The internal resistance of a practical current source is the high-value parallel resistance that dictates how much the output current drops as the load voltage increases. When you design or troubleshoot circuits relying on constant current—like LED drivers, battery charging circuits, or 4-20mA industrial sensor loops—this hidden parallel path is what determines whether your source acts "stiff" (unyielding) or "soft" (sagging under load). Understanding this parameter is the difference between a precision instrument and a drifting mess.
The Core Concept: Parallel Resistance and Source Stiffness
In textbook theory, an ideal current source delivers the exact same amperage regardless of the load connected to it. To achieve this mathematically, an ideal source must have an infinite internal resistance. In the real world, infinity doesn't exist on a workbench. Every practical current source—whether built from discrete transistors, an integrated circuit like the LM334, or a programmable bench supply—has a finite, measurable internal resistance wired in parallel with the ideal source.
This brings us to the most common point of confusion on the workbench: people constantly confuse the resistance of a current source with the internal resistance of a voltage source. A voltage source (like a 12V lead-acid battery or a bench supply in CV mode) has a series internal resistance that you want to be as close to 0Ω as possible to prevent voltage sag. A current source, conversely, has a parallel internal resistance that you want to be as close to ∞Ω as possible to prevent current leakage.
What it changes in a real circuit: This parallel resistance changes the actual current delivered to your load. As your load resistance increases, the voltage across the load must also increase to maintain the same current (Ohm's Law). That rising voltage pushes more current through the source's internal parallel resistance and away from the load.
The Water Analogy: Imagine a pump pushing a fixed 10 gallons per minute through a pipe, but the pipe has a microscopic pinhole leak (the internal resistance). As backpressure (load voltage) rises, more water escapes through the pinhole, leaving less for the nozzle (the load). The smaller the pinhole (higher resistance), the stiffer the source.
Real-World Values: Internal Resistance Across Source Types
Not all current sources are created equal. The "stiffness" of the source depends heavily on the topology and the components used. Below is a reference table of typical internal parallel resistance values you will encounter in actual designs and standard circuit topologies.
| Source Type / Topology | Typical Internal Parallel Resistance | Max Load Voltage (Compliance) | Stiffness Rating |
|---|---|---|---|
| Ideal Mathematical Source | ∞ Ω (Infinite) | ∞ V | Perfect |
| Discrete BJT Current Mirror (e.g., 2N3904 pair) | ~50kΩ to 100kΩ | VCC - 1.5V | Soft (Limited by Early Effect) |
| Integrated 2-Terminal Source (e.g., LM334 / LM334A) | ~10MΩ to 20MΩ | ~20V (depending on power dissipation) | Very Stiff |
| Industrial 4-20mA Transmitter (e.g., Rosemount 3051) | >100MΩ | Up to 45V (Loop powered) | Ultra-Stiff |
| Bench Power Supply (CC Mode, e.g., Rigol DP832) | >50MΩ (Dynamic, varies by range) | Up to rated max voltage (e.g., 30V) | Stiff (until compliance limit) |
Worked Numeric Example: Calculating Load Current Drop
Let's put numbers to the theory. Suppose you are using a practical current source configured to deliver 20.0 mA. Based on its datasheet, it has an internal parallel resistance ($R_p$) of 500 kΩ. We will calculate the actual current delivered to the load ($I_L$) under two different load conditions using the current divider rule: $I_L = I_{total} \times \frac{R_p}{R_p + R_L}$.
Scenario A: Low Resistance Load ($R_L = 100\Omega$)
With a 100Ω load, the voltage across the load is only 2V (if ideal). Let's see what the parallel leakage does.
- $I_L = 20\text{ mA} \times \frac{500,000}{500,000 + 100}$
- $I_L = 20\text{ mA} \times 0.9998$
- Actual Load Current = 19.996 mA
The error is a negligible 0.004 mA. The source is exceptionally stiff at this load.
Scenario B: High Resistance Load ($R_L = 5,000\Omega$)
Now we swap in a 5kΩ load. The ideal voltage would be 100V, but let's assume our source has the compliance voltage headroom to support it.
- $I_L = 20\text{ mA} \times \frac{500,000}{500,000 + 5,000}$
- $I_L = 20\text{ mA} \times 0.9901$
- Actual Load Current = 19.802 mA
The current has dropped by nearly 0.2 mA. While this might be acceptable for charging a battery, in a precision 4-20mA instrumentation loop, a 1% error (0.2mA out of 20mA) translates to a significant measurement offset on your PLC's analog input card.
Where You Meet This in Practice
You rarely sit down and calculate parallel leakage for a simple LED blinker, but in professional and industrial electronics, the resistance of a current source is a critical design parameter.
1. Industrial 4-20mA Sensor Loops
In process control, sensors transmit data by varying their current draw between 4mA and 20mA. The loop can have hundreds of feet of wire, plus the input impedance of the PLC's analog-to-digital converter. If the transmitter's internal parallel resistance isn't in the tens of megaohms, the changing voltage drop across the long wire run will cause the current to sag, resulting in false temperature or pressure readings at the control room. This is why industrial transmitters use active feedback loops with operational amplifiers to artificially boost their output impedance well past 100MΩ.
2. Transistor Biasing and Current Mirrors
In analog integrated circuit design (and discrete audio amplifiers), current mirrors are used to bias differential pairs. A basic two-transistor BJT mirror suffers from the "Early Effect," which effectively lowers its internal parallel resistance to perhaps 50kΩ. As the output voltage swings, the bias current shifts, causing distortion. Designers solve this by adding emitter degeneration resistors or using Wilson current mirror topologies to push that internal resistance into the megaohm range, ensuring the bias current remains rock-solid regardless of signal swings.
3. Lithium-Ion Battery Charging (Constant Current Phase)
During the CC phase of a Li-ion charge cycle, the charger acts as a current source. As the battery's state of charge (SoC) rises, its terminal voltage increases from ~3.0V to 4.2V. A cheap, poorly designed switching charger with low internal parallel resistance will actually deliver less current as the battery voltage rises, drastically increasing charge times. High-quality BMS and charge controllers (like those based on the TI bq24xxx series) use tight feedback loops to maintain a stiff current source right up until the 4.2V transition to Constant Voltage (CV) mode.
Compliance Voltage Limit: No matter how high the internal parallel resistance is, a practical current source will fail if the load resistance demands more voltage than the source can provide. This ceiling is called the compliance voltage. If you try to push 20mA through a 10kΩ load using a 12V-powered LM334, the source will saturate, the internal resistance effectively collapses, and the current will plummet. Always verify that $I_{set} \times R_{load(max)}$ is at least 2V to 3V below your supply rail.
Frequently Asked Questions
Can I measure a current source's internal resistance with my multimeter?
No. You cannot simply put a DMM in resistance mode across the terminals of a current source and get a valid reading. The internal parallel resistance is a dynamic, small-signal AC resistance (often denoted as $r_o$ in transistor models), not a static DC resistance. To measure it, you must apply a small AC signal or measure the DC current at two different load voltages and calculate the slope ($\Delta V / \Delta I$).
Why do schematics sometimes show a resistor in parallel with an ideal current source?
When engineers draw a Norton equivalent circuit, they explicitly draw an ideal current source in parallel with a resistor to model the real-world behavior of the physical component. That drawn resistor is the internal resistance we've been discussing. It's a modeling tool to remind you that as the voltage across the terminals rises, some current will bypass the load and flow through that internal resistor.
Does temperature affect the internal resistance of a current source?
Absolutely. In discrete transistor current mirrors, the output resistance is heavily dependent on the junction temperature and the Early voltage, which shifts with thermal runaway. In integrated solutions like the LM334, the internal trimming resistors and bandgap references have temperature coefficients (typically 50 to 100 ppm/°C). If your application operates in a harsh environment (e.g., an under-hood automotive sensor), you must account for thermal drift in both the set current and the effective source stiffness.
