# selection of transformer for flyback converter

I am designing a flyback converter. Now, necessary parameters are as stated -

Vin(voltage to primary) = 17-32V, Vout(voltage to secondary) = 12V, Iout = 3Amps.

Now, based on the above requirements, I calculated the following parameters for the flyback transformer -

Pout = 36W,

Assuming 80% efficiency, Pin = 45W.

Iaverage = Pin/Vin = 45/17 = 2.647Amps.

Iavg 0 Ipk*(Duty cycle)*0.5. Where D - 50% duty cycle. Hence-forth , Ipk(peak current) = 2.647/(0.5*0.5) = 10Amps.

So, V = LdI/dT => L = 17(ton)/dI = 8.02uH. The inductance o0f the primary coil.

Nr(ratios) = Nsec/Npri = 0.70588 => Nsec = (Npri)*(0.705). Now, with the above values, I tried selecting a flyback transformer.

I am looking at the transformer PA3856.006NL.

Now, is my approach correct ? If so, in the transformer I have show, where is the turns ratio specified ?

Also, what is the actual advantage of have a dual output in the secondary ? Does it have to do with the current on the output and its effects on the transformer saturation current.

• Um, PA3856.006NL is a transformer for a forward application (note "Forward Transformer" beneath the diagram and note the phase dots). They're not interchangeable. – Adam Lawrence Mar 21 '16 at 16:14
• Yes, my bad, its written forward transformer. But, the comment about dots dont stand as the same then will apply to the flyback ones shown in the same datasheet - productfinder.pulseeng.com/products/datasheets/P719.pdf Also, are flyback and forward transformers interchangeable in applications ? – Board-Man Mar 21 '16 at 16:23
• Absolutely NOT interchangeable! The construction and energy-handling requirements are totally different between forward and flyback transformers. – Adam Lawrence Mar 21 '16 at 16:26
• @AdamLawrence. Thank you for the advice. In the same family, there is a PA3855.006NL transformer(flyback). Now,, my calculated Nsecondary is 0.705w.r.t primary. or, Npri = 1.4184. What if I select a primary of larger turns, say 2.5 turns ? – Board-Man Mar 21 '16 at 16:41
• Also, what does 0.47 stand for in the primary ? Is it saturation current or turns in the coil ? – Board-Man Mar 21 '16 at 17:15

The PA3856.006NL can be used as a flyback transformer. Note 2 in the data sheet states this: -

1. For flyback topology applications, it is necessary to ensure that the transformer will not saturate in the application. The peak flux density (Bpk) should remain below 2700Gauss. To calculate the peak flux density use the following formula: Bpk (Gauss) = K1_Factor * Ipk(A)

Given that the peak primary current is calculated as 10 amps on the lowest input volts, Bpk (Gauss) = 41.7 x 10 = 417 gauss and this is well short of 2700 gauss.

Now, is my approach correct ?

Your approach does seem correct but you forgot to mention operating frequency however, i reasoned that the on time is 4.72 us if that helps.

where is the turns ratio specified ?

The turns ratio is 0.47:1 as shown by the little numbers on the picture of the transformer in your question.

Also, what is the actual advantage of have a dual output in the secondary ? Does it have to do with the current on the output and its effects on the transformer saturation current.

Dual output means you can wire them in series or in parallel. Saturation of the core is caused by primary current reaching too high a value.

• Saturation of the core is caused by primary current reaching too high a value. No? Saturation of the core is caused by having the core eat up too much voltage-time before it is magnetized in the other direction. You have to limit this voltage-time integral over a half-period. – Janka Apr 6 '18 at 13:13
• @Janka the two are totally inter-related. You can convert volt seconds into ampere turns and get the same answer. Do the math. Magnetic flux is caused by current (biot savart law) but can also be regarded as volt-seconds. No, I'm not wrong. Also look up core BH curves and see what causes saturation (ampere turns per metre = H) – Andy aka Apr 6 '18 at 13:25
• @Janka consider this: V = L.di/dt therefore V.dt = L.di or, if we reduced L to being proportional to turns squared you find that volt seconds per turn = ampere turns. – Andy aka Apr 6 '18 at 13:32
• This is a never-ending discussion, seen that at another place. The simplest way to calculate it is integrating the voltage curve over time during a half wave. That value has to be smaller than the Webers from the particular core datasheet (or Gauss/Tesla value times the cross section of the core.) – Janka Apr 6 '18 at 14:00
• Of course so why were you questioning my use of amps in the primary if you've seen this before (at another place). Sorry for asking but I get it in the head once or twice a year from @Neil and he just doesn't seem to comprehend that the two are the same and if I choose to use ampere turns or amps in describing a magnetic field then who can argue (or more importantly why should anyone argue?). – Andy aka Apr 6 '18 at 14:14