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Thread: SMPS Transformer secondary vs output voltage for high current

  1. #1

    SMPS Transformer secondary vs output voltage for high current

    Hello
    i have at hand 0-50v variable lab psu,i measure schottky diode output pwm which is after connect transformer secondary.i check waveform with oscilloscope i saw amplitude about about 80v and at max voltage 50v i can see duty cycle about %60

    when duty cycle %10 i can see output voltage is around 8v, so 80v amplitude /duty cycle = output voltage

    i would like to know for get 5v high current like 100-200A which amplitude value is better?

    100v %5 duty = 5v
    50v %10 duty = 5v
    30v %16 duty = 5v
    20v %25 duty = 5v
    10v %50 duty = 5v

    which value is better?if i can know so i can change secondary winding

  2. #2
    .... Silvio's Avatar
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    Hello Yokan, It does not work the way you are expecting. The duty cycle of the pwm varies not only to maintain the voltage but also to maintain the current drawn at any point. If say you have 5v and no load at the output the duty cycle will be very small like 5% for example but if you start loading the output voltage will drop and then the pulse width will grow larger to maintain the voltage. The more load you put on the larger pulse width it will be because the voltage tend to drop more as the load (Amps) are increased.

    I do not know what power the variable power supply can give but if you are aiming for 5v @ 200A then the power supply must be capable of 1200w or more (5v x 200A = 1000W) considering an efficiency of 80%. It also depend on the duty cycle of the load. For example if you want continuous load at 1000w then the transformer and the mosfets or transistors and also the heatsinks must be capable for this kind of load.

    Regarding the secondary peak voltage of the transformer I guess you will need 20v or so to maintain the 5v at that current you are mentioning. Your windings must be made of copper sheet not wire to get that current otherwise they will get very hot and cannot carry the current needed. The diodes for your secondary must also be very powerful. The output inductor must also be suited for the new current and must be large enough to carry the current needed. The feedback loop must also be changed for the new voltage.

    It is not so easy to change all these things as everything has to be considered, even the pcb traces need to be adequate to carry that kind of current and lastly you do not know how the power supply will behave because there will be a lot of stray inductance due to the high current involved and this might interfere with the operation of the low current paths controlling the PWM chip.

    I hope this helps you consider a little more before attempting to make this modification.

    Regards Silvio
    Last edited by Silvio; 06-09-2020 at 01:48 AM.

  3. #3
    Quote Originally Posted by Silvio View Post
    Hello Yokan, It does not work the way you are expecting. The duty cycle of the pwm varies not only to maintain the voltage but also to maintain the current drawn at any point. If say you have 5v and no load at the output the duty cycle will be very small like 5% for example but if you start loading the output voltage will drop and then the pulse width will grow larger to maintain the voltage. The more load you put on the larger pulse width it will be because the voltage tend to drop more as the load (Amps) are increased.

    I do not know what power the variable power supply can give but if you are aiming for 5v @ 200A then the power supply must be capable of 1200w or more (5v x 200A = 1000W) considering an efficiency of 80%. It also depend on the duty cycle of the load. For example if you want continuous load at 1000w then the transformer and the mosfets or transistors and also the heatsinks must be capable for this kind of load.

    Regarding the secondary peak voltage of the transformer I guess you will need 20v or so to maintain the 5v at that current you are mentioning. Your windings must be made of copper sheet not wire to get that current otherwise they will get very hot and cannot carry the current needed. The diodes for your secondary must also be very powerful. The output inductor must also be suited for the new current and must be large enough to carry the current needed. The feedback loop must also be changed for the new voltage.

    It is not so easy to change all these things as everything has to be considered, even the pcb traces need to be adequate to carry that kind of current and lastly you do not know how the power supply will behave because there will be a lot of stray inductance due to the high current involved and this might interfere with the operation of the low current paths controlling the PWM chip.

    I hope this helps you consider a little more before attempting to make this modification.

    Regards Silvio
    Thank you a lof Silvio

    i will try with 20v amplitude

  4. #4
    Hi
    Firstly sorry for bad english.I am new in this forum .I am learning SMPS details .
    I was watched @silvio 's Ferrite Transformer Claculation " title video in Youtube as link is
    https://www.youtube.com/watch?v=TLD6hUV64YU

    I think it have a calculation error about Primer winding number not 50 it is 65 spir with this values
    Meanwhile I was say in comments too in youtube ( Pathfinder63 is my comment)
    am I wrong or have another detail in this value .
    (Numerator values :155 volt and 100 million .Denominator values: 4, 33000(frequency) 1800 (Bmax average value) and 1,3 cm2 (Cross Section Area)
    I think this value for primer winding not cricital for transformer and all designs can work .

    Iwill continue calculate by excel .
    Regards

  5. #5
    .... Silvio's Avatar
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    Quote Originally Posted by Flatron View Post
    I think it have a calculation error about Primer winding number not 50 it is 65 spir with this values
    Meanwhile I was say in comments too in youtube ( Pathfinder63 is my comment)
    am I wrong or have another detail in this value
    I just re-worked the sum and in fact is 50turns 155x10^8 = 155,000,000,00

    4 x 33,000 x 1.3 x 1800 = 308,880,000

    If you divide the first number by the second using a calculator the sum comes to 50.181 Turns.

    Regards Silvio

  6. #6
    I was checked again and Iink is is related w was wrong .I don't multiplex with Cross Section area .So calculation is correct.But in some books 4 value was get 4.44 (I think it is related with sinus or square wave signal))

  7. #7
    .... Silvio's Avatar
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    The constant 4 is used for square wave while the constant 4.44 is used for sine wave. In smps which is hard switched is square wave.
    Regards Silvio

  8. #8
    I was watched all video of Mr Silvio.Have not any critical calculation process for this design by Excel.
    I am not sure for this core's energy transfer capability (350 watt) also seconder side so much like linear transformer's bridge filter side (Don't use any shotky diodes)
    Meanwhile I was learn as we must use 4.4 value for at 12:00 calculation for sinus signal ,(4 value for Square wave )

    Summary I will continue searching
    Regards

    Sorry
    Just see
    I was learn now Thank you Silvio
    Quote Originally Posted by Silvio View Post
    The constant 4 is used for square wave while the constant 4.44 is used for sine wave. In smps which is hard switched is square wave.
    Regards Silvio
    Last edited by Flatron; 07-14-2020 at 04:43 PM.

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