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I am designing a Discontinous conduction mode flyback converter, 100kHz using UC3844 IC with the spec. 220V,50Hz to 5V 5A. The issue is that after testing it for an hour, the transformer starts heating up(tried with different cores).

Ambient Temp.: 27degC

Transformer Spec: 1st try: EE20/10/6 Core, 0.5mm Gap, Al=100nH/turns2(EE20/10/6 Datasheet)

  • Turns Ratio: 21 to keep Vds(max) Low

  • Primary: 1.7mH, 30 SWG, 130 turns, 1 strand

  • Secondary: 5-6uH, 6 turns, 30 SWG, 7strands(I know, the diameter that i used is very low but could fit only this in the winding area and the strands have to be increased, for testing purpose designed this)

  • Auxiliary: 52uH, 14 turns,30SWG, 1 strand.

With this configuration, the transformer was heating upto 80deg C in just 30 min.

Form the calculations for the core loss, it was coming to 170mW. But Copper Losses were huge

Primary: Irms=0.4A, Resistance_dc=(total length)*Res. per length, Rp=(7.6*4*130)*0.221 mOhm=873mOhm, P=0.4*0.4*0.873*1.5=0.2W.

Secondary: Irms=9A, R_dc=(7.6*4*6/5)*0.221=8mOhm, P=9*9*.008*1.5=0.972W(This is the prob)

Seeing this with the thermal resistivity of 50K/W. Temp. would rise to almost 80 degC. So, it confirms with the problem.

Since i could not fit more dia. winding in this bobbin. Therefore, had to increase the core size.

2nd try: EE25/13/7 0.2mm gap Al=290nH/turns2 EE25/13/7

Same Inductance Parameters.

Primary: 77turns, 30 SWG, 1strand Secondary: 22SWG, 5 strand, 4turns.

But even then it is still rising to 65-70degC.

Currently working at 78% efficiency. What is happening in this case? Are my previous calculations correct? Is it because ac resistance has increased due to skin effect since operating at 100 Khz because of 22SWG wire? or something else?

Edit:

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3rd Try:

EE25/13/7 core: 0.4mm Gap, 187nH/turns2[earlier(340nh/turns2)]

Primary: 30SWG, 1 strand, 95 turns, 1.6mH

Secondary: 22SWG, 5strand, 5 turns, 5.2mH

Auxiliary: 35 SWG, 1 strand, 16turns.

Although the temp decreased but still hovering till 65degC.

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Circuit Diagram:

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  • 2
    \$\begingroup\$ We tend to design for more even split of dissipation. I'm not saying 50:50, but 1 watt copper and 170mW ferrite is a big ratio. A small increase in flux and reduction in turns would probably result in lower overall losses. Don't forget that the resistivity of copper increases by 10% every 25C, so your copper losses are underestimated if based on resistance measurements at room temperature, and then your RF resistance is higher than the DC value. \$\endgroup\$ – Neil_UK Nov 30 '16 at 7:37
  • \$\begingroup\$ Yes, You are correct. That is why next core that i tried, i used 22SWG copper wires 5strands in EE25/13/7. Still it was heating upto 70degC. Updated my 3rd try too above. \$\endgroup\$ – Div-lcr Nov 30 '16 at 11:18
  • \$\begingroup\$ I could not get the gap to go down, as then it would go towards saturation. Below 0.3mm, in EE25/13/6, Core would start saturating, that is why new design is for 0.4mm. \$\endgroup\$ – Div-lcr Nov 30 '16 at 11:34
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    \$\begingroup\$ 100 Kelvin-Hertz is a non-sensical unit for this question. \$\endgroup\$ – Olin Lathrop Nov 30 '16 at 11:47
  • \$\begingroup\$ What core material are you using? What flux swing did you calculate? \$\endgroup\$ – winny Nov 30 '16 at 11:57
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I think you have a borderline but significant core saturation problem. Using your 3rd example, the current into the primary rises at a rate of V/L where V is about 311 volts (rectified AC and smoothed) and L is 1.6 mH. So, in 5 us I would expect to see the current rise to about 1 amp.

This is based on the basic inductor formula of V = L.di/dt

5 us is the on time for a 50:50 duty cycle at a switching frequency of 100 kHz

The primary magneto motive force (MMF) is ampere turns or 1 x 95 At. But, to calculate the H field, we need the effective length of the core (57.5 mm in the data sheet linked in the question) so H = 1652 At/m.

An ungapped core would certainly be saturating but yours is gapped and has an effective permeability of around 170 compared to a permeability of around 1520 ungapped (again these were numbers I calculated from the data sheet you linked). The effect of gapping can be seen as reducing the H field so, your H field reduces to an equivalent value of around 185 At/m for an ungapped core. This allows us then to look at the published BH curve.

Now, if you look at the BH curve for N27 you will see this: -

enter image description here

On the two diagrams I've taken the liberty of drawing a red line that shows where the equivalent ungapped H field peak value is sitting (185 At/m). As can be seen on the left diagram (ambient of 25 degC) 185 At/m is starting to significantly saturate your core.

It's quite critical that a flyback transformer does not saturate very much.

So, as the core saturates the inductance tends to fall and instead of a linear rise in current per micro second you get a seemingly out of control rise like this: -

enter image description here

This may lead to a very significant rise in the peak H field and the core starts to get quite hot. But, you may say: -

So what, the controller will limit the current to that needed to store only the energy needed to pass to the secondary load

However, as the core saturates the inductance falls so what was sufficient current (for a given value of inductance and therefore the correct amount of energy based on E = \$I^2L/2\$), now needs to be more current.

Do you see the problem and this isn't even considering what happens when the core gets warm (see the graph on the right in the picture above). At 100 degC there is even more core saturation.

I think you are running into saturation problems.

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  • \$\begingroup\$ Thank you for the detailed response. So, the solution would be to increase the frequency and then test what happens or go for larger core(trade-off for small pcb). But still as you explained, I am limiting the current by using 1.3 ohm resistor, so that it does not let the transformer to go till saturation point, I am not able to understand how that is going to be a problem, if i don't take it to the point of saturation \$\endgroup\$ – Div-lcr Nov 30 '16 at 16:54
  • \$\begingroup\$ One more thing though, i noticed, since it is going for saturation at 1A, when i was using <0.83 ohm resistance, it's magnetizing inductance was taking up lot of energy for the snubber to dump. \$\endgroup\$ – Div-lcr Nov 30 '16 at 16:58
  • \$\begingroup\$ The CS threshold is typically 1.04 volts (figure 19 in the DS) and your 1.2 ohm sense resistor (as per your schematic) means the peak current is 867 mA. Given also that there is a 45 ns time delay in turning off current, that's another 10 mA (ish) rise before the MOSFET deactivates. You also have an RC filter that may add a little to the timing and give a little more current so, around about 900 mA is probably what your I(limit) is and you are still significantly into the saturation region. \$\endgroup\$ – Andy aka Nov 30 '16 at 17:50
  • \$\begingroup\$ I'll also point out that your snubber might as well be removed because with a 50 kohm resistor and 50 nF capacitor, that cap will never get discharged on a cyclic nature. Think about the time constant and the effect of the diode charging that capacitor up. The cap gets charged to some value and bearly discharges before the diode drops more energy into it. This means the cap keeps charging and charging and you might easily get failure of the MOSFET. \$\endgroup\$ – Andy aka Nov 30 '16 at 17:55
  • \$\begingroup\$ So, Probably i should go for higher frequency. Right? \$\endgroup\$ – Div-lcr Nov 30 '16 at 17:56

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