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Feb 7, 2022 at 14:24 comment added Neil_UK @JRVeale What's more interesting is if the 50 V point is stable, then why can it oscillate? The point is that it takes some series inductance in the line to make it oscillate. At some frequency, the series inductance increases the impedance of the line, so that when paralleled with the load's negative impedance, it can no longer pull it positive. Increasing the capacitance across the load can be described as either 'reducing the LC ratio', or 'reducing the resonant frequency so the impedance of the L is lower, allowing the line to make the total resistance positive'.
Feb 7, 2022 at 13:52 comment added Neil_UK @JRVeale When used to loadline graphs, one can tell from the slopes. Determine the dynamic resistance of a constant power load, either analytically or with a spreadsheet. That load will be in parallel ( ! not series, both V+ and GND are ground, as far as impedance goes) with the line resistance. At the 35 V point, the total resistance is -ve, which means unstable. At the 50 V point, the load is still -ve resistance, but paralleling with the line resistance brings it net positive, so it's stable. Try putting it into a SPICE simulator and use initial conditions to start it where required.
Feb 7, 2022 at 12:58 comment added JRVeale Thanks, yes, I can see the utility of the load line graph clearly. Could I ask for some detail on the bistability? The 35V solution is clearly less desirable than the 50V one, but why is unstable?
Feb 4, 2022 at 5:32 history edited Neil_UK CC BY-SA 4.0
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Feb 3, 2022 at 17:32 comment added Neil_UK It's worth understanding the style of a load line graph, because you'll see them everywhere for all sorts of purposes - biasing transistors, understanding how LED current changes with LED voltage, communicating power withstanding of transistors in data sheets, determining gain of active devices.
Feb 3, 2022 at 16:55 vote accept JRVeale
Feb 3, 2022 at 16:12 history edited Neil_UK CC BY-SA 4.0
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Feb 3, 2022 at 15:22 history answered Neil_UK CC BY-SA 4.0