How to Analyze an Example of Schematic: Thevenin Equivalent Walkthrough

Reading and reducing a complex circuit into a usable model is a foundational skill for any electrical hobbyist or engineering student. When you encounter an example of schematic that needs to interface with a sensitive load—like an analog-to-digital converter (ADC) or a high-impedance sensor—guessing the output behavior leads to noisy readings and stalled projects. Instead, we use circuit theorems to strip the network down to its bare essentials.

In this walkthrough, we will take a practical sensor interface schematic, reduce it to its Thevenin equivalent, verify the math, and use those results to select a specific, real-world op-amp part number for buffering.

The Problem Statement: A Sensor Interface Example of Schematic

Choosing the Right Method: Why Thevenin's Theorem Applies

Which theorem applies and why?
We use Thevenin's Theorem because we are analyzing the circuit from the perspective of a specific pair of output terminals (A and B) that will eventually connect to a load. Thevenin's theorem allows us to replace the entire resistor network and voltage source with a single equivalent voltage source ($V_{th}$) in series with a single equivalent resistor ($R_{th}$). This drastically simplifies load analysis.

The trap in this problem:
The most common mistake in this specific topology is calculating the open-circuit voltage at Node X, but then forgetting to include $R_3$ when calculating the Thevenin resistance. Because Terminal A is open-circuited during the $V_{th}$ calculation, no current flows through $R_3$, meaning there is no voltage drop across it. However, when calculating $R_{th}$, the independent source is turned off, and $R_3$ is physically in series with the parallel combination of $R_1$ and $R_2$ when looking back into the terminals.

Step-by-Step Algebraic Solution

Let's break down the algebra without skipping any steps.

Step 1: Calculate Thevenin Voltage ($V_{th}$)

$V_{th}$ is the open-circuit voltage across terminals A and B. Since the terminals are open, current through $R_3$ is zero ($I_{R3} = 0$). Therefore, the voltage at Terminal A equals the voltage at Node X ($V_A = V_X$).

Node X forms a simple voltage divider between $V_s$, $R_1$, and $R_2$:

• $V_X = V_s \times \frac{R_2}{R_1 + R_2}$
• $V_X = 3.3V \times \frac{4k\Omega}{2k\Omega + 4k\Omega}$
• $V_X = 3.3V \times \frac{4}{6}$
• $V_X = 3.3V \times 0.6667$
• $V_X = 2.2V$

$V_{th} = 2.2V$

Step 2: Calculate Thevenin Resistance ($R_{th}$)

To find $R_{th}$, we deactivate the independent voltage source (replace $V_s$ with a short circuit to ground) and look into terminals A and B.

1. With $V_s$ shorted to ground, $R_1$ is now connected between Node X and ground.
2. This places $R_1$ in parallel with $R_2$. Let's calculate $R_{1||2}$:
   • $R_{1||2} = \frac{R_1 \times R_2}{R_1 + R_2}$
   • $R_{1||2} = \frac{2k \times 4k}{2k + 4k}$
   • $R_{1||2} = \frac{8}{6} k\Omega = 1.333 k\Omega$
3. Looking in from Terminal A, $R_3$ is in series with this parallel combination:
   • $R_{th} = R_{1||2} + R_3$
   • $R_{th} = 1.333 k\Omega + 1 k\Omega$
   • $R_{th} = 2.333 k\Omega$ (or $2333 \Omega$)

Sanity Check and Independent Verification

Before moving to component selection, we must verify the math.

Order of Magnitude & Limits Check:

• Voltage: $V_{th}$ must be between 0V and 3.3V. At 2.2V, it sits correctly closer to 3.3V because $R_2$ (4k) is larger than $R_1$ (2k).
• Resistance: $R_{th}$ must be greater than $R_3$ (1k) alone, but less than $R_3$ plus either $R_1$ or $R_2$. Our result of 2.333kΩ fits perfectly between 1kΩ and 3kΩ.

How to verify the answer independently:
Attach a "test load" resistor ($R_L$) across terminals A and B equal to the calculated $R_{th}$ (2.333 kΩ). According to the voltage divider rule, the voltage across the load should drop to exactly half of $V_{th}$ (1.1V). If you build this on a breadboard and measure 1.1V across a 2.32kΩ test resistor, your Thevenin model is proven correct. For deeper theoretical verification, refer to Georgia State University's HyperPhysics guidelines on Thevenin equivalents.

Decision Path: Selecting the Buffer Op-Amp

We need to buffer this 2.2V signal for a 3.3V microcontroller ADC. The ADC requires a low-impedance source to charge its internal sample-and-hold capacitor quickly. Our $R_{th}$ of 2.333 kΩ is too high for direct connection to most 12-bit or 16-bit ADCs.

Design Constraints:

1. Supply: Must operate on a single 3.3V rail.
2. I/O: Must be Rail-to-Rail Input/Output (RRIO) to handle the 2.2V signal without clipping.
3. Input Bias Current ($I_b$): The error voltage is $V_{error} = I_b \times R_{th}$. To keep $V_{error} < 1mV$, we need $I_b < \frac{1mV}{2333\Omega} = 428 nA$.

Candidate Part	RRIO?	Max $I_b$ (nA)	$V_{error}$ at 2.33kΩ	Verdict
LM358	No	250 nA (but not RRIO)	N/A	Reject (Cannot reach 3.3V rail)
TLV2372	Yes	60,000 nA (60 µA)	140 mV	Reject ($I_b$ too high)
MCP6001	Yes	5 nA (Typical)	0.011 mV	Accept

Concrete Pick: Select the Microchip MCP6001T-I/OT (SOT-23-5 package). It operates down to 1.8V, features true RRIO, has a typical input bias current of just 1 pA (max 5 nA), and costs roughly $0.35 in single quantities. It easily satisfies the < 1mV error constraint and will drive the ADC's sample capacitor without issue.

Frequently Asked Questions

What if the example of schematic contains dependent sources?
If your schematic includes dependent sources (like a transistor model), you cannot simply turn them off to find $R_{th}$. Instead, you must apply a 1V test voltage source (or 1A test current source) at terminals A and B, calculate the resulting current (or voltage), and use Ohm's law ($R_{th} = V_{test} / I_{test}$) to find the equivalent resistance.

Does this method work for AC schematics?
Yes, but you must transition from resistance to impedance. $V_{th}$ and $R_{th}$ become phasors and complex impedances ($Z_{th}$). You will use complex algebra to combine resistors, capacitors, and inductors at your specific operating frequency before applying the same decision path for load matching.