Diving Deep Into Flyback Transformer Design

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  • Опубліковано 25 гру 2024

КОМЕНТАРІ • 41

  • @abidinakdag658
    @abidinakdag658 10 місяців тому

    Great presentation. Thank you so much for your sharing

  • @josephbaker9932
    @josephbaker9932 10 місяців тому +7

    Could you verify that the dots on the windings are correct?

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому +4

      Good catch! If you watch the other video on this you'll see that the windings work out correctly, that was my mistake while doing this on the board.

  • @rogersmith5183
    @rogersmith5183 16 днів тому

    Zach, @ 6:44 if I use your calculated 2.57Amps (Iout) and place it back in the formula to verify/determine the original Ispk, I get (2*Iout)/(1-D) = (2*2.57A)/(1-.3) = 5.14A/.7, Ispk = 7.34A Where did that come from?

    • @Zachariah-Peterson
      @Zachariah-Peterson 5 днів тому

      That value is not so important except for sizing discrete semiconductors (e.g., diodes) based on their peak pulse current rating. To protect the switcher on the primary side, the primary peak drain current needs to be below the switcher's aboslute maximum peak current rating. The UCC28881 has a peak drain current rating of 770 mA, and the 7.34 secondary peak current rating with 19.1 turns ratio gives a primary peak current of 385 mA, or 50% margin of safety on the primary side switcher.

    • @rogersmith5183
      @rogersmith5183 4 дні тому

      @@Zachariah-Peterson I will take another look. Thank you for taking the time to respond.

  • @Shampine1
    @Shampine1 6 місяців тому +1

    That's a great explanation, but I was a bit confused with how you got the secondary current to find the I out. Can you provide the formula you used ?

    • @Zachariah-Peterson
      @Zachariah-Peterson 5 місяців тому

      The secondary current is based on a target, or based on the allowed current limit in the primary side switcher. If you are targeting for example a 3 A current, then you have to size the switcher such that it can handle the peak primary current. For this example, we based the primary peak current on the specified peak current limit from the UCC series datasheet. Then using the turns ratio and duty cycle, you can calculate the secondary current.

  • @commonwombat-h6r
    @commonwombat-h6r 10 місяців тому

    a very good video! Thank you

  • @nulldev42
    @nulldev42 10 місяців тому +1

    Awesome video and I'm looking forward to the results from the build. One question that is a bit off-topic, what is the name of the music track that you use for your openings? I'd love to hear more of it. Thanks!

  • @asidesigner8542
    @asidesigner8542 10 місяців тому

    Zack, please show how to use calculators for it like ExcellentIT, thanks for sharing🎉

  • @Yorumcu63
    @Yorumcu63 7 місяців тому

    Great video

  • @alirezaabasi.
    @alirezaabasi. 10 місяців тому

    Hi great work 👍 I know my question is kind of out of subject but for high frequency puls transformer like 300kHz how can l do the calculating?

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому +1

      The same method applies. I did this for 62 kHz.

  • @hamidrezafallahian7752
    @hamidrezafallahian7752 9 місяців тому

    Hello, thank you very much for your good training🙏🙏🙏🙏🙏
    Sorry, I had a question
    How did you calculate the NP/NS ratio?

    • @Zachariah-Peterson
      @Zachariah-Peterson 9 місяців тому

      The Np/Ns ratio is calculated based on the duty cycle of the switcher, the input and output voltages, and the voltage of the diode. You can see the formula where it is calculated at 4:50.

    • @blakelight4378
      @blakelight4378 4 місяці тому

      ​@@Zachariah-Peterson Could you explain the proper use of the turns ratio formula shown at 4:50. My calculations come out to 13.534, and the given answer of 19.1 is confusing me as to whether or not this formula will give the proper turns ratio for a flyback?

    • @Zachariah-Peterson
      @Zachariah-Peterson 4 місяці тому

      @@blakelight4378 This is because you used 120 V AC as the input voltage. 120 V is an RMS voltage, once the input is rectified the peak voltage value is 169 V (assuming small diode loss in the rectification stage). This 169 V value is what you use to determine the turns ratio.

  • @mfbaba1467
    @mfbaba1467 6 місяців тому

    Dear zack, a 250 to 900 v input and 24, 18 V and 12 V multi output flyback gas been designed for powering up a pwm unit followed by auxiliary power needs. Switching frequency is 80 KhZ. For these extreme step down conversion ratio some times the switch is getting damaged, though it is equipped with RCD-R snubber. Any suggestion from your side to troubleshoot this concern.

    • @Zachariah-Peterson
      @Zachariah-Peterson 5 місяців тому

      I would suggest it is the peak current issue in the switch, you are clearly overstressing the switch. I would suggest also have an RC snubber in addition to the RCD clamp. Most reference designs only include the RCD clamp, but consider also using an RC snubber. If you google this you should find some example circuits showing where these are connected. You should also find some guides from Ray Ridley, he does a great job of explaining the usage of an RC snubber and the RCD snubber in the same design.

  • @teddyjamilonatefreire8797
    @teddyjamilonatefreire8797 10 місяців тому

    Hello dear Zach, how do you set your output current (Iout)?

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому +1

      There are two ways to go about this. First is to approach is to set the primary side peak/average currents based on what your driver or switcher can handle. Once you have that number you can determine the DC output. The other approach is to set the output current to a target value (you are using this as a design target), and then find a switcher/driver and a transformer turns ratio that can reach that value.

  • @sc0or
    @sc0or 3 місяці тому

    "Back" means dots are at opposite sides of the coils -)

  • @MohitSharma-eu9kl
    @MohitSharma-eu9kl 3 місяці тому

    How do you calculate secondary peak current ?

  • @nehayadav2862
    @nehayadav2862 10 місяців тому

    How you get the 19:1 turns ratio. I get the 13.53 After calculation.

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому

      At 9:26, you will see the inductances listed on the white board. Turns ratio is SQRT(820/2.25) = 19.1. That gives the 57:3 ratio I am targeting in the design.

  • @abbasamir4878
    @abbasamir4878 10 місяців тому

    how u get peak current value 770mA and duty cycle value 50% as these are not given in data sheet ?

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому +1

      On page 4 of the datasheet:
      Aboslute maximum positive drain current single pulse, pulse max duration 25 μs is 770 mA
      On page 6, the typical maximum duty cycle ranges from 45% to 55%, I so I listed 50% since that is the midpoint.

    • @abbasamir4878
      @abbasamir4878 10 місяців тому

      @@Zachariah-Peterson thanks

  • @victorlombardo2396
    @victorlombardo2396 10 місяців тому

    Great video! I assume that for the calculus of critical inductance Ls, the D you are using is 50,as is the critical value? Otherwise I don't obtain the same inductance value, as if I use a D of 30 the formula is Ls= (3.3+0.5)*(1-0.3)^2/2*2.57*62000 which gives an inductance of 5.84 uH

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому

      Yes that Ls value was taken at the maximum available duty cycle as that gives you the true upper limit for your conduction mode across the entire range of duty cycle. If it is in DCM with 50% duty cycle then it is also in DCM with 30% duty cycle, so you should choose an inductance value that is based on the maximum duty cycle since that is the true lower limit for this conduction mode. I should have clarified this a bit better.

    • @scottneels2628
      @scottneels2628 2 місяці тому

      Hi Victor, I'm trying to learn these calculations. Can you please explain where exactly the 2.57 was calculated from? I keep getting lower ?

  • @petersage5157
    @petersage5157 10 місяців тому

    I love that your design criteria are so much lower than the converter's maximum ratings. For power components, if you don't push them anywhere near their limits, they tend to last just a couple days longer than forever.
    Also, they're calling bobbins "coil formers" now?

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому

      Thanks, for power I try to get some derating, especially in this case where you might saturate the core and the vendor is not providing the best documentation on what the saturation flux/current will be. I used "coilformer" because that's what TDK calls it, so does Ferroxcube and a few others. But other vendors still call it a bobbin.

  • @SatyajitRoy2048
    @SatyajitRoy2048 10 місяців тому +1

    Dot represented for flyback is incorrect.

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому

      In haste on a whiteboard sometimes you put the dot in the wrong spot. But if you look at the design files for this design you'll see the circuit with the correct polarity.

  • @klauskragelund8883
    @klauskragelund8883 10 місяців тому

    I am pretty sure the formulas you are using are applied incorrectly. The formulas are for CCM, but it is stated that it will be running in DCM. In DCM it is not a volts seconds balance, it becomes an energy balance. If the load is reduced, the voltage goes to infinity.

    • @Zachariah-Peterson
      @Zachariah-Peterson 10 місяців тому +3

      The peak current equation, which was used for determining the secondary inductance limit, is for DCM. Rohm has a good article showing the formulas. Once you get the secondary inductance and the turns ratio you need to step the voltage, you can then get the primary inductance.