Spread-spectrum thing

May 08, 2020 20 Replies

If I have say two square waves, each 180 degrees out of phase. Say during period A I advance one slightly ahead in phase, and the second slightly back. and on period B I do the opposite.



If I sum them bridged-tied load into a resistor, I get a sub-harmonic at half the original period cuz I'm jittering forard and back over two cycles of the original period.



I'm thinking if I'm not wrong if I jitter them correctly I should be able to run the outputs thru a common mode? differential? choke and eliminate the sub-harmonic, that is make the sub-harmonic common mode and the original fundamental and higher harmonics differential mode. Or vice versa. Maybe I have to jitter them in the same direction. Or with different timing. Or something. Any ideas?


I don't think the voltage across the resistor has any sub-harmonics. It does have odd harmonics, whose amplitude will squirm around as the time shift changes.

The RMS voltage across the resistor of course goes down as the phases shift from 180 degrees.

What's the goal? EMI limits reduction? Wobble the frequency for that.

John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com

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Actually the idea wasn't "spread spectrum" exactly it was just a thought about reducing switch dissipation in an H-bridge arrangement you're going to filter anyway, - say your period is 8 units, instead of keeping top switch on for 4 units and bottom on for 4 units keep top switch on for 3 units, leading dead-center. then next cycle keep it on for 3 units, lagging. And vice versa on the bottom switch.

It does seem to reduce dissipation and the cycle to cycle variation gets mostly smoothed in the filter. Unfortunately it looks like if you then take the output single ended you get a sub-harmonic at half the fundamental which makes low-passing annoying.

Rather the long-term amplitude variation is smoothed but you end up with glitches around the zero-crossing from the sub-harmonic which is still within the pass-band of your filter.

AKA SED.

We get too many unspecified probems here. It becomes "20 questions" to figure what the guy wants. It's often not worth it to try to help.

John Larkin Highland Technology, Inc Science teaches us to doubt. Claude Bernard

It's ok I figured it out, so you don't feel you wasted your time this is great freeware:

All the filters and typologies you can shake a stick at!

The answer seems to be (I think) that into a properly-tuned resonant filter like a nth-order Chebyshev, with a differential choke as the first element, low-amplitude common-mode sub-harmonics well below the cutoff get wiped out they don't even "see" the first element

You would be best off thinking about this entirely in the time domain and draw out the two waveforms that you intend. I think you mean to have a reference square wave and a second rectangular wave with a fixed phase offset. But your description is far too vague to be sure.

Basically one of them cannot be a square wave.

Is this what you mean?

________________ ________________ |_______________| |_______________| _____________ _____________ |_______________| |__________________| |____

Difference between them. You really need to think hard about what you mean by "jittering the phase" of a rectangular wave over two periods of its square wave cousin. If you can draw out what you mean then you can easily compute its Fourier transform and then you have your answer.

Regards, Martin Brown

Isn?t it at half bridge arrangement you are talking about?

You always need deadtime

In your description first with 4 high side on and 4 low side on with 8 peri od you have 50% duty

On the other with 3 and 3, your duty cycle is 37%, and some funny freewheel ing in the deadtime

I can?t see a system where that would be any good?

Except if you were talking about a demand duty cycle in between the 3 and 4 and your manipulation/jitter is way faster than your system cross over

Changing duty cycle will change the harmonics from pure odd to combination of odd and even. I do not see half harmonics in that system?

Cheers

Klaus

Are you re-inventing the phase-shift full bridge switching convertor?

In the phase shift modulated SMPS the two legs of the bridge both switch a 50:50 square wave but the relative phase between them is varied to modulate power delivered to the load. It can be a very useful topology.

piglet

Please see linked files:

Waveforms:

Frequency is 3 MHz

The FFT of each phase:

There is a sub-harmonic at 1.5 MHz.

A differential-to-single-ended filter:

The FFT of the output (there's gain in the drive stage, driving the filter.) The sub-harmonic has been suppressed about 20 dB, even though it's within the pass-band of the filter:

It's something like that but rather in the case where the I'm experimenting with where the voltages are asymmetric on the bridge, one switch of each is tied to a higher voltage than the other, reducing dissipation by not keeping the ones switching higher voltages by not keeping them on as you usually would. In exchange for a bit of distortion

The two 10k and 2n2 are shorted by the wire to their right? So the output is basically from the center-tap of the dual 620nH?

piglet

Or put another way, distributing the energy-transfer between the fundamental and half the fundamental, then reject the sub-harmonic in the filter

Whoops! That's not right. But seems to be the way the sub-harmonic is rejected. Just remove the cap and resistors - this is just a one-pole filter with a differential choke on the input, lol. If I remove the connection the cap is in the path and acts as a high impedance to the sub-harmonic I think...don't work...

The other two poles should probably go after

Er, Two-pole, three pole? Something

Perhaps if you described what you were trying to do and explained what each of the coloured lines were intended to show it might be possible to make some sort of informed comment. Also increase the modulation to say

5% so that it is clear what is real and what is a plotting artefact.

There are no visible square waves in that plot there are two rectangular waves with ~3:1 mark space ratio in blue and green and a stepwise sine approximation in red.

The great thing about Fourier transform pairs is that they only care about the shape of the function. The actual frequency doesn't matter.

Regards, Martin Brown

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