AoNArt of Nanoelectronics
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02 · Foundational → intermediate

Extracting a Thévenin equivalent without fragile shortcuts

Separate open-circuit voltage, driving resistance and the test-source method instead of memorizing disconnected rules.

55–75 minFoundational → intermediate
In this lesson
  • Explain why open-circuit voltage and driving resistance are extracted from two different circuit states.
  • Compute the Thévenin voltage with the original independent sources still active.
  • Compute the Thévenin resistance by correctly zeroing independent sources.
  • Use a test source when dependent sources remain active.
  • Verify the equivalent with an independent loaded-circuit calculation.
The problem we are solving

We want to replace the network to the left of a load by only two numbers: a voltage and a resistance. The common mistake is to search for both numbers in the same modified circuit. Thévenin does not do that.

VTH asks what voltage remains when the port delivers no current. RTH asks how strongly port voltage changes when current is forced. One is an intercept; the other is a slope.

That distinction looks trivial on a divider, but it becomes essential with dependent sources, transistor stages and real instruments. The goal is not to memorize a short/open recipe; it is to know which property of the port is being identified at each step.

VTHthe intercept

Port voltage when the external current is zero.

RTHthe slope

Change in port voltage per unit current forced at the port.

open circuit → keep the sources activedriving resistance → zero independent sources only

02.1Two questions require two circuit states

Use a 12 V divider with R1 = 3 kΩ and R2 = 6 kΩ. To obtain VTH, remove the external load but leave the independent source active.

Figure 2.1 — Original network, load removed

The independent 12 V source is still active. Only the external load has been removed.

canonical circuit verified
Voltage divider one-port with load removed+−VS12 VR13 kΩR26 kΩVINPORTGND
port open: Iport = 0VTH = VOCinternal sources stay active

An open port means external port current is zero. It does not mean that the internal shunt resistor disappears.

VTH = VOC = 12 × 6/(3 + 6) = 8.00 V

02.2Why a zero-volt ideal source is a short circuit

An ideal voltage source imposes its terminal voltage. Setting its independent value to zero therefore imposes zero voltage between its terminals — exactly the electrical constraint of an ideal short. The dual statement holds for an ideal current source at 0 A: no current can flow through that branch, so it becomes an open circuit.

Figure 2.2 — Source-deactivated network seen from the port

Setting the ideal 12 V source to 0 V turns it into a short. R1 and R2 therefore both connect from the port to ground.

connectivity verified
Source-deactivated resistance seen from the portR13 kΩR26 kΩVOUTGND
0 V source → short circuitRTH = R1 ∥ R2do not open the voltage source
RTH = 3 kΩ ∥ 6 kΩ = 2.00 kΩ

The topology changed because one end of R1 is now tied to ground through the zero-volt source. Seen from the port, both resistors are shunts to ground.

02.3The test source measures the slope directly

For dependent-source networks, or simply as an independent check, zero the independent sources, keep controlled sources active and apply a test excitation at the port.

Figure 2.3 — Test-source method

A 1 V test source probes the slope of the deactivated one-port. The resulting current determines the driving-point resistance.

test-source topology verified
Test-source method for driving-point resistance+−Vt1 VR13 kΩR26 kΩPORTGND
Vt = 1 VIt = 0.5 mARTH = Vt / It
It = Vt/R1 + Vt/R2
= 1 V/3 kΩ + 1 V/6 kΩ = 0.500 mA
RTH = Vt/It = 1 V/0.500 mA = 2.00 kΩ

A 1 V test source is chosen for convenience. Any non-zero test amplitude gives the same ratio in a linear network.

02.4Reconnect the load and demand an independent check

Figure 2.4 — Final equivalent with the load reconnected

The original divider is replaced by an 8 V ideal source in series with 2 kΩ. RL = 4.7 kΩ must see the same operating point as in the full circuit.

equivalent circuit verified
Thévenin equivalent with load+−VTH8 VRTH2 kΩRL4.7 kΩVOUTGND
VTH = 8 VRTH = 2 kΩsame terminal law, same load point
VL = 8 × 4.7/(2 + 4.7) = 5.612 V
IL = 1.194 mA

Solving the original loaded divider gives the same 5.612 V. That equality of terminal behavior — not visual similarity — is the meaning of equivalence.

02.5Method selection

SituationNatural methodReason
Resistors + independent sourcesOpen circuit + source deactivationDriving resistance often reduces directly.
Dependent sourcesTest sourceControlled sources remain active.
Accessible linear black boxTerminal measurementsIdentify intercept and slope externally.
Nonlinear networkLocal linearizationOne fixed affine equivalent is not globally valid.

02.6Checkpoint before moving on

You should now be able to explain why VTH and RTH are extracted from different circuit states rather than reciting a short/open rule from memory.

Three self-check questions
  1. Why does an independent ideal voltage source set to zero become a short rather than an open circuit?
  2. Why must dependent sources remain active when finding RTH?
  3. If a 2 V test source produces 1 mA, what is the driving resistance? Would a 1 V test source change the ratio in a linear network?
Answers. Zero volts forces equipotential terminals; dependent sources encode internal coupling and therefore remain active; 2 V / 1 mA = 2 kΩ, and any non-zero test amplitude gives the same ratio while the network remains linear.
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