The inductors in parallel formula dictates that the equivalent inductance ($L_{eq}$) of parallel inductors (assuming zero mutual coupling) is the reciprocal of the sum of their reciprocals: $1/L_{eq} = 1/L_1 + 1/L_2 + ... + 1/L_n$. For two identical inductors, the equivalent inductance is simply half the value of one ($L/2$), while the current handling capacity and DC resistance (DCR) scale favorably.
While textbook problems treat inductors as ideal, bench reality involves saturation currents, thermal limits, and parasitic capacitance. This guide breaks down the topology, contrasts it with series configurations, and walks through a real-world power supply design using specific component values.
The Inductors in Parallel Formula and Topology Breakdown
In a parallel inductor topology, the components share the same voltage across their terminals but divide the current. Let's define the circuit nodes:
- Node A (Input/Switch Node): The common connection point where the incoming current splits.
- Node B (Output/Load Node): The common connection point where the currents recombine before reaching the load or filter capacitor.
Inductor L1 and Inductor L2 both connect directly between Node A and Node B. The governing math for the inductors in parallel formula with two components is:
$$L_{eq} = \frac{L_1 \times L_2}{L_1 + L_2}$$
This formula assumes the magnetic fields of L1 and L2 do not interact. If you place two unshielded inductors side-by-side on a PCB or breadboard, their magnetic flux will couple. Depending on the winding direction and physical orientation, this mutual inductance can cause $L_{eq}$ to be significantly higher or lower than the formula predicts. Always keep parallel inductors physically separated, orient them at 90-degree angles to one another, or use magnetically shielded parts (like molded ferrite types).
Why Parallel Over Series? (And Failure Mode Contrasts)
Why put inductors in parallel instead of series? In series, inductances add ($L_{eq} = L_1 + L_2$), but the current rating of the string is limited by the weakest component. If you series two inductors rated for 5A, your entire circuit is still capped at 5A.
Paralleling inductors is primarily a current-handling and DCR-reduction strategy. By splitting the load current, you delay core saturation and reduce $I^2R$ copper losses. However, this topology introduces distinct failure modes compared to a series string.
| Event / Fault | Parameter Change | Circuit Result & Consequence |
|---|---|---|
| L1 Opens (e.g., solder joint fails) | $L_1 \to \infty$ | $L_{eq}$ drops to exactly $L_2$. All load current shifts to L2. If L2 was sized for half the load, it will instantly saturate, causing massive current spikes and potential switch-node overvoltage. |
| L1 Shorts (e.g., internal winding insulation melts) | $L_1 \to 0$ (just DCR) | $L_{eq}$ collapses to near zero. The circuit loses all inductive filtering. In a switching regulator, this results in catastrophic overcurrent and immediate destruction of the MOSFET/diode. |
| L1 and L2 mismatched DCR | $I_{L1} \neq I_{L2}$ | Current divides inversely to DCR, not equally. The lower-DCR inductor hogs the current and hits thermal saturation ($I_{rms}$ limit) first, reducing the effective total current capacity. |
| Contrast: Series L1 Opens | $L_{string} \to \infty$ | The entire circuit path is broken. Load loses power completely, but no overcurrent/short-circuit event occurs. Safer fail-state for signal lines, worse for power delivery. |
Design Walkthrough: Sizing Real Inductors for a 12A Buck Converter
Let's apply the inductors in parallel formula to a real design problem. You are building a 12V-to-5V synchronous buck converter that needs to deliver 12A continuous. The controller requires a 4.7µH inductor.
Sourcing a single 4.7µH inductor with a saturation current ($I_{sat}$) above 15A (allowing for ripple) and low DCR often requires expensive, tall, through-hole toroids or custom SMD parts. Instead, we can parallel two standard, high-volume SMD inductors.
Selected Component: Bourns SRP1265A-100M
- Value: 10µH
- $I_{sat}$: 11A
- $I_{rms}$ (Thermal): 10.5A
- DCR: 12.5mΩ (typical)
The Math:
Using two 10µH inductors in parallel:
$L_{eq} = \frac{10 \times 10}{10 + 10} = 5\mu H$
A 5µH inductance is a perfectly acceptable substitute for 4.7µH in most modern current-mode controllers; it will slightly reduce the inductor ripple current, which actually improves efficiency and output voltage ripple.
Current and Thermal Scaling:
Theoretical $I_{sat}$ is $11A + 11A = 22A$. However, because of minor DCR mismatches and thermal coupling on the PCB, we apply a 20% derating factor.
Effective $I_{sat} \approx 17.6A$. This comfortably clears our 12A load requirement plus the ~3A peak ripple current.
Effective DCR = $12.5m\Omega / 2 = 6.25m\Omega$. This drastically cuts copper losses compared to a single, smaller inductor pushed to its limits.
How to Breadboard-Test Parallel Inductors Step-by-Step
Before soldering to a PCB, verify your parallel network on a breadboard using an LCR meter (like the DER EE DE-5000 or Keysight U1733C). Do not rely on a standard multimeter's resistance mode; you must measure reactance.
- Set the Test Frequency: Power inductors operate at high frequencies (e.g., 500kHz). Cheap LCR meters default to 100Hz or 1kHz. Set your meter to 100kHz if available. Measuring a power inductor at 100Hz will yield inaccurate inductance values due to core permeability variance.
- Zero the Leads: Short your test probes and press the "Zero" or "REL" button to null out the parasitic inductance of your test leads (usually around 0.5µH to 1.0µH, which is massive when measuring small parallel equivalents).
- Measure Individual Baselines: Measure L1 and L2 separately. Record the exact values. If L1 is 9.8µH and L2 is 10.2µH, your $L_{eq}$ will be 4.99µH, not exactly 5.0µH.
- Wire the Parallel Topology: Insert both inductors into the breadboard. Use thick, short jumper wires to tie both left legs together (Node A) and both right legs together (Node B). Keep the physical distance between the two inductor bodies at least 1 inch to minimize mutual coupling during the test.
- Measure $L_{eq}$: Probe Node A and Node B. Verify the reading matches your calculated inductors in parallel formula result within 5%.
- The "Wiggle" Test (Mutual Inductance Check): While probing, physically move one inductor closer to the other, or rotate it 90 degrees. If the $L_{eq}$ reading on the meter jumps by more than 2-3%, your physical layout is suffering from mutual inductance. You must adjust your PCB footprint to ensure orthogonal placement or increase spacing.
Frequently Asked Questions
Does the inductors in parallel formula work if the values are different?
Yes, the reciprocal formula ($1/L_{eq} = 1/L_1 + 1/L_2$) works for any values. However, in power supply design, paralleling mismatched inductors (e.g., a 10µH and a 22µH) is generally a bad idea. The smaller inductor (10µH) will have a lower impedance to high-frequency ripple current and will "hog" the AC ripple, potentially saturating long before the larger inductor reaches its thermal limit. Always parallel identical part numbers from the same manufacturing batch for predictable current sharing.
How does mutual inductance affect parallel inductors mathematically?
When magnetic coupling ($M$) is present, the formula changes. If the inductors are wound such that their flux aids each other, the equivalent inductance becomes $L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}$. If the flux opposes, it becomes $L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 + 2M}$. In practical PCB layout, you want $M$ to be as close to zero as possible so the standard formula holds true. For a deep dive on magnetics design, refer to Texas Instruments' magnetics design guides.
Can I parallel inductors to increase current rating in an AC line filter?
It is highly discouraged for AC line filtering (like common-mode chokes). In AC applications, paralleling inductors can create uneven current sharing due to slight differences in AC impedance and parasitic capacitance. Furthermore, if one inductor fails open in an AC safety filter, the remaining inductor will silently take the full line current, potentially overheating and causing a fire hazard without tripping a breaker. For AC line current scaling, it is always safer and more code-compliant to source a single, properly rated inductor or use a transformer topology.
