Pure sinewave SPWM inverter

CheeseToastie said:

Morning. Back again. Returning to,

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It's a bit difficult to separate out which bobbin is which but using minimum figures the available winding width, Ww, is C 26mm. The available winding height is (A-B)/2 or about 5.8mm.

We'll assume you are going to use 3mm margins either end for insulation. That reduces your winding width to 20mm. With 5 turns per primary and assuming you wind them side by side in one layer that is 10 turns total so each wire will have 2mm available to it.

That's going to cause problems with skin effect, you also have to worry about proximity effect...

http://www.ti.com/lit/ml/slup125/slup125.pdf

There is probably more comprehensive information available from elsewhere but the above gives you some idea as to what is going on. Otherwise you can 'cheat'.

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**broken link removed**

Software here,

www.genomerics.talktalk.net/software/roundwire/roundwire.zip

It was written in Delphi 4 and might fall over on newer versions of Windows latest operating systems. Should be fine under XP. You can see the underlying code in the roundedit.pas file which should be viewable in a text editor.

Available wire width is 2mm so make that up as a twisted rope of 7 strands. The overall diameter will be three times the wire diameter so your wire is 2/3 or 0.66mm.

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You can see in the above that AWG23 is selected. Insulated wire diameter is 0.63mm. L,S,H and T refer to the insulation thickness. Light, Single, Heavy and Thick. For 'high' voltage SMPS work you will generally be using Heavy.

The other stuff..

Top left are the input variables. Frequency is 60KHz which is your new transformer switching frequency. You IC clock will be at 120KHz. Layers deals with proximity or layer effect. In this case it is set to 0.5 because we are going to sandwich the primaries between the secondary, secondary wound in two sections. Temperature is your expected maximum transformer operating temperature. Strands, as per the above, is set to 7.

Calculated values, Single strand and Rope appear top middle and top right. DC and AC resistances per metre and the ratio. You cannot do better than 1 for the ratio and 1.5 is sometimes suggested as a 'target'. In this case the solution looks like a good one which is sort of unusual because you often find things do not fit together so well and you have to consider different solutions.

The bobbin data gives a value for the Mean Length of a Turn, MLT,

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60.8mm for the EER35. With 5 turns on the primary the winding length will be 5*60.8 or 0.304m so RDC = 0.0038R and RAC = 0.0039R. That ignores lead outs.

Previously we allowed for 2W core loss and 2W winding loss. Split the 2W winding loss between primaries and secondary and that gives 1W each. Then split that 1W for the primary between them for 0.5W each.

Without being rigorous assuming a 50% duty cycle current square pulse the RMS value is,

Irms = A.√ D = 0.7071 * A

Given the RAC/RDC ratio is one the whole wire will be utilised and we need not consider the DC or AC components but rather use the above as an absolute figure. Given RDC = 0.0038R for a loss of 0.5W the permissible RMS current is √(0.5/0.0038) or 11.5A giving an amplitude of 16A which will be the current draw from the battery.

Assuming the nominal battery voltage is 12V then the maximum power throughput for this transformer will be 192W.

Your burning question was how many secondary turns?

Given the end discharge voltage of a loaded lead acid battery is 10.5V

http://www.farnell.com/datasheets/19849.pdf

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With a desired Fclk of 120KHz the period is 8.333uS. Choose a discharge time of 1uS, right graph. This will be the 'dead time'. CT = 1nF RD = 150R. The remainder, 7.333uS becomes the charge time, left graph, which makes RT = 9K1. Maximum duty cycle is 7.33/8.33 or 88%

With a minimum input voltage of 10.5V the average at the transformer primary will be 0.88*10.5 = 9.24V You want 180V out so the required turns ratio is 180/9.24 or 19.48 which based on the 5 turn primaries makes the required number of secondary turns 97.4 so choose 100.

As before your available bobbin winding width after margins is 20mm. We'll do the secondary as two sections with two layers of 25 turns per layer sandwiching the primaries. Available width for the wire is 20/25 or 0.8mm. Again we use a rope but this time 7 strands of something close to 0.8/3 or 0.2667mm overall diameter.

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AWG31 fits the bill. Rope diameter will be 0.792 which, 25 turns, will fit the available winding width. The height for 4 layers will be 3.168mm and our primary occupied 1.89mm giving a total height for the windings alone of 5.058mm less than the available winding height of 5.8mm. That leaves adequate space for inter winding insulation.

You will end up with a 'power bulge' across the top of the bobbin as a result of the primary lead outs.

Again RAC/RDC is close to one with RDC = 0.080705 and RAC = 0.085073 MLT is 60.8mm so length of the secondary is about 6 metres giving RDC ~ RAC = 0.52R. This time secondary current approximates to a true square wave which is entirely composed of AC harmonics.

Irms = A

Given our limit on dissipation was set to 1W for the secondary A works out to be √1/0.52 or 1.39A. For the 180V out that's 250W. The numbers never match because you are dealing with discrete sizes of wire and have to fit them in to the available dimensions.

It's not very explicit but this represents the winding structure for your transformer.

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Now then.... about this 'voltage mode control' and flux walking. Whilst 'time is of the essence', smells like a final year project, would you at least consider using 'current mode control' instead? Assuming you do things right life, for this part at least, will become less stressful..

http://www2.microsemi.com/document-portal/doc_download/11139-sg1846-pdf

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Looks similar but different.

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Kryptone said:

From my power supply when I measure the current I am getting 0.1A at 12V and at the dc-dc converter stage I am getting 180Vdc at 0.1A therfore I guess thats why it the dc-dc converter can only light up a 20W bulb since 180V x 0.1A= 18W and the 20W bulb drops the 180Vdc to about 165-170Vdc when I connect the 180Vdc to the h-bridge I get about 24V-AC from a 14.7Vdc from power supply.

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CheeseToastie said:

Perhaps Mary Gannon from TI Ireland or Harwin with their new ACE award can join in on the discussion. Ooops did not think so. Why am I receiving their totally unrelated marketing spam?

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Kryptone said:

I am already getting 180Vdc across cap of LC filter in dc-dc converter. What is the difference between the SG1876 or the one in the diagram and the SG3525????

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Subject: Opt-in to Texas Instruments Product News and Technical Information

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Flux Walking

Faraday’s Law states that the flux through a winding is equal to the integral volt-seconds per turn. This requires that the voltage across any winding of any magnetic device must average zero over a period of time. The smallest dc voltage component in an applied ac waveform will slowly but inevitably “walk” the flux into saturation. In a low frequency mains transformer, the resistance of the primary winding is usually sufficient to control this problem. As a small dc voltage component pushes the flux slowly toward saturation, the magnetizing current becomes asymmetrical. The increasing dc component of the magnetizing current causes an IR drop in the winding which eventually cancels the dc voltage component of the drive waveform, hopefully well short of saturation.

In a high frequency switchmode power supply, a push-pull driver will theoretically apply equal and opposite volt-seconds to the windings during alternate switching periods, thus “resetting” the core (bringing the flux and the magnetizing current back to its starting point). But there are usually small volt-second asymmetries in the driving waveform due to inequalities in MOSFET RDSon or switching speeds.

The resulting small dc component will cause the flux to “walk”.

The high frequency transformer, with relatively few primary turns, has extremely low dc resistance, and the IR drop from the dc magnetizing current component is usually not sufficient to cancel the volt-second asymmetry until the core reaches saturation.

Flux walking is not a problem with the forward converter. When the switch turns off, the transformer magnetizing current causes the voltage to backswing, usually into a clamp. The reverse voltage causes the magnetizing current to decrease back to zero, from whence it started. The reverse volt-seconds will exactly equal the volt-seconds when the switch was ON.

Thus the forward converter automatically resets itself (assuming sufficient reset time is allowed, by limiting the maximum duty cycle).

The flux walking problem is a serious concern with any push-pull topology (bridge, half-bridge or push-pull CT), when using voltage mode control..

One solution is to put a small gap in series with the core. This will raise the magnetizing current so that the IR drop in the circuit resistances will be able to offset the dc asymmetry in the drive waveform.

But the increased magnetizing current represents increased energy in the mutual inductance which usually ends up in a snubber or clamp, increasing circuit losses.

A more elegant solution to the asymmetry problem is an automatic benefit of using current mode control (peak or average CMC). As the dc flux starts to walk in one direction due to volt-second drive asymmetry, the peak magnetizing current becomes progressively asymmetrical in alternate switching periods.

However, current mode control senses current and turns off the switches at the same peak current level in each switching period, so that ON times are alternately lengthened and shortened. The initial volt-second asymmetry is thereby corrected, peak magnetizing currents are approximately equal in both directions, and flux walking is minimized.

However, with the half-bridge topology this creates a new problem. When current mode control corrects the volt-second inequality by shortening and lengthening alternate pulse widths, an ampere-second (charge) inequality is created in alternate switching periods.

This is of no consequence in full bridge or push-pull center-tap circuits, but in the half-bridge,
the charge inequality causes the capacitor divider voltage to walk toward the positive or negative rail.

As the capacitor divider voltage moves away from the mid-point, the volt-second unbalance is made worse, resulting in further pulse width correction by the cur rent mode control. A runaway situation exists, and the voltage will walk (or run) to one of the rails.

This problem is corrected by adding a pair of diodes and a low-power winding to the transformer, as detailed in the Unitrode Applications Handbook.

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Kryptone said:

Can you tell me or calculate for me a suitable boot strap capacitor value for a DC bus of 180 V-dc please in order to get 110 VAC at the output of the inverter.

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CheeseToastie said:

You might wish to hunt elsewhere on the site for some of the application notes.. Otherwise conceptually you end up with one of these,

View attachment 104497

Your push-pull converter is now being, current fed, from a boost converter. The filtering inductor has been moved to the primary side, no need for the output inductor on the secondary side, and more importantly it is now in series with the battery.

As a result the battery no longer gets hammered with high frequency square amp currents and, assuming you keep things tight elsewhere, the pain might become less.

Of course you have now transferred that discontinuous current to the secondary side of the transformer but filtering becomes easier.

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CheeseToastie said:

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