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"?

  • \$\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


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.

  • \$\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\$ 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\$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.