In electrical control theory, an open system (more commonly called an open-loop system) is a configuration where the control action is entirely independent of the output. The most practical, bench-level example of an open system is a basic PWM-driven brushed DC motor without an encoder or tachometer for feedback. You set the input voltage, and the motor spins—but the controller has absolutely no idea what the shaft is actually doing.
Understanding how to model and calculate the behavior of open systems is a staple of engineering exams (like the FE or PE) and a critical debugging skill on the workbench. Below, we will walk through a classic exam problem, show every algebraic step, expose the common trap students fall into, and verify the result using power conservation.
The Practice Problem: Open-Loop DC Motor Under Load
A 24V brushed DC motor is driven by an open-loop PWM controller set to 100% duty cycle (effectively applying a constant terminal voltage Vt = 24V DC). The motor has an armature resistance Ra = 0.8 Ω and a torque/back-EMF constant Kt = Kv = 0.1 (in SI units: Nm/A and V/(rad/s)). Calculate the steady-state shaft speed (ω) in rad/s when a constant mechanical load torque of TL = 1.0 Nm is applied to the shaft.
Which theorem/method applies and why?
To solve this, we must use Kirchhoff’s Voltage Law (KVL) for the electrical armature loop and Newton’s Second Law for Rotation (torque balance) for the mechanical side. Because it is an open system, we cannot rely on a feedback controller to adjust Vt; the terminal voltage is fixed, meaning any mechanical load will directly alter the electrical current and resulting speed.
Step-by-Step Algebraic Solution
We will break the solution down into three distinct algebraic steps. No skipped math.
- Step 1: Find the Armature Current (Ia) using Torque Balance
In steady-state, the electromagnetic torque produced by the motor (Te) must exactly equal the applied mechanical load torque (TL).
Te = TL
Kt × Ia = 1.0 Nm
0.1 × Ia = 1.0
Ia = 10 A - Step 2: Find the Back-EMF (Eb) using KVL
Apply KVL around the armature circuit. The terminal voltage must equal the sum of the back-EMF and the voltage drop across the armature resistance.
Vt = Eb + (Ia × Ra)
24V = Eb + (10 A × 0.8 Ω)
24V = Eb + 8V
Eb = 24V - 8V
Eb = 16 V - Step 3: Calculate the Shaft Speed (ω)
The back-EMF is directly proportional to the shaft speed via the voltage constant Kv.
Eb = Kv × ω
16 V = 0.1 V/(rad/s) × ω
ω = 16 / 0.1
ω = 160 rad/s
Because this is an open system, the controller cannot detect if the motor stalls. If the load torque exceeds the motor's stall torque, the shaft stops (ω = 0), meaning Eb = 0. The current becomes Ia = Vt / Ra = 24V / 0.8Ω = 30A. The power dissipated as heat in the windings becomes I2R = 720W. Without closed-loop thermal or current feedback, the motor will rapidly melt its insulation and fail.
Sanity Check, The Exam Trap, and Independent Verification
The Exam Trap:
The most common mistake students make on this problem is calculating the no-load speed and assuming it remains constant. If TL = 0, then Ia = 0, Eb = 24V, and ω = 240 rad/s. Students see the 24V supply hasn't changed and incorrectly answer 240 rad/s. The defining flaw of an open system is that it lacks the feedback mechanism required to compensate for the IaRa voltage drop caused by the load. The speed must drop to generate the back-EMF difference that drives the required current.
Answer Sanity Check:
Our answer is 160 rad/s. Converting to RPM: 160 × (60 / 2π) ≈ 1528 RPM. A drop from a no-load speed of ~2292 RPM to 1528 RPM under a 1.0 Nm load is a highly realistic order of magnitude for a small 24V fractional-horsepower DC motor. The units (rad/s) match the SI constants provided.
Independent Verification (Power Conservation):
We can verify our answer by ensuring electrical power in equals mechanical power out plus resistive losses.
Pin = Vt × Ia = 24V × 10A = 240 W
Ploss = Ia2 × Ra = (10A)2 × 0.8Ω = 80 W
Pmech = TL × ω = 1.0 Nm × 160 rad/s = 160 W
Does Pin = Ploss + Pmech? Yes: 240 W = 80 W + 160 W. The math is bulletproof. For deeper reading on DC motor modeling, refer to the Electronics Tutorials DC Motor guide or MIT OpenCourseWare Dynamics and Control II.
| Criteria | Open System (Open-Loop) | Closed System (Closed-Loop) |
|---|---|---|
| Feedback Mechanism | None (Blind to output) | Encoder, tachometer, or current sensor |
| Speed Regulation under Load | Poor (Speed drops as load increases) | Excellent (Controller increases voltage to compensate) |
| Component Count & Cost | Low (Just a MOSFET and basic PWM) | High (Requires H-bridge, MCU, sensors, PID tuning) |
| Stability Risk | Inherently stable (cannot oscillate) | Can oscillate or hunt if PID gains are poorly tuned |
FAQ: Long-Tail Questions on Open Systems
What is an example of an open system in thermodynamics vs electrical engineering?
In thermodynamics, an open system (control volume) is one where both mass and energy cross the boundary—like a liquid cooling loop pumping water through an inverter cabinet. In electrical engineering and control theory, an open system refers to an open-loop control circuit where signal energy flows strictly from input to output without feedback. While the physics definitions differ, the core concept is the same: the system does not self-correct based on its own output state.
Why does an open-loop system fail to maintain constant speed under varying loads?
It fails because the input command (e.g., a fixed 24V DC supply) is completely decoupled from the physical reality of the shaft. As mechanical load increases, the motor slows down slightly, which reduces the back-EMF. This lower back-EMF allows more current to flow from the fixed voltage source, generating more torque to match the load. However, that speed drop is permanent as long as the load remains. Without a sensor to tell a controller to boost the terminal voltage above 24V, the system cannot recover the lost RPM.
Is a simple relay switching circuit considered an open system?
Yes. A standard electromechanical relay driving a resistive heater is a classic open system. The control signal (coil voltage) dictates the action (contacts closing), but the circuit has no mechanism to measure the actual temperature of the heater or the current flowing through the contacts to adjust its behavior. It simply executes the command blindly.
What happens to an open-loop DC motor if the mechanical load exceeds the stall torque?
The motor will stall (shaft speed drops to zero). Because it is an open system, the controller continues to apply full terminal voltage. With zero speed, back-EMF drops to zero, and the only thing limiting the current is the very low armature resistance. This results in a massive current spike (stall current), converting almost all input electrical power directly into heat (I2R losses) inside the motor windings, which will quickly lead to thermal destruction unless an external fuse or breaker interrupts the circuit.
