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I am reading about flyback converter design and it is often said that the flyback converter should operate either in DCM or CCM. However, I understand that flyback converters can operate in CCM at low input voltage (high load), which mode the transformer is designed for and switch to DCM at light load. The transition between the two modes is called BCM.

My question is: If I chose an inductance for CCM, how do I calculate the "critical" input voltage when the flyback operates in BCM? Thanks.

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  • \$\begingroup\$ For many applications, you don't, you simply need to deal with the change in behavior when you transition between DCM to CCM and vice versa, The BCM point will depend on both output voltage, output current and input voltage. \$\endgroup\$
    – winny
    Commented May 10, 2022 at 12:30
  • \$\begingroup\$ Let's say the output voltage and output current are the same. The load being constant, there must be an input voltage where the inductor core demagnetises, right? \$\endgroup\$
    – Ultra67
    Commented May 10, 2022 at 12:38
  • \$\begingroup\$ Correct. If all other parameters are fixed, only input voltage is left. Have you tried to simulate it? \$\endgroup\$
    – winny
    Commented May 10, 2022 at 12:52
  • \$\begingroup\$ Yes but I cannot find the theoretical calculation. When the input voltage increases the primary rms current decreases and at some point the flyback enters DCM. \$\endgroup\$
    – Ultra67
    Commented May 10, 2022 at 13:18
  • \$\begingroup\$ You can derive it from U = L di/dt. I don't have it on hand, but you should be able to find it. \$\endgroup\$
    – winny
    Commented May 10, 2022 at 13:20

1 Answer 1

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If I chose an inductance for CCM, how do I calculate the "critical" input voltage when the flyback operates in BCM?

As the input supply voltage rises, the converter will leave CCM and enter DCM. This is because the energy needed by the load per switching cycle can be fully delivered by the charge period of DCM.

That energy per cycle (for a constant load) needs to be kept constant. If it isn't kept constant then the output voltage across the load will vary and, of course, we don't want this to happen.

So, the formula that might help you is this: -

$$\text{Boundary duty cycle, }D_B = \dfrac{\frac{N_P}{N_S}\cdot V_{OUT}}{\frac{N_P}{N_S}\cdot V_{OUT}+V_{IN}}$$


An example of a 1:1 converter supplying 200 volts DC to a 600 Ω load (66.667 watts): -

enter image description here

When the input voltage is 273 volts, we are still just about in CCM. You can see this on the image above; the load power is 66.667 watts and the boundary power transfer limit that can be delivered is 66.624 watts.

This means it has to operate in CCM (to get that extra little bit of power to service the load). But, if the input supply voltage rises just a tad to 274 volts, the boundary power transfer limit rises to 66.830 watts i.e. fractionally above what is required by the load (66.667 watts) hence, to stabilize the output voltage, the converter must "drop" into DCM.

So, to calculate that boundary power limit you use this formula: -

$$\text{Boundary power limit, }P_{B} = \dfrac{V_{IN}^2\cdot D_B^2}{2\cdot L_P\cdot F_{SW}}$$

Using the two formulas listed above you should be able to combine them to get what you want. Each formula is derived on the example link from my basic website.

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