0
\$\begingroup\$

I'm trying to use the NCP1014 (65kHz) to make a flyback converter. I'm using coilcraft's transformer B0570-BL which is specifically designed for the NCP1014, with turns ratio 1:0.06 or 16.67:1. I intend to take in narrow european mains (actually, narrow Indian mains, 230VAC ~ 310VDC) and generate 1A@12V.

Using the 'textbook' flyback controller equations, I get a duty cycle requirement using that transformer of:

dctyp = N * Vout / Vin
      = 16.67 * 12.7 / 310 = 0.68

The datasheet of the NCP1014 uses its eq.24 on page 17 for operating duty cycle,

dc = Ip * Lp / Vin * Tsw
   = (450mA * 3.4mH) / (310V * 15.38us)
   = 0.321

The two duty cycles are very different. I understand why this is - the duty cycle for the NCP1014 is set only by the time it takes for primary current to ramp to 450mA. What I can't work out is how the output regulation will work for this, given that the required duty cycle is more than the cutoff.

Given the duty cycle the NCP1014 will cut off at, will the output have enough energy to sustain a stable 12V output? How does one make sure this will work for a given output voltage (12V, in this case)

I have attempted to work though the datasheet in nearly its entirety to the best of my understanding. I have put it up here, if it will help.

\$\endgroup\$
1
  • \$\begingroup\$ The definition of the duty ratio you give is for a continuous-conduction-operated (CCM) flyback converter. For the (high) input voltage you state, it is very likely the converter operates in discontinuous conduction mode (DCM) and the duty ratio definition for a given output voltage is different. The second formula states the maximum peak current of this switcher which is unlikely to be hit in normal operation. It is however a good starting point to see what maximum duty ratio you may expect in DCM. \$\endgroup\$ Apr 17, 2020 at 8:32

1 Answer 1

1
\$\begingroup\$

When operating in DCM (and that is what this design does), the voltage transfer is this: -

$$\dfrac{V_{OUT}}{V_{IN}} = N\cdot D\sqrt{\dfrac{R_L}{2\cdot F_{SW}\cdot L_P}}$$

Where: -

  • \$N\$ is the turns ratio as a fraction i.e. 1/16.67 (0.06) for your case
  • \$D\$ is the duty cycle percentage
  • \$R_L\$ is your load resistance on the secondary
  • \$F_{SW}\$ is your switching frequency i.e. 65 kHz
  • \$L_P\$ is your primary inductance

The important thing to note is that the voltage transfer is load dependent unlike when operating in CCM.

How does one make sure this will work for a given output voltage (12V, in this case)

Try plugging in the numbers into the above equation but do take account of what your maximum and minimum loading values are.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.