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i have been designing flyback converter of a 5Volt 5 ampere specifications output deriving from 220v ac mains. But i have been stuck at one point and can't seem to get out it. Efficiency is particularly very poor in this case getting around 49% when running at 25W output. Earlier, due to leakage inductance voltages were going around greater than 600v on the drain of the MOSFET(10n60) causing high stress on the FET.

So to clamp the voltage , introduced RCD clamp(50nF, 3-5k). I know its a very low value of resistance used causing excessive losses. But only at this value i was able to clamp the voltage down to 550v running at full load.(For stability purpose this is too high, but can't go down anymore without decreasing resistance). At this, MOSFET temperature does not increase to high levels.

Whether this problem is coming due to the leakage inductance too high for this circuit or some other problem?

Transformer specifications:

30swg, 61 turns primary, 1 strand(1.7mH)

22 swg, 3 turns secondary, 5 strands(4.6uH)

35 swg, 1 strand, auxillary, 10 turns( 100uH)

EE28 core used, interleaved windings.

Since, i am winding transformer by hand not by machine, already it is a problem to wind 22 swg wire that too 5 strands so that is why i am thinking it might be a problem of magnetizing inductance too high in this circuit.

Uc3844 ic with isolated feedback using 817 opto with tl431 reference.

Any help would be greatly appreciated.

Edit 1: Circuit for the following converter(FET used is 10n60 not IRF840)(earlier IRF840 was used)-

Schemiatic of the FLyback Converter

Edit 2: The solution I got was to increase the current sense resistor to limit the peak current thereby less energy stored per cycle causing resistance value used to almost negligible(50kohm,considering losses). Efficiency boosted up to 76% at 5A Load. Still trying to increase it up to 85 by optimizing the core and rectifier losses. Will write the results soon.

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  • \$\begingroup\$ If you want help post a circuit and magnetization inductance has nothing to do with the secondary wire type \$\endgroup\$
    – Andy aka
    Nov 3, 2016 at 13:42
  • \$\begingroup\$ Okay, i will post the circuit. I was thinking about it because since sandwiched winding is there, the upper primary and the lower primary has a certain distance between them causing leakage inductance to increase. Thank you for replying. \$\endgroup\$
    – Div-lcr
    Nov 3, 2016 at 13:44
  • \$\begingroup\$ A lot of flybacks will use zener clamps too. \$\endgroup\$
    – Andy aka
    Nov 3, 2016 at 13:48
  • \$\begingroup\$ I agree with @Andyaka about zener clamps. For such a low power Flyback, you can use diode-zener snubber instead of RCD (Example m.eet.com/media/1180028/fig1.jpg). Anyway, it's quite interesting to achieve under %50 efficiency. I designed a 32V/3A offline SMPS and it's efficiency is about %88 under full load even with an RCD clamp on tranny's primary. About transformer winding, I highly recommend you to wind the transformer with like this (From inner to outer) to reduce leakage inductance: "one layer pri -> one layer sec -> one layer pri -> one layer sec -> ... -> auxilliary". \$\endgroup\$ Nov 3, 2016 at 14:05
  • \$\begingroup\$ I have updated the post with the schematic.( Sorry about the image). Saved it from OrCad Capture. I agree with @RohatKılıç regarding the efficiency, it should be more than 80% efficiency since it is low output circuit. Considering the 3.7k used in clamp, losses increases many times fold. Without that, zener in FET activates causing high power losses in MOSFET. \$\endgroup\$
    – Div-lcr
    Nov 3, 2016 at 21:43

1 Answer 1

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Note that primary winding does affect input losses. Your transformer design seemed a bit odd to me (61 turns primary for a 40kHz offline converter seems quite low).

So I re-designed your transformer:

  • Parameters: Vin: 85-265Vac = 120-370VDC, Vout=5V, Iout=5A, eff: %85 so Pin=25/0.85=30W, f = 40kHz (from your UC2844 config: RT=4k and CT=10n), maximum duty: %45 at minimum Vin, Total transformer loss=%5Po=1.25W and total core loss = 1.25W/2=625mW, core:EE28 (Ae=85mm², Ve=43cm³).

From transformer equation, \$V_t=N\cdot A_e \cdot dB/dt\$.

For our needs, this equation turns to \$V_{inDC_{min}} = N_{p_{min}} \cdot A_e \cdot \Delta B/\Delta t_{max}\$ where \$\Delta t_{max}= 0.45 / 40kHz = 11\mu s\$. Assuming you're using 3F3 core, from 3F3 datasheet (Fig. 6), selected \$B_{pk}=80mT\$ for a core loss per volume of 625mW/43cm³=14mW/cm³. So \$\Delta B=2 \cdot B_{pk} = 160mT\$.

With these parameters, $$ N_{p_{min}}=\frac {120V\cdot 11\mu s}{85mm^2 \cdot 0.16T}=97 $$ So, if volts per turn is \$120/97=1.24\$ then number of secondary turns is \$N_s=5 \cdot 97 / 120 = 4\$. Supplying UC2844 with 15V is quite enough, so required number of turns for aux winding is \$N_a=15 \cdot 97 / 120 = 12\$.

For a current density of \$J=420A/cm^2=4.2A/mm^2\$, your wire selections are quite enough.

Besides, a few extras I can suggest:

  • Place a 1k resistor across opto's LED for proper biasing.
  • Although your feedback network is correct, I personally recommend you to connect FB pin and opto's emitter to GND, opto's collector to CMP. So you can get rid of R19, R20, R21 and C11.
  • D9 (Zener) is unnecessary since your auxiliary supply is regulated enough.
  • Instead of C10-R18 snubber network, place a, 200V zener or so.

If you're interested, take a look at one of my offline flyback designs (32V/3A) and ST's application note.

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  • \$\begingroup\$ Thank you for the redesign, i will test it and will tell you the results soon. For the added parts, i will apply it after the transformer design. \$\endgroup\$
    – Div-lcr
    Nov 4, 2016 at 12:38
  • \$\begingroup\$ One more problem that was coming up related to start-up of the pwm ic. At no-load, the auxiliary winding is not able to supply enough current for the ic supply and thus ic turns off and turns on again via Rstart-up resistor. this keeps on going on and on and because of this not able to have regulated voltage. At load though, the auxiliary winding combination is working as it is supposed to be. I have increased no. of turns even but to no avail. Currently 16 turns auxiliary 35swg 1strand is designed. \$\endgroup\$
    – Div-lcr
    Nov 7, 2016 at 5:18
  • \$\begingroup\$ @Div-lcr It's not related to number of turns. It's all related to flyback action itself. Flyback action cannot be initiated and completed, because there's no load. Remember how flyback converters work: Energy stored in primary winding will be transfer to the "load". No load, no transfer, no flyback action. The solution is putting a "dummy load" in the converter. EDIT: Seems like you opened a new question for this. I'm writing the same answer there. \$\endgroup\$ Nov 7, 2016 at 6:10
  • \$\begingroup\$ Thank you for answering. Sorry for again opening a new question.(My mind became restless :)). yes. that is the problem. So in all flyback operations, we have to put a dummy load, right? So, is that why they specify minimum amount of output current in the design? \$\endgroup\$
    – Div-lcr
    Nov 7, 2016 at 6:16
  • \$\begingroup\$ Yes, all of the switching converters need a dummy load. It actually depends but %2-5 (or even %10) of nominal output "current" would be enough. Try and find. EDIT: Don't delete that question. It can be very useful for others. \$\endgroup\$ Nov 7, 2016 at 6:18

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