Current mirror output current, finite β effects

Current mirror circuit

A current mirror copies a reference current to an output branch using two matched transistors with equal VBE. Q1 is diode-connected (collector shorted to base).

  +VCC                 IO (from load)
    |                    |
    R                    |
    | IREF               |
    +------+             |
    |      |             |
    C1     |             C2
  Q1  B----+-------------B  Q2
    E1   (Q1: C tied to B)   E2
    |                    |
   GND                  GND

Expression for output current

Q1 and Q2 are matched and have the same VBE, so IC1 = IC2 = IC and IB1 = IB2 = IC/β.

IREF = (VCC - VBE)/R
KCL at the collector of Q1:
IREF = IC1 + IB1 + IB2 = IC + 2IC/β
IO   = IC2 = IREF / (1 + 2/β) = IREF·β/(β + 2)

For β → ∞, IO = IREF = (VCC − VBE)/R.

Effect of finite β

Part of IREF is used to supply the base currents of both transistors, so IO is smaller than IREF:

Error = (IREF - IO)/IREF = 2/(β + 2)
β = 100  → IO = 0.980 IREF (2% error)
β = 50   → IO = 0.962 IREF (3.8% error)

The error also changes with temperature because β does. It is reduced by a buffered mirror (emitter-follower Q3 supplies the base currents, IO = IREF/(1 + 2/β²)) or a Wilson mirror.

Effect of finite output resistance

Because of the Early effect, IC2 depends on VCE2. Q1 has VCE1 = VBE, but VCE2 is set by the load:

IO = IREF/(1 + 2/β) · [1 + (VCE2 - VBE)/VA]
Output resistance  Ro = ΔVCE2/ΔIO = ro = VA/IO

So IO is not constant; it rises as the output voltage rises, and the source is not ideal (ideal Ro = ∞). For VA = 100 V and IO = 1 mA, ro = 100 kΩ, so a 10 V change in VCE2 changes IO by about 10%. Widlar, cascode and Wilson sources raise Ro to reduce this effect.