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As part of my pure sine wave inverter, I have a DC - DC subsystem in which a signal stepped up through a HF transformer is put through an uncontrolled diode rectifier to get a DC output which I am o feed into my PWM inverter (not shown in circuit).

I am using a 12V to -12V square wave as input to the transformer and getting the required 475 to -475 V wave output, but the moment I connect a diode rectifier to it, the output of my transformer sinks to a range of 4 to -4 V on the secondary.

I am using a threshold voltage of 1.7 V and a diode resistance of 0.57 ohms.

  • Rpri = 2.48 mOhms
  • Rsec = 0.411 Ohms
  • Lpri = 1.909 uH
  • Lsec = 2.98 mH
  • Lm = 27.51 uH
  • Npri = 2
  • Nsec = 79
  • SWG 19 wire (1mm diameter) used on primary coils
  • SWG 24 (0.5 mm diameter) for secondary

MOSFET ON STATE resistance = 0.015 Ohms

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    \$\begingroup\$ What is there to limit your Peak Current ? \$\endgroup\$ – Autistic Jul 23 '16 at 23:43
  • \$\begingroup\$ You connect ONLY the diode rectifier and voltage drops ? No load on the rectifier? \$\endgroup\$ – Marla Jul 23 '16 at 23:52
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    \$\begingroup\$ Ignoring the immediate problem that this does not have an actual question, your schematic does not seem to match your description. I see a rectifier and a load. \$\endgroup\$ – pipe Jul 24 '16 at 0:59
  • \$\begingroup\$ well the load is the equivilant of what the PWM inverter to be connected to the rectifier will represent. i donno if that makes sense? \$\endgroup\$ – A Tee Jul 24 '16 at 12:58
  • \$\begingroup\$ please properly format your question. The formatting buttons above the editing field are your friends. A bullet list would be appropriate to list properties of a transformer \$\endgroup\$ – Marcus Müller Jul 24 '16 at 13:05
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Well, obviously, the HV supply you've built can't source enough current!

Hence, the voltage drops as soon as you connect a load.

I don't know the specifications (winding thickness, winding count, core material, transformer topology) of your transformer, but that's where I'd start looking.

Also, I don't know the effective resistance of your MOSFETs in ON state; you might also consider the fact that these device are frequency dependent, too!

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  • \$\begingroup\$ i can list out the transformer specs for you: \$\endgroup\$ – A Tee Jul 24 '16 at 12:59
  • \$\begingroup\$ Rpri = 2.48 mOhms/ Rsec = 0.411 Ohms/ Lpri = 1.909 uH/ Lsec = 2.98 mH/ Lm = 27.51 uH/ Npri = 2 and Nsec = 79. SWG 19 wire (1mm diameter) used on primary coils and SWG 24 (0.5 mm diameter0 for secondary \$\endgroup\$ – A Tee Jul 24 '16 at 13:03
  • \$\begingroup\$ im sorry, im new here, hence im making silly mistakes, please bear with me \$\endgroup\$ – A Tee Jul 24 '16 at 13:04
  • \$\begingroup\$ please edit your question. Address the concern made in this and the other, very good answer in your question. \$\endgroup\$ – Marcus Müller Jul 24 '16 at 13:05
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So, you're putting +/- 475 volts into a bridge rectifier, followed by a 596 ohm resistor. In a perfect world, you'll get about 950 volts out of the rectifer, and then your trouble starts. 596 ohms? Really? (And where in the world did that extra .07 ohms come from, anyways? Are you truly measuring resistance to 5 significant figures?) Apply Ohm's Law. 950 / 596 is 1.6 amps! 1.6 amps times 950 volts is 1500 watts!

Have you really got a 1.5 kW transformer, and a 1.5 kW resistor? You need to rethink your design. From your trace, it's apparent that you're driving this at about 60 kHz, and you should be aware the designing a transformer for these currents and frequencies is not simple. And it's pretty clear that you've done something wrong.

Oh yes, and you have not specified the diodes in your rectifier. You're not using 1N400x types, I hope.

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  • \$\begingroup\$ wait, when i rectify a voltage of 475, am i not supposed to get a DC voltage of roughly around that area? I used the formula Vripple = Iload/f.C to get my cap value, with my V ripple being 1% of 425 V required by the PWM inverter as an input. The resistance R = 596 Ohms i got from R = V^2/P with The V representing the DC Bus voltage into the PWM inverter (out of the rectifier) and P representing the total watt output required by the system (300 W) \$\endgroup\$ – A Tee Jul 24 '16 at 13:13
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So it's just a simulation isn't it?

All your secondary circuit, bridge, filter, load is floating, i.e. it has no reference to 0V node.

Most simulators are not happy with this.


So I didn't read problem's data carefully enough and I apologize for this.

The matter is much simpler: it cannot work.

Switches: a 500W system run from 12V would take over 40A. Just rds(on) and primary windings resistance score up to 18mohm: this alone drops nearly 1V from 12V and dissipates over 30W statically only. Then add switching losses Moving up to 24V or 48V would be a very good choice unless really impossible.

In any case such an inverter will take several MOS for each switch and for sure transformer primary winding will not be made of 1mm copper wire. Wound copper stabs or several parallel windings are mandatory.

Talking of transformer: so far it is no clear to me wether we are talking of measured or simulated data but inductances tell that coupling between primary and secondary is very loose.

With Lpri=3mH, Lsec=1.9uH and perfect k=1 coupling Lm should be around 75uH. If you have 27uH it means k=0.35 or so and hence around 2.6mH of series inductance.

Around 1kohm reactance at 60kHz.


First you need a much better transformer, but IMHO inverter design is also a far cry from a good working circuit.

I swapped Lpri and Lsec but it's just the same, they are very loosely coupled, they cannot transfer much energy each other. How have you designed it? What is the core size and shape? Where have you got those inductances?

Simulators are toghether the most usefull and the most dangerous to any field of engineering.

I am afraid that used with no background experience, sensibilty to results and experimental work will not give any good results. By the way designing SMPSU transformer is one of the most demanding tasks. It is made of educated guesses, trials and errors back and forth electric, magnetic and thermal design at the same time.

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  • \$\begingroup\$ Hey, thank you for replying. it means a lot. I am using a N87 ferrite core with Kcu = 0.25, Psp = 200mw/cm3, Bc_max = 200mT, J = 4.924 A/mm2 and the relatios b = 0.735a, d = 1.13a and h = 2.35a. do you have any advice for me as to how i can fix this issue im facing @carloc?? \$\endgroup\$ – A Tee Jul 26 '16 at 15:07
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So I didn't read problem's data carefully enough and I apologize for this.

The matter is much simpler: it cannot work.

Switches: a 500W system run from 12V would take over 40A. Just rds(on) and primary windings resistance score up to 18mohm: this alone drops nearly 1V from 12V and dissipates over 30W statically only. Then add switching losses Moving up to 24V or 48V would be a very good choice unless really impossible.

In any case such an inverter will take several MOS for each switch and for sure transformer primary winding will not be made of 1mm copper wire. Wound copper stabs or several parallel windings are mandatory.

Talking of transformer: so far it is no clear to me wether we are talking of measured or simulated data but inductances tell that coupling between primary and secondary is very loose.

With Lpri=3mH, Lsec=1.9uH and perfect k=1 coupling Lm should be around 75uH. If you have 27uH it means k=0.35 or so and hence around 2.6mH of series inductance.

Around 1kohm reactance at 60kHz.

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  • \$\begingroup\$ This isn't a forum - it's a Q&A site and the answers float up and down with votes and user's sort order. You would have been warned of this when you posted your second answer. Merge the two of them and delete one to keep tidy. \$\endgroup\$ – Transistor Jul 24 '16 at 18:38
  • \$\begingroup\$ Hello carloc, i forgot to mention that the primary windings are stranded with 7 strands to make up the 1 mm. The values i used in my transformer are those that i calculated. it is not Lpri but rather Lsec that is 3mH and Lpri is 1.9uH. I do not have any option but to use a 12 V source and in essence what i want to do is make a 240 V sine wave inverter. Can you explain a mbit more simply why this will not work and what i can do to get it to work? \$\endgroup\$ – A Tee Jul 25 '16 at 11:53

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