From: Grant on
On Sun, 25 Jul 2010 16:00:54 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote:

>On Mon, 26 Jul 2010 08:49:12 +1000, Grant <omg(a)grrr.id.au> wrote:
>
>
>>>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.
>>
>>But then one may consider delivering a constant current into / from
>>a cap as not being efficient, since not a tuned circuit? Can you
>>really have this both ways?
>
>Sure. The sim works with no lossy parts at all, just one L and one C
>and the switches. An actual implementation could have only minor
>losses.
>
>This *is* a resonant circuit. The switches just time-warp it in big
>jumps so that we only use the segments of the sine wave that appeal to
>us. Just skip over the parts you don't like.

It's amazing what a tuned circuit can do, I remember driving big
inductive loop (think doorway size) as tuned circuit with a
bullet proof RS485 diff. driver chip and a seeing a whopping
great signal in the loop, tuned with roughly binary value sequence
caps with an 8way dip switch.

Grant.
>
>John
>
From: Tim Williams on
"Grant" <omg(a)grrr.id.au> wrote in message news:jtfp46pvifns5kho0up43joc6bfo926gep(a)4ax.com...
> PP, whazzat? Mind blank.

Push-pull. Instead of driving both ends of the load, as H bridge, or just one end, as half bridge, you drive a tapped winding from each side. You can only pull down on the winding from either end, so transformer action fills in the rest: a pull on one end looks like a push from the other.

Tim

--
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From: Grant on
On Sun, 25 Jul 2010 21:22:35 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote:

>"Grant" <omg(a)grrr.id.au> wrote in message news:jtfp46pvifns5kho0up43joc6bfo926gep(a)4ax.com...
>> PP, whazzat? Mind blank.
>
>Push-pull. Instead of driving both ends of the load, as H bridge, or just one end, as half bridge, you drive a tapped winding from each side. You can only pull down on the winding from either end, so transformer action fills in the rest: a pull on one end looks like a push from the other.

Yeah, know what push-pull is :)

But PP recently for me is polypropylene film caps.

Rarely see push-pull transformer these days, more common to see half or
full bridge drive, to utilise copper all the time instead of half the
time?

Though centre-tap output windings are common, to save on expensive
diodes, so copper utilisation perhaps a poor argument?

Grant.
From: Winfield Hill on
markp wrote...
>
> On Jul 21, "markp" wrote:
>
>>> 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.
>
> Oh dear! Just to be clear, by energy retrieval I meant the stored energy
> in the capacitor when charged needs to be recovered back when discharging
> so the cycle can repeat and process is efficient, much like a resonant LC
> oscillator but with triangle waves instead of sine waves. See my post to
> Jim.

It's a lot of fuss to save a few watts. For 2uF, 138.5Vpp and 200Hz
I calculate you need a roughly 110mA square-wave drive current.

A resonant inductor would be 0.32H, which is a high inductance, and
to insure a linear ramp, rather than a sine wave, you'd need a much
higher inductance than that (did you say how much nonlinearity you
can tolerate?). What's more, the H-bridge involved must be made
from floating bidirectional switches. Ouch. So it appears any
full-cycle energy-storage idea is going to be very painful.

OTOH, a +/-110mA class-D chopper current-source drive with +/-70V
compliance would be relatively easy, using components created for the
high-power audio market. Class D also uses energy storage you know.

But you said you'd not like chopper noise, so what's so bad about
less than 8 watts of dissipation in a simple linear circuit?


--
Thanks,
- Win
From: Jim Thompson on
On 26 Jul 2010 05:35:02 -0700, Winfield Hill
<Winfield_member(a)newsguy.com> wrote:

>markp wrote...
>>
>> On Jul 21, "markp" wrote:
>>
>>>> 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.
>>
>> Oh dear! Just to be clear, by energy retrieval I meant the stored energy
>> in the capacitor when charged needs to be recovered back when discharging
>> so the cycle can repeat and process is efficient, much like a resonant LC
>> oscillator but with triangle waves instead of sine waves. See my post to
>> Jim.
>
> It's a lot of fuss to save a few watts. For 2uF, 138.5Vpp and 200Hz
> I calculate you need a roughly 110mA square-wave drive current.
>
> A resonant inductor would be 0.32H, which is a high inductance, and
> to insure a linear ramp, rather than a sine wave, you'd need a much
> higher inductance than that (did you say how much nonlinearity you
> can tolerate?). What's more, the H-bridge involved must be made
> from floating bidirectional switches. Ouch. So it appears any
> full-cycle energy-storage idea is going to be very painful.
>
> OTOH, a +/-110mA class-D chopper current-source drive with +/-70V
> compliance would be relatively easy, using components created for the
> high-power audio market. Class D also uses energy storage you know.
>
> But you said you'd not like chopper noise, so what's so bad about
> less than 8 watts of dissipation in a simple linear circuit?

Agreed!

Look back at...

Message-ID: <pshp46dqpnrd688pdia6hlm25iqvat7nl0(a)4ax.com>

From the OP's loose description (he says a few amps, which I took as
2A), I got 36uF :-(

...Jim Thompson
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