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What is the formula for average current with step-up PWM circuit (and about 30 kHz frequency and 300 uH inductor) that charges battery at 72 volts, and source is 24 volts, relative to duty cycle? I tested this with CPU PWM signal and noticed that current begins to raise at some critical level very quickly, which caused a problem with controlling, because my PWM signal was based on 1 MHz clock and could not be adjusted finely enough. I'm trying to use continuous mode, although I noticed that discontinuous mode makes it easier to control the average current (by adjusting gap between known charge/discharge pulses)

The TI doc has the basic step up schematic, and Vin and Vout can be assumed to be fixed at 72 and 24V:

enter image description here

The components can be assumed to be ideal, and since the current will be infinite after some critical point, where D > 1-(24/72) = 0.66, I'm wondering if there is function f for inductor current = f(D)? Anyway I'm interested in duty cycle range that is close to the maximum value, and probably will change PWM clock to 16 MHz to get more resolution. (16 MHz / 30 kHz = 533 time steps)

The problem with low resolution PWM is that I get for example about 500 mA average current with pwm value = 20, and 4 amps when pwm value = 21. So not very good control system. And would like to be able to calculate the currents for different D that are possible.

The inductor ripple current formula is rather close, but it does not say anything about what the final value of current will be after lots of cycles.

The infinite current can be understood for example by thinking what happens when duty cycle = 1, or switch is on always.

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  • \$\begingroup\$ ti.com/lit/an/slva372d/… \$\endgroup\$
    – Antonio51
    Nov 29, 2022 at 9:20
  • \$\begingroup\$ Show the schematic please. \$\endgroup\$
    – Andy aka
    Nov 29, 2022 at 9:43
  • \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$
    – Community Bot
    Nov 29, 2022 at 10:41

1 Answer 1

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Tried simulating with LTSpice, and noticed that inductor current goes to infinity if duty cycle is too high, and this means that continuous mode does not work without fast feedback loop. So current can be increased by using smaller value of inductance, and then in discontinuous mode the average current can be controlled rather linearly by slowly changing PWM duty cycle, and there is no need for fast feedback. With microcontroller it is easy to measure current and voltage with A/D converter, but this is rather slow. (but still might be possible to do continuous mode...)

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    \$\begingroup\$ Zenner - Hi, This answer isn't clear to me. Either: (a) You correctly wrote it as an answer, because it really does answer your original question (even though it doesn't seem to do that to me). In that case, please clearly state that this is the final solution, and come back in 2 days to "accept" this answer (or another one, if a better one has been posted by then), to close the whole topic. (Self-answers can only be accepted when 48 hours has elapsed since asking the question.) \$\endgroup\$
    – SamGibson
    Nov 30, 2022 at 10:17
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    \$\begingroup\$ [continued] Or: (b) You mistakenly wrote this as an answer, when in fact you still want more responses. In which case, this is an update and not an answer, and must be "edited into" your original question e.g. click "Edit" under the question, add this new information at the bottom, then delete this "answer". Which applies here, (a) or (b)? Thanks. (As the OP, you would only write an answer if you solved the problem on your own and the topic can be closed.) \$\endgroup\$
    – SamGibson
    Nov 30, 2022 at 10:17
  • \$\begingroup\$ Yes, it is answer. I did not understand what happens when inductor does not reach zero current in the cycle, i.e. when it works in continuous mode. In discontinuous mode inductor current always starts from zero and reaches zero at end of cycle, and switch on-time determines the current (as long as input and output voltages remain the same). \$\endgroup\$
    – Zenner
    Nov 30, 2022 at 17:28
  • \$\begingroup\$ Zenner - Hi, "Yes, it is answer." Ok, thanks for explaining. In this case, please come back and accept your answer (or a different one, if one gets posted and you prefer it) in order to effectively mark the topic as closed, when 48 hours has elapsed since you asked the question. That means after 08.47 UTC on 01 December. Then I can remove all these meta comments. Thanks. \$\endgroup\$
    – SamGibson
    Dec 1, 2022 at 1:48

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