Thevenin resistance with dependent sources is the equivalent output resistance of a linear network, calculated by applying an external test source to the open terminals because internal dependent sources cannot simply be zeroed out like independent ones. In a real circuit or installation, dependent sources—such as the transconductance in a MOSFET or the current gain in a BJT—actively alter the output impedance based on internal voltages or currents. This means the Thevenin resistance can be significantly lower, higher, or even negative compared to the physical resistors sitting on your breadboard. Beginners frequently confuse dependent (controlled) sources with independent sources, mistakenly trying to "turn off" a dependent voltage source by shorting it, which instantly invalidates the circuit's governing equations and leads to completely wrong bench measurements.
The Core Problem: Why Standard Source Transformation Fails
When you first learn Thevenin's theorem, the rule for finding the Thevenin resistance ($R_{th}$) is straightforward: turn off all independent sources (short voltage sources, open current sources) and calculate the equivalent resistance looking into the terminals using series and parallel combinations. This works perfectly for purely resistive networks.
However, dependent sources break this rule. A dependent source's output is mathematically tied to a controlling variable ($v_x$ or $i_x$) elsewhere in the circuit. If you attempt to "turn off" a dependent source, you destroy the very mechanism that defines the circuit's behavior. Instead of passive resistance, the dependent source injects or absorbs current dynamically, effectively acting as an active component that shifts the equivalent resistance.
| Source Type | Schematic Symbol | Action for $V_{oc}$ Calculation | Action for $R_{th}$ Calculation |
|---|---|---|---|
| Independent Voltage | Circle with +/- | Leave Active | Short Circuit (0V) |
| Independent Current | Circle with arrow | Leave Active | Open Circuit (0A) |
| Dependent Voltage | Diamond with +/- | Leave Active | Leave Active (Solve via Test Source) |
| Dependent Current | Diamond with arrow | Leave Active | Leave Active (Solve via Test Source) |
The Two Valid Methods for Finding Rth
Because you cannot use simple series/parallel reductions, you must rely on one of two fundamental network analysis techniques to extract the Thevenin resistance when dependent sources are present.
Method 1: The Test Source Method
This is generally the fastest method for circuits containing only dependent sources (after independent sources are turned off). You deactivate all independent sources, apply a hypothetical test voltage ($V_t$) or test current ($I_t$) to the open terminals, and use Kirchhoff's laws to solve for the resulting current or voltage. The Thevenin resistance is simply the ratio: $R_{th} = V_t / I_t$. Applying a $1V$ test source or $1A$ test source makes the math trivial, as the denominator or numerator becomes 1.
Method 2: The Short-Circuit Current Method
If the circuit is complex and applying a test source results in a messy system of equations, you can fall back on the definition of the Thevenin equivalent. First, calculate the open-circuit voltage ($V_{oc}$) at the terminals with all sources active. Next, short the terminals and calculate the short-circuit current ($I_{sc}$). The resistance is then $R_{th} = V_{oc} / I_{sc}$. This method requires solving the circuit twice, but it avoids the mental gymnastics of redefining controlling variables for a test source.
Worked Numeric Example: Test Source Method in Action
Let's walk through a concrete numeric example that mimics the small-signal output impedance of a common-source MOSFET amplifier. We want to find the Thevenin resistance at terminals A-B.
The Circuit Setup:
- Branch 1: A $1k\Omega$ resistor connected from Node A to ground (Node B). This represents a biasing resistor after a DC voltage source has been shorted.
- Branch 2: A $5k\Omega$ resistor connected from Node A to ground. This represents the transistor's output resistance ($r_o$).
- Branch 3: A voltage-controlled current source (VCCS) pointing down from Node A to ground, with a value of $0.5 \text{ mA/V} \cdot v_x$. This represents the transistor's transconductance ($g_m$).
- Controlling Variable: $v_x$ is defined as the voltage at Node A with respect to Node B.
Step 1: Apply the Test Source
Because all independent sources have already been turned off (which is why Branch 1 is just a resistor to ground), we apply a test voltage $V_t$ at terminals A-B. Therefore, the voltage at Node A is exactly $V_t$, which means our controlling variable $v_x = V_t$.
Step 2: Write the KCL Equation
We sum the currents leaving Node A and set them equal to the test current $I_t$ entering the node from our source:
$$I_t = I_{R1} + I_{R2} + I_{VCCS}$$
$$I_t = \frac{V_t}{1000} + \frac{V_t}{5000} + (0.0005 \cdot v_x)$$
Step 3: Substitute and Solve
Substitute $v_x = V_t$ into the equation:
$$I_t = V_t(0.001) + V_t(0.0002) + V_t(0.0005)$$
$$I_t = V_t(0.0017)$$
Step 4: Calculate $R_{th}$
$$R_{th} = \frac{V_t}{I_t} = \frac{1}{0.0017} = \mathbf{588.2 \Omega}$$
Where You Meet This in Practice
You rarely sit down with a schematic full of diamond-shaped symbols outside of a university exam, but the physics they represent are everywhere in practical electronics. According to foundational circuit theory resources like MIT OpenCourseWare, mastering this concept is the bridge between basic DC analysis and real-world AC signal design.
- Transistor Amplifier Output Impedance: When designing an audio preamp using a 2N3904 BJT or a BS170 MOSFET, the small-signal model relies entirely on dependent sources ($g_m v_{be}$ or $g_m v_{gs}$). Calculating the $R_{th}$ at the drain or collector tells you the amplifier's output impedance, which is critical for impedance matching to the next stage.
- Op-Amp Output Stages: While an ideal op-amp has zero output resistance, real op-amps like the TL072 or NE5532 have a physical output resistance (often around $50\Omega$ to $100\Omega$). When you apply negative feedback, the dependent nature of the op-amp's open-loop gain drives the effective Thevenin resistance down to milliohms.
- Active Filters and Synthetic Inductors: Circuits like the Antoniou gyrator use op-amps and dependent source behavior to simulate massive inductors using only small capacitors and resistors. The Thevenin resistance calculations dictate the Q-factor and insertion loss of the filter.
For deeper reading on how these equivalent circuits map to physical semiconductor datasheets, Electronics Tutorials provides excellent visual breakdowns of source transformations.
Common Confusions and Troubleshooting
Can Thevenin resistance be negative?
Yes. In circuits with dependent sources (like certain oscillator topologies or negative-impedance converters), the dependent source can push current back into the test source rather than drawing it. This results in a negative $R_{th}$. On the bench, you won't measure a "negative ohm" directly with a multimeter, but you will observe the circuit exhibiting instability, latching up, or oscillating when connected to a capacitive load.
What if the controlling variable is outside the Thevenin boundary?
If $v_x$ or $i_x$ is located in the part of the circuit you are "cutting off" to form the Thevenin equivalent, you must express that controlling variable in terms of the variables inside your Thevenin boundary before solving. If you cannot mathematically link the external variable to the terminals A-B, the circuit cannot be reduced to a simple Thevenin equivalent from that perspective.
Why does my SPICE simulation show a different $R_{th}$ than my hand calculation?
Hand calculations assume ideal dependent sources. SPICE models for transistors include parasitic capacitances, body diodes, and non-linear resistance (like the Early effect in BJTs). Ensure your SPICE simulation is running an AC small-signal analysis (.ac) at a low frequency (e.g., 10 Hz) to approximate the DC Thevenin resistance, rather than a transient (.tran) sweep which captures dynamic charging effects.
