finding the value of C

Discussion in 'Homework Help' started by led23, Apr 27, 2009.

  1. led23

    Thread Starter New Member

    Mar 10, 2009

    using this diagram i can't figure out how to find the value of C. Can anyone help to lead me to a formula to solve this problem?
  2. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    No answer is possible, unless the question specifies the amount of ripple.

  3. led23

    Thread Starter New Member

    Mar 10, 2009
    I forgot to mention the value of the capacitor c , to have a DC voltage across the load equal to 9.5v
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    I don't think this is possible.

    Even with zero capacitance the average DC value will be at least 12.3 V assuming some realistic forward voltage drop in the rectifier diodes.

    Check your original problem description.
  5. led23

    Thread Starter New Member

    Mar 10, 2009
    the only other part of this I am to assume that they are ideal diodes.
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    OK. I did originally assume the primary to secondary ratio was 5:1 for each winding. That's probably incorrect so ...

    Let me go through the circuit behavior in a little detail.

    The primary is connected to a 60Hz voltage supply with a peak of 100V.

    The primary to secondary ratio is stated as 5:1. Assuming this means the total secondary winding is 1/5th of the primary or 20V peak, then the secondary half voltage will be 10V peak. That will then "work" for the expected 9.5V DC load voltage. So on that basis disregard my earlier post.

    To solve your problem requires several assumptions - given there is no other data such as KL7AJ suggests.

    I would then try this approach.

    1. Assume ideal diodes. Hence rectified wave Vmax = 10V.
    2. The DC voltage = (Vmax+Vmin)/2 = 9.5V
    3. Vmin= 9V
    4. The rectified 60Hz sine wave goes from zero to Vmin [= 9V] in 2.97ms
    5. So the capacitor discharge time Δt = 8.33/2 + 2.97 = 7.14ms
    6. Assuming I load is 'constant' = 9.5/330 = 28.8mA
    7. Capacitor voltage droop ΔVc= 1v {from Vmax to Vmin}
    8. For this ΔVc at 28.8mA drain for Δt=7.14ms we require C=IxΔt/ΔVc
    9. C=205.5uF

    Hope that makes sense.
  7. led23

    Thread Starter New Member

    Mar 10, 2009
    I have finally had time to look at this and there is something that bothers me.
    the Time period of a 60 Hz sin wave is 1/60= 1.6 mS, so I am confused how you came up with 2.93? then you say the capacitor discharge time = 8.33 how am I to come up with the number.Sorry for the questions, i am just trying to understand.
  8. Papabravo


    Feb 24, 2006
    Off by a decimal place. 1/60 th of a second is 16.6666... milliseconds
  9. jvjtech


    Jan 26, 2008

    For single-phase AC, if the transformer is center-tapped, then two diodes back-to-back (i.e. anodes-to-anode or cathode-to-cathode) can form a full-wave rectifier.

    My text follows: Refering to the graphic of the recifier using a center tapped transformer on the wiki page you will see the waveform of rectified AC. Without a load the capacitor will charge up to the peak voltage and remain at that level. The load will load down both the transformer (through the diodes) and the capacitor. The capacitor will discharge through the load during the time the voltage less than the AC peak. This discharge formula is given by the RC time constant of RL and C. As the frequency is 60 Hz and the rectifier is a full wave rectifier the time we are dealing is 1/2 of 1/60 or 8.3 ms. The ripple, or the amount the capacitor is allowed to discharge, is given by the target of 9.5 V at the load. t_n_k showed that Vmax is 10V and calculated Vmin at 9V to get Vavg at 9.5V. Regards. JJ
  10. bertus


    Apr 5, 2008

    Read the attached PDF. That will make things more clear.