Why does duty cycle seem relevant when calculating inductor current in boost converter circuit?

Question

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%?

Answer 1

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.