Solenoid Work Notes

Latching solenoid principle: how they work, and the drive cost nobody budgets for

2 October 2026

A latching solenoid holds on a permanent magnet and moves on a pulse, so holding current is zero. The energy saving scales with hold time: a 12 V, 1 A, 50 ms pulse costs 0.6 J, against 16 Wh for the same load held continuously for eight hours. The bill arrives as a bipolar driver and a capacitor sized from C = 2E/V squared.

Why this happens

A conventional solenoid holds by current. A latching one holds by permanent magnet. The plunger is driven across by a pulse, the pulse is removed, and the plunger stays where it was put. Holding force comes from the magnet’s flux rather than from the winding, so holding current is zero and the coil carries current only while the state is changing.

Most magnetic latching designs use two permanent magnets with the armature between them. Each magnet supplies flux through the armature along a closed path, and the armature sits stably at either end. A pulse of one polarity shifts the flux balance so the armature crosses over; a pulse of the opposite polarity sends it back. The coil has to beat the difference between the two magnet circuits, not the full holding force. That is why the pulse can be short.

The energy advantage is entirely a function of hold time. This is the part that gets argued about, and the arithmetic settles it quickly.

CaseEnergy per operationEnergy per day at 20 operations
Latching: 12 V × 1 A × 50 ms0.6 J12 J (3.3 mWh)
Standard coil: 2 W held for 2 s4 J80 J (22 mWh)
Standard coil: 2 W held for 8 h57,600 J576,000 J (160 Wh)

Read the table as arithmetic rather than as a claim, because the arithmetic is what decides the design. Hold for two seconds and latching is a modest improvement: 12 J against 80 J across 20 operations, a factor of 6.7. Hold for eight hours and a single operation costs 0.6 J with the latch against 57,600 J with the coil, a factor of 96,000. The comparison is a straight multiplication by hold time, so the first question to ask is how long the state has to persist, and everything else follows from the answer. Battery-powered locks are the obvious case, and the inrush accounting for those is set out in battery life for door lock solenoids.

The bill arrives in three places, and all three are on the driver side.

The first is polarity. A latching coil has to be driven in both directions, which means an H-bridge or a pair of devices with a changeover rather than a single low-side MOSFET. Four switching elements where there was one, and the bridge has to carry the full pulse current, 1 A in this example, while holding off the supply plus whatever the coil’s flyback adds, so a 30 V device on a 12 V rail is normal practice rather than extravagance.

The second is energy storage. The pulse is short, the current is high, and most supplies cannot deliver it on demand, so a reservoir capacitor does the work. Sizing it is one line: C = 2E/V². For 0.6 J at 12 V that is 2 × 0.6 / 144, which is 8,300 µF, and the stored-energy expression gives the same answer from the other direction, since ½CV² at 8,300 µF and 12 V is 0.60 J. Only about half of that is usable, because what remains at 70 % of nominal voltage is 0.30 J, so the capacitor is normally specified at roughly twice the minimum. It then sets the shortest interval between operations. Charging 8,300 µF through a source resistance of 1 kΩ gives a time constant of 8.3 s, and five time constants for a full charge is about 42 s. A mechanism that has to fire twice in ten seconds will not do it from that supply, and this constraint is easy to miss because it never appears on a schematic.

The third is the magnet’s own temperature behaviour. Neodymium magnets lose about 0.11 to 0.12 % of their flux density per kelvin, so a magnet 80 K above ambient holds with roughly 9 % less flux. Because holding force varies with the square of flux density in the gap, that 9 % becomes about 17 % less force. Samarium cobalt runs at roughly a third of that coefficient, 0.03 to 0.04 %/K, for a comparable loss of about 6 % over the same span, at materially higher cost. A latch specified at room temperature and installed in a warm enclosure has already spent part of its margin.

There is a fourth cost that rarely appears on a bill of materials, and it is a property of the winding rather than of the driver. What heats a coil is average power, so a winding rated 3 W continuously will absorb 300 W at 1 % duty. A 50 ms pulse repeated every 5 s is exactly 1 % duty, which is why a latching coil can be driven at currents that would destroy a continuously rated winding of the same size. The rating only holds while the duty holds, though. Stretch the pulse to 100 ms at the same repetition rate and the allowable peak falls to 150 W; drive two operations a second and the duty is 10 %, where the same coil may dissipate only 30 W and will not survive the current it was designed around.

Check these in order

1. How long must the state persist? The crossover is easy to compute and usually lands far lower than people expect. A 0.6 J pulse holds the same energy as 0.3 s of running a 2 W coil, so any hold longer than about a third of a second already favours the latch on energy. Past a few seconds the advantage is large; past minutes it is decisive. Below a third of a second a latching design costs you a bridge and a capacitor for nothing.

2. What happens when power fails during the stroke? A latch stays where it was last put. That is either exactly the fail-safe behaviour you want or the exact opposite. For a lock, staying latched with no power is usually the requirement. For a valve that must vent on power loss, it is a hazard. Decide which you need and write it down, because the magnet circuit cannot be changed later without changing the whole actuator.

3. Can the mechanism tolerate a force pulse rather than a controlled force? A latching solenoid delivers a short, high-force stroke. It does not produce a defined force profile across the stroke, and it cannot be made to. The pulse is typically 20 to 100 ms, over which the plunger either crosses or does not; there is no intermediate operating point to modulate. If the load needs progressive force or position control, this is not the right device.

4. Cost the driver at concept stage. The bridge, the reservoir capacitor, the charge-time budget and the extra board area. Four switching devices and an 8,300 µF capacitor are not expensive individually, but they take several times the board area of a single low-side drive and they add a part that has to be derated for temperature. Designers who compare only the coil price find that the saving disappears, and it usually surfaces once the enclosure drawing is done rather than before.

5. Measure set and release energy separately. The release pulse is working against whatever the load is doing, so the two directions are rarely equal. Measure both at the temperature extremes as well, because three things move at once when it gets cold: the lubricant thickens, the seal stiffens, and the magnet gets stronger, since a negative temperature coefficient works in that direction. A mechanism that releases cleanly at 20 °C can refuse at −20 °C with the same capacitor. Size for the larger of the two energies, and remember that the load is often absent on the bench.

6. Check the pulse from both sides. There is a lower bound and an upper bound. Too little and the plunger will not cross reliably. Too much and the reverse field can exceed the magnet’s intrinsic coercivity, after which the magnet is permanently weaker. Grades differ by a factor of two or more in this respect: a standard N grade carries an intrinsic coercivity around 955 kA/m and an SH grade around 1,590 kA/m, so a pulse that is safe in one is not automatically safe in the other. The loss is not recoverable by re-magnetising in the field, and it usually surfaces as a holding-force complaint after a few thousand operations rather than as an immediate failure.

7. Confirm the pulse is being terminated. Set and release are edge events. A firmware path that leaves the drive on after the plunger has moved turns a pulse-rated coil into a continuous-duty load it was never wound for. A winding that is content at 300 W for 1 % of the time is not content at 300 W continuously, and the difference shows up within seconds rather than over a shift. The same failure appears in standard solenoids driven from a controller that never releases, covered in coils that will not release.

What actually to change

SymptomWhat to changeWhy not the obvious thing
Battery life below the calculationConfirm the pulse is actually terminated, then measure its widthRewinding the coil misses a firmware fault that costs nothing to fix
Will not release reliablyRaise release-pulse energy, then inspect the guided surface for stictionA bigger capacitor alone treats friction as an energy problem
Loses holding force when warmCheck the magnet grade and its temperature coefficientAdding holding current defeats the design’s entire purpose
Second operation fails if commanded too soonSize the reservoir from C = 2E/V² and the source resistanceA larger supply does not help if the bottleneck is the charge resistor
Holds correctly but draw is higher than expectedLook for a holding current left on after the pulseThe coil is wound for pulse duty and will not survive it
Force varies between unitsMeasure friction on the plunger, not flux in the gapFriction scatter normally exceeds magnetic scatter
Unit buzzes when it should be staticSomething is still being driven, or the armature is not seating fullyA latch has no AC holding mode to hum in, so this is mechanical

When it IS the harder problem

The load can back-drive the plunger. A latch resists a static load well because the magnet’s flux holds the armature against its seat. A shock or a sustained vibration can walk it off, and the margin that decides this is the detent force rather than the holding force. A factor of two between the detent force and the worst-case back-driving force is a sensible starting point, and it has to hold at the temperature extreme rather than at room temperature.

The reverse pulse is close to the magnet’s coercivity limit. Then the design is living at the edge of a permanent failure mode, and the pulse energy has to be controlled rather than just provided. Constant-current drive instead of a capacitor discharge is the usual answer, at the cost of more driver complexity.

The operating temperature is high. Magnet loss is partly reversible and partly not. Past the maximum operating temperature the loss becomes permanent, and that limit depends on the circuit as much as on the material: for neodymium it runs from around 80 °C for a standard N grade to 150 °C or more for an SH grade, and the permeance coefficient of the surrounding circuit moves it further. This is the point at which material cost genuinely has to rise, and it is a specification decision rather than a design one.

The supply is AC. A latching coil needs a defined polarity for each direction, so an AC supply has to be rectified and switched, which changes the driver architecture rather than just its components. The wider differences between the two supply types are covered in AC versus DC solenoids.

The mechanism has to report its state. A latch is bistable but not self-sensing. Nothing in the magnetic circuit tells the controller which position the plunger is in, so an unverified assumption about state is the usual source of intermittent faults in field returns. If a position signal is needed, it has to be designed in as a separate element.

The principle is simple enough to explain in a paragraph, and that is why the driver cost gets missed. A latching actuator is a magnet, a pulse and a capacitor, and the engineering is in the energy budget rather than in the magnetics. Work the hold-time arithmetic first, size the capacitor from the pulse energy, and check the reverse pulse against the magnet’s limit. Those three numbers decide whether the design works.

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Frequently asked

Does a latching solenoid use any current while it is holding?
No. The holding force comes from a permanent magnet, so once the pulse has ended the coil carries nothing and the plunger stays where it was put. That is the whole point of the design, and it is why a battery-powered lock can run for a year on cells that would flatten in days with a continuously energised coil.
How much energy does one operation take?
Multiply voltage, current and pulse width. A 12 V coil drawing 1 A for 50 ms uses 12 times 1 times 0.05, which is 0.6 J. Twenty operations a day is 12 J, or about 3.3 mWh. Compare that with the same load held at 2 W for eight hours, which costs 57,600 J or 16 Wh, and the factor between them is 96,000.
Why does a latching solenoid need a capacitor?
Because the pulse is short and the current is high, and most supplies cannot deliver it on demand. The reservoir is sized from C = 2E/V squared, so 0.6 J at 12 V needs about 8300 µF. That capacitor then sets how often the mechanism can fire: charging 8300 µF through 1 kΩ takes about 42 s to reach five time constants.
Can you drive a latching solenoid with a single MOSFET?
No, because the polarity has to reverse. You need an H-bridge, or a pair of devices with a changeover, or a capacitor that is switched in opposite senses for set and release. That is at least four switching elements where a standard solenoid needs one, and it is the cost that tends to surface after the enclosure drawing is finished.
What happens if you leave the pulse on?
The coil is wound for pulse duty, so it has little copper mass and a small thermal time constant. Held continuously it reaches its insulation limit in seconds to minutes rather than hours. Set and release are both edge events, and the firmware has to treat them that way.