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Understanding the specification of flyback Converter

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Advanced Member level 1
Oct 6, 2006
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The following are the design spec of a flyback converter given in the attached file (at page 13).

VIN: 28 V
Output 1: 5 V
Full Load Current: 10 A
Circuit Topology: Flyback, Continuous Mode
Switching Freq, fs: 100 kHz
Desired Duty Cycle: 0.5 at 28V input
Max Ripple Current (delta)lpp: 5 A @ 32V (secondary)
Peak Short-circuit Current: 25A (secondary)
Secondary Inductance, L: 6.8 uH
Max Loss (absolute): 2.0 W
Max Temperature Rise: 40°C
Cooling Method: Natural Convection

I need explanation of the Max Ripple Current, (delta)lpp: 5 A @ 32V (secondary). How ripple currents are defined in flyback converter and on which factors they depend?


  • Flyback.pdf
    2.1 MB · Views: 108

Output caps and low ESR in the regulator loop cause current spikes which is limited by selection of parts and total ESR of each.
Caps are rated by ESR and max ripple current at say 10kHz this affects temp. rise.
Coil, Rs and FET RdsOn are other parameters that affect ripple current as well as external loads and current limits.
You see that the primary and secondary currents of a flyback converter are trapezoidal (CCM) respectively triangle-shaped (DCM). Ripple current is simple the difference of minimum to maximum current of this waveform (designated Imin and Imax in the TI seminar paper).

The current waveform parameters can be calcuated by basic inductor circuit relations, they depend on secondary voltage, inductance, primary and secondary on-time.
So someone is asking you to make a 50W flyback with 96% efficiency? Not feasible at all, regardless of ripple involved.
So someone is asking you to make a 50W flyback with 96% efficiency? Not feasible at all, regardless of ripple involved.
The seminar example is talking about total transformer losses (core + copper), which sounds feasible.
Assuming that also doesn't include losses induced by leakage inductance, then I agree.
A 50W resonant flyback with mosfet rectifier is just about 94-95% efficient, 96 very tricky to get....
Thanks for reply.

In the attached file at page 8 Ferrite Core calculation are given as follows. These following lines are confusing me

The minimum length of the airgap for the core is then:

Ig =0.4*Pi Lpri*10^8/Ac*Bmax^2

Ig = 0.046 cm or 18 mils

An airgap that produces an AL of 100 mH/1000T is larger than this, so that is what is used. The number of turns needed to produce the required primary inductance is:

N = 1000Sqrt(Lpri/Al) = 74 turns


  • Very Wide Input Voltage Range, Off-Line Flyback Switching Power Supply.pdf
    326.6 KB · Views: 62

AL = Ae Uo / Lgap, N^2 x AL = Lpri

It is necessary to have a Lpri and a peak current large enough to get the desired power, so Preq = 0.5 Lpri Ipk^2 freq. for a fully discontinuous design also di/dt = V/L so L must be low enough at the min Vin to ramp the current up to the desired settings, by setting L = v.dt / di and putting this into the power equation you can solve for Ipk and L exactly (there is only one solution). From L and Ipk you can get the gap size, AL and Npri, easy....

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Bmax should be <160mT for 100kHz for a low loss efficient design.
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