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Quantity referred to as 'self-capacitance

  1. Apr 10, 2009 #1
    Hi all!

    I've got a question about the quantity referred to as 'self-capacitance'. From what I can gather, self-capacitance is the amount of charge necessary to generate a 1 Volt E-field between an arbitrary surface and a ground plane @ infinity. This doesn't make sense to me, but that's why I'm hoping someone here can help me understand it!

    Typically when people talk about capacitance, it means 'mutual capacitance', the amount of charge @ the surface of the conductor/volt between two conducting surfaces, which makes alot more sense to me...

    How would one measure the 'self-capacitance' of a conductor of arbitrary geometry????
     
  2. jcsd
  3. Apr 11, 2009 #2
    Re: Self-capacitace

    Actually you're right! Capacitance can exist only between two conductors. However, the concept of self-capacitance is purely theoretical, in that the other plate is considered to be Earth...... as earth is a very large conductor and hence is considered to be at zero potential w.r.t a charged conductor.

    However, in pure theoretical sense the other plate is considered to be located at infinity, but that becomes quite an abstract assumption. Hence, Earth is considered as the other plate or conductor.

    Regards,
    Shahvir
     
  4. Apr 11, 2009 #3
    Re: Self-capacitace

    If you use a 1-meg resistance in a real circuit, like in the feedback loop of a high frequency op amp, then the 1 meg resistor should be modelled in SPICE like two 500 k resistors in series, with a shunt 10 pF capacitor to ground in the middle. The shunt capacitance is called the self or distributed capacitance. If you use an air-coil inductance at a very high frequency, it will cross over from being inductive to being capacitive, due to internal turn-to-turn self capacitance.
     
  5. Apr 12, 2009 #4
    Re: Self-capacitace

    This is a good analogy :smile:
     
  6. Apr 13, 2009 #5
    Re: Self-capacitace

    Ok, but I still have a problem!

    How can I measure the self-capacitance of a small length of wire?
     
  7. Apr 13, 2009 #6
    Re: Self-capacitace

    Suppose you have a foot of wire that has a distributed capacitance of 10 picoFarads to the surroundings. If you touch it with a 1000 volt DC source, you will charge the wire to 10 nanoCoulombs. Then if you touch the wire to a charge integrator amplifier you can measure the charge on the wire.

    I attach a sketch of a charge integrator Linear Technology LTC 6084) or equiv. op amp with a 1 picoamp typ leakage current. The 10 nano Coulombs will flow onto the integrating amplifier therough the input resistor and give you 10 volts out (full scale). It will leak off at a 1 pA (1 picoCoulomb per second) rate. Use reset button to zero the capacitor through a 10 meg resistor. Change values of components as necessary.
     

    Attached Files:

  8. Apr 19, 2009 #7

    vk6kro

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    Re: Self-capacitace

    If you have a coil of wire and you measure the impedance of the coil, it should rise uniformly according to the formula for inductive reactance which is
    XL = 2 pi F L

    But when you do this there is always a peak in the impedance which then drops steadily as the frequency is increased.
    This is due to the capacitance between the windings of the coil. It can be reduced by winding the coil with a space between the winding turns. Doing this raises the self resonant frequency of the coil at the expense of making the coil bigger for a given inductance.

    This capacitance is the self capacitance of the coil.
     
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