Designing a switch-mode power supply requires moving quickly between abstract circuit theory and concrete component selection. When you are at the bench prototyping a buck converter, you need to calculate the required inductance, verify the ripple current, and pick a physical inductor that will not saturate under peak load. While SPICE simulators are great for final validation, the TI-30XS MultiView scientific calculator online emulator (or its physical counterpart) remains the fastest tool for iterative component sizing. Its MathPrint engine renders stacked fractions and scientific notation exactly as they appear in datasheets, eliminating the parenthesis-counting errors common on basic calculators.

The Core Inductor Sizing Formula and Symbol Definitions

To maintain Continuous Conduction Mode (CCM) and limit output voltage ripple, the inductance $L$ must be sized relative to your input voltage, output voltage, switching frequency, and acceptable ripple current. The governing equation for a standard synchronous or asynchronous buck converter is:

$L = \frac{(V_{in} - V_{out}) \times V_{out}}{V_{in} \times f_{sw} \times \Delta I_L}$

Table 1: Formula Symbol Definitions and Standard Units
Symbol Parameter Standard SI Unit Typical Datasheet Unit
$L$ Inductance Henrys (H) Microhenrys ($\mu H$)
$V_{in}$ Nominal Input Voltage Volts (V) Volts (V)
$V_{out}$ Target Output Voltage Volts (V) Volts (V)
$f_{sw}$ Switching Frequency Hertz (Hz) Kilohertz (kHz) or Megahertz (MHz)
$\Delta I_L$ Peak-to-Peak Inductor Ripple Current Amperes (A) Amperes (A)

When This Formula Applies (and When It Breaks)

This derivation assumes the converter is operating in Continuous Conduction Mode (CCM), meaning the inductor current never falls to zero during the switching cycle. It also assumes ideal components (zero switch resistance, zero diode forward voltage drop). For quick bench prototyping, these assumptions hold up well; for production-level efficiency mapping, you must account for duty cycle losses.

Unit Mistakes That Break the Math:
The most common failure when using any calculator for power electronics is mixing base SI units with datasheet prefixes. If you enter $f_{sw}$ as 500 (meaning 500 kHz) instead of 500,000 Hz, your calculated $L$ will be off by a factor of 1,000. The TI-30XS MultiView's dedicated x10^n key (located on the bottom right of the keypad) is specifically designed to handle these magnitude shifts cleanly without requiring nested parentheses for scientific notation.

Realistic Answer Magnitude: For modern DC-DC buck converters operating between 500 kHz and 2 MHz with load currents from 1A to 5A, a realistic $L$ magnitude falls between $1.0 \mu H$ and $47 \mu H$ ($1.0 \times 10^{-6}$ to $47 \times 10^{-6}$ H). If your calculator spits out $0.04 H$ (40,000 $\mu H$), you have almost certainly forgotten to convert your switching frequency from kHz to Hz.

Rearranged Forms for Reverse Engineering

When debugging an existing board or selecting a switching frequency for a fixed inductor, you need to isolate different variables. Here are the algebraically rearranged forms:

  • Solve for Ripple Current ($\Delta I_L$):
    $\Delta I_L = \frac{(V_{in} - V_{out}) \times V_{out}}{V_{in} \times f_{sw} \times L}$
  • Solve for Switching Frequency ($f_{sw}$):
    $f_{sw} = \frac{(V_{in} - V_{out}) \times V_{out}}{V_{in} \times L \times \Delta I_L}$
  • Solve for Duty Cycle ($D$) equivalent form:
    Since $D = \frac{V_{out}}{V_{in}}$, the formula simplifies to $L = \frac{V_{out} \times (1 - D)}{f_{sw} \times \Delta I_L}$

Worked Examples Using the TI-30XS MultiView Keystrokes

Let us run through two real-world sizing problems, tracking units at every step and utilizing the specific interface features of the TI-30XS MultiView online emulator.

Problem 1: 12V to 5V Point-of-Load Regulator

Given: $V_{in} = 12V$, $V_{out} = 5V$, $f_{sw} = 500 \text{ kHz}$, Maximum Load Current $I_{out} = 2A$. We will target a 30% ripple ratio, meaning $\Delta I_L = 2A \times 0.30 = 0.6A$.

  1. Set up the numerator: $(12 - 5) \times 5 = 7 \times 5 = 35 \text{ V}^2$
  2. Set up the denominator: $12 \times (500 \times 10^3) \times 0.6 = 3,600,000 \text{ V}\cdot\text{Hz}\cdot\text{A}$
  3. TI-30XS Keystrokes: Press ( 1 2 - 5 ) × 5 ÷ ( 1 2 × 5 x10^n 3 × . 6 ) enter.
  4. Display Result: The MathPrint screen shows the stacked fraction, resolving to $9.7222... \times 10^{-6}$.
  5. Unit Conversion: $9.72 \times 10^{-6} \text{ H} = 9.72 \mu H$.

Problem 2: 24V to 3.3V High-Frequency Rail

Given: $V_{in} = 24V$, $V_{out} = 3.3V$, $f_{sw} = 1.2 \text{ MHz}$, $I_{out} = 3A$. Target 40% ripple ratio, so $\Delta I_L = 3A \times 0.40 = 1.2A$.

  1. Numerator: $(24 - 3.3) \times 3.3 = 20.7 \times 3.3 = 68.31 \text{ V}^2$
  2. Denominator: $24 \times (1.2 \times 10^6) \times 1.2 = 34,560,000 \text{ V}\cdot\text{Hz}\cdot\text{A}$
  3. TI-30XS Keystrokes: Press ( 2 4 - 3 . 3 ) × 3 . 3 ÷ ( 2 4 × 1 . 2 x10^n 6 × 1 . 2 ) enter.
  4. Display Result: $1.9765... \times 10^{-6}$.
  5. Unit Conversion: $1.976 \times 10^{-6} \text{ H} = 1.98 \mu H$.

Pro-Tip: If your emulator displays the result as an exact fraction instead of a decimal, press the S<->D button (Symbol to Decimal) located above the enter key to toggle it into a usable engineering format.

Decision Path: From Calculated Value to Concrete Part Number

Calculating $L$ is only half the job. You must now select a physical inductor that meets the inductance requirement while surviving the peak current without saturating. The peak current ($I_{peak}$) is defined as $I_{out} + \frac{\Delta I_L}{2}$. For Problem 1 ($9.72 \mu H$, $I_{out} = 2A$, $\Delta I_L = 0.6A$), $I_{peak} = 2A + 0.3A = 2.3A$.

Use the following decision matrix to lock in your final component choice. You can cross-reference these parameters using the Coilcraft Power Inductor Finder or TI Power Management portals.

Table 2: Inductor Selection Decision Path (Applied to Problem 1)
Condition / Calculation Action Required Resulting Parameter
Calculated $L = 9.72 \mu H$ Round UP to the nearest standard E12 series value to ensure ripple stays below the 30% target. $L_{selected} = 10 \mu H$
Calculated $I_{peak} = 2.3A$ Apply a 20% safety margin to prevent core saturation during transient load spikes. $I_{sat(min)} = 2.76A$
Calculated $I_{out(RMS)} = 2.0A$ Ensure the inductor's thermal RMS current rating ($I_{rms}$) exceeds the continuous load. $I_{rms(min)} = 2.0A$
Form Factor Constraint Select a shielded drum core to minimize EMI radiation in mixed-signal PCB layouts. Shielded SMD Topology
Final Cross-Reference Match $10 \mu H$, $I_{sat} > 2.76A$, Shielded. Select: Coilcraft MSS1210-103

By following this exact sequence, you terminate the design phase with a concrete, orderable part number: the Coilcraft MSS1210-103. This specific inductor provides $10 \mu H$ inductance, a saturation current ($I_{sat}$) of 4.2A (safely above our 2.76A requirement), and an RMS current rating of 3.9A, ensuring your buck converter will run cool and stable without unexpected magnetic saturation tripping your controller's overcurrent protection.