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While working on my first voltage regulator circuit I got stuck at calculating the inductor value... The IC is LM5143A-Q1 from Texas Instruments, and according to the datasheet, the formulas for calculating the required inductance are these: enter image description here

I'm not sure what I_L(peak), I_OUT, and ∆I_L mean, nor how we can find them... I assume I_OUT is the maximum output current that it will need to provide, but don't know what the other two are. How can we find the 3 variables mentioned above? Are there any formulas?

At mouser, most inductors only include values like max DC current/resistance and saturation current (other than their actual inductance). If it helps, the other factors in the L_O Equation are known. V_OUT = 7.4V, V_IN =50V, F_SW (Swithcing frequency) = 100kHz

Also, what do they mean by "choose a buck inductance such that the inductor ripple current, ΔI_L, is between 30% to 50% of the maximum DC output current at nominal input voltage"?

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  • \$\begingroup\$ Your current ripple is a trade off between inductor size and transistor+capacitor cost/size. With hypothetical zero ESR and ESL capacitor, you can choose as high current ripple as you want. 30-50 % gives realistic trade-offs. Have you tried to simulate your circuit with realistic output capacitor and tried to vary the inductance to build Your understanding? \$\endgroup\$
    – winny
    Commented Jul 22, 2023 at 17:28
  • \$\begingroup\$ @winny I haven't tried that. But shouldn't that formula give an estimated inductance value, and from there on just pick an inductor with a higher maximum dc current than what we need (and high saturation current)? This is my first SMD board attempt and I'm not sure how to find any of the I_L(peak), I_OUT, and ∆I_L to find the other, and then integrate it into the equation 15. I suppose we could find I_L(peak) and I_OUT? \$\endgroup\$
    – Mito
    Commented Jul 22, 2023 at 17:59
  • \$\begingroup\$ Note that higher current ripple in the inductor also increases AC and core losses in the inductor which can affect efficiency. It's all a tradeoff. \$\endgroup\$
    – John D
    Commented Jul 22, 2023 at 18:43
  • \$\begingroup\$ The formula is already given, but you need to learn and understand the tradeoff or you’ll end up with cooked capacitors or huge chokes. I_OUT you must know since it’s a design parameter. \$\endgroup\$
    – winny
    Commented Jul 22, 2023 at 19:27
  • \$\begingroup\$ @winny I understood the formula, but what are the tradeoffs I need to know? I_OUT is known, 35A. Higher ripple current = lower efficiency? By ripple current you mean ΔI_L, right? And if so, how can we reduce it? Also, I figured out the inductance to be 6 micro Henry in the answer below \$\endgroup\$
    – Mito
    Commented Jul 22, 2023 at 19:47

1 Answer 1

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ΔIL "is between 30 to 50%" of IOUT.

IOUT is the steady-state output current.

IL(pk) is the peak, or half the delta above the average.

Δ is a symbol used to denote an overall change in a parameter. Inductor current cycles up and down around the steady-state output (that's how it works, it's being switched). ΔIL is the peak-to-peak range of that current. The amounts can all be solved from the inductor equation \$V = L \dfrac{dI}{dt}\$, taking dees as deltas during each phase of the switching waveform (square-wave voltage causes triangle-wave current).

Note that choosing a lower ripple fraction also degrades current limiting ability (i.e., PWM depends on VOUT as well as peak current), which is probably why they include a separate current-limiting function (which will however be susceptible to chaotic behavior because it's simply a no-slope peak current threshold). Presumably that limit will only be used transiently, so the instability doesn't have a big impact on power dissipation or emissions.

Permissible ripple fraction depends upon slope compensation factor, see §9.3.13.

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  • \$\begingroup\$ So that was it! If ΔIL is I_OUT * 30% (or anything in 30-50 range), then the value of IL(pk) is known too! In our case, I_OUT will be 35A, Input voltage 50, output 7.4, so ΔIL= 35*30/100 = 10.5! Knowing this, we can calculate L_O using equation (15) and it will be: 7.4/(10.5*100*10^3)*(50-7.4/50) = 6 micro Henry Inductor! Is this right? Did I get it? \$\endgroup\$
    – Mito
    Commented Jul 22, 2023 at 19:05
  • \$\begingroup\$ That is correct. \$\endgroup\$
    – Tim Williams
    Commented Jul 22, 2023 at 19:56
  • \$\begingroup\$ could you clarify me if high ripple current is bad? If yes, then how can it be lowered. And also, when picking an inductor I should get one with a saturation current higher than that of the IL(pk), right? From what I understood, saturation is the point where inductance lowers, which is not good as far as I know? \$\endgroup\$
    – Mito
    Commented Jul 22, 2023 at 20:04
  • \$\begingroup\$ "Bad" in what sense? Yes, saturation should generally be above Ipk. \$\endgroup\$
    – Tim Williams
    Commented Jul 22, 2023 at 20:12
  • \$\begingroup\$ Bad in the sense it will affect the circuit the buck converter powers. Like burn an IC or capacitor. From what I understand ripple current is a variation of the current in a sine wave-style, which (maybe) is not good? \$\endgroup\$
    – Mito
    Commented Jul 22, 2023 at 20:16

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