Definition: A Widlar current source is a two-transistor circuit that uses an emitter degeneration resistor on the output device to generate a highly stable, low-magnitude output current from a much larger reference current.
What this topology fundamentally changes in circuit design is the ability to generate microamp-level bias currents without requiring impractically large resistors or mismatched transistor geometries. In a standard basic current mirror, the output current is forced to equal the reference current. If you need a 10 µA bias current from a 10V supply using a basic mirror, you must place a 1 MΩ resistor in the reference leg. On a silicon integrated circuit, a 1 MΩ resistor consumes massive die area; on a discrete breadboard, 1 MΩ resistors are highly susceptible to noise and leakage. The Widlar current source topology solves this by allowing a 1 mA reference current to be mathematically scaled down to 10 µA using standard, low-value resistors and identically sized transistors.
The most common confusion with this circuit is mixing it up with a Wilson current mirror. While a Wilson mirror adds a third transistor to drastically increase output impedance and reduce Early effect errors, it does not inherently scale the current ratio. The Widlar modification is specifically about current scaling via the $V_{BE}$ mismatch, whereas the Wilson modification is about output impedance enhancement.
The Core Mechanism: Scaling Current Without Giant Resistors
In a basic BJT current mirror, two identical transistors share the same base-emitter voltage ($V_{BE}$), forcing their collector currents to be equal. The Widlar modification breaks this symmetry intentionally. By inserting a resistor ($R_E$) in the emitter of the output transistor ($Q_2$), the base-emitter voltage of $Q_2$ is forced to be lower than that of the diode-connected reference transistor ($Q_1$).
The voltage dropped across $R_E$ is exactly equal to the difference in $V_{BE}$ between the two transistors. Because the bipolar transistor's collector current relates exponentially to its $V_{BE}$, a very small difference in $V_{BE}$ (typically between 50 mV and 150 mV) corresponds to a massive logarithmic ratio in current. This allows $Q_2$ to pass a tiny fraction of $Q_1$'s current while keeping $R_E$ in the practical 5 kΩ to 50 kΩ range.
Worked Numeric Example: Designing a 10 µA Source
Let’s design a discrete Widlar source to generate exactly 10 µA ($I_{out}$) from a 1 mA reference current ($I_{ref}$). We will assume an ambient temperature of 27°C (300K), which gives a thermal voltage ($V_T$) of approximately 25.85 mV. We also assume $Q_1$ and $Q_2$ are perfectly matched (identical saturation currents, $I_S$).
The governing equation for the Widlar emitter resistor is:
$$I_{out} \times R_E = V_T \times \ln\left(\frac{I_{ref}}{I_{out}}\right)$$
- Calculate the $V_{BE}$ delta: $\Delta V_{BE} = 25.85\text{ mV} \times \ln(1\text{ mA} / 10\text{ \mu A})$
- Solve the natural log: $\ln(100) \approx 4.605$
- Multiply by $V_T$: $25.85\text{ mV} \times 4.605 = 119.04\text{ mV}$
- Solve for $R_E$: $R_E = 119.04\text{ mV} / 10\text{ \mu A} = 11,904\text{ \Omega}$
Notice that to get a 100:1 current reduction, we only needed an 11.8 kΩ resistor. If we had used a basic mirror to get 10 µA from a 10V rail, we would have needed a 1 MΩ reference resistor. The Widlar topology saves silicon area in ICs and avoids high-impedance noise pickup in discrete builds.
Where You Meet the Widlar Current Source in Practice
You interact with this circuit every time you use a classic operational amplifier. The most famous implementation is inside the LM741 op-amp. The input differential pair of the 741 requires a highly stable tail current of roughly 19 µA to set its transconductance and slew rate.
Inside the 741 die, transistors Q10 and Q11 form a Widlar source. Q11 is diode-connected and fed by the main bias network (providing roughly 730 µA). Q10 has a 5 kΩ resistor in its emitter. If we run the math: $26\text{ mV} \times \ln(730\text{ \mu A} / I_{out}) = I_{out} \times 5\text{ k\Omega}$. Solving this iteratively yields an $I_{out}$ of approximately 18.9 µA. This precise microamp tail current is what gives the 741 its predictable input bias characteristics.
In modern discrete designs, you will use this topology when biasing high-impedance sensors. For example, biasing a photodiode in photoconductive mode or setting the quiescent current for a low-power piezo preamplifier often requires exactly 5 µA to 20 µA. A Widlar source powered by a low-noise LDO provides a vastly cleaner, lower-impedance bias point than a simple high-value resistor tied to the positive rail.
Decision Tree: Which Current Mirror Topology to Pick
Choosing the right current mirror depends on your headroom, scaling needs, and output impedance requirements. Use this matrix to terminate your design choice.
| Topology | Current Scaling | Output Impedance | Voltage Headroom Required | Best Use Case |
|---|---|---|---|---|
| Basic Mirror | 1:1 only | Low ($r_o$) | Low ($V_{CE(sat)}$) | Simple level shifting, matched 1:1 biasing where Early effect is negligible. |
| Widlar Source | High (10:1 to 1000:1) | Medium (boosted by $R_E$) | Medium ($V_{CE(sat)} + I_{out}R_E$) | Generating microamp bias currents from milliamp references. |
| Wilson Mirror | 1:1 only | Very High ($\beta r_o / 2$) | High ($2 \times V_{BE}$) | Active loads in high-gain amplifier stages where output impedance dictates gain. |
| Cascode Mirror | 1:1 (or scaled) | Extremely High ($g_m r_o^2$) | Very High ($V_{CE(sat)} + V_{BE}$) | Precision current sources in low-voltage IC design (using wide-swing variants). |
Practical Build Notes and Discrete Failure Modes
When moving this circuit from SPICE simulation to a physical breadboard or PCB, the assumption that $Q_1$ and $Q_2$ are perfectly matched falls apart. The governing equation actually includes the mismatch in base-emitter voltages ($\Delta V_{BE(mismatch)}$):
$$I_{out} \times R_E = V_T \times \ln\left(\frac{I_{ref}}{I_{out}}\right) \pm \Delta V_{BE(mismatch)}$$
If you use two random 2N3904 transistors from different manufacturing batches, their $V_{BE}$ might differ by 5 mV to 10 mV at the same current. In our 10 µA design, a mere 2 mV mismatch translates to a current error of $2\text{ mV} / 11.8\text{ k\Omega} = 169\text{ nA}$. While 169 nA sounds small, it represents a 1.6% error on a 10 µA source. If you are biasing a sensitive transimpedance amplifier, this error will manifest as a massive DC offset at the output.
The Fix: Never use discrete, single-packaged transistors for a precision Widlar source. Always use a monolithic dual matched transistor pair like the BCM847 matched dual NPN or the MAT12. Because both dies are fabricated on the same silicon substrate, their $V_{BE}$ mismatch is typically guaranteed to be less than 1 mV, and their thermal coupling is practically perfect.
Thermal Runaway Risk: The output transistor ($Q_2$) in a Widlar source has a negative temperature coefficient regarding its $V_{BE}$. As $Q_2$ heats up, its $V_{BE}$ drops, which forces more voltage across $R_E$, increasing $I_{out}$, which causes more heating. Because $I_{out}$ is in the microamp range, self-heating is rarely an issue. However, if $Q_2$ is placed near a hot power transistor on the PCB, the ambient heat will shift your bias current. Keep the matched pair thermally isolated from heat sinks and power stages.
Frequently Asked Questions
Can I use MOSFETs instead of BJTs for a Widlar current source?
Yes, but the math changes entirely. BJTs rely on the exponential Shockley diode equation, meaning a tiny $\Delta V_{BE}$ creates a massive current ratio. MOSFETs operate on a square-law (in saturation) or linear (in subthreshold) relationship. To get a 100:1 current scaling with MOSFETs, you typically just scale the $W/L$ (width-to-length) ratio of the transistor geometries on the silicon die rather than using a source degeneration resistor. In discrete MOSFET designs, matching $V_{GS(th)}$ is notoriously poor, making discrete MOSFET Widlar-style sources highly inaccurate.
Does the Early effect ruin the output impedance of a Widlar source?
Actually, the emitter resistor $R_E$ improves the output impedance compared to a basic mirror. The output impedance of a Widlar source is approximately $R_{out} \approx r_o \times (1 + g_m R_E)$, where $r_o$ is the transistor's intrinsic Early resistance. The local negative feedback provided by $R_E$ makes the output current much stiffer against changes in the output voltage ($V_{CE}$ of $Q_2$) than a basic diode-connected mirror.
What happens if my reference current ($I_{ref}$) drifts?
The Widlar source is a ratio-based circuit, not an absolute regulator. If your reference current drifts by 5% due to supply voltage ripple or a drifting reference resistor, your output microamp current will also drift by roughly 5%. For absolute precision, $I_{ref}$ must be generated by a dedicated bandgap reference IC or a precision voltage reference driving a stable resistor, rather than just a simple resistor tied to an unregulated $V_{CC}$ rail.
