Superposition Theorem for Linear Circuits

Every linear circuit with more than one independent source can, in principle, be solved directly with node or mesh analysis. But when sources have different frequencies, different types (DC bias plus AC signal), or when you want to isolate the contribution of one particular source for design or troubleshooting, solving the whole system at once is wasteful and obscures what's happening. Superposition lets you break the problem into single-source sub-circuits, solve each with simple series/parallel and divider reasoning, and add the results. It's the theoretical justification for the common engineering habit of analyzing "DC operating point" and "AC signal" separately in amplifier design.

Statement of the Theorem

In any linear circuit containing multiple independent sources, the response (voltage or current) at any element equals the algebraic sum of the responses caused by each independent source acting alone, with all other independent sources set to zero.

"Set to zero" means:

• Voltage sources → replaced by a short circuit (0 V, zero internal impedance for the ideal case)
• Current sources → replaced by an open circuit (0 A)

Dependent (controlled) sources are never zeroed out — they stay in every sub-circuit exactly as they are, because their value depends on a variable elsewhere in the circuit that superposition is not allowed to strip away. This is the single most common mistake when applying the theorem to transistor small-signal models.

Why It Works: Linearity

Superposition is a direct consequence of linearity. A circuit made only of linear elements — resistors, capacitors, inductors, linear dependent sources — obeys:

Homogeneity:  f(k·x) = k·f(x)
Additivity:   f(x₁ + x₂) = f(x₁) + f(x₂)

Kirchhoff's laws (KCL, KVL) combined with linear V-I relationships (V = IR, I = C·dV/dt, V = L·dI/dt) produce a system of linear equations. Solving that system for multiple simultaneous inputs is mathematically identical to solving it once per input and summing — this is exactly what linear algebra guarantees for a linear system Ax = b, where b is a sum of independent excitation vectors.

This is why superposition fails for nonlinear elements — diodes, transistors in large-signal operation, saturating inductors. If f(x) = x², then f(x₁+x₂) ≠ f(x₁) + f(x₂), so a circuit with a diode cannot have its total current found by summing the diode currents from each source acting alone.

Procedure

1. Identify all independent sources.
2. For each source in turn: zero all other independent voltage sources (short) and current sources (open); keep all dependent sources and all passive components unchanged.
3. Solve the resulting single-source circuit for the desired voltage or current using any method (series/parallel reduction, divider rules, node/mesh analysis).
4. Repeat for every independent source.
5. Algebraically sum all the partial responses to get the total.

Step 5 requires care with sign conventions — define a single reference polarity/direction for the quantity of interest before starting, and keep it consistent across all sub-circuits.

Worked Example: Two Sources, One Resistor

        R1=4Ω        R2=2Ω
  A ●───/\/\───●───/\/\───● B
              node X       |
                          R3=6Ω
                           |
  Vs=12V (A–gnd)         gnd
  Is=3A  (into node X, from gnd)

Circuit: a 12 V source connects through R1 = 4 Ω to node X. A 3 A current source injects current into node X from ground. R2 = 2 Ω connects node X onward to node B, and R3 = 6 Ω connects node X to ground. Node B is left open (no load), so no current flows through R2, meaning the current through R2 is irrelevant here — but let's instead find the voltage at node X, V_X, which is the more informative target since R2 carries no current with B open.

Find V_X by superposition.

Step 1 — Voltage source alone (Is opened): With the current source removed (open circuit), the circuit is just Vs = 12 V through R1 = 4 Ω into R3 = 6 Ω to ground (R2 carries no current since B is open, so it doesn't load node X).

Voltage divider from Vs across R1 and R3:

V_X1 = Vs · R3 / (R1 + R3) = 12 · 6 / (4 + 6) = 12 · 0.6 = 7.2 V

Step 2 — Current source alone (Vs shorted): With Vs replaced by a short, R1 goes from node X to ground (since A is now grounded). So R1 (4 Ω) is in parallel with R3 (6 Ω) at node X, fed by the 3 A source.

R1 ∥ R3 = (4·6)/(4+6) = 24/10 = 2.4 Ω
V_X2 = Is · (R1 ∥ R3) = 3 · 2.4 = 7.2 V

Step 3 — Superpose:

V_X = V_X1 + V_X2 = 7.2 + 7.2 = 14.4 V

Verification by direct nodal analysis: KCL at node X (currents leaving node X sum to zero), with A = 12 V fixed:

(V_X − 12)/4 + V_X/6 − 3 = 0

Multiply through by 12:

3(V_X − 12) + 2·V_X − 36 = 0
3V_X − 36 + 2V_X − 36 = 0
5V_X = 72
V_X = 14.4 V

Matches the superposition result exactly. ✓

Application to AC/DC Mixed Circuits

Superposition is the formal basis for splitting amplifier analysis into DC bias and AC small-signal halves:

1. Solve the DC circuit (all AC sources set to zero, capacitors treated as open at DC) to find the operating point (Q-point).
2. Solve the AC small-signal circuit (DC supply zeroed to AC ground, capacitors treated as short if large enough at the signal frequency, transistor replaced by its small-signal linear model) to find gain and impedance.
3. The total instantaneous voltage/current at any node is the DC value plus the AC value — a direct application of superposition, valid because the small-signal model is a linearized approximation of the transistor around the Q-point.

This only holds as an approximation: real transistors are nonlinear, and the small-signal model is valid only for signal excursions small enough that the linearization holds. Superposition breaks down for large-signal swings (clipping, slewing) precisely because the underlying device is no longer behaving linearly.

Superposition is also the natural tool for circuits with sources at different frequencies (e.g., a DC supply plus multiple AC signals or harmonics) — since impedances (Z = R, jωL, 1/jωC) depend on frequency, each source's sub-circuit must be solved at its own frequency, and the results summed as phasors only if they share the same frequency; responses at different frequencies must be summed as time-domain waveforms, not combined as phasors, because phasors are only meaningful for a single frequency.

Power Is the Exception

Superposition does not apply directly to power. Power is proportional to the square of voltage or current (P = V²/R, P = I²R), which is a nonlinear relationship. You cannot compute the power dissipated in a resistor by finding the power due to each source separately and adding them — cross terms are lost.

Correct:   P_total = (V1 + V2)² / R = V1²/R + 2·V1·V2/R + V2²/R
Incorrect: P_total ≠ V1²/R + V2²/R   (misses the cross term 2·V1·V2/R)

The safe procedure: use superposition to find the total voltage or current first, then compute power from that total.

Practical Value

• Circuit troubleshooting: isolating which source (signal vs. supply noise vs. crosstalk) contributes how much to a measured voltage at a test point.
• Amplifier design: separating DC bias analysis from AC gain analysis, as described above.
• EMI/noise analysis: computing the contribution of an interfering source independently of the wanted signal, since both are linear excitations on the same network.
• Simplifying hand analysis: converting a circuit with N sources into N simpler single-source problems, each solvable by divider rules instead of full matrix solutions.

Key Takeaways

• Superposition states that the response in a linear circuit with multiple independent sources equals the sum of responses from each source acting alone, with other independent sources zeroed (voltage sources shorted, current sources opened).
• Dependent sources are never zeroed — they remain active in every sub-circuit.
• The theorem is a direct consequence of linearity and fails for any nonlinear element (diodes, saturated transistors, saturated inductors).
• It underlies the standard practice of separating DC bias analysis from AC small-signal analysis in amplifier design.
• Power cannot be superposed directly because P = V²/R is nonlinear — sum voltages or currents first, then compute power from the total.
• Always verify a superposition result against a direct nodal or mesh solution, especially when sign conventions for multiple sources could be mixed up.