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From: John Larkin on 24 Jul 2010 15:35
On Fri, 23 Jul 2010 22:27:20 -0700 (PDT), Bill Sloman
<bill.sloman(a)ieee.org> wrote:

>On Jul 24, 1:53�am, John Larkin
><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
>> On Thu, 22 Jul 2010 15:04:46 +0100, "markp" <map.nos...(a)f2s.com>
>> wrote:
>>
>>
>>
>> >Hi All,
>>
>> >If I had a Baxandall class D resonant oscillator, would it be possible by
>> >modulating the current input to the drive circuit to produce a rounded off
>> >triangle like waveform?
>>
>> >BillSloman'sexcellent work
>> >http://home.planet.nl/~sloma000/Baxandall%20parallel-resonant%20Class...
>> >found that the odd harmonic distortion was caused mainly by the AC ripple
>> >current flowing through the source inductor (and hence the driving
>> >windings). This presumably causes a perturbation in the dB/dt of the flux
>> >which modifies the output waveform.
>>
>> >So would it be possible (in theory) by controlling this drive current more
>> >accurately to produce a rounded off triangle waveform without losing all the
>> >efficiencies and advantages of a resonant class D oscillator?
>>
>> >Mark.
>>
>> Doesn't this work?
>>
>> ftp://jjlarkin.lmi.net/Triangle_Cap.JPG
>>
>> What's interesting is that, once it's all going, the power supply can
>> be cranked down to zero and you can make the triangle forever, for
>> free, since the ideal circuit is lossless. The slopes are technically
>> segments of sine waves, not linear bits, so there will be some small
>> curvature, less as L gets bigger. Given a real inductor, simple tweaks
>> could make the slopes straight.
>
>Nice - and a much neater solution than mine. I am fixated on the
>Baxandall circuit, which lead me to a much more complicated (and less
>good) solution.
>
>I'm wondering if you could get away with putting a second winding on
>your inductor, return one end to Vcc/2, and drive the second winding
>with the buffered (and offset by Vcc/2) voltage appearing at the
>switch end of the inductor, thus making up the voltage drive lost as
>the capacitor charges up

Something like that should work. Or just add a modulated linear
current source at the top of the h-bridge, to add a little correction
current, so you don't have to buy such a big inductor.

There may be a passive network that improves triangle linearity, too,
at least at one frequency. Possibly a parallel LC tank in series with
the main L, to raise the effective current-source impedance at 2F, the
major ripple current frequency. Yeah, that might work.

It's interesting that the triangle sides are in fact s-curves,
steepest in the middle. That's because they are actually little slices
of a sine wave, straddling the zero crossing. I actually drew the
sawtooth waveform in my sketch with such a little s-curve, but I
didn't do that consciously. This circuit has some beautiful voltage
and current waveforms.

Run Fred's sim and change L to about 0.85 H. It becomes a sinewave
generator, sort of Baxandall-like.

John

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