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Half wave rectifier with ideal transformer

  1. Jun 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine the rms values of the currents of primary and secondary windings of transformer in the network
    circuit diagram. Transformer is perfect, with ratio 6: 1,and negligible
    AC component of magnetizing current, is connected to the sinus supply network U1 = 220 V,
    f = 50 Hz.Ld = 40 mH, R = 0.5 Ω

    2. Relevant equations


    3. The attempt at a solution
    power_eletronics.png

    My questions is how did we get square wave forms for the currents i1 and i2 , it really confuses me( I found waveforms in the book) .Shouldn't current i1 be a sine current with 90 angle delay behind voltage source.
     
  2. jcsd
  3. Jun 12, 2016 #2

    cnh1995

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    Is because the inductance is very large compared to the resistance. If you calculate the time constant L/R, it comes out to be 80ms, which means the sinusoidal steady state will be reached in 400ms (5*time constant), which takes 20 cycles of input voltage. Here, at the beginning of every cycle, a new transient starts, hence the inductor never really attains a steady state. The transient exponential behavior of the current can be approximated to a square wave since rate of change of current is small due to high time constant. Also, current through Rd can be approximated to constant dc.
     
    Last edited: Jun 12, 2016
  4. Jun 12, 2016 #3
    But the how do you explain current i1?
     
  5. Jun 12, 2016 #4
  6. Jun 13, 2016 #5

    cnh1995

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    I1 is reflection of i2. So, when i2 becomes 0, i1 should also be zero.
     
    Last edited: Jun 13, 2016
  7. Jun 15, 2016 #6
    Thank you for your answer , transformers are really my weak spot :/
     
  8. Jun 22, 2016 #7

    rude man

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    This circuit is very hard to analyze not because of the transformer but because the diodes introduce nonlinear behavior. This problem is best (probaly solely) solved with software like SPICE.
     
  9. Jun 25, 2016 #8
    Yes nonilinearity of the diodes produces really non intuitive waveforms
     
  10. Jun 25, 2016 #9

    jim hardy

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    The academic way is to write differential equations and solve for i(t) , which defines both the startup transient and steady state.

    Steady state is a lot easier

    i think in pictures and figure out the formulas
    here's an approach to get you started

    examine the circuit
    can we assume ideal diodes too ?
    What do we know? I wrote it in brown.
    homwrkhalfwave1.jpg

    Ideal diodes allow no forward voltage drop so we get at junction of the diodes perfect half wave rectified 20Vac and i assumed RMS.
    But since it's now DC (lumpy but unidirectional) we should switch to average
    average of a halfwave rectified sinewave is 0.318Vpeak, see http://my.ece.ucsb.edu/York/Bobsclass/2B/Extras/Half-Wave Rectifier.pdf
    and Vpeak is VRMS X √2

    20 X √2 V 0.318 = 8.99 (to slide rule accuracy of 3 figures)

    and that'll be the average voltage at junction of diodes and Ld.

    Now an inductor cannot oppose DC beyond a few time constants so the average voltage at other side of Ld must also be 8.99 volts
    meaning average current through Ld and Rd must be by Ohm's law 8.99 /0.5 = 18.0 amps
    During positive half cycles the current increases to a value above average
    during negative half cycles it decreases to a value below average
    with a time constant L/r = ##\frac {0.04} {0.5} ## = 0.08 sec

    homwrkhalfwave2.jpg

    Between positive half cycles , while current is decreasing, , i = imax et/0.08
    in 30 miliseconds , just a little longer than 1/2 cycle, it'll decrease to Imin = ~68% of whatever Imax was, 32% decrease
    so we can estimate that current swings between Imax and Imin , roughly Iavg ±16% , as sketched above.

    Of course Ohm's Law dictates voltage across Rd must do exactly the same.

    .......................................................
    Why the circuit works:
    During positive half cycles the inductor absorbs energy from the supply and transformer
    during negative half cycles the inductor delivers some of its absorbed energy to Rd

    Of course at startup, current will start from zero and increase during every positive half cycle, after several cycles approaching this steady state sawtooth looking waveform , ±16% about average .

    .....................................................

    Now you have an intuitive :"Sanity Check" method to compare against whatever Pspice (or your differential equation) tells you.
    That's how i always had to flush out my algebra mistakes, and later on when QBasic and computers replaced my slide rule, my programming errors .
    ..............................................................................................................................

    Your homework circuit is the basis of many kinds of regulators,
    both simple ones like we just analyzed and
    complex ones where they use switching transistors instead of diodes, so called "Switch Mode Power Supplies" ...

    So it's important that you get accustomed to working it in your head .

    Doubtless your math is fifty years more fresh than mine so i wouldn't presume to lecture you there.
    I offer this "simple thinking" approach to complement your math skill. Maybe it'll help you spot an impossible result before it propagates through an entire derivation.

    Hope it helps.

    old jim
     
    Last edited: Jun 25, 2016
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