Why use reduced-order models in coupled problems?

Coupling passes one physics' results to another repeatedly and over long times. Computing semiconductor temperature at every step of a circuit simulation with a detailed thermal model is too slow. Here a 200-DOF thermal model was reduced to three states, and over a 300 s coupled run the peak temperature differed by 0.02 K.

How is the ROM built?

Here the detailed model's temperature response to a constant heat step was computed and fitted with a sum of three exponentials (a Foster network): thermal resistances 0.061, 0.446 and 5.943 K/W with time constants 0.0029, 0.68 and 32.1 s. Thermal responses are often well represented by a few time constants like this.

Where does ROM error grow?

Outside the conditions used for fitting. Here at 0.01 s the full model gives 0.060 K/W and the ROM 0.067 K/W — over 10% off. Millisecond pulses need more short-time-constant terms. A change in cooling changes the time constants themselves, requiring a new ROM.

Reduced Order Model (ROM) for Coupled Problems

Category: Coupled Analysis – Multiphysics | Revised 2026-10-01

CAE visualization for reduced order multiphysics theory - technical simulation diagram

Reduced-Order Models (ROM) for Coupled Problems

Theory: representing response with a few time constants

Overview

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If a ROM loses accuracy, why use it in coupled analysis?

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Because in coupling, speed matters as much as accuracy. An inverter circuit simulation, for example, takes hundreds of thousands of fine time steps to follow switching, and at each step semiconductor temperature must be found and temperature-dependent loss fed back. Dropping a detailed finite-element thermal model in directly never finishes. So a ROM reproduces only the input (heat) to output (temperature) relationship with few variables. Heat conduction is linear and its response is usually set by a few time constants, so a well-built ROM with a handful of variables matches the full model closely. What matters is checking the range over which it is valid.

Reduction to a Foster network

$$ Z_{th}(t) = \sum_{i=1}^{3} R_i\left(1 - e^{-t/\tau_i}\right),\qquad T_j(t) = T_a + \sum_i \theta_i,\quad \tau_i\frac{d\theta_i}{dt} = R_i P - \theta_i $$

The full model stacks 0.3 mm silicon, 0.1 mm solder, 3 mm copper, 0.1 mm thermal grease and 20 mm aluminium over 1 cm², cooled at the bottom with 2,000 W/m²K — a 1D model with 200 DOF, 6.45 K/W thermal resistance and slowest time constant 32.1 s. Three terms were fitted to its step response.

Coupling with temperature-dependent loss

$$ P(T_j) = P_0\,[1 + 0.006\,(T_j - 25)],\qquad \Delta T_{ss} = \frac{R_{th}P_0}{1 - 0.006\,R_{th}P_0} $$

P₀ = 10 W, ambient 25 °C. Loss rises 0.6% per kelvin (representing on-resistance temperature dependence).

Coffee Break Trivia

Foster and Cauer

Representing a given response with networks of resistors, capacitors and the like grew out of 1920s telephone-circuit research. Foster in 1924 and Cauer in 1926 presented different network forms. The Foster network expresses response as a sum of exponentials and is easy to fit; the Cauer ladder maps naturally to layer stacks. Century-old telephone theory is now used daily for thermal ROMs of power semiconductors.

Worked examples

Example 1: fitting a three-stage Foster network

Stage	Thermal resistance	Time constant	Mainly represents
1	0.061 K/W	0.0029 s	Die and solder
2	0.446 K/W	0.68 s	Copper base and grease
3	5.943 K/W	32.1 s	Heatsink and cooling

Time	Full model	ROM
0.01 s	0.060 K/W	0.067 K/W
0.1 s	0.146 K/W	0.140 K/W
1 s	0.580 K/W	0.587 K/W
10 s	2.101 K/W	2.098 K/W
100 s	6.185 K/W	6.186 K/W

Overall maximum fit error is 0.008 K/W (0.1% of thermal resistance), but at 0.01 s the relative error exceeds 10%.

Example 2: electro-thermal coupled run

Case	Full (200 DOF)	ROM (3 states)
5 s on / 5 s off for 300 s, peak	71.79℃	71.76℃
Continuous, steady (coupled)	130.2℃	130.2℃
Continuous, steady (constant loss)	89.5℃	89.5℃

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Coupling raises steady temperature from 89.5 to 130.2 °C. With loss rising only 0.6% per K, why so different?

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Because of positive feedback: higher temperature raises loss, which raises temperature further. With constant loss the rise is thermal resistance × loss = 64.5 K; with coupling it becomes 64.5/(1 − 0.006×64.5) = 105 K. As the 0.39 in the denominator approaches 1, temperature climbs steeply, and at 1 it never settles (thermal runaway). Constant-loss thermal calculations therefore underestimate risk in electro-thermal problems. This is exactly where ROMs help: exchanging temperature and loss every step inside a circuit simulation adds almost no run time. But a ROM assumes linear conduction, so if conductivity changes strongly with temperature or cooling conditions change, verify with the full model or rebuild the ROM per condition.

Modelling workflow

Compute step responses for each heat source with the detailed thermal model.
Fit sums of exponentials and confirm the time-constant range covers the application time scales.
With several heat sources, include mutual influence (mutual thermal impedance).
Embed in the circuit simulation and exchange temperature and loss.
Check against the full model for representative cases.

Coffee Break Trivia

“A prediction that missed on short pulses”

In a motor drive, ROM-predicted die temperature during millisecond high-current motor starts came out nearly 10 K below measurement. The ROM had been fitted to second-scale response, its shortest time constant longer than a few milliseconds, so it couldn't capture rapid heating within the die. Rebuilding with two additional short-time-constant terms brought start-up temperatures into agreement with measurement.

Common mistakes

Mistakes and fixes

Mistake	Effect	Fix
Fitting long times only	Short-pulse temperature wrong	Fit over application time range
Constant loss	Temperature underestimated	Couple with temperature
Single heat source only	Neighbour heating missed	Include mutual impedance
Reusing ROM after cooling change	Time constants mismatch	Rebuild per condition
Using Foster node temperatures	No physical meaning	Use input–output only

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