A Wilson current mirror is a three-terminal circuit (Fig. 1) that accepts an input current at the input terminal and provides a "mirrored" current source or sink output at the output terminal. The mirrored current is a precise copy of the input current.

It may be used as a Wilson current source by applying a constant bias current to the input branch as in Fig. 2. The circuit is named after George R. Wilson, an integrated circuit design engineer who worked for Tektronix. Wilson devised this configuration in 1967 when he and Barrie Gilbert challenged each other to find an improved current mirror overnight that would use only three transistors. Wilson won the challenge.

There are three principal metrics of how well a current mirror will perform as part of a larger circuit.

An approximate analysis due to Gilbert shows how the Wilson current mirror works and why its static error should be very low. Transistors Q₁ and Q₂ in Fig. 1 are a matched pair sharing the same emitter and base potentials and therefore have ${\displaystyle \scriptstyle i_{C1}~=~i_{C2}}$ and ${\displaystyle \scriptstyle i_{B1}~=~i_{B2}}$. This is a simple two-transistor current mirror with ${\displaystyle \scriptstyle i_{E3}}$ as its input and ${\displaystyle \scriptstyle i_{C1}}$ as its output. When a current ${\displaystyle \scriptstyle i_{\text{in}}}$ is applied to the input node (the connection between the base of Q₃ and collector of Q₁), the voltage from that node to ground begins to increase. As it exceeds the voltage required to bias the emitter-base junction of Q₃, Q₃ acts as an emitter follower or common collector amplifier and the base voltage of Q₁ and Q₂ begins to rise. As this base voltage increases, current begins to flow in the collector of Q₁. All increases in voltage and current stop when the sum of the collector current of Q₁ and base current of Q₃ exactly balance ${\displaystyle \scriptstyle i_{\text{in}}}$.^{[dubious – discuss]} Under this condition all three transistors have nearly equal collector currents and therefore approximately equal base currents. Let ${\displaystyle \scriptstyle i_{B}~=~i_{B1}~=~i_{B2}~\approx ~i_{B3}}$. Then the collector current of Q₁ is ${\displaystyle \scriptstyle i_{\text{in}}\,-\,i_{B}}$; the collector current of Q₂ is exactly equal to that of Q₁ so the emitter current of Q₃ is ${\displaystyle \scriptstyle i_{E3}~=~i_{C2}\,+\,2i_{B}~=~i_{\text{in}}\,-\,i_{B}\,+\,2i_{B}~=~i_{\text{in}}\,+\,i_{B}}$. The collector current of Q₃ is its emitter current minus the base current so ${\displaystyle \scriptstyle i_{\text{out}}~=~i_{\text{in}}\,+\,i_{B}\,-\,i_{B}~=~i_{\text{in}}}$. In this approximation, the static error is zero.

A more exact formal analysis shows the expected static error. We assume:

Therefore, ${\displaystyle \scriptstyle i_{C1}~=~i_{C2}~\equiv ~i_{C}}$ and ${\displaystyle \scriptstyle i_{B1}~=~i_{B2}~\equiv ~i_{B}}$. The base current of Q₃ is given by, ${\displaystyle \scriptstyle i_{B3}~=~{\frac {i_{C3}}{\beta }}}$ and the emitter current by,

From the sum of currents at the node shared by the emitter of Q₃, the collector of Q₂ and the bases of Q₁ and Q₂, the emitter current of Q₃ must be:

Equating the expressions for ${\displaystyle \scriptstyle i_{E3}}$ in (1) and (2) gives: