Inductor Peak Current Check
Ipeak = Idc + V × Δt / (2 × L)
Quick answer
This calculator solves ΔI = V × Δt / L, Ipeak = Idc + ΔI/2 for Current Ripple. With Applied Voltage 12 V, On-Time 100 µs, Inductance 100 µH, the result is 12 A.
Calculator
Formula
Description
In a switching converter, the inductor current ramps up during the on-time and ramps down during the off-time. The current ripple ΔI depends on the applied voltage, on-time, and inductance. The peak current is the DC component plus half the ripple: Ipeak = Idc + ΔI/2. If Ipeak exceeds the inductor saturation current, the inductance drops dramatically, causing current to spike and potentially damaging the converter. This is the most critical check in power inductor selection.
Variables
- ΔI — Peak-to-peak current ripple (A)
- V — Voltage across inductor during on-time (V)
- Δt — Switch on-time (s)
- L — Inductance (H)
Practical Notes
The forward result returns ΔI (the full ripple). To get Ipeak, add Idc + ΔI/2. Choose an inductor with saturation current rating at least 20% above Ipeak at maximum operating temperature. Ferrite inductors lose saturation current with temperature, while powdered iron types are more stable but have lower initial permeability.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| dI | Current Ripple | A | pA, nA, µA, mA, A, kA |
| V | Applied Voltage | V | pV, nV, µV, mV, V, kV |
| dt | On-Time | µs | ps, ns, µs, ms, s, min, h |
| L | Inductance | µH | nH, µH, mH, H |
Step by step
- 1. What you know: Applied Voltage 12 V, On-Time 100 µs, Inductance 100 µH
- 2. The relation: ΔI = V × Δt / L, Ipeak = Idc + ΔI/2
- 3. Substitute: V = 12 V, dt = 100 µs, L = 100 µH
- 4. Result: Current Ripple = 12 A
Related quantities from the same inputs
- P = V × I = 144 W
Worked examples
Minimum range
Given: Applied Voltage 0.33 V, On-Time 100 ns, Inductance 100 nH
Result: Current Ripple = 0.33 A
Low range
Given: Applied Voltage 3.3 V, On-Time 1 µs, Inductance 1 µH
Result: Current Ripple = 3.3 A
Typical
Given: Applied Voltage 12 V, On-Time 100 µs, Inductance 100 µH
Result: Current Ripple = 12 A
High range
Given: Applied Voltage 0.23 kV, On-Time 0.5 ms, Inductance 0.47 mH
Result: Current Ripple = 0.2447 kA
Maximum range
Given: Applied Voltage 2.3 kV, On-Time 5 ms, Inductance 4.7 mH
Result: Current Ripple = 2.447 kA
How Current Ripple changes with Applied Voltage
| Applied Voltage (input) | Current Ripple (result) |
|---|---|
| 1.2 V | 1.2 A |
| 3 V | 3 A |
| 6 V | 6 A |
| 9 V | 9 A |
| 12 V | 12 A |
| 18 V | 18 A |
| 24 V | 24 A |
| 48 V | 48 A |
| 84 V | 84 A |
| 120 V | 120 A |
The same result in other units
| Unit | Value |
|---|---|
| mA | 12000 |
| A | 12 |
| kA | 0.012 |
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<p><a href="https://vbatpower.com/formulas/inductor-saturation-check/">Inductor Peak Current Check</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Inductor Peak Current Check?
Use ΔI = V × Δt / L, Ipeak = Idc + ΔI/2. With Applied Voltage 12 V, On-Time 100 µs, Inductance 100 µH, Current Ripple works out to 12 A.
What is the formula for Inductor Peak Current Check?
The formula is ΔI = V × Δt / L, Ipeak = Idc + ΔI/2. In a switching converter, the inductor current ramps up during the on-time and ramps down during the off-time.
What units does Inductor Peak Current Check use?
Current Ripple is shown in mA, A, kA. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
Related Concepts
All Inductors formulas →Need more features?
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