Full-Wave rectifier with center-tap transformer problem

Discussion in 'Homework Help' started by PsySc0rpi0n, Jul 4, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I'm starting a new problem.

    As the title says it's a Full-Wave rectifier with center-tap transformer problem.
    Teacher is asking us to plot the waveform at the rectifier entrance, the waveform across the load and the current waveform at the diodes.

    Check the attached circuit/screen.

    The info we have from teacher is:
    230V
    1kHz
    10:1 ratio.
    Vdiodes = 0.7V
    R_load = 500Ω

    The first thing I did was to find the amplitude of the rectifier input waveform as:

    (230√2)/(10*2) = 16.26V

    Then the voltage drop across the load will have Vd drop so:

    V_out = 16.26 - 0.7 = 15.56V

    Then I need to find at what time the diodes starts conducting using the following formula:

    Vin*sin (ω*t1) = Vd
    ⇔16.26*sin (2*π*1000*t1) = 0.7
    ⇔sin (2000*π*t1) ≅ 0.0431
    ⇔2000*π*t1 = arcsin (0.0431)
    ⇔t1 ≅ 6.8538 μS

    Then t2 = 0.5mS - 6.854μS = 0.4931mS

    Please someone help me confirming the calcs with LTSpice!

    Thanks
     
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  2. crutschow

    Expert

    Mar 14, 2008
    13,003
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    That looks about right as compared to the simulation.

    Note that, in the simulation, the sine voltage source is specified by the peak voltage so you need to multiply 230 by 1.414 giving 325.2Vpk to get 230Vrms.

    Why don't you do your own simulation?
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I think I was forgetting that detail about red values in LTSpice. I tried the simulation but probably because of that I was not being able to understand the results.
     
  4. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Hi Psy,
    Recall what we said about RMS and Peak, LTS expects Peak in the Sine setup.
    In the Sine set up window use curly braces around the Amplitude(V) value ie: {230*1.414} , this gives Vpeak in.

    NOTE:
    You say V_out = 16.26 - 0.7 = 15.56V, how can that be.? which way around are the rectifier diodes.???

    E
     
    Last edited: Jul 5, 2015
  5. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,184
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    Morning.

    Yes, I'm correcting the Vpeak value!

    You mean, Vout = 16.26 + 0.7 =16.96V???

    The plot shows that Vout is around 15.56V so I thought the math was correct!

    Apart from the problem, I'm having some problems running the simulation. I'm getting some weird small peaks in the Vout and Dx current waveforms. But the very same circuit that someone posted in Yahoo groups that I have downloaded is working perfectly. Wonder why would this be! 2.1 is the circuit from Yahoo Groups, the 2.h is the one I built!
     
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  6. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Hi psy,
    The diodes are connected with their Cathodes to the transformer and the Anodes connect into the load resistor, so which half of the full cycle will they be conducting , so what polarity will the Vout be,??


    EDIT:
    Also consider as there is no smoothing capacitor, you define your Vout result either as Vpeak out or RMS
     
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ahh I think I got it.

    Vout = -16.26 + 0.7V = -15.56V

    Didn't understood your EDIT:. I'vealso edited my last post!
     
  8. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Your weird peaks are due to the .tran setting, remove the max time set, see image.

    Who chose the transformer indcutance values.?
     
  9. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Consider if you read some else's document that Vout was -15.56v

    How would you know that Vout was Peak, RMS or Average.? I would suggest Vout = -15.56V peak.

    Also have you allowed for the 10:1 transformer ratio, that the coupling factor is only 0.98 not 1.0.??
     
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Well, the 0.98 factor was told me to use it because it would be more realistic! I'm still not very sure about how to calculate correct inductances to have the correct ration from primary to secondary!

    And yes, you're right! I should say which value is that of -15.56V. If I say that 230V is RMS, then the peak would be 230*√2. So, as I've done -(230*√2)/(10*2) + 0.7 would also be a peak value that it's what we need to plot the graph!

    About the Maximum Timestep, I removed it and it has no influence in the plot. Also, the .asc file that I've downloaded from Yahoo groups also has that value there.
     
  11. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    The coupling factor of 0.98 indicates a loss of 2% in the transformer ratio.

    Change the 0.98 to 0.49 and rerun the sim, what Vpeak do you get on the secondary.?

    EDIT:
    Study this link.
    http://www.linear.com/solutions/5092
     
  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,184
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    If I change the 0.98 to 0.49 the plot goes a bit weird, I guess!

    Anyway, that's details I can't afford to solve now! My teacher want's to know if we know the math and the theoretic concepts and not if we know how to use LTSpice, although I would like to learn more about LTSpice.

    Now I need to know how can I calculate the current across the diodes. Can I find it by means of the voltage across R_load divided by the resistance of that load? -15.56/500 = -31.12mA???

    About the link on ratios and inductances relationships, I tried to find out how to calculate the correct inductances for let's say a primary coil of 10H but I'm not sure if it is correct (for a coupling with no losses)

    I found that those 2 secondary coils should be using that link's formula:

    L1/L2 = (N1/N2)^2
    10/L2 = (10/1)^2
    L2 = 50mH

    But L2 wil become actually L21 and L22, so L21 = 25mH and L22 = 25mH.
     
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  13. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Assuming both the diodes have identical specifications.
    One diode will conduct on one half cycle and the other diode on other half cycle.

    Which current has the tutor asked you to calculate.?
    eg:
    the total peak current thru the load.?
    the Average or RMS thru the load or thru each diode.?

    BTW: its incorrect to say the current across the diodes.
    [voltage across the diode or current flow through the diodes]
     
    Last edited: Jul 5, 2015
  14. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Yes they are both identical. Further, when the teacher gives us the circuit, the only thing we know is the specs he gives . We do not solve problems for a specific model of diode.

    He do not asks for a specific current. So we assume the peak values, I guess.

    Ok, I'll have that in mind in the future!
     
  15. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, as far as I can tell, looks like I can calculate diode's current by dividing the voltage across the load by the load. I got 31.12mA and LTSpice also says something similar! So I assume I can do it like that.

    The following question is what should be the correct capacitor to place in parallel with the load to achieve a ripple not greater than 10% of it's output voltage! Here's how I did this one:

    10% of Vout = 10%*15.56V = 1.556V

    Vripple = Iout DC/(f_out*C)

    Fout = 2*Fin --> because this is a full wave rectifier.

    So

    Vripple = Iout DC/(2*f_in*C)

    C = Iout DC/(Vripple*2*f_in)
    C = 31.12/(1.556*2*1000) = 10μF

    LTSpice says that Vripple is 1.2567V with a capacitor of 10μF. Hope this is acceptable to say my calcs are correct!


    By the way, the articrafts in the simulation, I found out to be due to the frequency. For a lower frequency as 50Hz, for instance, the articraft is not visible.
     
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  16. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    I think there as many ripple capacitor formula's as there are engineers.

    I find that Q =CV, ie: C= (I * t)/Vripple gives an answer close enough for most practical applications.
    There are so many variables in the charge and discharge of the smoothing capacitor.

    The capacitance tolerance range of electrolytic capacitors can be in the order of +/-30%.

    As a check on your answer, C= (I * t)/Vrip, using your values.

    10% of Vpeak is 1.55v
    f = 1kHz , so t= 500uSecs ; note full wave rectification, so half a cycle.
    Ipeak = 31.1mA

    So, Csmoothing = (31.1mA * 500uS)/ 1.55v = ~10uF
     
    PsySc0rpi0n likes this.
  17. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Great. Nice...

    Now I was asked to plot the current flowing through the diodes and output voltage but I'm going to leave that to discuss with my teacher tomorrow or Wednesday as I have a question for the current.

    LTSpice plots the current (with the capacitor) flowing through the diodes as a positive current. But shouldn't it be negative as is Vout waveform?
     
  18. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    The diode current is negative, it is LTS that shows it in that sense.

    If you want to plot the current it the opposite sense, multiply the D1 and D2 current (* -1)
     
  19. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, that's what I wanted to know. I'm going to start a new thread for another problem!

    Thanks Eric.
     
  20. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi Psy,
    Why are the asc files for this thread being posted on the Yahoo LTS User group, by Helmut and yourself.??
    E
     
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