+ Post New Thread
Page 2 of 2 FirstFirst 1 2
Results 21 to 25 of 25
  1. #21
    Advanced Member level 5
    Points: 14,284, Level: 28

    Join Date
    Apr 2014
    Posts
    2,239
    Helped
    890 / 890
    Points
    14,284
    Level
    28

    Re: Calculate Capacitance Value from Smith Chart

    Quote Originally Posted by BradtheRad View Post
    I think my simulation is hard to argue with too.
    Here is the mistake in interpretation of your voltage divider experiment:

    You expect the output voltage to be half the total voltage, because R and X are equal.

    But that's wrong. The voltage divider gives R divided by total impedance, and that total impedance magnitude is sqrt(R2 + X2) in this case.
    Last edited by volker@muehlhaus; 19th October 2019 at 09:04.



    •   AltAdvertisement

        
       

  2. #22
    Super Moderator
    Points: 53,103, Level: 56

    Join Date
    Apr 2011
    Location
    Minneapolis, Minnesota, USA
    Posts
    12,982
    Helped
    2585 / 2585
    Points
    53,103
    Level
    56

    Re: Calculate Capacitance Value from Smith Chart

    Quote Originally Posted by volker@muehlhaus View Post
    that total impedance magnitude is sqrt(R2 + X2)
    And yes, I pointed out that formula entering the picture in my post #17. So let's see if things are clearing up. I compared more variations to see what happens.

    At left, equal voltages appear across the 10pF and 77 ohm. As you state, they are not 1/2 source voltage.

    The other layouts are 2 series capacitors, or 2 series resistors. Voltages are 1/2 source voltage.

    Click image for larger version. 

Name:	compare series layouts RC CC 10pF  RR 70 ohms.png 
Views:	2 
Size:	40.2 KB 
ID:	156115

    I believe the RC arrangement above yields 3 dB drop at the rolloff frequency. Output is 0.707 x source voltage, whether taken across C or R.



    •   AltAdvertisement

        
       

  3. #23
    Advanced Member level 5
    Points: 11,565, Level: 25

    Join Date
    Aug 2015
    Location
    Melbourne
    Posts
    1,881
    Helped
    695 / 695
    Points
    11,565
    Level
    25

    Re: Calculate Capacitance Value from Smith Chart

    at the 3dB freq the volts reduce to 0.707 across the C only ....



    •   AltAdvertisement

        
       

  4. #24
    Advanced Member level 5
    Points: 14,284, Level: 28

    Join Date
    Apr 2014
    Posts
    2,239
    Helped
    890 / 890
    Points
    14,284
    Level
    28

    Re: Calculate Capacitance Value from Smith Chart

    Quote Originally Posted by Easy peasy View Post
    at the 3dB freq the volts reduce to 0.707 across the C only ....
    No, Brad ist correct: the magnitude of R and X are equal in his left testcase, and current is equal, so indeed there is 0.7V across each of the components. But these voltages have a 90° phase offset, so that the total voltage magnitude is sqrt (0.7V2 +0.7V2) = 1V



  5. #25
    Advanced Member level 5
    Points: 11,565, Level: 25

    Join Date
    Aug 2015
    Location
    Melbourne
    Posts
    1,881
    Helped
    695 / 695
    Points
    11,565
    Level
    25

    Re: Calculate Capacitance Value from Smith Chart

    apologies - I must be so used to dealing with L & C, certainly the freq most quoted is when R = 1/ ( 2 pi f C ) so the impedances are the same at that freq.

    - - - Updated - - -

    as the freq goes up the volts decrease across the C - I was stuck on that thought too



--[[ ]]--