DC-DC Flyback magnetics selection (esp, inductance)

When choosing magnetics for a flyback type DC-DC converter with Vin=36-57V Imax=400 mA DC to 3,3V, with a Pmax=12,95W. Is the inductance of the pulse transformer calculated as below..? If not, how does one calculate it ..?

L_s = (V_ut*T_s*(1-d_min)) / (2*I_s*I_r)

L_s - Inductance on secondary [H] V_ut - Output voltage [V] T_s - Cycle time [s] d_min - Minimum duty cycle I_s - Power through secondary [A] I_r - Allowed ripple [A]

L_p = sqrt(L_s^2 / (N_s/N_p) )

L_p - Inductance primary [H] L_s - Inductance secondary [H] N_s - Turns secondary N_p - Turns primary

Reply to
kgll8ss
Loading thread data ...

If it is a flyback then, by definition, there is no secondary current whilst the primary is 'charging'. The primary inductance is give by: L=N^2.AL

Reply to
RHRRC

Ofcourse current in secondary is only present when current in primary is removed. But rather when the current is present in the different windings that they follow the formulas pointed out. I'm looking howto find out which inductance to look for rather than how the inductance of a specific transformer coil is calculated.

Reply to
kgll8ss

This is an oddly arranged formula with some ambiguous terms. I_s and I_r would have to be defined in peak or rms values in a particular circuit branch. The fact that a maximum discharge period is used doesn't make much sense, except that it would include the worst-case flux excursion in an IET ie continuous current application. There is really no such thing as a minimum duty cycle in a flyback circuit, until a lot of other parameters have been rigged.You should maybe quote the text and page number if you want further advice on its application.

The secondary inductance determines the slope of the output current waveform during energy discharge. As L reduces, the closer the circuit runs towards CET or discontinuous operation, the lower the voltage stress during the charging period and the higher the peak to average current will be in output components and windings. The discharge period is minimum (at low line - hence T_s(1-Dmax).

When power transfer alone is considered, secondary and primary inductance are not inherently dependant due to decoupling in the power transfer.

The primary inductance can store the required power if it can be induced to sufficient peak current to satisfy the formula:

W = Lmax x Iminpk^2 x f / 2

W = required power Watts Lmax = maximum permissible primay inductance Henries Iminpk = minimum primary current required Amps f 0perating frequency in Hertz

(from e = L.I^2.f/2 energy formula)

This is most difficult at low line

Vmin = Lmax x Ipk x f / Dmax

Vmin = minimum primary voltage during the storage period.Volts Lmax = inductance limit to achieve necessary current. Henries f = operating frequency Hertz Dmax = maximum duty cycle before enforced limiting

(from v = L.di/dt induction formula)

Note that both formulas can be stated in terms of either L or Ipk. Through substitution, a quadratic equation is possible for fixed power, voltage, frequency and duty cycle for the primary winding.

RL

Reply to
legg

Go to

formatting link
and look for a flyback assistance software that they have there. Is really good.

I use the N67 material from Epcos.

I recomend you to read about "Gap" in flyback ferrite transformer design.

Reply to
steven.cano

I'm pretty well acquainted with most versions of this kind of SW.

The OP has not ventured to question the best methods for establishing gap or turns on his transformer. This would be the next logical question.

Using SW-assisted magnetic design GUI's can produce rubbish, if the basics behind them aren't understood by the end-user. This is particularly true if the software is formulated around a specific application IC or schematic. This is one of the reasons why I asked the OP where his original formula came from.

With a basic understanding of the flyback transformer's application and some simple, universally applicable relationships, you'd likely not need the GUI. It looks better in your lab notes as well, being traceable, checkable and reusable.

RL

Reply to
legg

RL, I got the formula from a degree thesis written about DC-DC in the

300W class. What I'm trying to determine is the proper parameters for the transformer magnetics in a flyback converter. Not how to build the transformer magnetics. So parameters like L [Henries], N [Ratio], Ipk [Ampere] etc.. are the one's I'm seeking. I found the energy principle lately a good way to analyze this lately.

Another puzzle is how to analyze the input filter choke, there's a inductor with 10uH in series and then a 26,7 uF capacitor (48V). I calculate I need only about 1uF. So is the inductor choke there to handle the transient surge that the capacitor can't handle due ESR ..?, because there's simple not enough energy in the inductor to take the load of charging the transformer primary.

Reply to
kgll8ss

Could you please quote the Author's name, the thesis title and year, and the institution holding it on record?

There are increasingly severe restrictions affecting the simple flyback topology at higher power levels, although I have used it with some modifications, and seen similar work, above 3KW.

A study of higher power flyback applications that does not include a physical iteration of the magnetic structure runs the risk of not being taken seriously.

Minimization of core and copper losses, with rationalization of the resulting stress affects on the surrounding components, would need to consider, at least in passing, the physical turns count of the winding and the flux densities anticipated in the core.

If the self-resonant frequency of the filter components or parasitics is anywhere near the operating frequency of the converter, there will be undamped resonances that can interfere with normal or efficient operation. Input filter inductors are not normally configured to enter into the conversion process itself, in the flyback topology.

At 48V, a 300W flyback will be drawing large pulsating currents that a

1uF supply decoupling component might have difficulty handling, hence a larger value.

The fact that the component is specified with 3 significant figures of accuracy in capacitance is an indication that it's value does not originate in a practical situation. Power electronic components are seldom available in tolerances better than 5% at room temperature.

Theseseses often make interesting reading. Their value is often not in the actual practical results recorded - in fact the best may describe a complete experimental disaster and point to important errors in innitial assumptions or quoted references.

RL

Reply to
legg

The first paper gives the inductor for the secondary in equation 2.1, but it seems to be half-forward topology (now that I looked again), I had trouble finding the original document. But here it is: ftp://ftp.elteknik.chalmers.se/Publications/MSc/Karlsson&RinnemoMSc.pdf (in Swedish)

Here's another document I looked at for hints (ignore the CAN stuff):

formatting link
(in English)

My application is Uin between 36-57V, 48V DC nominal. Max 400mA. No

300W :) Output 3,3V-12V depending on what turns ratio and voltage divider I choose. Topology: isolated flyback transformer. So I'm looking for how to figure out the acceptable range for the primary inductor Lp in [Henry]. Secondary inductance is also of interest ofcourse.
Reply to
kgll8ss

Well, I recommend you either study unfamiliar topics in a language that you have no trouble with or stick to a topic you're familiar with in an unfamiliar languages.

This paragraph preceding equation 2.1 states that a relationship will exist between the primary and secondary inductances. It also assumes, given the 300V intermediate bus being maintained by this battery-backup port, that the secondary current will vary between 2.2 and 3A peak and that the 300uF capacitor, which just happens already to by available, can handle the stress.

Equation2.1 simply states the expected relationship between the primary and secondary inductances that the preceding paragraph assumes, however the symbol 'r' in the original equation is intended to be a dimensionless decimal number symbolizing a ratio of rms to peak ripple current.

The equation predicts that the peak to average current ratios will change as the inductance ratios or duty cycle limitations change. It's not a cookbook formula to estimate either current or inductance.

Because of the decoupling nature of the flyback circuit, the output voltage is not closely dependant upon the turns ratio. You should be able to make one transformer cover this output voltage range, unless transformer size or component cost is critical.

A flyback circuit capable of producing 12V at a specific power level, will likely produce a 3V3 output at the same power level, providing that the rectifier current rating, output capacitor ripple current rating and secondary winding copper cross-sectional area are suitable. Flybacks are not the first choice for lower voltage circuits due to the high peak to average ratio, which predicts a tendency to higher rms losses in copper or other resistive circuit elements.

The 2A output at 12V will be less stressful than the 3V3 output at 7A.

I'm listing some web references that you might find useful:

SEM300 topic8 slup072 Switching Power Supply Design Review - 60 Watt Flyback Regulator Raoji Patel and Glenn Fritz.

formatting link

SEM400 topic2 slup076 Filter Inductor and Flyback Transformer Design for Switching Power Supplies Lloyd Dixon Jr.

formatting link

Section 5 slup127 Inductor and Flyback transformer design

formatting link

There are a lot of other archived app notes from the TI-Unitrode seminars at their web site.

formatting link

notably the magnetics design handbook

formatting link

RL

Reply to
legg

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.