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The Control Loop Cookbook from TI explained about sub-harmonic oscillation as below.

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I am really confused by the part "cross over each other at their points of intersection".

This is probably due to my English language. If two waveforms cross over then they should cross at their intersection points, don't they?

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    \$\begingroup\$ I think they might mean that the non-crossing intersection is where the waves are just "touching" each other without one wave to "cross" to the "other side" of the second one (like a tangent point). \$\endgroup\$ – Eugene Sh. Mar 8 '18 at 18:57
  • \$\begingroup\$ Also your title says subharmonic instability in average current mode control. Subharmonic instability for the most part happens only in peak current mode control, not average. \$\endgroup\$ – John D Mar 8 '18 at 19:20
  • \$\begingroup\$ @JohnD: if you read that article and also the one in the link below, the subharmonic oscillation does happen in average current mode control. ti.com/lit/an/slua079/slua079.pdf \$\endgroup\$ – anhnha Mar 8 '18 at 19:25
  • \$\begingroup\$ In each case it is about a condition that creates instability, such as too light of a load, a sudden change in load. They show samples of properly stabilized loops. The 'I' in PID is so important. \$\endgroup\$ – Sparky256 Mar 8 '18 at 20:33
  • \$\begingroup\$ Yeah, it can happen, (hence my "for the most part" comment) but you don't need slope compensation like you do with peak current mode. If your compensator gain is correct you don't have to worry about it. \$\endgroup\$ – John D Mar 8 '18 at 20:49
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Here's a SIMPLIS simulation of an average current mode loop similar to this figure:

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From this TI app note

Note that the compensator inverts the sensed inductor current, so the down-slope becomes the up-slope in the figure.

If the inverted downslope of the inductor current is equal or just less than the ramp slope there is optimal gain because any additional gain will result in subharmonic instability. This looks like the following. The red trace is the sensed inverted inductor current, the green is the ramp and the blue is the switch. You can see the regular duty cycle with no subharmonic oscillation by looking at the switch.

enter image description here

Since this is average current mode we want the loop to respond to the average current, not the peak current. Once the inverted downslope of the current has a slope that's greater than the ramp you get the situation below. The extra crossing of the ramp now changes the duty cycle in an unintended way. The inverted inductor up-slope is always well above the comparator trip point so it's only the downslope that can cause this kind of trouble.

enter image description here

In practice subharmonic instability in average current mode control isn't nearly as troublesome as peak current mode control. There's no need to add a compensating ramp or avoid duty cycles greater than 50% (or even less). If you do see it in average current mode control you can just raise the value of the R1 resistor lowering the compensator gain a bit. (Doing the math to be sure you have a robust compensation.) (Actually it's not a big deal in peak current mode either as most controllers have built-in slope compensation these days which makes it very easy to use.)

Here's a more extreme example with higher compensator gain (Again, the sensed inductor current is inverted so inductor current ramp up is a down slope here):

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And back to stability:

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  • \$\begingroup\$ The inverted inductor up-slope is always well above the comparator trip point so it's only the downslope that can cause this kind of trouble.: I don't get this. What is comparator trip point in this case? Also, did you simulate it? If so, could you add the output waveform in two cases so we can see the subharmonic oscillation? \$\endgroup\$ – anhnha Mar 14 '18 at 16:27
  • \$\begingroup\$ @anhnha The green and red waveforms are the two inputs to the comparator. The comparator trips as the sensed current in red crosses the ramp in green. The waveforms shown are from simulation, and you can see the oscillation in the switch waveform which is the output of the comparator. I can add the current waveform later if that's not clear. Though it just ramps up during the switch on time and down during the switch off time. Not that when the red waveform is on a downslope there's no chance of it crossing the green waveform except for the small portion at the top. \$\endgroup\$ – John D Mar 14 '18 at 16:47
  • \$\begingroup\$ Is there a case the red-waveform downslope doesn't touch the ramp? It think it could happen when the sensed current slope is too small. Also, why the sensed inductor current up-slope in red goes down near the end of period? why doesn't it goes up as a straight line? \$\endgroup\$ – anhnha Mar 14 '18 at 17:07
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    \$\begingroup\$ Remember this is a control loop, so the loop will force the red sensed current to cross the ramp and generate whatever duty cycle is necessary to get the programmed current value. If the waveform didn't cross the ramp you would have either 0% or 100% duty cycle causing no current or max possible current. Because of the compensation, which is an integrator followed by a zero then finally a pole, there is some phase delay and filtration of the signal causing the rounding and phase offset you see. \$\endgroup\$ – John D Mar 14 '18 at 17:35
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    \$\begingroup\$ @anhnha Done. You might want to delete your comment with your email address now, otherwise you may get unwanted SPAM. \$\endgroup\$ – John D Mar 14 '18 at 19:26

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