coupled-inductor capacitance - the dirty underbelly

Coupled inductors are attractive devices, easily made and inexpensive with bifilar windings, and all the rage. They're quite useful for SEPIC and Cuk converters, for balanced-output and zeta converters, and for fly-buck and for enhanced boost tapped inductors: get a free extra 2x output.

But these parts have a dirty underbelly, one you're not told about on the datasheets and that's rarely mentioned. I'm talking about their high winding-winding capacitance. We know they're bifilar wound, so it should come as no surprise that the winding pairs have capacitance, assuming you think about it.

Here's measured data, for Coilcraft's MSD1583- series, fairly powerful 15x15mm parts with 3 to 6 amp ratings.

. -223ME 22uH 170pF . -333ME 33uH 144pF . -473ME 47uH 160pF . -683ME 68uH 177pF . -104KE 100uH 203pF

I'm guessing the measured capacitances are higher than you thought. It's high, and not particularly correlated to the inductance value, which changes over a 5x range with these parts. A smaller sized part, the 12x12mm MSD1278 series, doesn't give you much relief, its 47uH version measures 126pF. In even smaller parts, like Bourns' 7x7mm SRF0703 series, I measured 180pF. No relief there.

It's thought that high capacitance doesn't matter with some configurations, like SEPIC, because the pins are flying together, and the pin-pin voltage doesn't change. But that's wrong, because the high capacitance can lead to rather high internal resonant circulating currents, and unexpected ac-resistance losses. In fact, some manufacturers such as CoilCraft are offering loosely- coupled degraded coupling coefficients, like the MSC1278, with k = 0.8 instead of >0.98, to fight this phenomena. Note: MSC rather than MSD in the p/n.

Furthermore, the high capacitance can be a killer in tapped and series inductor applications. For example, John Larkin's cute HV boost circuit with its 2x output multiplier (switch the tap to ground, then let it fly back to half the output voltage) ran out of steam at lower maximum voltages than he expected. The E =

0.5 L I^2 energy stored in L gets used up charging the hidden stealth E = 0.5 C V^2. Here the high C and V^2 is killing you at 500 volts, and less energy than intended is available to charge the output capacitors each cycle.
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Reply to
Winfield Hill
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He expected?

The E =

I kept the voltage boost down, and multiplied, partly because of capacitances and partly because I was concerned about the breakdown voltage of bifalar windings. Plus, I needed the intermediate voltages. I did breadboard it, and efficiency was OK.

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The hottest thing was the transformer, core and/or copper loss. On the PCB layout, all that copper is mostly to cool the transformer.

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High ratios in boost converters are difficult to achieve, as the L*I^2 energy gets gobbled up by C*V^2.

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John Larkin         Highland Technology, Inc 
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Reply to
John Larkin

You see the same capacitance effect in HV transformers. the Interlayer capacitance can be a killer as you go up in frequency. Some Z wind the layers, or split the winding across a split bobbin, you can also stack the layers so that there is a defined voltage between layers. There is a way to approximate the capacitance if you know the winding details.

Cheers

Reply to
Martin Riddle

Den fredag den 2. september 2016 kl. 00.02.34 UTC+2 skrev Winfield Hill:

an possibly reduce how many different part you need, because you can use them series, parallel or just one of the windings

-Lasse

Reply to
Lasse Langwadt Christensen

The mystery generalizes away when you consider the transformer as a transmission line component instead -- which it is, to a much more fundamental degree.

The capacitance is distributed along the winding, which as you say, can lead to lots of reactive power (which usually ends up wasted).

One trivially simple trick, that's often missed: putting the secondary rectifier diode on the "wrong" side of the winding. For some converters (usually flyback), this can transform the induced common mode voltage from worse-case to almost zero!

Old fashioned CRT flyback transformers took advantage of this, following one layer (of particularly fine wire, so that a single layer accumulates a thousand turns or so) with a diode, and stacking enough layers (and diodes) to get the required output voltage (~30kV typical). Winding voltage does not couple into the inter-layer capacitance, so the parasitic capacitance doesn't scale with voltage. It's the only way to build a transformer, for such high voltages, that does anything at deflection speeds up to 100kHz!

A multilayer transformer isn't much of a transmission line, but the principle still applies. The further apart two wires are, the higher the transmission line impedance: and thus, the higher the leakage inductance and lower the coupling capacitance. This works best when your circuit impedance (usually the ratio of p-p switch voltage to pk switch current) is near the TL impedance, and higher circuit impedances are affected disproportionately by capacitance, and lower by leakage inductance.

(Note: every time I say "inductance" or "capacitance", I mean that impedance which is the low-frequency approximation to TL properties. Real physics is TLs, period: there's no such thing as a true R, L or C! We engineers approximate with L or C, when we can, because it's easier. The fractional-wavelength rule applies.)

Tim

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Seven Transistor Labs, LLC 
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Reply to
Tim Williams

Of course resonant converters exploit the capacitance, and Baxandall is rumoured to have invented the Class-D oscillator to get a kV or so out of 24V rails to drive photomultipliers.

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I've even recently added my own version of an imagined photomultiplier to my web-site. The circuit diagram is well down the page

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The .asc file hasn't come out as well as I would have liked. I may get it into more useful shape today.

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Bill Sloman, Sydney
Reply to
bill.sloman

Sorry, it's rejected by the sci.electronics.design usenet server system.

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Reply to
Winfield Hill

Got it via Supernews but not ES.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Phil Hobbs

John, let me remind you that your original switching MOSFET was a 1.5kV part. Clearly you started with thoughts of going higher than 325V in the first stage.

I wonder if most of your heat might not be in the MOSFET. which shares the same copper pad? The FDD7N20 has Rds(on) = 1 ohm max when it's hot, and I^2 R = 9 watts at the end of the ramp. By contrast the inductor resistances are low.

More history. Your early drawing showed a 100uH inductor, which would charge to about 1A in the maximum 80% duty = 4us on-time for the LTC3803. The charged inductor energy would be about 50uJ. At 325V the CV^2 scene robs about 15uJ of that, and then all of it at slightly higher voltages. I remember you commenting on a wimpy voltage rise at higher voltages in your early tests.

We can see how you changed the inductor to 33uH, so it'd ramp up to 3A, and added the 3:1 divider on the 0.1-ohm sense resistor to allow this 3A. That increased the inductor energy to 150uJ, successfully combating capacitive losses. But now you were forced to power a 30W converter, where you had hoped for a nice 10W converter.

Exactly, but consider, an ordinary inductor might have had 1/2 to 1/3 as much CV^2 loss.

It'n not that I'm down on coupled inductors, I love them. But I think the hidden inter- winding capacitance can be a serious issue in some configurations.

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    - Win
Reply to
Winfield Hill

What;d it say?

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 Thanks, 
    - Win
Reply to
Winfield Hill

Actually, no. I needed those 2SK4177 fets in the output amps, so I figured I'd use one in the converter too. We like to keep our BOM line count down. But the 2SK efficiency was terrible, so I went with the smaller FDD7N20, which worked much better. Lower C and especially lower Rds-on.

I always needed the other supplies, +325 and -500. The dual inductor isn't breakdown voltage rated, another reason to keep the boost voltage down. And the SOT23 dual diodes are rated at 400.

The inductor is indeed the hottest part. Core loss?

The converter makes as much power as the load needs. It's about 71% efficient, which is good enough for this app.

Actually, the bifalar wire-wire voltage, that drives most of the capacitance, is only half of the peak boost voltage. So that's 1/4 the energy, seems to me.

This works, tweaked by Spice and the breadboard, and it was a minor consideration on this board. It uses mostly parts that we stock, and wasn't worth a lot of time to optimize any further.

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John Larkin         Highland Technology, Inc 
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Reply to
John Larkin

OK, that sounds like a good point.

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 Thanks, 
    - Win
Reply to
Winfield Hill

It showed the two windings broken up into many sections, with little capacitors going across like rungs of a ladder, to make the point about the capacitance being distributed rather than lumped at the ends.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Phil Hobbs

I guess noting the ringing frequency is a way to approximate the capacitance. Or just the unloaded flyback voltage. Or SRF. The fet drain capacitance wastes a little energy too.

I should test one of those dual inductors for breakdown voltage.

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John Larkin         Highland Technology, Inc 
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John Larkin

Sadly, breakdown-voltage tests aren't good for much, unless you conduct really long tests. A manufacturer may test a part to 5kV for one minute, but then rate it for 660V for 10 years.

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 Thanks, 
    - Win
Reply to
Winfield Hill

That's where trick #2 comes in, Faraday. Either a shield foil if you have a pot core or EE-core bobbin. Or as I use a lot on toroids and "double-barrels" sans bobbin, a coax cable. One of the windings is made with coax and its shield is grounded. Of course only on one side to prevent sparks and molten soldering from flying.

A shield winding (only one end grounded) can also work but not as efficiently and it is not suited for multi-filar coils.

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Regards, Joerg 

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Reply to
Joerg

Try

Thanks Phil, I couldn't describe it well, but you did. There are many ways the capacitance can be (and is) distributed, including across the same winding. The capacitors shown are especially those in the bifilar pair even before the transformer is wound. They join two points that move together, so the capacitance should not matter.

Different styles of winding will have capacitance between points further apart on the same coil as well as on the other coil. Guitar pickup enthusiasts worry about close-winding vs scramble-winding, and I'm sure that it matters.

I'm not sure how you could measure the "degree of distribution" of the interwinding capacitance. Perhaps there just aren't enough ports to even approximate it.

Any thoughts?

Clifford Heath.

Reply to
Clifford Heath

Yup, years ago I made a multiple floating gate supply for a full-bridge switch driver, using quadrafilar wire. It sure made winding the transformer easy. But, it blew up the driver for the supply. Only THEN did I measure the capacitance, it was something over 300 pF, IIRC. I then used a split bobbin and put the swinging windings on the other side of the bobbin, and it worked great. Random-wound coils on the same side of the bobbin were also vastly lower capacitance, under 100 pF for sure.

Jon

Reply to
Jon Elson

Thanks Tim, for your valuable remarks. I played around with some flyback configurations, but didn't find one where moving the diode was helpful.

Some time ago I wrote here about a tracking bipolar power-supply smps I made using coupled inductors: inverting-mode neg output, pos side going along for the ride (I got nearly 0.1% tracking, even though only one side was regulated). I started with the pos diode on the output side, but moved it to the "wrong" input side, so I could add a zeta-converter capacitor (I found the zeta capacitor unnecessary, that task was handled by the high coil coupling). What I had not realized was the dramatic reduction I've been enjoying in capacitive losses. Thanks!

I'd like to understand this, it's hard to see what the diode was doing. Keeping local voltages down?

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 Thanks, 
    - Win
Reply to
Winfield Hill

The diode makes sure that the switching node on the primary is opposite the switching node on the secondary, so the relative dV/dt from primary to secondary is reduced which reduces the capacitive CM currents

It all depends on the dot notation on the windings layout

Cheers

Klaus

Reply to
klaus.kragelund

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