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Parasitic values

Discussion in 'Electronic Design' started by Steve Hill, May 22, 2004.

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  1. Steve Hill

    Steve Hill Guest

    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.
     
  2. 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.
     
  3. Randy Yates

    Randy Yates Guest

    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
     
  4. Genome

    Genome Guest

    | 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
     
  5. 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.
     
  6. Genome

    Genome Guest

    | 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
     
  7. Genome

    Genome Guest

    |
    | | | 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
     
  8. 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.
     
  9. James Meyer

    James Meyer Guest

    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
     
  10. Fred Bartoli

    Fred Bartoli Guest


    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.
     
  11. John Larkin

    John Larkin Guest

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


    John
     
  12. Genome

    Genome Guest

    | 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
     
  13. James Meyer

    James Meyer Guest

    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
     
  14. Randy Yates

    Randy Yates Guest

    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)
     
  15. Randy Yates

    Randy Yates Guest

    X_C = R*Z_T/X_L

    would be easier.
     
  16. KR Williams

    KR Williams Guest

    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.
     
  17. Jim Thompson

    Jim Thompson Guest

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

    Best regards,
    Spehro Pefhany
     
  19. Don

    Don Guest

    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.
     
  20. Fred Bloggs

    Fred Bloggs Guest

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