Chapter 2 — Companion models: discretizing R, L, and C

In Chapter 1 we saw the strategy: discretize time, then turn the network into a resistive circuit the computer can solve once per step. This chapter does the first half rigorously. We derive the companion model (also called the discrete or Norton-equivalent model) of the three lumped elements — resistor, inductor, capacitor — using the trapezoidal rule. By the end you will be able to compute, by hand, the very numbers the simulator produces.

Learning objectives

• State why inductors and capacitors cannot be handled by ordinary nodal analysis directly.
• Apply the trapezoidal rule to derive the companion conductance and history-current source of L and C.
• Compute the companion parameters for given element and time-step values.
• Hand-iterate the first steps of an R-L or R-C circuit and match the simulator's output.
• Explain, qualitatively, why the trapezoidal rule is the EMT workhorse.

2.1 The obstacle: two of our elements are differential

Nodal analysis (which we build in Chapter 3) loves elements whose current is an instantaneous function of voltage — i.e. conductances. The resistor obliges:

$$i_R(t)=\frac{1}{R}{v}_R(t)=G{v}_R(t).$$

The inductor and capacitor do not. Their defining laws involve a derivative:

$$v_L(t)=L\frac{d{i}_L}{dt},i_C(t)=C\frac{d{v}_C}{dt}.$$

The current now depends on the rate of change, not the present value alone. We cannot stamp a derivative into a linear algebraic system. The fix is to replace the derivative with a difference using a numerical integration rule, evaluated over one time step of length $\mathrm{\Delta }t$.

2.2 The trapezoidal rule

Suppose a quantity $x(t)$ obeys $\dot{x}=f(t)$, and we know $x$ at the previous step $t-\mathrm{\Delta }t$. The exact update is

$$x(t)=x(t-\mathrm{\Delta }t)+{\int }_{t-\mathrm{\Delta }t}^{t}f(\tau )d\tau .$$

The trapezoidal rule approximates the integral by the area of the trapezoid under $f$ between the two endpoints:

$${\int }_{t-\mathrm{\Delta }t}^{t}f(\tau )d\tau \approx \frac{\mathrm{\Delta }t}{2}(f(t)+f(t-\mathrm{\Delta }t)).$$

It is implicit (the unknown endpoint $f(t)$ appears on the right), second-order accurate (the error per step shrinks like $(\mathrm{\Delta }t{)}^3$, and globally like $(\mathrm{\Delta }t{)}^2$), and A-stable (it will not blow up for a stable system, no matter how large $\mathrm{\Delta }t$ is). Those three properties are exactly what an EMT integrator needs. We will apply it to each reactive element.

2.3 Companion model of the inductor

Start from $v_L=Ld{i}_L/dt$, i.e. $d{i}_L/dt=v_L/L$. Integrate over one step and apply the trapezoidal rule:

$$i_L(t)=i_L(t-\mathrm{\Delta }t)+\frac{\mathrm{\Delta }t}{2L}(v_L(t)+v_L(t-\mathrm{\Delta }t)).$$

Group the terms into "depends on the present unknown $v_L(t)$" and "known from the previous step":

$$\boxed{{\displaystyle i_L(t)=G_L{v}_L(t)+I_L^{\text{hist}}}},G_L=\frac{\mathrm{\Delta }t}{2L},$$$$I_L^{\text{hist}}=i_L(t-\mathrm{\Delta }t)+G_L{v}_L(t-\mathrm{\Delta }t).$$

Read this as a circuit: the inductor, over one step, behaves like a conductance $G_L=\mathrm{\Delta }t/2L$ in parallel with a current source $I_L^{\text{hist}}$ whose value is fixed by the previous step's current and voltage. That parallel combination is the inductor's companion model. Once $I_L^{\text{hist}}$ is computed (it is just a number at solve time), the inductor is indistinguishable from a resistor-plus-source — and those, nodal analysis handles happily.

Companion model of the inductor: over one step it behaves as a conductance $G_L=\mathrm{\Delta }t/2L$ in parallel with the history current source $I_L^{\mathrm{hist}}=i_L(t-\mathrm{\Delta }t)+G_L{v}_L(t-\mathrm{\Delta }t)$.

Note the physics baked into $G_L=\mathrm{\Delta }t/2L$: a large inductance gives a tiny conductance (the inductor barely lets current change in one step), which is why an inductor looks almost like an open circuit at the first instant.

2.4 Companion model of the capacitor

Start from $i_C=Cd{v}_C/dt$, i.e. $d{v}_C/dt=i_C/C$. Trapezoidal integration of $v_C$ gives

$$v_C(t)=v_C(t-\mathrm{\Delta }t)+\frac{\mathrm{\Delta }t}{2C}(i_C(t)+i_C(t-\mathrm{\Delta }t)).$$

Solve for the present current $i_C(t)$:

$$\boxed{{\displaystyle i_C(t)=G_C{v}_C(t)+I_C^{\text{hist}}}},G_C=\frac{2C}{\mathrm{\Delta }t},$$$$I_C^{\text{hist}}=-(G_C{v}_C(t-\mathrm{\Delta }t)+i_C(t-\mathrm{\Delta }t)).$$

Again a conductance $G_C=2C/\mathrm{\Delta }t$ in parallel with a known history source. Here a large capacitance gives a large conductance (the capacitor easily passes current in one step) — the dual of the inductor, exactly as intuition demands: a capacitor looks almost like a short circuit to a sudden change.

Companion model of the capacitor: a conductance $G_C=2C/\mathrm{\Delta }t$ in parallel with the history source $I_C^{\mathrm{hist}}=-(G_C{v}_C(t-\mathrm{\Delta }t)+i_C(t-\mathrm{\Delta }t))$. It is the exact dual of the inductor.

2.5 The resistor (for completeness)

No derivative, no history term:

$$i_R(t)=G{v}_R(t),G=\frac{1}{R}.$$

2.6 The three companion models at a glance

Element	Conductance $G_{eq}$	History source $I^{\text{hist}}$ (known from $t-\mathrm{\Delta }t$)
Resistor $R$	$1/R$	— (none)
Inductor $L$	$\mathrm{\Delta }t/2L$	$i(t-\mathrm{\Delta }t)+G_{L}v(t-\mathrm{\Delta }t)$
Capacitor $C$	$2C/\mathrm{\Delta }t$	$-(G_{C}v(t-\mathrm{\Delta }t)+i(t-\mathrm{\Delta }t))$

The history source is the memory of the element: it carries stored energy forward from one step to the next. Because it depends only on already-known quantities, it is a constant by the time the present step's network is solved.

2.7 Worked example: the first steps of the R-L step response

Let us reproduce, by hand, what the simulator does on the R-L step response sample ($V=10\text{V}$ DC, $R=10\mathrm{\Omega }$, $L=0.1\text{H}$, $\mathrm{\Delta }t=50\mu \text{s}$). The inductor's lower terminal is grounded, so the inductor voltage equals the mid-node voltage, $v_L=V_{\text{mid}}$.

The R-L step-response circuit: an ideal $10\mathrm{V}$ source, series $R=10\mathrm{\Omega }$, and $L=0.1\mathrm{H}$ whose lower terminal is grounded, so $v_L=V_{\mathrm{mid}}$.

Companion conductance.

$$G_L=\frac{\mathrm{\Delta }t}{2L}=\frac{50\times {10}^{-6}}{2(0.1)}=2.5\times {10}^{-4}\text{S}.$$

Step 1 ($t=\mathrm{\Delta }t$). The simulator starts every branch current and the history term at zero, so $I_L^{\text{hist}}=0$. Kirchhoff's current law at the mid-node says the current through $R$ equals the inductor current:

$$\frac{V-V_{\text{mid}}}{R}=G_L{V}_{\text{mid}}+I_L^{\text{hist}}.$$$$\frac{10-V_{\text{mid}}}{10}=2.5\times {10}^{-4}{V}_{\text{mid}}+0\Rightarrow V_{\text{mid}}=\frac{1}{0.10025}=9.975\text{V},$$$$i_L(\mathrm{\Delta }t)=G_L{V}_{\text{mid}}=2.5\times {10}^{-4}(9.975)=2.494\times {10}^{-3}\text{A}.$$

Now open the sample, run it, and read the very first sample on the scope (or in the exported CSV): $V_{\text{mid}}=9.975\text{V}$, $I_L=0.002494\text{A}$. Our hand calculation matches the simulator to four significant figures — because we performed exactly the arithmetic the engine performs.

Two physical sanity checks fall out immediately:

• The inductor current is almost zero at the first instant — the inductor resists the sudden change, behaving like a near-open circuit (its companion conductance $G_L$ is tiny).
• Almost the entire source voltage appears across the inductor ($V_{\text{mid}}\approx 10\text{V}$), again because it is momentarily open.

Onward in time. Each step updates $I_L^{\text{hist}}$ from the previous step and re-solves. Marched forward, the current traces the exponential rise $i_L(t)=(V/R)(1-e^{-t/\tau })$ with time constant

$$\tau =\frac{L}{R}=\frac{0.1}{10}=0.01\text{s}=10\text{ms}.$$

At $t=\tau$ the analytic value is $i_L=(1)(1-e^{-1})=0.632\text{A}$; the simulator reports $0.631\text{A}$ — agreement to about a tenth of a percent. The two values are not identical because the engine seeds its history from zero current at $t=0$ (a one-step initialization choice) and because trapezoidal integration carries its own small, second-order error. Both effects are tiny here and shrink as you reduce $\mathrm{\Delta }t$.

2.8 Lab: confirm the companion model

Run both first-order samples and verify the theory you just derived.

Open the R-L step response in app →

Open the R-C charging circuit in app →

For the R-L circuit:

1. Confirm the final current settles near $V/R=1\text{A}$.
2. Find the time at which $i_L$ reaches $0.632\text{A}$ — it should be one time constant, ~10 ms.
3. Read the first scope sample and check it against the hand calc above.

The R-C charging circuit: an ideal $10\mathrm{V}$ source, series $R=1\mathrm{k}\mathrm{\Omega }$, and $C=10\mu \mathrm{F}$ to ground; the capacitor-node voltage is $V_{\mathrm{cap}}$.

For the R-C circuit ($R=1\text{k}\mathrm{\Omega }$, $C=10\mu \text{F}$), repeat the derivation yourself (Problem 2.3) and confirm:

1. $G_C=2C/\mathrm{\Delta }t=0.4\text{S}$.
2. The first sample is $V_{\text{cap}}=0.0249\text{V}$, $i_C=0.009975\text{A}$.
3. $V_{\text{cap}}$ reaches $63\mathrm{\%}$ of $10\text{V}$ (i.e. 6.32 V) after $\tau =RC=10\text{ms}$, and the charging current starts at $V/R=10\text{mA}$ and decays to zero.

2.9 Why the trapezoidal rule?

EMT engines overwhelmingly use the trapezoidal rule because it is:

• A-stable — large steps cannot make a stable circuit's simulation diverge (an explicit rule like forward Euler can).
• Second-order accurate — halving $\mathrm{\Delta }t$ quarters the error, cheap accuracy.
• Symmetric, so for a fixed $\mathrm{\Delta }t$ the companion conductances $G_L$, $G_C$ are constant and the system matrix can be factorized once and reused (Chapter 3).

Its one notorious weakness is a tendency to produce small sustained numerical oscillations right after an abrupt event (for example, opening a switch that carries inductor current). Production engines damp these with techniques such as critical-damping adjustment or a brief switch to backward Euler around discontinuities; we return to this in Module 2 and the appendices.

2.10 Summary

• Resistors are conductances; inductors and capacitors are differential and need discretizing.
• The trapezoidal rule turns each reactive element into a conductance in parallel with a known history current source — its companion model: $G_L=\mathrm{\Delta }t/2L$ and $G_C=2C/\mathrm{\Delta }t$.
• The history source encodes the element's stored energy and is a known constant when the present step is solved.
• Hand-iterating the companion recursion reproduces the simulator's output exactly; over a time constant it tracks the analytic exponential to a fraction of a percent.

2.11 Problems

Problem 2.1. A $5\text{mH}$ inductor is simulated with $\mathrm{\Delta }t=20\mu \text{s}$. Compute its companion conductance $G_L$. Repeat for $\mathrm{\Delta }t=50\mu \text{s}$. What happens to $G_L$ as you refine the step?

Solution 2.1

$G_L=\mathrm{\Delta }t/2L$. For $\mathrm{\Delta }t=20\mu \text{s}$: $G_L=20\times {10}^{-6}/(2\cdot 5\times {10}^{-3})=2\times {10}^{-3}\text{S}$. For $\mathrm{\Delta }t=50\mu \text{s}$: $G_L=50\times {10}^{-6}/0.01=5\times {10}^{-3}\text{S}$. $G_L$ is proportional to $\mathrm{\Delta }t$, so a smaller step gives a smaller companion conductance — the inductor looks "more open" over a shorter interval, consistent with $v=Ldi/dt$.

Problem 2.2. A $100\mu \text{F}$ capacitor is simulated with $\mathrm{\Delta }t=50\mu \text{s}$. Compute $G_C$. If at step $n-1$ the capacitor had $v=4\text{V}$ and $i=0.2\text{A}$, compute the history source $I_C^{\text{hist}}$ to be used at step $n$.

Solution 2.2

$G_C=2C/\mathrm{\Delta }t=2(100\times {10}^{-6})/(50\times {10}^{-6})=4\text{S}$. $I_C^{\text{hist}}=-(G_{C}v(t-\mathrm{\Delta }t)+i(t-\mathrm{\Delta }t))=-(4\cdot 4+0.2)=-16.2\text{A}$.

Problem 2.3. Derive the single-node update equation for the R-C charging circuit (ideal source $V$, series $R$, capacitor $C$ to ground) using the capacitor companion model, then compute $V_{\text{cap}}$ and $i_C$ at the first step ($V=10$, $R=1000$, $C=10\mu \text{F}$, $\mathrm{\Delta }t=50\mu \text{s}$, zero initial history).

Solution 2.3

KCL at the capacitor node: $(V-V_{\text{cap}})/R=i_C=G_C{V}_{\text{cap}}+I_C^{\text{hist}}$. With $G_C=2C/\mathrm{\Delta }t=0.4\text{S}$ and $I_C^{\text{hist}}=0$ at step 1:

$$\frac{10-V_{\text{cap}}}{1000}=0.4V_{\text{cap}}\Rightarrow 0.01=(0.001+0.4)V_{\text{cap}}\Rightarrow V_{\text{cap}}=\frac{0.01}{0.401}=0.02494\text{V}.$$$$i_C=G_C{V}_{\text{cap}}=0.4(0.02494)=0.009975\text{A}.$$

These match the simulator's first sample exactly. Note $i_C$ starts near $V/R=10\text{mA}$ — the capacitor looks momentarily like a short.

Problem 2.4. Trapezoidal integration is second-order accurate. If a study has unacceptable integration error at $\mathrm{\Delta }t=40\mu \text{s}$, roughly what step would cut the error by a factor of 16, and what is the cost?

Solution 2.4

Global error scales like $(\mathrm{\Delta }t{)}^2$. To cut error by 16 you need $(\mathrm{\Delta }t{)}^2\to (\mathrm{\Delta }t{)}^2/16$, i.e. $\mathrm{\Delta }t\to \mathrm{\Delta }t/4=10\mu \text{s}$. The cost is 4× as many time steps (4× the run time), since the number of steps scales like $1/\mathrm{\Delta }t$.

Problem 2.5. Explain physically why $G_C=2C/\mathrm{\Delta }t$ grows as the step shrinks, while $G_L=\mathrm{\Delta }t/2L$ shrinks. Relate each to how the element responds to a fast change.

Solution 2.5

Over a very short interval a capacitor can absorb a large current for a small voltage change ($i=Cdv/dt$ is large when $dt$ is small), so it looks like a large conductance — hence $G_C\propto 1/\mathrm{\Delta }t$. An inductor resists current change over a short interval ($v=Ldi/dt$ is large for small $dt$), so very little current flows for a given voltage: it looks like a small conductance — hence $G_L\propto \mathrm{\Delta }t$. The two are duals.

2.12 References

• H. W. Dommel, Electromagnetic Transients Program (EMTP) Theory Book, Bonneville Power Administration — the companion-model derivations for lumped elements.
• J. Arrillaga and N. R. Watson, Power Systems Electromagnetic Transients Simulation, IET Power and Energy Series 39 — numerical integrator substitution and the discretization of R, L, C.

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