Driving capacitive loads with an H bridge
in [Design]
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From: Fred Bartoli on 23 Jul 2010 15:19 Jim Thompson a �crit : > On Fri, 23 Jul 2010 18:46:13 +0200, Fred Bartoli <" "> wrote: > >> Jim Thompson a �crit : >>> On Thu, 22 Jul 2010 21:56:35 -0500, legg <legg(a)nospam.magma.ca> wrote: >>> >>>> On Thu, 22 Jul 2010 00:53:47 +0100, "markp" <map.nospam(a)f2s.com> >>>> wrote: >>>> >>>> <snip> >>>>>>> I'm not looking for a design as such, just a possible architecture. So far >>>>>>> the info you've been given is a triangle waveform across the cap, bipolar >>>>>>> so >>>>>>> no DC component, >40V AC(true RMS), a few hundred Hertz and a question >>>>>>> whether an H-bridge architecture could do it. I'm not sure how much more >>>>>>> information you want...:) >>>>>>> >>>>>>> Mark >>>>>>> >>>>>> A single ended load would require a single ended source - a >>>>>> half-bridge. >>>>>> >>>>>> A triangle wave is generated by a constant current, reversing, but >>>>>> residual DC can only be limited by the accuracy of the modulator. >>>>>> >>>>>> Driving purely inductive or purely capacitive loads is as efficient as >>>>>> driving a short circuit, no matter what the drive method. >>>>>> >>>>>> RL >>>>> Come again? I can create a pair of constant current sources with transistors >>>>> and resistors and alternately charge and discharge the capacitor with them. >>>>> But that's not efficient... >>>>> >>>>> Mark. >>>>> >>>> Neither is any other method of driving a short circuit. >>>> >>>> Given a large enough and perfect enough inductor, and ideal switches, >>>> a short or a capacitor can be driven efficiently by reversing the >>>> inductor's polarity at the waveform peak. That would require 4 >>>> switches. >>>> >>>> Given imperfect and realistically sized components, a half bridge can >>>> reverse it's output inductor current in a finite time period while >>>> supplying approximately constant current of the correct polarity, with >>>> a modest ripple component and reasonable losses. >>>> >>>> RL >>> The OP (markp) implies this is a _real_ situation. How do you meet >>> his requirement "...it has to be efficient, i.e. some kind of energy >>> retrieval" ?? >>> >>> ...Jim Thompson >> >> ROFL! >> >> You'd better not write such nonsense when you intend to blast everybody >> that don't fit your taste here... >> >> I take it that you didn't think much when writing this, which anyway >> shouldn't be with the posture you choose to display, but even, >> understanding this doesn't take much thinking, so? > > Where's the "energy retrieval"? > > An H-bridge driven from an inductor doesn't "retrieve". > > I see nada of substance in your comments. > > Another "Bloviator" ?:-) > > ...Jim Thompson 2.9uJ loss for 360uJ transfered to the cap and "retrieved" per cycle. That's 0.8% loss. How do you call that? Version 4 SHEET 1 1072 680 WIRE 96 -16 -48 -16 WIRE 336 -16 176 -16 WIRE 448 -16 336 -16 WIRE 576 -16 448 -16 WIRE 336 48 336 -16 WIRE 576 48 576 -16 WIRE 992 48 912 48 WIRE 288 64 272 64 WIRE 640 64 624 64 WIRE 288 112 240 112 WIRE 672 112 624 112 WIRE -48 160 -48 -16 WIRE 912 176 912 48 WIRE 336 192 336 128 WIRE 416 192 336 192 WIRE 576 192 576 128 WIRE 576 192 480 192 WIRE 864 192 576 192 WIRE 336 240 336 192 WIRE 864 240 336 240 WIRE 336 272 336 240 WIRE 576 272 576 192 WIRE 288 288 272 288 WIRE 640 288 624 288 WIRE 64 336 64 304 WIRE 176 336 176 304 WIRE -48 432 -48 240 WIRE 64 432 64 416 WIRE 64 432 -48 432 WIRE 176 432 176 416 WIRE 176 432 64 432 WIRE 240 432 240 112 WIRE 240 432 176 432 WIRE 288 432 288 336 WIRE 288 432 240 432 WIRE 336 432 336 352 WIRE 336 432 288 432 WIRE 576 432 576 352 WIRE 576 432 336 432 WIRE 624 432 624 336 WIRE 624 432 576 432 WIRE 672 432 672 112 WIRE 672 432 624 432 WIRE 912 432 912 256 WIRE 912 432 672 432 WIRE -48 512 -48 432 FLAG -48 512 0 FLAG 64 304 A FLAG 272 64 A FLAG 640 288 A FLAG 176 304 B FLAG 272 288 B FLAG 640 64 B FLAG 992 48 Vcap FLAG 448 -16 Sup SYMBOL voltage -48 144 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value 0.022 SYMBOL sw 336 368 M180 WINDOW 0 32 15 Left 0 WINDOW 3 32 44 Left 0 SYMATTR InstName S1 SYMBOL sw 336 144 M180 WINDOW 0 32 15 Left 0 WINDOW 3 32 44 Left 0 SYMATTR InstName S2 SYMBOL voltage 64 320 R0 WINDOW 0 -53 5 Left 0 WINDOW 3 -242 110 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(1 0 0 1E-6 1E-6 .004999 .01) SYMBOL sw 576 144 R180 WINDOW 0 32 15 Left 0 WINDOW 3 32 44 Left 0 SYMATTR InstName S3 SYMBOL sw 576 368 R180 WINDOW 0 32 15 Left 0 WINDOW 3 32 44 Left 0 SYMATTR InstName S4 SYMBOL cap 416 208 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName C2 SYMATTR Value 3e-6 SYMBOL ind 80 0 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 5 56 VBottom 0 SYMATTR InstName L2 SYMATTR Value 10 SYMATTR SpiceLine Ipk=12m Rser=1 SYMBOL voltage 176 320 R0 WINDOW 0 -53 5 Left 0 WINDOW 3 -242 110 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V3 SYMATTR Value PULSE(0 1 0 1E-6 1E-6 .004999 .01) SYMBOL e 912 160 R0 SYMATTR InstName E1 SYMATTR Value 1 TEXT -40 480 Left 0 !.model SW SW(Ron=0.1 Roff=1E6 Vt=0.5 Vh=0) TEXT -32 456 Left 0 !.tran 0 25 24.9 1m Same apply for RL's half bridge, which is nothing more than an unloaded synchronous buck. The real question is that the OP's 40V peak at 1kHz on a 3uF cap is only 10W dissipation. Is it worth the trouble? BTW, Bloviator yourself. -- Thanks, Fred.
From: whit3rd on 23 Jul 2010 16:09 On Jul 23, 9:58 am, "markp" <map.nos...(a)f2s.com> wrote: > "John Fields" <jfie...(a)austininstruments.com> wrote in message > > How much power are you willing to lose? > > 2. Change the inductance of the choke with respect to time so its > > increasing and decreasing reactance does the current limiting. > > Now that's interesting. But, not interesting enough. There are sync circuits for CRT deflection that use a permanent magnet to so alter an inductor and get a sawtooth, but it's not symmetric (and DC zero output was a requirement). > > Either way is going to cost you power, so what efficiency would you > > _like_ to see? > > Something like 85% would be nice. Best solution I see is to resonate (make your C part of an LC tank) at the fundamental, and drive (transformer-coupled, low impedance winding in series with C) according to feedback to 'fill in' the sinewave on the inductor with the difference signal to get a triangle on the C. I'm still fond of the figured-pole generator solution, but it's not something you could get off the shelf.
From: Jim Thompson on 23 Jul 2010 18:44 On Fri, 23 Jul 2010 21:19:32 +0200, Fred Bartoli <" "> wrote: >Jim Thompson a �crit : >> On Fri, 23 Jul 2010 18:46:13 +0200, Fred Bartoli <" "> wrote: >> >>> Jim Thompson a �crit : >>>> On Thu, 22 Jul 2010 21:56:35 -0500, legg <legg(a)nospam.magma.ca> wrote: >>>> >>>>> On Thu, 22 Jul 2010 00:53:47 +0100, "markp" <map.nospam(a)f2s.com> >>>>> wrote: >>>>> >>>>> <snip> >>>>>>>> I'm not looking for a design as such, just a possible architecture. So far >>>>>>>> the info you've been given is a triangle waveform across the cap, bipolar >>>>>>>> so >>>>>>>> no DC component, >40V AC(true RMS), a few hundred Hertz and a question >>>>>>>> whether an H-bridge architecture could do it. I'm not sure how much more >>>>>>>> information you want...:) >>>>>>>> >>>>>>>> Mark >>>>>>>> >>>>>>> A single ended load would require a single ended source - a >>>>>>> half-bridge. >>>>>>> >>>>>>> A triangle wave is generated by a constant current, reversing, but >>>>>>> residual DC can only be limited by the accuracy of the modulator. >>>>>>> >>>>>>> Driving purely inductive or purely capacitive loads is as efficient as >>>>>>> driving a short circuit, no matter what the drive method. >>>>>>> >>>>>>> RL >>>>>> Come again? I can create a pair of constant current sources with transistors >>>>>> and resistors and alternately charge and discharge the capacitor with them. >>>>>> But that's not efficient... >>>>>> >>>>>> Mark. >>>>>> >>>>> Neither is any other method of driving a short circuit. >>>>> >>>>> Given a large enough and perfect enough inductor, and ideal switches, >>>>> a short or a capacitor can be driven efficiently by reversing the >>>>> inductor's polarity at the waveform peak. That would require 4 >>>>> switches. >>>>> >>>>> Given imperfect and realistically sized components, a half bridge can >>>>> reverse it's output inductor current in a finite time period while >>>>> supplying approximately constant current of the correct polarity, with >>>>> a modest ripple component and reasonable losses. >>>>> >>>>> RL >>>> The OP (markp) implies this is a _real_ situation. How do you meet >>>> his requirement "...it has to be efficient, i.e. some kind of energy >>>> retrieval" ?? >>>> >>>> ...Jim Thompson >>> >>> ROFL! >>> >>> You'd better not write such nonsense when you intend to blast everybody >>> that don't fit your taste here... >>> >>> I take it that you didn't think much when writing this, which anyway >>> shouldn't be with the posture you choose to display, but even, >>> understanding this doesn't take much thinking, so? >> >> Where's the "energy retrieval"? >> >> An H-bridge driven from an inductor doesn't "retrieve". >> >> I see nada of substance in your comments. >> >> Another "Bloviator" ?:-) >> >> ...Jim Thompson > >2.9uJ loss for 360uJ transfered to the cap and "retrieved" per cycle. >That's 0.8% loss. > >How do you call that? > Same way you spelled it... "loss". Build me one ;-) ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Spice is like a sports car... Only as good as the person behind the wheel.
From: ehsjr on 23 Jul 2010 23:38 Jim Thompson wrote: > On Fri, 23 Jul 2010 15:35:30 +1000, Grant <omg(a)grrr.id.au> wrote: > > >>On Thu, 22 Jul 2010 19:11:06 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: >> >> >>>On Fri, 23 Jul 2010 12:04:46 +1000, Grant <omg(a)grrr.id.au> wrote: >>> >>> >>>>On Thu, 22 Jul 2010 21:56:35 -0500, legg <legg(a)nospam.magma.ca> wrote: >>>> >>>> >>>>>On Thu, 22 Jul 2010 00:53:47 +0100, "markp" <map.nospam(a)f2s.com> >>>>>wrote: >>>>> >>>>><snip> >>>>> >>>>>>>>I'm not looking for a design as such, just a possible architecture. So far >>>>>>>>the info you've been given is a triangle waveform across the cap, bipolar >>>>>>>>so >>>>>>>>no DC component, >40V AC(true RMS), a few hundred Hertz and a question >>>>>>>>whether an H-bridge architecture could do it. I'm not sure how much more >>>>>>>>information you want...:) >>>>>>>> >>>>>>>>Mark >>>>>>>> >>>>>>> >>>>>>>A single ended load would require a single ended source - a >>>>>>>half-bridge. >>>>>>> >>>>>>>A triangle wave is generated by a constant current, reversing, but >>>>>>>residual DC can only be limited by the accuracy of the modulator. >>>>>>> >>>>>>>Driving purely inductive or purely capacitive loads is as efficient as >>>>>>>driving a short circuit, no matter what the drive method. >>>>>>> >>>>>>>RL >>>>>> >>>>>>Come again? I can create a pair of constant current sources with transistors >>>>>>and resistors and alternately charge and discharge the capacitor with them. >>>>>>But that's not efficient... >>>>>> >>>>>>Mark. >>>>>> >>>>> >>>>>Neither is any other method of driving a short circuit. >>>>> >>>>>Given a large enough and perfect enough inductor, and ideal switches, >>>>>a short or a capacitor can be driven efficiently by reversing the >>>>>inductor's polarity at the waveform peak. That would require 4 >>>>>switches. >>>>> >>>>>Given imperfect and realistically sized components, a half bridge can >>>>>reverse it's output inductor current in a finite time period while >>>>>supplying approximately constant current of the correct polarity, with >>>>>a modest ripple component and reasonable losses. >>>> >>>>Wouldn't it be easier to run the H bridge as switching current >>>>direction only, with current through the bridge controlled by a >>>>separate switching regulator? >>> >>>(*) >>> >>> >>>>Saves talking about ginormous inductors, for starters. Might >>>>even be buildable ;) >>>> >>>>Grant. >>> >>>That would meet the "shape" requirement, but I still ponder what does >>>"...it has to be efficient, i.e. some kind of energy retrieval" mean? >> >>Dunno, may be this is a piezoelectric thingy that wants to squish >>the current back out when being relaxed at a controlled rate? >> >>Rotate the bridge 90' so the capacitor voltage see-saws and the >>charge doesn't fall out? ;^) >> >>>* You'd need some kind of loop to keep it "centered" also. >> >>Yeah, that too, unless that thing is taken rail to rail, self >>centering? >> >>Grant. > > > Yep, You do need to be careful to keep the charge from falling off the > plates ;-) > > ...Jim Thompson That does seem to be happening around here lately. Maybe you could saw the capacitor open and put some Coulomb Glue on the plates. :-) Ed
From: John Larkin on 25 Jul 2010 13:38
On Wed, 21 Jul 2010 19:12:10 -0700 (PDT), amark <amarkpalmer(a)gmail.com> wrote: >On Jul 21, 9:48�pm, "markp" <map.nos...(a)f2s.com> wrote: >> Hi All, >> >> I need to drive capacitor with a triangle wave with no DC across the >> capacitor (i.e. a symmetrical bipolar drive but triangular) but it has to be >> efficient, i.e some kind of energy retrieval. >> >> Is it possible to use standard H bridge circuits to do this? Does anyone >> have any links or app notes? >> >> Thanks! >> Mark. > >There is a classic circuit which is an op-amp integrator, followed by >an inverting comparator whose o/p is fed back to the integrator. >What's an H bridge? The four switches here ftp://jjlarkin.lmi.net/Triangle_Cap.JPG make an h-bridge. The a-a and b-b switch pairs are alternately turned on. So the power source can be connected to the load (the capacitor here) in one direction or the other. Or the two bottom switches could be turned on to short the load, or all four turned off to open it. The switches are usually transistors of some sort, and fully integrated h-bridge chips are common. H-bridges are commonly used to drive motors and speakers, using pulse-width modulation to control how much drive goes into the load. The power source would ususlly be a voltage, not a current like in my circuit. They allow you to, theoretically, make a 100% efficient amplifier. This particular circuit can drive a triangle wave into the capacitive load with, theoretically, zero power required after startup. In practise, it would be far more efficient than a linear amplifier. Hmmm, you can also turn on all four switches in an h-bridge. John |