A voltage source is an active circuit element that maintains a specific potential difference across its terminals regardless of the current drawn by the load, up to its physical limits. In a real circuit or installation, it changes the mathematical hierarchy of the system: voltage becomes the independent variable, and the load’s resistance dictates the current flow via Ohm’s Law, rather than the circuit dictating the voltage. Beginners commonly confuse it with a current source (which maintains constant current regardless of voltage) or mistakenly assume all voltage sources are "ideal" with zero internal resistance.
Think of it like a municipal water tower: the height of the water (voltage) creates a constant pressure at the tap, regardless of whether you open the valve a little or a lot (the load), until the supply pipes max out their flow capacity. But as any seasoned maker or electrician knows, real-world pipes have friction, and real-world voltage sources have internal resistance.
The Ideal vs. Real Voltage Source (And Why Internal Resistance Matters)
In textbook circuit theory, an ideal voltage source has zero internal resistance. If you short-circuit it, it delivers infinite current. If you draw 100A from a 12V ideal source, the terminals remain at exactly 12.000V. This is a mathematical fiction used to simplify Kirchhoff's Voltage Law (KVL) calculations.
A real voltage source always possesses some internal resistance (Rint), which acts as a resistor in series with the ideal source. As current increases, the voltage drop across this internal resistance increases, causing the terminal voltage to sag. According to SparkFun's Ohm's Law tutorials, the terminal voltage ($V_{term}$) is calculated as:
V_term = V_source - (I_load × R_int)
Imagine a 12V lead-acid battery with an internal resistance of 0.05Ω. You connect a 1.2Ω heater element.
• Ideal assumption: I = 12V / 1.2Ω = 10A.
• Real calculation: Total resistance = 1.2Ω + 0.05Ω = 1.25Ω. Real current = 12V / 1.25Ω = 9.6A.
• Terminal Voltage: 12V - (9.6A × 0.05Ω) = 11.52V.
That 0.48V of sag is the energy lost as heat inside the battery itself.
Where You Meet This in Practice
You interact with real voltage sources every time you power a project or wire a branch circuit. Here is how they manifest on the bench and the jobsite:
- Bench Power Supplies (e.g., Rigol DP832): These operate in Constant Voltage (CV) mode up to a set current limit. If your circuit tries to pull more current than the limit, the supply switches to Constant Current (CC) mode, and the voltage collapses to protect the output transistors.
- Lithium Batteries (18650 / LiFePO4): A high-quality 18650 cell (like the Sony VTC6) has an Rint of roughly 12mΩ when fully charged, but this can spike above 50mΩ as it ages or drops below 3.2V, causing severe voltage sag under high-drain loads like drone motors.
- The Mains Grid: Your wall outlet is a massive 120V/230V AC voltage source. However, the transformer impedance and the AWG of your branch circuit wiring create a combined source impedance. This is why the lights dim momentarily when a 15A table saw motor starts—the source impedance causes a temporary voltage drop.
Real-World Scenario: The 12V Compressor Voltage Sag
To see why treating a real voltage source as an ideal one leads to failure, let's look at a common off-grid DIY mistake involving a 12V DC compressor fridge.
- The Setup: A hobbyist wires a 12V 60W compressor fridge to a 12V 100Ah AGM battery using 10 feet of 16 AWG copper wire. The battery is fully charged at 12.8V.
- The Numbers: The fridge draws a steady 5A while running. However, the compressor motor has a startup inrush (Locked Rotor Amperage) of 25A for about 0.5 seconds. The 16 AWG wire has a resistance of 0.004016Ω/ft. For a 20-foot round trip, wire resistance is 0.08Ω. The AGM battery's Rint is roughly 0.02Ω. Total series source resistance = 0.10Ω.
- The Outcome: When the compressor tries to start, it pulls 25A. The voltage drop across the source resistance is 25A × 0.10Ω = 2.5V. The voltage actually reaching the fridge terminals drops to 10.3V (12.8V - 2.5V).
- What Went Wrong: Most 12V compressor fridges have a Low Voltage Disconnect (LVD) set at 10.5V to protect the battery from deep discharge. Because the terminal voltage sagged to 10.3V during inrush, the fridge's internal controller thought the battery was dead and shut off. It then tried to restart 30 seconds later, sagged again, and entered an endless locked-rotor loop, eventually burning out the start relay.
The Fix: Upgrade the wiring to 10 AWG (reducing wire resistance to ~0.02Ω) and ensure the battery is a low-impedance LiFePO4 chemistry rather than AGM. This keeps the inrush sag under 1.0V, keeping the terminal voltage above the 10.5V LVD threshold.
Voltage Source vs. Current Source: Clearing the Confusion
As detailed in Analog Devices' application notes, confusing a voltage source with a current source will result in destroyed components. A current source varies its output voltage to force a specific current through a load, whereas a voltage source varies its output current to maintain a specific voltage.
| Feature | Voltage Source (e.g., USB Charger, Battery) | Current Source (e.g., LED Driver, 4-20mA Loop) |
|---|---|---|
| Independent Variable | Voltage (V) | Current (I) |
| Dependent Variable | Current (dictated by load resistance) | Voltage (dictated by load resistance) |
| Open Circuit Behavior | Outputs rated voltage, 0A current. | Voltage rails to maximum compliance limit to try and push current. |
| Short Circuit Behavior | Current spikes to maximum capacity (trips protection). | Outputs 0V, maintains rated current safely. |
| Common Application | Powering microcontrollers (ESP32, Arduino), logic ICs, motors. | Driving high-power LEDs, industrial sensor loops, battery charging (CC phase). |
Frequently Asked Questions
Can a voltage source absorb power instead of delivering it?
Yes. A real voltage source can absorb power if current is forced backward through it by a higher potential. This is exactly what happens when you charge a battery or when a DC motor undergoes regenerative braking, feeding energy back into the supply bus. If the source cannot absorb this energy (like a standard linear bench supply), the bus voltage will rise uncontrollably unless a bleeder resistor or active clamp is used.
Why does my bench power supply show a lower voltage than I set when I connect my circuit?
Your bench supply is likely hitting its current limit and transitioning from Constant Voltage (CV) to Constant Current (CC) mode. Alternatively, you are measuring the voltage at the load end of thin jumper wires. The supply might be outputting exactly 5.00V at its binding posts, but the resistance of 22 AWG breadboard wires carrying 500mA will drop the voltage to ~4.7V at the microcontroller's VCC pin.
Is the electrical grid a voltage source or a current source?
The grid is a voltage source. The utility maintains a nominal 120V/230V RMS at the service entrance. However, during extreme peak demand or brownouts, the sheer magnitude of the collective load causes voltage sag across the utility's transformer impedance, dropping your wall voltage to 110V or lower. It remains a voltage source, just one with a highly variable and sometimes significant internal impedance.
