200V @ 100mA from 3.2V

Ooohh, I want part numbers!

--
 Thanks, 
    - Win
Reply to
Winfield Hill
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Here's one:

DigiKey 732-4485-1-ND, Wurth 760871113

It's complicated by the fact that it has two 5V windings -- I'm assuming that I could just run them in parallel.

DigiKey has a bunch of Wurth flyback transformers, but if you start cross- referencing part numbers you find that Wurth calls them out as custom. Given that the one such that I drilled all the way down was referenced as "TI custom" I don't know if they have chip manufacturer support behind them and will be there for a while or if they're evanescent.

I found the above part number by starting with the Wurth catalog, finding a stock transformer that I liked, and then working backwards.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

Yes, 3.2V is more or less the minimum voltage to expect from a LiPo cell.

And yes, it takes a big cell that won't run for long at that power level. What the customer wants is what the customer wants...

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

does tend to turn into a low resonant frequency for the transformer, but yo u can help that a bit by breaking the secondary winding into banks.

with low resonant frequency transformers in photomultiplier supplies. Baxan dall's paper doesn't mention it, but one of my bosses who had worked for Ba xandall at the time said that it was the motivation.

cathode back-light driver as described in Linear Technology's application notes AN45, AN49, AN51, AN55, AN61, AN65). Jim Williams did describes the i nverter as a "current driven Royer" which is isn't, but Peter Baxandall had published the circuit long before that (in a UK journal that isn't easy to get hold of and no American has ever read) and there are suggestions that Jim Williams got the circuit from somebody in the UK who hadn't bothered to tell him where it came from.

Can you show us a schematics for the 200V version?

m wound

solution.

It's far less complicated than the Baxandal, and is straightforward needing no special magnetics

Standard magnetics, if it fits, is way cheaper than any custom magnetics

Cheers

Klaus

Reply to
klaus.kragelund

If you go to trouble of finding your local coil-winding shop, you can get small quantities of special purpose transformers wound onto stock formers without too much trouble or expense.

They won't be as cheap as something that's been mass-produced, but they do tend to be cheaper than cobbling together enough mass-produced wound parts to create the necessary complexity.

--
Bill Sloman, Sydney
Reply to
Bill Sloman

e:

:

o does tend to turn into a low resonant frequency for the transformer, but you can help that a bit by breaking the secondary winding into banks.

e with low resonant frequency transformers in photomultiplier supplies. Bax andall's paper doesn't mention it, but one of my bosses who had worked for Baxandall at the time said that it was the motivation.

ld cathode back-light driver as described in Linear Technology's applicatio n notes AN45, AN49, AN51, AN55, AN61, AN65). Jim Williams did describes the inverter as a "current driven Royer" which is isn't, but Peter Baxandall h ad published the circuit long before that (in a UK journal that isn't easy to get hold of and no American has ever read) and there are suggestions tha t Jim Williams got the circuit from somebody in the UK who hadn't bothered to tell him where it came from.

tom wound

t solution.

ng no special magnetics.

The Baxandall class-D oscillator is pretty simple.

Mass production does beat low volume production, but standard magnetics are just wound on automated version of the same machines that your local coil, winder uses, so "way cheaper" overstates the case.

Custom magnetics comes out with the minimum copper and ferrite that you can use to get the result you want, and that is usually cheaper than something comp licated enough to work with off-the-shelf parts.

Hand-winding does cost money, but neither copper nor ferrite are cheap.

--
Bill Sloman, Sydney
Reply to
Bill Sloman

If your target audience is physicists learning just enough electronics to i nstrument their own experiments, this might be a valid objection.

Sci.electronics.design does address a group who mostly can design their own custom transformers, and get them wound, even if most of them would prefer not to. There are quite a few situations where a custom transformer is the right answer, and acting as if every custom transformer needs to buried at cross-roads with a stake through its magnetic axis isn't a constructive re action.

--
Bill Sloman, Sydney
Reply to
Bill Sloman

The 23:1 ratio is nice. If you had 9 to 10V you could step it up to 200V with a push-pull driver. Or use a low-V boost first, to create say 12V.

--
 Thanks, 
    - Win
Reply to
Winfield Hill

Sometimes it's not only what the customer wants, but also what the customer needs, though one has to be careful not to confuse the two.

--
Bill Sloman, Sydney
Reply to
Bill Sloman

That's nice Win. For high boost ratios the peak currents get pretty high.

Have you considered a dual-winding inductor? They're cheap, and common. Putting the windings in series and driving the center-tap doubles the L's discharging duty-cycle, cutting peak currents in half.

For example, for an ordinary 8x boost stage, 100uH, Vin=3.2V and 1A ripple current (for convenience), on-time would be dt = L di/E, (100uH * 1A)/3.2V = 31uS, discharge time would be (100uH * 1A)/((8-1)*3.2V) = 4.5uS.

Discharge duty-cycle = 4.5uS / (31uS + 4.5uS) = 13%. i.avg(L) has to be 8x avg output current.

With the tapped scheme, charging time for a dual 100uH inductor is still 31uS, and discharge time is (200uH * 1A) / ((8-1) * 3.2V) = 9uS.

Discharge duty is 9uS / (31uS + 9uS) = 22.5%, which cuts the required i.avg(L) to 4.4 x Iout.avg.

Spreading the windings' copper across two coils doubles each coil's d.c.r., so the net effect is that i^2*R loss in the inductor is halved, and conduction loss in the power switch is cut by a factor of four.

Cheers, James Arthur

Reply to
dagmargoodboat

I wonder about the implications of producing an intermediate voltage then discharging that through the transformer again to get 200v.

NT

Reply to
tabbypurr

Essentially you just use an off-the-shelf CCFL transformer and drive it bang-bang style, using two transistors. Cheap. It is customary to have the primary side resonant which reduces EMI worries. To regulate, one can adjust the input voltage supply which takes the power devices out of the high voltage realm.

They are normally meant to supply 600V (and 1200V peaks for striking) so this method is very suited to generate 200V from a low supply voltage. Just not in Tim's case because CCFL transformers aren't widely available at the >20W power level he needs.

Of course, there will probably come a day where CCFL transformers become harder to obatain because the backlight on monitors migrates from fluorescent to LEDs which don't need such high voltages.

Yup. But it depends on quantity. I had one design where we prescribed a custom inductor on purpose because it cost less than off-the-shelf (tens of thousands per year).

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

You can use a tapped inductor and auxiliary diode to get what the old TV engineers called "B+ boost".

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

Just as with my suggested two-stage boost, once you go to a stage and charge a big capacitor and create in essence a new higher DC voltage, next stage's boost is a new unrelated converter. But, in my example, the FET boost switches shared the same controller and feedback control loop. They are intertwined and the intermediate voltage will be determined by the L1 L2 inductance ratios.

--
 Thanks, 
    - Win
Reply to
Winfield Hill

I'm not clear why you say that. It can be, or can reuse some of the first s tage. Yours effectively reuses/shares the control system, mine reuses the t ransformer. I think it would mean switching the load cap.

Just in case the circuit isn't clear, The initial pulse is transformed & du mped into C1. A 2nd tr then dumps that charge back into the transformer pri mary, the secondary of which then charges output cap C2. Thus C1 is disconn ected during the 2nd transformation.

NT

Reply to
tabbypurr

On Thursday, 21 January 2016 00:15:53 UTC+11, Winfield Hill wrote:

Like I said, funny sort of simplicity.

Version 4 SHEET 1 1544 736 WIRE 768 -32 -160 -32 WIRE 896 -32 768 -32 WIRE 896 0 896 -32 WIRE -160 96 -160 -32 WIRE -112 96 -160 96 WIRE 608 96 -32 96 WIRE 1040 96 608 96 WIRE 1040 144 1040 96 WIRE -128 192 -160 192 WIRE -32 192 -128 192 WIRE 128 192 48 192 WIRE 256 192 128 192 WIRE 368 192 336 192 WIRE 896 240 896 64 WIRE 1040 240 1040 208 WIRE 1040 240 896 240 WIRE -368 256 -464 256 WIRE -208 256 -288 256 WIRE 128 256 128 192 WIRE 128 256 -208 256 WIRE 1040 272 1040 240 WIRE -464 336 -464 256 WIRE 768 384 768 -32 WIRE 1040 400 1040 352 WIRE 1120 400 1040 400 WIRE 1232 400 1200 400 WIRE 1376 400 1312 400 WIRE 1392 400 1376 400 WIRE 1488 400 1392 400 WIRE 1040 416 1040 400 WIRE 608 432 608 96 WIRE -160 448 -160 192 WIRE -16 448 -160 448 WIRE 144 480 -48 480 WIRE 304 480 224 480 WIRE 368 480 368 192 WIRE 368 480 304 480 WIRE -160 496 -160 448 WIRE 368 496 368 480 WIRE 1392 496 1392 400 WIRE 1488 496 1488 400 WIRE 1040 528 1040 496 WIRE -48 576 -48 480 WIRE -48 576 -112 576 WIRE -16 576 -16 448 WIRE 176 576 -16 576 WIRE 320 576 256 576 WIRE -464 640 -464 416 WIRE -160 640 -160 592 WIRE -160 640 -464 640 WIRE 368 640 368 592 WIRE 368 640 -160 640 WIRE 608 640 608 496 WIRE 608 640 368 640 WIRE 768 640 768 448 WIRE 768 640 608 640 WIRE 1040 640 1040 592 WIRE 1040 640 768 640 WIRE 1392 640 1392 560 WIRE 1392 640 1040 640 WIRE 1488 640 1488 576 WIRE 1488 640 1392 640 WIRE -464 672 -464 640 FLAG -208 256 Vct FLAG 1376 400 Vout+ FLAG -128 192 tank- FLAG 304 480 tank+ FLAG -464 672 0 SYMBOL ind2 -48 208 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 4 56 VBottom 2 SYMATTR InstName L1 SYMATTR Value 0.054m SYMATTR Type ind SYMATTR SpiceLine Rser=0.00012 SYMBOL ind2 240 208 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 4 56 VBottom 2 SYMATTR InstName L2 SYMATTR Value 0.054m SYMATTR Type ind SYMATTR SpiceLine Rser=0.00012 SYMBOL nmos 320 496 R0 SYMATTR InstName M1 SYMATTR Value FDS6680A SYMBOL nmos -112 496 M0 SYMATTR InstName M2 SYMATTR Value FDS6680A SYMBOL ind -384 272 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 5 56 VBottom 2 SYMATTR InstName L3 SYMATTR Value 0.0775m SYMATTR SpiceLine Rser=0.0001 Cpar=100p SYMBOL voltage -464 320 R0 WINDOW 123 0 0 Left 2 WINDOW 39 24 132 Left 2 SYMATTR SpiceLine Rser=0.001 SYMATTR InstName V1 SYMATTR Value 3.2 SYMBOL res 272 560 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 820 SYMBOL res 240 464 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R2 SYMATTR Value 820 SYMBOL ind2 -128 112 R270 WINDOW 0 44 45 VTop 2 WINDOW 3 5 56 VBottom 2 SYMATTR InstName L4 SYMATTR Value 250m SYMATTR Type ind SYMATTR SpiceLine Rser=7.2 Cpar=100pF SYMBOL schottky 880 0 R0 SYMATTR InstName D2 SYMATTR Value UPSC600 SYMATTR Description Diode SYMATTR Type diode SYMBOL res 1472 480 R0 SYMATTR InstName R4 SYMATTR Value 1k SYMBOL cap 1024 528 R0 SYMATTR InstName C1

SYMBOL ind 1024 256 R0 SYMATTR InstName L5

SYMBOL schottky 624 496 R180 WINDOW 0 24 64 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D1 SYMATTR Value UPSC600 SYMATTR Description Diode SYMATTR Type diode SYMBOL schottky 784 448 R180 WINDOW 0 24 64 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D4 SYMATTR Value UPSC600 SYMATTR Description Diode SYMATTR Type diode SYMBOL schottky 1024 144 R0 SYMATTR InstName D3 SYMATTR Value UPSC600 SYMATTR Description Diode SYMATTR Type diode SYMBOL res 1024 400 R0 SYMATTR InstName R3 SYMATTR Value 22 SYMBOL ind 1104 416 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 5 56 VBottom 2 SYMATTR InstName L8

SYMBOL res 1328 384 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R5 SYMATTR Value 22 SYMBOL cap 1376 496 R0 SYMATTR InstName C2

TEXT -400 664 Left 2 !.tran 0 20m 1u 10n startup TEXT -448 8 Left 2 !K1 L1 L2 0.9995 TEXT -448 40 Left 2 !K2 L1 L4 0.9995 TEXT -448 72 Left 2 !K3 L2 L4 0.9995

is a Class-D version.

The big resistors at R1 and R2 are there to fool LTSpice into not making th e circuit take off at a few MHz. The N87 EPCOS ferrite that I've got in min d for the inductors would probably do that naturally - it's lossy enough th at MHz oscillations would be killed by the core. I could probably get LTSpi ce to think the same way by adding in a fifth linked inductor - a 6uH singl e turn coil with a damping resistance across it - but I haven't got a clue what the damping resistance might be.

L1, L2 and L4 are 3 turns, 3 turns and 204 turns on an EPCOS ungapped RM14 N87 core - B65887E0000R087 which you can buy ex-stock from Farnell/element

14 against order number 2355135.

L3 would be 22 turns of wire on a gapped EPCOS RM14 core - B65887E9160A087

- for which the order number is 1781866. It would be running close to satur ation.

The 56uH inductors at L5 and L8 are also Farnell parts, order number 312487

  1. The FDS6680A D-Mos transistors came out of LTSpice. The worst case threshol d voltage is just low enough to guarantee that the circuit will always turn on, and the maximum acceptable continuous drain current (at 12A) just low enough to work in the circuit, where each of M1 and M2 carries 12A for 50% of the time.

Something a bit bigger might be a better choice.

The circuit isn't optimised - it wouldn't be worth spending time on that un til one had built a real circuit and seen what it did in real life - but th e Spice simulation works well enough to suggest that that might be worth do ing.

The RM14 core is big and non-cheap, but you need that much winding window t o keep the resistive losses in the copper windings respectably low.

--
Bill Sloman, Sydney
Reply to
Bill Sloman

I'm not sure that I get your point, but let me say, of course we assume that C1 is a big enough capacitor that its voltage is not significantly affected by one cycle of L2 or T2 charging C2.

--
 Thanks, 
    - Win
Reply to
Winfield Hill

As I suspected a misunderstanding. C1 is discharged into the transformer more or less completely after each pulse that charges it. There is no L2. Maybe I'll get time to draw it later.

NT

Reply to
tabbypurr

Yes, it appears we're talking about two different circuits. Let me say, it doesn't take that long to draw an ASCII circuit. I like to use an editor that allows typing anywhere on the page. I used AB-Edit,

formatting link
which is free. Start with a row of periods, etc, on the left-hand edge, this sets up the line for different types of viewers, and with something on the lines, AB-Edit lets you type anywhere. Then ctrl-A_ctrl-C ctrl-V copy-paste the finished drawing into your usenet message field.

--
 Thanks, 
    - Win
Reply to
Winfield Hill

This was from several years ago, he built a 10 watt converter from an off the shelf transformer run "backwards" and harvested the energy that would be lost from the large leakage inductance. The link 404s but I believe I have the schematic saved here somewhere.

formatting link

RIP Vlad.

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Reply to
bitrex

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