Copper resistance and temperature: predicting hot-state solenoid force from a cold measurement

Copper resistance rises 0.393 % per kelvin, so a winding that settles 70 K above ambient carries 27.5 % more resistance than it did cold. At fixed voltage that leaves 78 % of the current and about 61 % of the force, because force follows current squared while the magnetic circuit is unsaturated. The same ratio works in reverse: a resistance measurement is a thermometer for the winding, accurate to roughly 2 K, and it costs nothing but a multimeter.

Why this happens

The mechanism is simple enough to carry in your head, and it explains a large share of field failures that get attributed to the wrong cause.

Resistance and current. Ohm’s law at fixed voltage says current falls in proportion to resistance. A 27.5 % rise in resistance is a current of 1 divided by 1.275, which is 0.784. The coil still has the same number of turns, so ampere-turns fall by the same 21.6 %.

Current and force. Force at a pole face goes with flux density squared, and flux density is proportional to ampere-turns while the iron is below saturation. So force falls with the square of the current ratio. 0.784 squared is 0.615. A coil that produced 10 N cold produces 6.2 N hot, and nothing has moved or worn.

Power falls, which sounds like good news and is not. At fixed voltage, power is voltage squared over resistance, so a 27.5 % resistance rise reduces dissipation by 21.6 %. The coil runs cooler than a constant-current design would, which is why constant-voltage drive is partly self-limiting. It also means the temperature stabilises at a point where the force has already fallen, and that equilibrium point is what the machine lives with.

The same physics makes resistance a thermometer. Rearranging the ratio gives temperature directly. Take a cold reading at a known ambient, run the coil to equilibrium, take a hot reading immediately after switching off, and solve. A coil reading 4.9 ohms at 20 °C that reads 6.2 ohms hot sits at 87 °C, a rise of 67 K. The same arithmetic in ratio form uses 234.5 °C as the inferred zero-resistance point of copper, and agrees to within a fraction of a kelvin over the working range.

Why the surface reads low. The winding generates the heat, the insulation and the bobbin wall sit between the copper and the outside, and the housing is cooler than both. A thermocouple bonded to the outside of the coil sees the cold end of that gradient and typically reads 10 to 20 K below the mean winding temperature. The resistance method integrates the whole winding, including the inner layers that have the longest thermal path, so it reports the number that matters for insulation life. The limits it should be compared against are in coil temperature limits by insulation class.

Thermal time constant is minutes, not seconds. A small solenoid winding typically reaches 63 % of its final rise in 5 to 15 minutes, depending on mass and conduction path. That is why a 30 second bench test shows no problem and a 30 minute run in a cabinet shows a failure: the unit has not arrived at its operating point yet during the short test. The pattern of a unit that works cold and fails warm is covered in diagnosing thermal failure.

Aluminium and copper-clad aluminium behave differently, and worse. Aluminium has a coefficient of about 0.00403 per kelvin and roughly 1.6 times the resistivity of copper, so a winding wound with clad aluminium has both a higher cold resistance and a steeper rise. The temperature coefficient difference is small; the resistivity difference is not, and it is the one that shows up as a hot force shortfall. If a supplier will not state the conductor, the resistance measurement will tell you which metal you have once you know the turns count.

Where the linear approximation stops. The 0.393 % figure is defined at 20 °C and drifts slightly at higher temperature. Over 20 °C to 150 °C the ratio form stays within about 1 %, which is 1 K, and that is well inside the uncertainty of the measurement. Extrapolating far below freezing, or using the coefficient across a range that includes a phase change in the insulation, is where it becomes unreliable.

Check these in order

1. Record the cold resistance with the ambient beside it. A resistance alone is not a baseline, because the same part reads differently in a cold workshop and a warm office. Note both. A 4.9 ohm reading means 87 °C if it was taken at 20 °C and 84 °C if it was taken at 25 °C, so the ambient is part of the measurement rather than a note about it.

2. Measure with four wires, or account for the leads. A two-wire meter includes the lead resistance, which for a 0.5 ohm coil and 0.3 ohm of leads is a 60 % error. Zero the meter on a shorted pair of probes, or use a four-wire measurement. On a low-resistance coil this step matters more than the meter’s accuracy class.

3. Take the hot reading within seconds of switching off. The winding starts cooling immediately, and a delay of a minute on a small coil can lose several kelvin. Better still, measure current and voltage while energised and divide, which avoids the cooling question entirely and costs one extra meter.

4. Compare the result against the insulation class limit, not against a temperature. Class B allows 80 K of rise, class F allows 105 K and class H allows 125 K, measured by resistance over the ambient. A winding at 67 K of rise in a 40 °C cabinet is at 107 °C, well inside class B. The same 67 K in a 70 °C cabinet is at 137 °C and outside it. Ambient is half the calculation.

5. Check whether the rise is a design point or a symptom. A rise that matches the thermal calculation is normal. A rise that is double what the copper loss predicts means heat is arriving from somewhere else, usually the valve body or the machine rather than the coil. Measuring with the unit on the bench and again installed separates the two.

6. Use the ratio to predict force before the test rig exists. Compute the expected hot resistance from the intended stabilised temperature, derive the current, and square the ratio. That number is the force the customer will see after ten minutes of running, and it is the number worth putting in a specification.

What actually to change

If the hot force is short, decide whether to fix the heat or the winding. A cooler winding recovers force without changing the electrical design, and the usual bottleneck is conduction from the coil to the housing rather than convection from the housing to air. Potting compound, a tighter bobbin-to-shell fit, or a metal end cap can each remove 10 to 20 K. A redesigned winding delivers more cold force but more heat as well, and the equilibrium may not move much.

If the application is intermittent, use the thermal mass. A coil that is energised for 2 seconds in every 30 has a duty cycle low enough that the winding barely warms, and the cold force is the relevant figure. Specifying a continuously rated coil for that application overpays in copper and size. The duty classification is in duty cycle and percent ED.

If force at temperature is critical, specify it at temperature. A requirement written as 10 N at 20 °C is a different requirement from 10 N at 85 °C, and the second one costs more. Stating the temperature turns an argument about test conditions into a design input, which is the same discipline that applies to the force-stroke curve.

If the machine runs on a supply with wide tolerance, combine the two losses. A 10 % low line costs 19 % of the force, and a 70 K rise costs 39 %. Together they leave about half. Designing to the nominal condition and hoping the tolerances do not coincide is how a product passes qualification and fails in a customer’s installation.

When it IS the harder problem

When the force loss exceeds what the resistance explains. A 27.5 % resistance rise explains a 39 % force loss. A 60 % loss at the same current means the magnetic circuit changed, not the winding. The candidates are a bobbin creeping under load, a spring relaxing, or joints settling, and the diagnostic is to measure the assembled gap hot and cold. The sequence is in force dropping after thermal cycling.

When the winding is hot but the current is normal. Constant-current drive holds ampere-turns steady as resistance rises, so the force does not fall with temperature. If the current is steady and the force still falls, the heat is affecting the magnetic circuit directly, which happens through differential expansion rather than through the winding. This is common where an aluminium housing carries a steel plunger.

When the resistance itself drifts over time at constant temperature. Copper does not age in a way that changes resistance by a measurable amount, so a resistance that climbs over weeks means the winding is shorting turn to turn or the insulation is degrading. That is an insulation failure investigation rather than a thermal one, and it belongs with insulation testing rather than with the temperature calculation.

When the customer measures hot resistance and reports a fault. A coil that reads 6.2 ohms instead of 4.9 is not faulty, it is warm. Explaining the ratio before replacing the part saves a return, and it is worth putting the expected hot resistance on the drawing so the number can be checked rather than argued about.

FAQ

What is the resistance ratio for a 100 K rise? 1 plus 0.00393 times 100, which is 1.393. A 10 ohm coil becomes 13.9 ohms. In force terms at fixed voltage the current falls to 0.718 and the force to 0.516, so a little over half. The same calculation with the 234.5 °C ratio form gives 1.395, which is within the measurement uncertainty.

Can I use this to estimate the force without a test rig? Yes, to about 10 %. Measure cold resistance and ambient, estimate the stabilised temperature from the intended duty and mounting, compute the hot resistance, derive the current from the supply voltage, and square the current ratio against the cold state. Multiply by the cold force from the datasheet curve at the working gap. The two largest uncertainties are the temperature estimate and whether the datasheet force was measured at the same gap.

Why does the current not fall as much as the resistance rises? It does, at fixed voltage. The confusion arises with constant-current drivers, where the supply raises its output voltage to hold current, so the resistance rise shows up as a voltage rise rather than a current fall. The winding still gets hotter and the insulation still ages; the force just does not fall with it. That is one reason constant-current drive is used for force-critical applications and one reason the driver costs more.

Does the surrounding metal change the answer? It changes the temperature, not the coefficient. A coil in a steel housing conducts heat away better than one in free air, so it stabilises lower, and the force holds up better as a result. The resistance ratio remains the correct way to read the winding temperature in either case, provided the hot reading is taken before the unit cools.