transformer magnetizing inductance vs magnetizing current

I have 2-switch forward converter operating at 83khz frequency. Load current must be 10A maximum D=0.5 I2=Iload*sqrt(D) I2=7.1A transformer turns ratio is 2:1, so I1=3.55A

The voltage applied to primary is 300V (220*sqrt2)results in value ner 300V. I have ETD59 core. lets calculate E per turn: Assume that flux swing is 0.2T (Br is 0.066T, Bmax is 0.266T, so dB is 0.266T) Lets calculate E per turn: E(1 turn)=dФ/dt=deltaB*S(core)*f/D

D max for forward converter is 0.5. OK

So, E per turn looks like 12.21 V. Ok.

Lets calculate primary turns: 300/12.21=24.57 turns of primary (assume Np=26 turns). OK

The transformer must be 2:1, So there must be 13 turns of secondary (Ns=13)

The next part is the part that i dont understand: magnetizing current depends on value of magnetizing inductance, right? And magnetizing inductanse is Al*Np^2 (Al for ETD59 core is 5300+-20%). OK.

If i assume that we have the worst case, Al(worst)=4240nH/turn So, calculating magnetizing inductance gives me Lm=Al(worst)*Np^2=2.86mH Is this value ok? I have also found a formula for calculating minimum value of magnetizing (primary) inductance:

Here, Drmin is minimum Duty ratio, Vsmax - maximum supply voltage, fs is a freqeuncy and delta ilmag is a swing of current flowing through magnetizing inductance. Is value of magnetizing inductance so necessary? If i have a transformer which has 26 turns of primary and 13 turns of secondary, is magnetizing current a concern? Is it a good idea to increase the number of turns to get higher value of inductance? This should reduce magnetizing current... Where am i wrong? If i assume that Drmin is 0.15 (i just assume - i am not sure), Vsmax=400V, f-83kHz and delta i(lmag)=0.355A (its 10 percent of rms primary current), it gives me Lmag=2mH So, is it ok? If i would done calculation i described before for a transformer which operated at 100kHz of frequency, I get 16 turns of primary and 8 turns of secondary... So, if i calculate value of magnetizing current the same way (it is MINIMUM inductance required?) i get inductance value L(100kHz)=Al(worst)*16*16=1mH So, it looks like according to formula for Lmagmin, this value isn enoght to provide 0.355ma of magnetizing current Im totally stuck. And does the value of magnetizing current affects core saturation? If it is MINIMUM required value of inductance, more is better?

• Try thinking of it as for any voltage fed converter, it's the voltage-time area which exites and thus saturates the core and your magnitizing inductance is a separate problem for you to provide the current. In real life, chaging the number of turns changes both the peak flux density and the magnitizing inductance, but you will just confuse yourself to see them as coupled to each other. Also, you have way too many questions. Narrow it down! – winny Mar 19 '18 at 15:42

1 Answer

If the primary inductance is 2.86 mH and you apply 400 volts for 6 us (50% of the period of 83 kHz) then the magnetization current will rise to 400 x 6 us/ 2860 uH = 0.839 amps over that period.

The above calculation is a re-arrangement of V = L di/dt.

Given that you have 26 turns, that's a peak ampere-turns (aka magneto motive force) of 21.8. Given that the effective length of an ETD59 core-set is 139 mm, then the H-field will be 21.8/0.139 = 157 ampere-turns per metre.

If you are using N27 material, the peak flux will be close to saturation: -

I've drawn the red line roughly indicating where the H field is 157 At/m

If you doubled the turns, the inductance would be 4 times greater and the current would be a quarter of what it previously was after 6 us. However, turns have doubled so, in effect, there is only a net reduction in H-field of 2 but this would be better positioned to give significantly less saturation.

See also THIS question and answer for a similar situation to yours.

• 6us? If DC=50%, it must be 3 us? Maybe i still dont get it... – silkyre6xtenz Mar 19 '18 at 16:17
• The period of 83 kHz is about 12 us so 50% is 6 us. My answer explains what the magnetization current will rise to in 6 us. Current x turns divided by length of the core gives you the H-field and the BH graph converts H-field to flux density and allows you to see how close to saturation you are. – Andy aka Mar 19 '18 at 16:18
• Right. thank you. So, for future: my calculations for number of turns (assuming flux swing) were right, right?) But i also have to look for maximum ampere-turns? – silkyre6xtenz Mar 19 '18 at 16:20
• They appear to be correct i.e. L = $A_L\cdot N^2$ – Andy aka Mar 19 '18 at 16:22
• @silkyre6xtenz you've asked ten questions now and I don't know if you realize that you should consider formally accepting the best answer to a given question and that upvoting any good answer is a nice thing to do. Call it a small payment for help. – Andy aka Mar 19 '18 at 16:26