Circuit Operation
There are three principal metrics of how well a current mirror will perform as part of a larger circuit. The first measure is the static error, the difference between the input and output currents expressed as a fraction of the input current. Minimizing this difference is critical in such applications of a current mirror as the differential to single-ended output signal conversion in a differential amplifier stage because this difference controls the common mode and power supply rejection ratios. The second measure is the output impedance of the current source or equivalently its inverse, the output conductance. This impedance affects stage gain when a current source is used as an active load and affects common mode gain when the source provides the tail current of a differential pair. The last metric is the pair of minimum voltages from the common terminal, usually a power rail connection, to the input and output terminals that are required for proper operation of the circuit. These voltages affect the headroom to the power supply rails that is available for the circuitry in which the current mirror is embedded.
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 and . This is a simple two-transistor current mirror with as its input and as its output. When a current 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 . Under this condition all three transistors have nearly equal collector currents and therefore approximately equal base currents. Let . Then the collector current of Q1 is ; the collector current of Q2 is exactly equal to that of Q1 so the emitter current of Q3 is . The collector current of Q3 is its emitter current minus the base current so . In this approximation, the static error is zero.
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