Superposition
Real circuits often contain more than one independent source. A transceiver power supply must handle both a charging voltage and a battery terminal voltage simultaneously. An amplifier input stage has both a DC bias supply and an AC signal source. A two-source bridge circuit carries current from two different voltage references at once. Superposition is the theorem that makes multi-source circuits tractable: rather than dealing with all sources simultaneously using a system of equations, you consider each source acting alone and sum the results. The principle is valid because the circuits are linear — and it works for both voltage sources and current sources.
Superposition: analyse each source independently (all other sources deactivated), then algebraically add the individual responses to find the total.
View LargerThe Superposition Principle
Deactivating a source means replacing it with its internal resistance — for an ideal source, that is:
- An ideal voltage source is deactivated by replacing it with a short circuit (ideal voltage source has zero internal resistance → a short).
- An ideal current source is deactivated by replacing it with an open circuit (ideal current source has infinite internal resistance → an open).
The principle follows from the mathematical property of linearity. In a linear circuit, every voltage and current is a linear function of all the source values. If the source values are V1, V2, I1, …, then any response x is:
x = a₁V1 + a₂V2 + b₁I1 + …
where a₁, a₂, b₁ are constants determined by the circuit topology. Setting all sources to zero except V1 gives x due to V1 = a₁V1. The total response is the sum of all the individual contributions. This is superposition — not an approximation, but an exact consequence of linearity.
An important note: dependent sources are never deactivated. Only independent sources (batteries, independent current sources) are turned off one at a time. Dependent sources (controlled by a circuit variable) remain active in every sub-circuit because they represent the inherent behaviour of a component, not an externally applied source.
Procedure for Applying Superposition
- Identify all independent sources in the circuit.
- For each source in turn:
- Deactivate all other independent sources (short voltage sources, open current sources).
- Keep all dependent sources active.
- Analyse the resulting simplified circuit using any method: series-parallel reduction, voltage divider, KVL, KCL, Thevenin, etc.
- Find the voltage or current of interest for this source alone. Note the polarity carefully.
- Sum all the individual contributions algebraically (paying attention to sign and polarity). If two contributions produce voltage in the same direction at a node, they add. If they oppose each other, the smaller is subtracted from the larger.
Superposition doubles (or multiplies) the amount of analysis needed — one pass per source — but each sub-circuit is simpler because it has only one active source. For two-source circuits, this is often much simpler than setting up simultaneous equations.
Worked Examples
Example 1 — Two sources aiding (same polarity at the node)
Sub-circuit 1 — V1 active, V2 shorted:
With V2 shorted, R2 and R3 are both connected from node A to ground (R2's far end is now at the V2 ground, which is the same ground). They are in parallel:
R23 = R2 ∥ R3 = (6000 × 12000)/(6000 + 12000) = 72,000,000/18,000 = 4000 Ω
VA1 = V1 × R23/(R1 + R23) = 10 × 4000/(4000 + 4000) = 5.00 V
Sub-circuit 2 — V2 active, V1 shorted:
With V1 shorted, R1 and R3 are both connected from node A to ground. They are in parallel:
R13 = R1 ∥ R3 = (4000 × 12000)/(4000 + 12000) = 48,000,000/16,000 = 3000 Ω
VA2 = V2 × R13/(R2 + R13) = 6 × 3000/(6000 + 3000) = 2.00 V
Both contributions push node A positive (both sources have + terminal toward A):
VA = VA1 + VA2 = 5.00 + 2.00 = 7.00 V
Verification — KCL at node A:
(7 − 10)/4000 + (7 − 6)/6000 + 7/12000 = −0.750 + 0.167 + 0.583 = 0.000 mA ✓
Example 2 — Two sources opposing (one source "fights" the other)
Sub-circuit 1 — V1 active (V2 shorted):
R2 ∥ R3 = (2000 × 6000)/(2000 + 6000) = 12,000,000/8000 = 1500 Ω
VA1 = 12 × 1500/(3000 + 1500) = 12 × 1/3 = +4.00 V (V1 raises node A)
Sub-circuit 2 — V2 active (V1 shorted):
With V1 shorted, R1 connects from A to ground. R1 ∥ R3 = (3000 × 6000)/(3000 + 6000) = 18,000,000/9000 = 2000 Ω
Because V2's positive terminal faces away from node A (toward ground), V2 drives current from A through R2 toward V2+ — it pulls A down:
VA2 = −4 × 2000/(2000 + 2000) = −4 × 0.5 = −2.00 V (V2 pulls A down)
Total — algebraic sum (contributions oppose each other):
VA = VA1 + VA2 = 4.00 + (−2.00) = 2.00 V
Key lesson: When sources oppose, one contribution is negative. The sign of each contribution must be tracked carefully — always check which direction each source pushes the node voltage before adding.
Example 3 — Mixed sources: one voltage source, one current source
Sub-circuit 1 — V1 active, current source I1 opened:
With I1 opened, no current can flow through the branch that contained I1 (open circuit). The circuit reduces to a voltage divider: V1 → R1 → node A → R2 → ground.
VA1 = V1 × R2/(R1 + R2) = 10 × 4000/(2000 + 4000) = 10 × 2/3 = 6.667 V
Sub-circuit 2 — I1 active, V1 shorted:
With V1 shorted, R1's left end is at ground. Both R1 and R2 are now connected from node A to ground (in parallel). I1 = 3 mA flows into node A and splits between R1 and R2.
Rparallel = R1 ∥ R2 = (2000 × 4000)/(2000 + 4000) = 8,000,000/6000 = 1333 Ω
VA2 = I1 × Rparallel = 3 mA × 1333 Ω = 4.000 V
Total (both contributions push A positive):
VA = VA1 + VA2 = 6.667 + 4.000 = 10.667 V
Verification — KCL at node A:
(VA − V1)/R1 + VA/R2 − I1 = 0
(10.667 − 10)/2000 + 10.667/4000 − 0.003
= 0.000333 + 0.002667 − 0.003 = 0 ✓
Limitations: Power and Non-Linear Circuits
Superposition has two critical limitations that every student must understand:
Limitation 1: Superposition does not apply to power
Power is a non-linear function of voltage and current (P = V²/R = I²R). You cannot find the total power in a resistor by summing the individual powers from each source. Here is a concrete numerical demonstration:
Why superposition fails for power: the cross-term 2VA1VA2/R is missing when you add individual powers. Always find total voltage or current first via superposition, then calculate power from the combined value.
View LargerCircuit: V1 = 6 V through R1 = 1 kΩ to node A; V2 = 4 V through R2 = 1 kΩ to node A; R3 = 1 kΩ from A to ground.
Step 1 — Find VA correctly using superposition:
Sub-circuit 1 (V1 active, V2 shorted): R2 ∥ R3 = 500 Ω; VA1 = 6 × 500/1500 = 2.00 V
Sub-circuit 2 (V2 active, V1 shorted): R1 ∥ R3 = 500 Ω; VA2 = 4 × 500/1500 = 1.333 V
VA,total = 2.00 + 1.333 = 3.333 V
Step 2 — Correct total power in R3:
PR3,correct = VA,total² / R3 = (3.333)² / 1000 = 11.11 / 1000 = 11.11 mW
Step 3 — Incorrect method: summing individual powers:
PR3 from V1 alone: VA1² / R3 = (2.00)² / 1000 = 4.00 mW
PR3 from V2 alone: VA2² / R3 = (1.333)² / 1000 = 1.78 mW
Incorrect sum: 4.00 + 1.78 = 5.78 mW — wrong by 5.33 mW!
Where is the missing power?
The cross-term: 2 × VA1 × VA2 / R3 = 2 × 2.00 × 1.333 / 1000 = 5.33 mW. This cross-term comes from (VA1 + VA2)² = VA1² + 2VA1VA2 + VA2². The 2VA1VA2 term is the power that superposition of voltages cannot account for.
Rule: Always use superposition to find the total voltage or current first, then calculate power: P = Vtotal² / R or P = Itotal² × R.
Limitation 2: Superposition requires a linear circuit
Diodes, transistors operating in saturation or cut-off, and any other non-linear element invalidate superposition. In non-linear circuits, the response to a sum of inputs is not the sum of individual responses. This is why intermodulation distortion occurs in non-linear RF amplifiers: superposition fails, and the output contains products and sums of the input frequencies — not just the individual frequencies. Intermodulation is the practical consequence of the superposition principle breaking down in overloaded or poorly designed amplifier stages.
Superposition Calculator
Two-Source Superposition Calculator
Topology: V1 through R1 to node A; V2 through R2 to node A; R3 from node A to ground. The calculator applies superposition to find VA, verifies with KCL, and shows the correct power in R3 versus the incorrect sum-of-powers.
⚖ Experiment: Verify Superposition with Two Supplies
Build a two-source resistor network, measure node voltage with both sources active, then with each source separately. Verify the total equals the sum of the individual contributions.
- Two batteries: 9 V and 4.5 V (or two different voltage sources)
- R1: 10 kΩ; R2: 10 kΩ; R3: 10 kΩ
- Breadboard and jumper wires
- Digital multimeter
- Build the circuit: 9 V battery through R1 to node A; 4.5 V battery through R2 to node A; R3 from node A to ground. Both battery positive terminals face node A (sources aid each other).
- Measure VA with both sources active. Record VA(total).
- Replace the 4.5 V battery with a wire (short circuit — wire across its terminals). Measure VA with only the 9 V active. Record VA(9V only).
- Restore the 4.5 V battery and replace the 9 V battery with a wire. Measure VA with only the 4.5 V active. Record VA(4.5V only).
- Verify: VA(total) ≈ VA(9V only) + VA(4.5V only).
- Calculate the correct power in R3: P = VA(total)² / R3. Then calculate the incorrect sum: P9V alone + P4.5V alone. Compare the two values to see the power error in action.
For three equal 10 kΩ resistors: VA due to each source = Vsource/3. VA(9V only) = 3.0 V; VA(4.5V only) = 1.5 V; VA(total) = 4.5 V. Correct PR3 = 4.5²/10000 = 2.025 mW. Wrong sum: 9/1000 + 2.25/10000 = 0.9 + 0.225 = 1.125 mW — nearly half the correct answer. The cross-term 2 × 3.0 × 1.5 / 10000 = 0.9 mW is missing. Measurements should agree with voltage predictions within 5%.
Superposition in Ham Radio
DC bias and AC signal coexistence
In every linear amplifier, a DC bias voltage and an AC signal are simultaneously present at the transistor's base or gate. Superposition is why this works: the DC supply establishes the operating point (Q-point), and the AC signal independently produces the signal swing around that Q-point. The coupling capacitors separate the contributions physically — they block DC from the signal source while passing AC. Superposition is the mathematical foundation for analysing DC and AC conditions separately in amplifier design. The DC analysis sets the operating point; the small-signal AC analysis (using the linearised transistor model) gives the gain. Both analyses are valid simultaneously because the transistor's operation is approximately linear around the Q-point.
Interference from multiple signal sources
When two stations transmit simultaneously on adjacent frequencies, the receiver's linear front-end responds to both signals according to superposition — the output is the sum of the two independent signal contributions. This is ideal behaviour. When the receiver's non-linear regions are reached (a saturated LNA, a mixer in compression), superposition fails and intermodulation products appear. The third-order intercept point (IP3) and the 1 dB compression point quantify the degree to which superposition holds across input power levels — the engineer's measure of how well the amplifier preserves linear behaviour across signal combinations.
Power supply ripple and supply rejection
A power supply delivers DC to a transceiver but also carries AC ripple (at 100 Hz or 120 Hz for line voltage rectifiers). By superposition, the transceiver's response to the supply can be analysed as a DC response plus an independent AC ripple response. The power supply rejection ratio (PSRR) of each amplifier stage quantifies how much the AC ripple contributes to the output relative to the desired signal. A PSRR of 60 dB means 1 mV of ripple on the supply produces only 1 µV of noise at the output. Designing for high PSRR requires that the AC contribution (by superposition) be far smaller than the signal contribution at every stage.
Mixing action: a deliberate violation of superposition
A mixer multiplies the RF signal by the local oscillator — a fundamentally non-linear operation. Superposition therefore does not apply: the output contains the sum and difference frequencies (the IF), not just the individual input frequencies. This is by design. The non-linearity of the mixer creates the frequency translation that is the heart of the superheterodyne architecture. Superposition is intentionally violated in mixers, while being carefully preserved everywhere else in the signal chain where linearity and dynamic range are paramount.
Frequently Asked Questions
Why can't I use superposition to calculate power?
Power is proportional to the square of voltage (or current): P = V²/R. Squaring is a non-linear operation. The square of a sum is not the sum of the squares: (V1 + V2)² = V1² + 2V1V2 + V2². The cross-term 2V1V2 is the power that superposition misses. To find total power correctly: first find the total voltage using superposition, then calculate power from that combined value: P = Vtotal²/R. Summing individual powers (P1 + P2) always gives the wrong answer unless V1 and V2 are orthogonal (as in AC circuits where they are at different frequencies — a special case).
Does superposition apply to dependent sources?
Dependent sources must remain active in every sub-circuit — they are never deactivated. Only independent sources are turned off one at a time. A dependent voltage source (such as the controlled source in a transistor's small-signal model) depends on a circuit variable; deactivating it would remove the transistor's effect from the analysis entirely. When applying superposition to a circuit with dependent sources, keep the dependent sources active in every sub-circuit and deactivate only the independent sources in turn. The dependent source's value will change in each sub-circuit as the circuit variables change.
Is superposition ever faster than node analysis or mesh analysis?
Yes — particularly when there are only two sources and the circuit simplifies greatly when each acts alone. Superposition is often the fastest method when: (1) you have exactly two sources; (2) each sub-circuit reduces to a simple voltage divider or Ohm's Law application; or (3) you need to understand each source's individual contribution (e.g. to design a filter that attenuates one source while passing another). For circuits with three or more sources, node or mesh analysis handling all sources simultaneously is usually faster. Superposition also has pedagogical value: it shows exactly how much each source contributes to the total response.
What does it mean physically for voltages to "add" in superposition?
Each source independently drives currents through the network. Those currents produce voltage drops across each resistor. The total voltage across any resistor is the algebraic sum of all the voltage drops produced by the independent current contributions. Physically, electrons in the conductor respond to the total electric field, which is the vector sum of the fields from all sources — this is the electromagnetic basis of superposition. In a linear medium (a metallic conductor at normal operating conditions), the fields add independently, making superposition an exact result rather than an approximation.
What happens when two sources completely oppose each other?
If two sources produce equal and opposite contributions to the node voltage or branch current, the total is zero. This is the principle behind cancellation noise-cancelling circuits, common-mode rejection in differential amplifiers, and balanced transmission lines. In a balanced line carrying a signal, the desired signal adds (both wires carry the signal in opposite senses, but the differential amplifier subtracts the voltages and doubles the signal). Noise entering both wires equally is a common-mode signal — the amplifier subtracts these equal contributions and they cancel to zero. This is superposition and cancellation working together to reject interference.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.