full wave rectification using center tap transformer

Discussion in 'General Electronics Chat' started by PG1995, Apr 16, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    I'm beginner in this stuff so please your reply simple. Thanks :)

    I did an experiment on full wave rectification using center tap transformer. The transformer read: Input 220V, Output 12V, 12V. I used two diodes and 1 micro farad capacitor. As you see capacitor's capacitance is very little so it didn't effect the overall output because I checked the output without using it and there was not much difference. The peaks had amplitude of 18V. I connected oscilloscope across the resistor. Have a look on the the given diagram:

    http://img233.imageshack.us/img233/4901/cttfullwaverectificatio.jpg

    RMS vs. Peak
    RMS = peak / √2
    12 ~= 18/√2

    Which means my output was correct. Because the transformer had 12V across each of its secondaries.

    I used the formula, ripple factor=IL/FC. IL was almost 12.09 mA. I used the value of F as 60 Hz which was the frequency of AC source and the voltage of AC was 220V.

    I think the instructor gave us the wrong formula because here on Wikipedia it seems Wiki is saying something different:
    http://en.wikipedia.org/wiki/Ripple_...-domain_ripple

    What do you say, what is the real error? Please help me with this. Thanks.
     
    Last edited: Apr 16, 2011
  2. RRITESH KAKKAR

    Senior Member

    Jun 29, 2010
    2,831
    89
    Firstly, your image is not there, and there must be high value capacitor lor large current as 1uf is nothing for any rectifier. you can also use 78xx or other voltage regulator ic for something like pure D.C.
     
  3. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,754
    760
    Do not double post.
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
  5. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi R!f

    Where did I double post? Would you please tell me? Thanks.
     
  6. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    I = C(ΔV/ΔT) rearrange to get:

    ΔV = I(ΔT/C)

    T = 1/F for 1/2 wave ckt, or 1/2F for full wave ckt therefore ripple voltage =


    ΔV = I/2FC
     
  7. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thanks a lot, Jaguarjoe. It would be very nice of you if you could help me a bit more. Please don't forget I'm a beginner.

    What is ΔV?

    What is ΔT?
     
  8. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Δ, or delta, means change, so ΔT is the change in time from the beginning of a particular event to the end of that event. ΔV is voltage change and so on and so forth.
     
  9. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Once again, thanks, Jaguarjoe.

    Okay, delta stands for change. ΔV stands for voltage change. But which voltage? In the experiment I did on full wave rectification the peak voltage on the oscilloscope was 18V. Is the ΔV = (18V - 0V)?

    Is the current, I, the one which runs through the load resistor? In my experiment it was 12.09 mA.

    What is ΔT? Is it the period of the wave shown on the oscilloscope?

    Could you please also tell me what "ckt" stands for?

    I'm very much grateful to you for all this guidance.

    Regards
    PG
     
  10. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    ΔV could be 18v if you were figuring something out that wanted the full change in output voltage from 0 to 18v. But in this case you are only interested in the AC (alternating current) ripple portion of the output. Here, ΔV is the change in ripple voltage from 0v AC at the bottom of the ripple waveform to its value at the top of its waveform. This AC ripple voltage rides along the peaks of the power supply output.

    ΔT is the change in time for one wave (its period).

    ckt is/was my abbreviation for circuit.
     
    PG1995 likes this.
  11. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    Your capacitor was so small that the expression ΔV = I/2FC did not really apply. This is an approximate formula which can only give a reasonable answer on the assumption that I is approximately constant. The load current is only approximately constant if a large reservoir capacitor is used, so that the ripple is a small proportion of the total.


    Let's try a few values, assuming in each case I=10mA, F=60Hz.

    C = 1μF gives ΔV = 83V, which is clearly ridiculous - this is more than the total voltage.
    C = 10μF gives ΔV = 8.3V, which is still nearly half the peak value.
    C = 100μF gives ΔV = 0.83V, a bit more sensible.
    C = 1000μF gives ΔV = 0.083V, which is pretty smooth.
     
    PG1995 likes this.
  12. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    Please have a look on the attached diagram. I think ΔV = V2 - V1. And period is (V2 - V3). Please correct me if I'm wrong. Thanks.
     
  13. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Correct on both counts!
     
    PG1995 likes this.
  14. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thanks a lot, Januarjoe.
     
Loading...