First-Order Circuits

The two forms of excitation deserve their names. In the natural response the circuit is free of independent sources; whatever energy was stored in the capacitor or inductor drives a current that the resistors gradually dissipate, and the circuit relaxes to rest. A source-free circuit may still contain dependent sources — "source-free" means no independent sources. In the forced response an independent source, usually DC here, drives the circuit to a new condition.

The remarkable thing — established in Section 7.8 — is that all four cases collapse into a single formula requiring three numbers. Work through the derivations first, because the formula means little without them, but do not lose sight of where this is heading.

A source-free RC circuit arises whenever a charged capacitor is suddenly disconnected from its supply and left connected to a resistance. The energy stored in the capacitor is released into the resistor and dissipated as heat.

Derivation. Let the initial capacitor voltage be

Apply KCL at the top node. The capacitor current and the resistor current must sum to zero:

This is the first-order differential equation the chapter is named for. Divide by \(C\) and separate the variables:

Integrating both sides, with \(\ln A\) as the constant of integration:

The constant follows from the initial condition. At \(t = 0\) the exponential is unity, so \(A = V_{0}\), giving the natural response:

The voltage decays exponentially from its initial value towards zero. Note that we could write down the initial condition immediately only because capacitor voltage is continuous — the property established in Section 6.3. Without it there would be no way to connect the circuit's state before the switching to its state after.

Everything else in the circuit follows at once:

Observe that the power decays at twice the rate of the voltage, since it depends on the square.

The rate at which the response decays is set entirely by the product \(RC\), called the time constant and denoted \(\tau\). Its definition is worth stating precisely:

The time constant is the time required for the response to decay to \(1/e\) — that is, 36.8 % — of its initial value. Setting \(t = \tau\) confirms it:

Despite being a product of ohms and farads, \(\tau\) has units of seconds — as it must, since it appears in the exponent alongside \(t\). A quick check: \(\Omega \cdot \mathrm{F} = (\mathrm{V/A})(\mathrm{C/V}) = \mathrm{C/A} = \mathrm{s}\).

This table justifies a rule used throughout engineering: after five time constants the response is within 1 % of its final value, and the circuit is taken as fully settled. The exponential never mathematically reaches zero, but after \(5\tau\) the remaining difference is smaller than the tolerance of any component in the circuit.

There is also a neat geometrical reading. The tangent to the curve at \(t = 0\) has slope \(-V_{0}/\tau\), so if the response continued to fall at its initial rate it would reach zero exactly at \(t = \tau\). This gives a practical way to measure \(\tau\) from an oscilloscope trace.

The dual situation is an inductor carrying current, suddenly left to discharge into a resistance. Everything runs parallel to Section 7.2, with current in place of voltage.

Derivation. Let the initial inductor current be

Apply KVL around the loop. The inductor voltage and the resistor voltage must sum to zero:

Separating variables and integrating exactly as before:

Note the time constant carefully. For the RC circuit it was a product, \(RC\); here it is a quotient, \(L/R\). A large resistance makes an RC circuit slow but an RL circuit fast. Confusing the two is among the most frequent errors in this chapter, and the units settle it: \(\mathrm{H}/\Omega = (\mathrm{V\cdot s/A})/(\mathrm{V/A}) = \mathrm{s}\) \(\checkmark\).

Again the initial condition could be carried across the switching instant only because inductor current is continuous. The associated quantities follow:

Before turning to step responses we need a compact notation for switching. Singularity functions are discontinuous functions, or functions with discontinuous derivatives, that describe abrupt events. Three of them matter here, and each is the derivative of the next.

Now a DC source is suddenly applied to an RC circuit — the switch closes rather than opens. The response is called the step response.

Derivation. With a source \(V_{s}\) applied through \(R\) at \(t = 0\) and an initial capacitor voltage \(V_{0}\), KCL gives

This differs from the source-free equation only by the constant on the right. Separating variables:

At \(t = 0\) the capacitor voltage is \(V_{0}\), so \(A = V_{0} - V_{s}\) and

For the common case of an initially uncharged capacitor, \(V_{0} = 0\) and this reduces to the familiar charging curve:

Note the asymmetry: the voltage rises exponentially towards \(V_{s}\) while the current falls exponentially from \(V_{s}/R\) towards zero. At the first instant the uncharged capacitor behaves like a short circuit; after a long time it behaves like an open circuit.

The RL step response follows by duality, with current replacing voltage throughout. Applying KVL after the switch closes:

Solving exactly as before, with initial current \(I_{0}\) and final current \(V_{s}/R\):

For an inductor initially at rest, \(I_{0} = 0\) and

Again the roles are exchanged relative to the RC case: here it is the current that rises and the voltage that decays. At the first instant an inductor initially at rest behaves like an open circuit — it refuses to let current start — and after a long time it behaves like a short circuit.

Look back at the four results derived so far. Every one of them has the same shape, and that shape can be stated once and applied to all of them:

where \(x\) is capacitor voltage or inductor current — the continuous variable — and the three ingredients are:

No differential equation need ever be solved again for a first-order circuit. Find three numbers using techniques from earlier chapters, and substitute.

Delayed switching. If the switch operates at \(t = t_{0}\) rather than at the origin, the response is simply shifted:

Other quantities. The formula applies directly only to capacitor voltage and inductor current, because only those are guaranteed continuous. To find any other voltage or current, first obtain the continuous variable, then work outwards using Ohm's and Kirchhoff's laws. Applying the formula directly to, say, a resistor current can give the wrong initial value, because that current may well jump at \(t=0\).

An op-amp circuit containing a single capacitor is still a first-order circuit, and the universal formula still applies. The only new work is finding the effective time constant, and here the two golden rules of Chapter 5 do it for us.

Consider an inverting amplifier in which the feedback resistor \(R_{f}\) has a capacitor \(C\) in parallel with it, driven through \(R_{1}\) by a step of height \(V_{s}\).

Because the inverting terminal is a virtual ground, the capacitor sees exactly \(R_{f}\) across it — nothing else in the circuit is connected to both its terminals. Hence

The initial value is zero if the capacitor starts uncharged, since the output must equal the (zero) capacitor voltage at that instant. The final value is the ordinary inverting-amplifier gain, because in DC steady state the capacitor is an open circuit and only \(R_{f}\) remains:

Substituting into the universal formula:

The circuit behaves as an amplifier at DC and as an integrator at high frequencies, which makes it a practical compromise between the two: unlike the pure integrator of Section 6.9, its DC gain is finite, so input offsets do not accumulate without limit.

The exponential charging of a capacitor gives a simple and reliable way of measuring time. A capacitor charges through a resistor until its voltage reaches a threshold, at which point a switching device — a neon lamp, a comparator, or the trigger input of a 555 timer — fires. The delay is set by \(\tau = RC\) and the threshold fraction:

For an initially uncharged capacitor charging towards \(V_{s}\) and triggering at half the supply, the delay is \(RC\ln 2 = 0.693\,RC\) — a number that appears in every 555 timer design.

The strength of the method is that the threshold is a fraction of the supply, so the delay is independent of supply voltage. Its weakness is that \(R\) and \(C\) both have wide tolerances and drift with temperature, so RC timing is fine for a windscreen-wiper interval and useless for a clock.

• A first-order circuit contains one storage element and is described by a first-order differential equation. There are four cases: RC or RL, source-free or step-excited.

• The source-free RC response is \(v = V_{0}e^{-t/\tau}\) with \(\tau = RC\); the source-free RL response is \(i = I_{0}e^{-t/\tau}\) with \(\tau = L/R\).

• The time constant is the time to decay to 36.8 % of the initial value; after \(5\tau\) the response is within 1 % of its final value.

• Singularity functions describe switching: \(u(t)\), its derivative \(\delta(t)\), and its integral \(r(t)\). The impulse has unit area and the sifting property.

• A complete response splits either into natural plus forced, or into transient plus steady-state.

• The universal formula \(x(t) = x(\infty)+[x(0)-x(\infty)]e^{-t/\tau}\) solves every first-order circuit from three numbers.

• It applies to capacitor voltage and inductor current, the continuous variables; other quantities are found from them afterwards.

• \(R_{Th}\) is the Thévenin resistance seen by the storage element after switching.

• RC timing underlies delay circuits; RL switching requires freewheeling protection.

• Using \(\tau = RC\) for an RL circuit, or \(L/R\) for an RC one. The RC constant is a product, the RL constant a quotient. Check the units if in doubt.

• Computing \(R_{Th}\) from the circuit before switching. The time constant belongs to the circuit that exists after the switch operates. This is the single most common error in the chapter.

• Applying the universal formula to a resistor current or voltage. It is valid only for capacitor voltage and inductor current, because only those are continuous. Find the continuous variable first, then work outwards.

• Assuming the initial value of the response is zero. It is zero only if the element started with no stored energy. Always check the steady state before switching.

• Forgetting that \(x(\infty)\) may be non-zero in a "discharge" problem, or zero in a "charge" one. Compute it; do not assume from the wording.

• Treating a capacitor as an open circuit during the transient. That substitution is valid only in DC steady state — before \(t=0\) and after \(5\tau\), not in between.

• Forgetting the time shift for delayed switching. If the switch acts at \(t_{0}\), the exponent is \(-(t-t_{0})/\tau\), not \(-t/\tau\).

• Expecting inductor voltage or capacitor current to be continuous. They jump routinely, and the jump is often the most interesting feature of the problem.

• Mixing units in the time constant. \(\mathrm{k}\Omega \times \mu\mathrm{F}\) gives milliseconds — convenient, but only if you know it. When unsure, convert everything to ohms and farads.

• Reading \(5\tau\) as an exact settling time. It is a convention meaning "within 1 %", not a mathematical endpoint. The exponential never actually reaches its final value.

Work these before opening the answers. Quote results to four significant figures with correct units.

1. P7.1 A \(5~\mu\mathrm{F}\) capacitor charged to 100 V discharges through \(20~\mathrm{k}\Omega\). Find \(\tau\), \(v(t)\), and \(v\) at \(t = 0.2~\mathrm{s}\).

   Show answer

   \[\tau = \left(20\times10^{3}\right)\left(5\times10^{-6}\right) = 0.1~\mathrm{s}\]

   \[v(t) = 100e^{-10t}~\mathrm{V}, \qquad v(0.2) = 100e^{-2} = 13.53~\mathrm{V}\]

2. P7.2 How long does a first-order response take to decay to 10 % of its initial value, expressed in time constants?

   Show answer

   \[e^{-t/\tau} = 0.1 \;\Longrightarrow\; \frac{t}{\tau} = \ln 10 = 2.303\]

   So \(t = 2.303\tau\) — a little over two time constants.

3. P7.3 A 4 H inductor carrying 3 A discharges through \(20~\Omega\). Find \(\tau\), \(i(t)\), and the initial stored energy.

   Show answer

   \[\tau = \frac{4}{20} = 0.2~\mathrm{s}, \qquad i(t) = 3e^{-5t}~\mathrm{A}\]

   \[w(0) = \tfrac{1}{2}(4)(9) = 18.00~\mathrm{J}\]

4. P7.4 Express a 5 V pulse lasting from \(t = 1~\mathrm{s}\) to \(t = 4~\mathrm{s}\) using step functions, and evaluate \(\int_{-\infty}^{\infty}(4t+1)\delta(t-3)\,dt\).

   Show answer

   \[v(t) = 5\left[u(t-1) - u(t-4)\right]~\mathrm{V}\]

   By the sifting property, \(\int (4t+1)\delta(t-3)\,dt = 4(3)+1 = 13.00\).

5. P7.5 An uncharged capacitor is connected at \(t=0\) to a source such that \(v(\infty) = 50~\mathrm{V}\) and \(\tau = 0.5~\mathrm{s}\). Find \(v(t)\) and \(v(1)\).

   Show answer

   \[v(t) = 50 + (0-50)e^{-2t} = 50\left(1-e^{-2t}\right)~\mathrm{V}\]

   \[v(1) = 50\left(1-e^{-2}\right) = 50(0.8647) = 43.23~\mathrm{V}\]

6. P7.6 An inductor initially at rest reaches a final current of 4 A with \(\tau = 0.25~\mathrm{s}\). Find \(i(t)\) and \(i(0.5)\).

   Show answer

   \[i(t) = 4\left(1-e^{-4t}\right)~\mathrm{A}, \qquad i(0.5) = 4\left(1-e^{-2}\right) = 3.459~\mathrm{A}\]

7. P7.7 A \(100~\mu\mathrm{F}\) capacitor is connected across a network whose Thévenin resistance is \(20~\Omega\) in parallel with \(30~\Omega\). Find \(\tau\) and the settling time.

   Show answer

   \[R_{Th} = \frac{(20)(30)}{50} = 12~\Omega, \qquad \tau = (12)\left(100\times10^{-6}\right) = 1.200~\mathrm{ms}\]

   Settling time \(5\tau = 6.000~\mathrm{ms}\).

8. P7.8 An op-amp inverting stage has \(R_{1} = 10~\mathrm{k}\Omega\), \(R_{f} = 100~\mathrm{k}\Omega\) and \(C = 1~\mu\mathrm{F}\) across \(R_{f}\). A 0.5 V step is applied. Find \(\tau\), the final output, and \(v_{o}(t)\).

   Show answer

   \[\tau = \left(100\times10^{3}\right)\left(1\times10^{-6}\right) = 0.1~\mathrm{s}\]

   \[v_{o}(\infty) = -\frac{100}{10}(0.5) = -5~\mathrm{V}, \qquad v_{o}(t) = -5\left(1-e^{-10t}\right)~\mathrm{V}\]

9. P7.9 How many time constants are needed for a response to reach 99 % of its final value?

   Show answer

   \[1 - e^{-n} = 0.99 \;\Longrightarrow\; e^{-n} = 0.01 \;\Longrightarrow\; n = \ln 100 = 4.605\]

   Slightly fewer than five — which is why the \(5\tau\) convention is a safe rounding.

10. P7.10 A capacitor at 40 V is switched at \(t = 0\) into a circuit whose final value is 10 V, with \(\tau = 0.4~\mathrm{s}\). Write \(v(t)\), identify the transient and steady-state parts, and find when \(v = 20~\mathrm{V}\).

    Show answer

    \[v(t) = 10 + (40-10)e^{-2.5t} = 10 + 30e^{-2.5t}~\mathrm{V}\]

    Steady-state part 10 V; transient part \(30e^{-2.5t}\) V.

    \[20 = 10 + 30e^{-2.5t} \;\Longrightarrow\; e^{-2.5t} = \tfrac{1}{3} \;\Longrightarrow\; t = \frac{\ln 3}{2.5} = 0.4394~\mathrm{s}\]