the same as solving C for XC=127R and F=1GHz.. 1p25?
|
| Not at all. If the inductor looks purely resistive, it is the
| inductance and capacitance that have canceled each other, leaving
| whatever series resistance the inductor had as the only remaining
| visible impedance. The value of that resistance really isn't involved
| in the calculation of that capacitance. If no inductance remains,
| then XC=XL.
|
| --
| John Popelish

Aha!, Trick question, like Randy says the resistance is irrelevant.

Thanks

DNA

 

|
| | | Genome wrote:
| | >
| | > John Popelish wrote:
| |
| | > | 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.
| | >
| | > Is that the same as solving C for XC=127R and F=1GHz.. 1p25?
| |
| | Not at all. If the inductor looks purely resistive, it is the
| | inductance and capacitance that have canceled each other, leaving
| | whatever series resistance the inductor had as the only remaining
| | visible impedance. The value of that resistance really isn't
involved
| | in the calculation of that capacitance. If no inductance remains,
| | then XC=XL.
| |
| | --
| | John Popelish
|
| Aha!, Trick question, like Randy says the resistance is irrelevant.
|
| Thanks
|
| DNA
|
|

And you said it as well.

Thanks again

DNA

 

That's how I see it. The resistance value (and to 3 digits of
precision, too) is misdirection. I am watching to see if someone else
interprets the problem differently.

 

Post ALL the questions. We'll give you the answers and you'll ace the
test!

For this one: The only way the inductor can look like a resistor is if
it is resonant at the applied frequency. A resonant circuit will have inductive
and capacitive elements with identical impedances, of opposite signs, of course.
Calculate the impedance of the inductor at the applied frequency and then
calculate a value for the parasitic capacitance that will have the same
impedance at the same frequency. That resistance value given in the question is
either a red herring or the thickness of your skull.

Perhaps you should switch your career goals over to something like
business administration or sales. "Do you want fries with that order, Sir?"

Jim

 

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,
Fred.

 

Is it? Not for an exact solution, I think. But close enough for the
typical homework problem.


John

 

| On 22 May 2004 06:45:46 -0700, (Steve Hill)
posted this:
|
| >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]
| >
| >
| >
| >Thanks all.
|
| Post ALL the questions. We'll give you the answers and you'll ace the
| test!
|
| For this one: The only way the inductor can look like a resistor is
if
| it is resonant at the applied frequency. A resonant circuit will have
inductive
| and capacitive elements with identical impedances, of opposite signs,
of course.
| Calculate the impedance of the inductor at the applied frequency and
then
| calculate a value for the parasitic capacitance that will have the
same
| impedance at the same frequency. That resistance value given in the
question is
| either a red herring or the thickness of your skull.
|
| Perhaps you should switch your career goals over to something like
| business administration or sales. "Do you want fries with that order,
Sir?"
|
| Jim
|

Well.... I got it wrong......

By the way the correct question is. "Do you want chips with that?"

DNA

 

I assumed that he'd be too embarrased to stay in the UK.

A sailor was berating a new recruit for using the wrong terminology.
"It's not the floor, it's the deck. And that's not the ceiling, it's the
overhead. The next time I hear you screw something like that up, I'll throw you
through that little round window over there!"

Jim

 

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)

 

X_C = R*Z_T/X_L

would be easier.

 

My first night in Taxyourtwoshits we couldn't find a damned place
to get a drink! What a piss-poor place that wuz. Yes, we
eventually found Cambridge (we were staying in Burlington) and a
nice local hole-in-wall.

 

[snip]

Naaah, It's Massa2shits ;-)

By being a student there, and getting an arrest warrant for working on
Sunday and then dodging the fine, I grew to hate the place... damned
church/police state.

Been back there now on business many times, but they still know how to
irritate... went into a hotel restaurant wearing a $1000 Sakowitz
suede jacket, with no tie, and was told I wasn't properly dressed...
what scum... ship 'em to Iraq.

Or just cut 'em off and let 'em float out to sea.

I affectionately refer to Massa2shits as the asshole of the nation ;-)

...Jim Thompson

 

You're probably thinking of Athol, MA. ;-)

Best regards,
Spehro Pefhany

 

Ah, it's just because the place has been settled since 1620. There is alot of
"old things" around. The "Blue Laws" are over, you can now work on Sundays,
and all stores are open. Burlington must be a "dry" town. Some towns allow
selling booze in stores, some towns stores and bars and restuarants, some are
dry.
The hotel must not have liked your looks, most such places offer to loan a tie
in such situations. Was it the Ritz?
Actually, Massachusetts is not that heavily taxed anymore. We even have had
two republican governors in a row. No tax on food or clothes either. The worse
place I've been to was Arkansas.

 

HOW CRASS!!!!!!!!!! You were NOT properly dressed.