Current Mirror

Part of The Transistor

A fundamental analog circuit that copies a reference current into one or more output branches with high accuracy.

Why This Matters

The current mirror is one of the most important circuit building blocks in analog electronics. Given a reference current, it produces an equal (or scaled) current in a separate branch, independent of that branch’s voltage. This enables biasing multiple circuit stages from a single reference, matching currents in differential pairs, and building precision current sources for test equipment and sensors.

Discrete transistor circuits in a rebuilding context use current mirrors extensively: to bias audio amplifiers, to provide constant-current loads that increase amplifier gain, and to maintain stable operating points over temperature. The current mirror is also the basis of the current source, which replaces resistive collector loads with dramatically higher effective impedance, enabling much higher voltage gain from a single transistor stage.

Understanding current mirrors requires grasping the fundamental relationship: two transistors with identical VBE carry identical collector currents regardless of their collector voltages (within the active region). This is a direct consequence of the exponential I-V relationship of the base-emitter junction.

Basic Current Mirror

The simplest current mirror uses two transistors:

Reference transistor (Q1): Connected with collector tied to base (diode-connected). Base and collector are at the same voltage. The reference current IREF flows through Q1 (supplied by a resistor from Vcc, or any current source).

Mirror transistor (Q2): Base connected to the base of Q1 (same VBE). Collector is the current output.

Operation: Because Q1 and Q2 have the same VBE (their bases are connected), and if both transistors are identical (same IS, same hFE, at same temperature), they carry the same collector current:

IC1 = IC2 = IS × e^(VBE/VT)

The current in Q2’s collector mirrors the current in Q1’s collector, hence “current mirror.”

Design: IREF = (Vcc - VBE) / R

For Vcc = 9V, VBE = 0.7V, desired IREF = 1 mA: R = (9 - 0.7) / 0.001 = 8.3 kΩ → use 8.2 kΩ. IOUT ≈ IREF = 1 mA, flowing through any load connected to Q2’s collector.

Error analysis: In the basic mirror, each transistor’s base draws base current IB = IC/hFE. Q1’s collector current is IREF minus two base currents (one for Q1, one for Q2). So the actual IC1 = IREF - 2×IB. Q2 mirrors IC1, not IREF:

IOUT = IC2 = IC1 = IREF - 2 × (IREF / hFE) ≈ IREF × (1 - 2/hFE)

For hFE = 100: IOUT = IREF × 0.98 (2% error). For hFE = 20: 10% error. For precision mirrors, this error must be reduced — the Widlar mirror or Wilson mirror provide corrections.

Widlar Current Mirror

The Widlar mirror produces an output current smaller than the reference, using an emitter resistor in the output transistor.

Circuit: same as basic mirror, but add RE2 in series with Q2’s emitter. Q1 remains diode-connected with emitter to ground.

The emitter resistor RE2 creates a voltage VE2 = IC2 × RE2, which reduces VBE2 relative to VBE1. A lower VBE means less collector current. The output current satisfies:

VBE1 - VBE2 = IC2 × RE2 VT × ln(IC2/IC1) = -IC2 × RE2 (since VBE2 < VBE1)

This transcendental equation is solved numerically or graphically for IC2.

For IREF = 1 mA and RE2 = 26 kΩ: IOUT ≈ 10 µA (100:1 current reduction). The Widlar mirror is ideal when a small reference current is needed from a circuit that can only practically generate a much larger reference.

Wilson Current Mirror

The Wilson mirror dramatically reduces the effect of finite hFE:

Circuit: three transistors. Q1 diode-connected to ground. Q2 diode-connected in feedback path. Q3 is the output transistor. The feedback ensures that Q3’s base sees correct voltage to mirror the reference.

Error for Wilson mirror: IOUT = IREF × (1 - 2/hFE²) — the error is divided by hFE again. For hFE = 100: error is 0.02% instead of 2%. For precision analog circuits where 1% matching is needed, the Wilson mirror is essential.

This represents the progression from simple to more sophisticated: basic mirror (2% error at hFE=100), Widlar (still 2% but enables current scaling), Wilson (0.02% error). Choose the complexity level that matches your application’s precision requirement.

Current Mirror as Constant Current Load

In a common-emitter amplifier, the voltage gain is Av = gm × (RC || ro). With a resistive RC, ro >> RC and Av ≈ gm × RC.

Replace RC with a current mirror output as the load: the current mirror maintains constant IC regardless of VCE (within active region). The effective load resistance is ro of the mirror transistor — typically 50-100 kΩ. Now Av = gm × (ro_mirror || ro_transistor) ≈ gm × (50 kΩ) = 0.038 × 50,000 = 1900 at IC = 1 mA.

Replacing a 4.7 kΩ resistor with a current mirror load increases voltage gain from ~180 to ~1900 — a 10× improvement from the same transistor. This is why operational amplifier first stages use current mirror loads: they achieve very high gain in a single stage.

PNP current mirror as load for NPN: The PNP current mirror provides the load for the NPN differential pair in most op-amp input stages. The PNP mirror’s high output impedance lets the NPN transistors develop large signal swing without the gain-limiting effect of a resistive load.

Practical Construction and Matching

For a basic current mirror to work well, Q1 and Q2 must be matched:

• Same type: both NPN or both PNP — the basic mirror is single-polarity
• Similar IS: both from same batch, same die if possible. IS varies with die area and process
• Same temperature: thermal coupling is critical. Mount on same heatsink or use a dual transistor package (two matched transistors in one package), which are physically adjacent.

Testing a current mirror:

1. Build the reference leg: Q1 (diode-connected), series resistor R from Vcc to Q1 collector.
2. Connect Q2 base to Q1 base. Q2 collector is the output.
3. Connect 100 Ω × 100V = 10 kΩ resistor from Vcc to Q2 collector (provides load and limits voltage).
4. Measure: (a) voltage across R (compute IREF), (b) voltage at Q2 collector versus Vcc (compute IOUT from 10 kΩ resistor).
5. IOUT / IREF should be within 5-10% for a basic mirror with hFE = 100-200 transistors.

Scaling: A current mirror with Q2 having 2× the emitter area of Q1 provides IOUT = 2 × IREF. In discrete circuits, replace Q2 with two transistors connected in parallel — effectively doubling the emitter area. Use this to create biasing networks where one reference current sets multiple output stages at scaled currents.