Driving capacitive loads with an H bridge
in [Design]
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From: Jim Thompson on 22 Jul 2010 21:59 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 -- | 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: Grant on 22 Jul 2010 22:04 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.
From: Jim Thompson on 22 Jul 2010 22:11 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? * You'd need some kind of loop to keep it "centered" also. ...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: Grant on 23 Jul 2010 01:35 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.
From: markp on 23 Jul 2010 05:57
"John Fields" <jfields(a)austininstruments.com> wrote in message news:66ch469elfa2i4k2l2lqeknf4mtqlasq09(a)4ax.com... > On Thu, 22 Jul 2010 18:05:46 +0100, "markp" <map.nospam(a)f2s.com> > wrote: > >> >>"John Fields" <jfields(a)austininstruments.com> wrote in message >>news:vbtg46ditdftdgl5hg3i06b84oj0irdhoc(a)4ax.com... >>> On Thu, 22 Jul 2010 10:02:23 +0100, "markp" <map.nospam(a)f2s.com> >>> wrote: >>> >>> >>>>Sorry Jim, what I meant was if you drive a triangle wave across a >>>>capacitor >>>>then the capacitor will store energy when it's charged up, but when it >>>>is >>>>discharged the energy has to be recovered back (retrieved) so it can be >>>>used >>>>again in the next cycle. A parallel resonant LC circuit does just this >>>>by >>>>shifting the stored energy from the capacitor to the inductor and back >>>>again, so for ideal components no power is needed to sustain >>>>oscillation. >>>>An >>>>H brigde would do it by temporarily storing energy in the inductor part >>>>and >>>>dumping that energy back to the DC supply reservoir. >>> >>> --- >>> Here's six circuits: >>> >>> On the top, an "H" bridge driving a capacitor, a series resonant >>> circuit, and a parallel resonant circuit, and on the bottom a bipolar >>> half-bridge driving the same circuits. >>> >>> Does it look like any of them will do what you want? >>> >>> JF >>> >> >>Did you forget to include the link? > > --- > Nope, the circuit list:-( > > > Version 4 > SHEET 1 960 680 > WIRE -272 -224 -336 -224 > WIRE -144 -224 -208 -224 > WIRE 16 -224 0 -224 > WIRE 112 -224 96 -224 > WIRE 192 -224 176 -224 > WIRE 384 -224 384 -320 > WIRE 384 -224 336 -224 > WIRE 400 -224 384 -224 > WIRE 464 -224 464 -320 > WIRE 528 -224 464 -224 > WIRE -336 -176 -336 -224 > WIRE -144 -176 -144 -224 > WIRE 0 -176 0 -224 > WIRE 192 -176 192 -224 > WIRE 336 -176 336 -224 > WIRE 528 -176 528 -224 > WIRE -336 -48 -336 -96 > WIRE -144 -48 -144 -96 > WIRE -144 -48 -336 -48 > WIRE 0 -48 0 -96 > WIRE 0 -48 -144 -48 > WIRE 192 -48 192 -96 > WIRE 192 -48 0 -48 > WIRE 336 -48 336 -96 > WIRE 336 -48 192 -48 > WIRE 528 -48 528 -96 > WIRE 528 -48 336 -48 > WIRE -336 16 -336 -48 > WIRE -160 96 -336 96 > WIRE 80 96 16 96 > WIRE 192 96 160 96 > WIRE 480 96 352 96 > WIRE 544 96 480 96 > WIRE -336 144 -336 96 > WIRE 16 144 16 96 > WIRE 352 144 352 96 > WIRE 480 144 480 96 > WIRE -160 160 -160 96 > WIRE 192 160 192 96 > WIRE 544 160 544 96 > WIRE -336 272 -336 224 > WIRE -160 272 -160 224 > WIRE -160 272 -336 272 > WIRE 16 272 16 224 > WIRE 16 272 -160 272 > WIRE 192 272 192 224 > WIRE 192 272 16 272 > WIRE 352 272 352 224 > WIRE 352 272 192 272 > WIRE 480 272 480 224 > WIRE 480 272 352 272 > WIRE 544 272 544 224 > WIRE 544 272 480 272 > WIRE -336 352 -336 272 > FLAG -336 352 0 > FLAG -336 16 0 > SYMBOL voltage 16 128 R0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMATTR InstName V1 > SYMBOL ind 496 240 R180 > WINDOW 0 36 80 Left 0 > WINDOW 3 36 40 Left 0 > SYMATTR InstName L1 > SYMATTR Value .845 > SYMBOL ind 64 112 R270 > WINDOW 0 32 56 VTop 0 > WINDOW 3 5 56 VBottom 0 > SYMATTR InstName L2 > SYMATTR Value .845 > SYMBOL cap 176 160 R0 > SYMATTR InstName C2 > SYMATTR Value 3e-6 > SYMBOL cap 528 160 R0 > SYMATTR InstName C1 > SYMATTR Value 3e-6 > SYMBOL voltage 352 128 R0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMATTR InstName V2 > SYMBOL voltage -336 128 R0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMATTR InstName V3 > SYMBOL cap -176 160 R0 > SYMATTR InstName C3 > SYMATTR Value 3e-6 > SYMBOL cap -208 -240 R90 > WINDOW 0 0 32 VBottom 0 > WINDOW 3 32 32 VTop 0 > SYMATTR InstName C4 > SYMATTR Value 3e-6 > SYMBOL voltage -336 -192 R0 > WINDOW 0 -42 0 Left 0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR InstName V4 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMBOL voltage -144 -80 R180 > WINDOW 0 -43 116 Left 0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR InstName V5 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMBOL cap 112 -208 R270 > WINDOW 0 32 32 VTop 0 > WINDOW 3 0 32 VBottom 0 > SYMATTR InstName C5 > SYMATTR Value 3e-6 > SYMBOL voltage 0 -192 R0 > WINDOW 0 -42 -2 Left 0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR InstName V6 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMBOL voltage 192 -80 R180 > WINDOW 0 -44 115 Left 0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR InstName V7 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMBOL cap 464 -240 R90 > WINDOW 0 0 32 VBottom 0 > WINDOW 3 32 32 VTop 0 > SYMATTR InstName C6 > SYMATTR Value 3e-6 > SYMBOL voltage 336 -192 R0 > WINDOW 0 -44 -3 Left 0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR InstName V8 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMBOL voltage 528 -80 R180 > WINDOW 0 -38 106 Left 0 > WINDOW 3 24 104 Invisible 0 > WINDOW 123 0 0 Left 0 > WINDOW 39 0 0 Left 0 > SYMATTR InstName V9 > SYMATTR Value PULSE(-35 35 0 .005 .005 0 .01) > SYMBOL ind 368 -304 R270 > WINDOW 0 32 56 VTop 0 > WINDOW 3 5 56 VBottom 0 > SYMATTR InstName L3 > SYMATTR Value .845 > SYMBOL ind 0 -208 R270 > WINDOW 0 32 56 VTop 0 > WINDOW 3 5 56 VBottom 0 > SYMATTR InstName L4 > SYMATTR Value .845 > TEXT -320 304 Left 0 !.tran .1 > > JF > Thanks for that. Yes, one of these is going to solve it - my preference at the moment is not actually an H bridge for noise reasons but rather a resonant type oscillator like the one in the top right of the schematic, if I can make the waveform triangular. That actually *might* be possible by using a class D oscillator and forcing additional current in and out of the drive windings, not sure yet. The question is how the driving sources are implemented. For example the top right (a capacitor driven by two voltage sources either side) would need to have a current source as well, and that current source mustn't be resistive so that the energy used to charge the capacitor in one direction is recovered from the capacitor and not thrown away by resistive losses. Mark. |