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I've been working hard to understand boost converters and the role of inductors in the boost converter circuit. I found a pretty good list of basic calculations from Texas Instruments and I started doing some back of the napkin calculations for the TPS61232.

My logic so far has been to say, roughly:

If I need 5V at 1.2A, that's 6W of output. If I get 80% efficiency I'll need 7.5W of input at 2.8V worst case . . . so I'll be pulling 2.7A from the source.

As I read through that TI document, I noticed that D and (1 - D) show up pretty often in these calculations.

So, I computed D for my circuit:

$$ D = 1 - \frac{V_{in(min)} * η }{V_{out}} = 1 - \frac{2.8V * .8}{5V} = .552 $$

Then, I read about computing the ripple current: $$ ΔI_{L} = \frac{V_{in(min)} * D}{f_{s} * L} = \frac{2.8V * .552}{2.0MHz * 1uH} = .7728A $$

The datasheet says the input voltage ripple is ±200 mV and the TI document says you can usually estimate inductor ripple to be 20% - 40% of the output current. My calculation at 1.2A is .7728A which is about 60%; that seems kind of high, but I can't tell that I'm doing something wrong. Maybe it's by design? Maybe it's because they're more optimistic about their 90% efficiency? Or, maybe it's based on their output current rating of 2.1A?

In any case, I wanted to know how much DC current the inductor needed to be rated for at various output currents, so I tried to come up with a formula. I saw that the I(max out) formula uses ΔI(L)/2. I assumed that's because the ripple is half above and half below V(in)?

So, I decided that something along these lines is probably pretty close:

$$ I_{L} = \frac{ΔI_{L}}{2} + \frac{V_{out} * I_{out}}{V_{in(min)} * η} $$

As I was factoring I(out) out, I realized that my formula could be expressed as to:

$$ I_{L} = \frac{ΔI_{L}}{2} + \frac{I_{out}}{1 - D} $$

So, I thought, "hey, there's that duty cycle again. Why does it keep showing up?" What is a duty cycle and why does it seem so important to switch mode power supply circuits?

Based on this formula, I looked at the effects of output current on the inductor current and came up with these:

$$ I_{L} = .3875A + \frac{1.2A}{.448} \approx 3.1A $$ $$ I_{L} = .3875A + \frac{0.75A}{.448} \approx 2.1A $$

Am I even in the ballpark here? If I am, how much should I derate inductors? I mean, if I'm looking for 3.1A in the inductor, should I look for a thermal and saturation DC rating greater than, say, 130% of 3.1A? 200%?

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  • \$\begingroup\$ A transistor is switched rapidly on and off to either charge the inductor or let it freewheel into the output cap/load. Consider the switching frequency fixed. Duty cycle is the amount of on vs. off time within one switching cycle. A longer on time allows a larger current to build up in the inductor. \$\endgroup\$ – Unimportant Feb 11 at 7:46
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    \$\begingroup\$ Duty cycle affects effective peak input current because the input switch is only on for D of the time, so to achieve the average current it needs to draw more current when it is on to make up for when it is off. eg if a switch was on 25% of the time and the mean current across the cycle was 1.2A, then, because it had only 25% of the cycle to draw all the current involved then it needs to draw 1/25% = 1/D = 4 x as much = 4.8A for the portion of the cycle when it IS on. | Related. A converter has t_on and t_off. During toff the energy is transferred to the load. The on and off periods are ... \$\endgroup\$ – Russell McMahon Oct 19 at 6:24
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    \$\begingroup\$ ... related to the in and out voltages for a single coil. Fot a transformer the turns ratio gets involved. And EFFECTIVE Vout includes a factor related to efficiency 0 < Z < 1. Ton:Toff ~~= Vin:Vout/Z . If the output is boosted by a transformer ratio of N then Ton/Toff ~= Vin/(Vout/Z x N) (as the N stepup makes the Vout "look" lower. \$\endgroup\$ – Russell McMahon Oct 19 at 6:28
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Stop! You will need some incredible luck if you are going to get something usable by shuffling formulas without understanding them.

Start by learnig how inductors actually work. You must both qualitatively and also in numbers understand how a coil stores magnetic energy, when its current gradually grows after DC voltage is applied to it. You should also understand that the current never stops immediately, but gradually, the coil will generate as high voltage as needed to let the current flow despite you turned the switch off. Start by reading this:

How does the inductor ''really'' induce voltage?

Then learn the induction law and try to calculate some current growth rates with known dc voltage and inductance. Calculate also how inductor current decreases, when the generated inductive peak is supplied to known existing DC voltage.

Then you can learn how flyback type switchmode power supplies really work. The duty cycle is a trivial and natural quantity in repeating pulse systems to describe how big part of the time something is on, typically the switching transistor.

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  • \$\begingroup\$ I’m not positive this addresses my question. I think it would be helpful if you indicated at what point my current understanding becomes inaccurate and which formulas I appear to have misunderstood. It would also be helpful if you included specific examples of the calculations you recommend that I make. Thank you very much! \$\endgroup\$ – D. Patrick Feb 11 at 7:25
  • \$\begingroup\$ "at what point my current understanding becomes inaccurate and which formulas I appear to have misunderstood" ... for example here: "So, I thought, "hey, there's that duty cycle again. Why does it keep showing up?" What is a duty cycle and why does it seem so important to switch mode power supply circuits?" user287001 is showing the steps how to understand this duty cycle, which is quite crucial. \$\endgroup\$ – Huisman Feb 11 at 7:33
  • \$\begingroup\$ So my calculations are all wrong? Or they’re all right, but just by luck? I’m not sure I don’t understand the formulas. I just didn’t understand why duty cycle keeps showing up as a ratio of voltages when I would expect it to be a function of at least frequency. However, I think that I’ve found the answer elsewhere. It appears that if you apply KVL to the loop during the on phase and solve for VL, and do the same for the off phase, you can substitute those values into VL = L(di/dt). Since the currents are equal in both phases, you integrate and reduce giving you D as a function of voltage. \$\endgroup\$ – D. Patrick Feb 11 at 7:53
  • \$\begingroup\$ @D.Patrick I didn't recalculate them. You can well do algebraic and calculus manipulations ok. But there's allways a possiblity to start from wrong formulas or continue without noticing that in different formulas the same symbol means different things. I have in my childhood calculated with velocity and acceleration: s=vt, v=at => s=at^2 and got angry when the teacher said "you do better by learning the basics at first". Duty cycle as a surprising variable suggested that you have skipped something before starting to apply equations from a handbook. I admit that it's not my business. \$\endgroup\$ – user287001 Feb 11 at 8:05
  • \$\begingroup\$ Yeah, I think I didn’t know what to ask or how to ask it. When I read another TI white-paper, I realized that time was factored out of the duty cycle calculation and it makes a lot more sense now how the duty cycle can be calculated as a ratio of voltages and why all of the formulas use that ratio. \$\endgroup\$ – D. Patrick Feb 11 at 8:11

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