Input impedance vs imput capacitance

Discussion in 'General Electronics Chat' started by anhnha, Dec 26, 2015.

  1. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    If a circuit has high input impedance, does it mean that the circuit also has high input capacitance?
    Is there a correlation between them?
     
  2. #12

    Expert

    Nov 30, 2010
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    No.
     
  3. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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  4. panic mode

    Senior Member

    Oct 10, 2011
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    you need to consider that circuit as three stages: signal source (producing Vin), interconnect (medium, aka transmission line), load (some VLSI device that reads Vout as input signal).
    load is not represented as a component (impedance). but if we have all things same (interconnect) and ONLY change impedance at Vout, then resitance R of interconnect does not matter (much smaller than your huge Z=Rout which is near infinity) but capacitance of the interconnect itself becomes significant factor.
     
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  5. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    Thanks, that makes sense. Actually, I don't understand the part "high input impedance of the receiver gives rise to high input capacitance".
    I just made a draft calculation. It seems that input capacitance doesn't change.
    I am having a problem uploading image on the site so I uploaded here:
    http://i.imgur.com/FKbWVKd.jpg

    Rint: interconnect resistance
    Rout: resistance of output network
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    Check the units. Capacitance is a real number with units of Farads. Impedance is a complex number whose magnitude has units of Ohms and whose phase is expressed in degrees or radians. How could they be correlated? One is frequency dependent and one is not, again how could the be correlated?
     
  7. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    Actually, I didn't intend to mean correlation that way (maybe that is because of my English). I think a better work is "result in". I was confused if a high input impedance results in high input capacitance.
     
  8. #12

    Expert

    Nov 30, 2010
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    I think the instruction page is wrong, and that would explain why it is difficult to understand.
    I have designed high input impedance circuits with only picofarads of capacitance (because low capacitance was an important part of making the circuit work properly). The fact that high input impedance, "gives rise to" high input capacitance is not (a fact).
     
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  9. sailorjoe

    Member

    Jun 4, 2013
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    That's right, the instructions are imprecise. What they're trying to say is that, no matter what you do, there is always at least a little bit of capacitance at the input to a signal receiver, the part with ideally infinite impedance. In the circuit, R would be infinite. But since C is always there, the large R increases the time constant of R times C. So the high input impedance doesn't really "create" the high capacitance, but it magnifies the effect of the capacitance.
     
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  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Try exam this non-inverter amplifier with open loop gain Aol = 10V/V; and Aol = 100V/V;
    And Cin = 10nF ; Rin = 10kΩ and Rs = 1kΩ;

    aa.PNG

    For Aol = 1V/V the -3dB is at F ≈ 0.16/RsCin ≈ 16kHz and Rin ≈ 10kΩ
    But as Aol increasing the Rin also increases and -3dB drops as if Cin was reduced.
    For example for Aol = 10V/V -3dB is at F ≈ 0.16/(Rs*Cin/2) ≈ 32kHz and Rin ≈ 2*Rin = 20kΩ
    This "2" is a (1 + Aol * R1/(R1+R2))
    And as Aol increase to infinity (ideal opamp) Rin and -3dB rise to infinity.
     
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