When you place inductors in parallel, the total inductance decreases while the overall current-handling capability increases. For two identical inductors, the equivalent inductance ($L_{eq}$) is exactly half the value of a single unit, but the saturation current rating doubles. This topology is a staple in high-current switching power supplies where a single inductor with the required inductance and current rating would be physically too large or prohibitively expensive.
However, unlike parallel resistors, parallel inductors introduce hidden physical traps: DC resistance (DCR) mismatch and mutual inductance. If you ignore these, your circuit will suffer from unequal current sharing, localized thermal runaway, and catastrophic MOSFET failure. This guide breaks down the exact node behavior, failure extremes, and a real-world buck converter design using off-the-shelf components.
The Parallel Inductor Topology: Node Labels and Core Behavior
In a standard parallel inductor configuration, the components share two common electrical nodes. Let's define the topology:
- Node A (Input/Switch Node): The common connection point tied to the upstream voltage source or switching MOSFET drain.
- Node B (Output/Load Node): The common connection point tied to the downstream load and output filter capacitors.
The governing equation for inductance in parallel (assuming zero mutual coupling) mirrors the parallel resistor formula:
$$ \frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} + ... + \frac{1}{L_n} $$
For two inductors, this simplifies to the product-over-sum formula: $L_{eq} = (L_1 \times L_2) / (L_1 + L_2)$. If $L_1 = L_2 = 10\mu H$, then $L_{eq} = 5\mu H$.
Placing inductors in series increases total inductance ($L_{eq} = L_1 + L_2$) but limits your maximum continuous current to the saturation rating of the weakest inductor in the chain. Placing them in parallel sacrifices inductance but splits the ripple and DC current across multiple magnetic cores, effectively multiplying your current handling. In high-power DC-DC converters, current capacity is usually the bottleneck, making parallel the superior choice.
For deeper theoretical background on magnetic field interactions in these configurations, refer to the foundational guides on inductors in parallel at Electronics Tutorials.
Behavior Matrix and Extreme Failure Modes
Understanding how the circuit reacts when a single element drifts or fails is critical for designing reliable power stages. The table below maps parameter shifts and extreme fault conditions.
| Condition / Fault | Effect on Total Inductance ($L_{eq}$) | Effect on Current Distribution | System-Level Consequence |
|---|---|---|---|
| Nominal (Matched) | Decreases (e.g., $L/2$) | Split 50/50 | Optimal ripple filtering and thermal balance. |
| One Inductor Opens | Increases (Approaches value of remaining unit) | 100% shifts to remaining unit | Remaining inductor saturates from overcurrent; output voltage ripple spikes massively. |
| One Inductor Shorts | Drops to near zero ($\approx 0H$) | Massive spike through shorted path | Upstream switching MOSFET experiences shoot-through and explodes; input fuse blows. |
| DCR Mismatch | Unchanged (theoretically) | Lower-DCR unit hogs DC current | Lower-DCR unit hits thermal saturation first, shifting load dynamically until failure. |
If a solder joint fractures on Node A of $L_1$, $L_1$ becomes an open circuit. The total inductance instantly jumps from $5\mu H$ back to $10\mu H$. More dangerously, $L_2$ must now carry the entire load current. If $L_2$ was sized to carry only 50% of the load, its core will saturate. Once saturated, its inductance collapses, turning it into a low-value resistor, which will rapidly destroy your switching IC.
Design Walkthrough: Sizing Parallel Inductors for a 5A Buck Converter
Let's design the output filter for a 12V-to-5V synchronous buck converter delivering 5A continuous load current. We need a target inductance of roughly $4.7\mu H$ to maintain a 30% ripple current ratio.
Sourcing a single $4.7\mu H$ inductor rated for 7A+ (accounting for the 5A DC load plus ripple peak) requires a massive, expensive component like the Coilcraft XEL6060 series. Instead, we can use inductance in parallel with two smaller, cheaper, and more readily available parts.
Component Selection
We will use two Wurth Elektronik 744774210 shielded power inductors.
- Single Unit Specs: $10\mu H$, 3.2A $I_{RMS}$, 4.5A $I_{sat}$, DCR = $46m\Omega$.
- Parallel Combination: $5\mu H$ total inductance (close enough to our $4.7\mu H$ target), 6.4A $I_{RMS}$ combined, 9.0A combined saturation threshold.
Because the Wurth parts are laser-trimmed and manufactured in the same batch, their DCR ($46m\Omega$) will match within 1-2%. This ensures the 5A DC load splits evenly (2.5A each), keeping both cores well below their thermal limits. For more on selecting magnetics for switching regulators, review Analog Devices' technical articles on inductor selection.
PCB Layout Rule
When routing Node A and Node B on your PCB, ensure the copper traces feeding each inductor are symmetrical in length and width. A 10mm difference in trace length adds parasitic resistance and inductance to one leg, ruining the current sharing you just calculated.
Breadboard Testing: Step-by-Step Verification
Before committing to a PCB, you should verify the parallel inductance on a breadboard using an LCR meter (like the DER EE DE-5000 or a benchtop Keysight U1733C). However, you must avoid mutual inductance—where the magnetic field of one inductor couples into the other, invalidating the standard parallel formula.
- Measure Baselines: Measure $L_1$ and $L_2$ individually at your converter's switching frequency (e.g., 100 kHz). Record both inductance and DCR.
- Physical Placement: Plug the inductors into the breadboard at least 2 inches apart. If space is constrained, place them at a 90-degree orthogonal angle to each other. Never place them side-by-side with their magnetic axes aligned.
- Wire the Nodes: Use 22 AWG solid core wire to jumper the left legs together (Node A) and the right legs together (Node B).
- Measure Combined: Place your LCR meter probes across Node A and Node B. Set the meter to series-equivalent mode (Ls) at 100 kHz.
- Verify the Math: If $L_1 = 10.1\mu H$ and $L_2 = 9.9\mu H$, your meter should read approximately $4.97\mu H$. If the reading is significantly higher (e.g., $6\mu H$), your inductors are too close together and are exhibiting positive mutual coupling. Move them further apart or rotate one 90 degrees.
Frequently Asked Questions
Does mutual inductance change the parallel inductance formula?
Yes, drastically. The standard formula assumes the coupling coefficient ($k$) is zero. If two inductors are placed close together and their magnetic fields interact, the total inductance becomes $L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 \pm 2M}$, where $M$ is the mutual inductance. If the fields aid each other, total inductance drops less than expected; if they oppose, it drops more. In power supply design, this unpredictability causes loop instability. Always space shielded inductors apart or use orthogonal placement to force $k \approx 0$.
Can I parallel inductors with completely different values?
You can mathematically, but it is a poor engineering practice for power delivery. If you parallel a $10\mu H$ inductor with a $2\mu H$ inductor, the equivalent inductance is $1.66\mu H$. However, the $2\mu H$ inductor will have a significantly lower DCR and will hog the majority of the DC current, saturating prematurely. Furthermore, the ripple current will not split evenly; the lower-value inductor will experience a much higher $di/dt$ ripple, leading to disproportionate core losses and heating. Always use matched values from the same manufacturer and BOM line.
Why did my parallel inductors overheat in my switching regulator?
If your inductors are overheating despite being rated for the combined current, check for high-frequency AC proximity effects and DCR mismatch. At switching frequencies above 500 kHz, skin effect and proximity effect in the inductor windings increase the effective AC resistance ($R_{AC}$). If one inductor is physically closer to the switching node's high $dv/dt$ copper pour, it may suffer higher capacitive coupling and core losses. Measure the temperature of each inductor individually with a thermocouple; if one is 15°C hotter than the other, your PCB layout has asymmetric parasitic impedances.
