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Widlar emitter resistance for 10 μA

Short answer

Requirements: matched, same-temperature, equal-VBE, active-region, high-β transistors. IO≠IREF due to finite β, Early effect, mismatch. Using RE = (VT/IO)·ln(IREF/IO) with VT=26mV, IO=10μA, IREF=1mA, RE ≈ 11.97 kΩ.

What are the basic requirements for current mirror circuit? Why is the output current of simple current mirror circuit not exactly equal to the input reference current? Determine the emitter resistance, RE for Widlar current source which supplies 10 μA from a reference current of 1 mA.

Basic requirements for a current mirror

  • Matched transistors: Q1 and Q2 must have the same IS (same VBE–IC curve), same β and same Early voltage. This is easy in an IC because both are made in the same process steps, side by side.
  • Same temperature: both transistors must be at the same temperature (close together on one chip), so that VBE changes track each other.
  • Equal VBE: the bases are tied together and the emitters are tied together (or returned to the same rail), so VBE1 = VBE2 and therefore IC2 = IC1.
  • Output transistor in the active region: Q2 must have VCE2 > VCE(sat), i.e. the load must leave enough voltage across Q2.
  • High β: so that the base currents taken from the reference branch are negligible.

Why the output current is not exactly equal to IREF

  • Finite β: IREF supplies IC1 plus both base currents, so IO = IREF/(1 + 2/β), slightly less than IREF.
  • Early effect: VCE2 is larger than VCE1 = VBE, so IO = IC1(1 + (VCE2 − VBE)/VA), slightly more than IC1 and dependent on the load voltage.
  • Mismatch in IS (area, doping) and temperature between Q1 and Q2.

Emitter resistance for the Widlar source

Given IREF = 1 mA, IO = 10 µA. Take VT = 26 mV (room temperature, about 27 °C) and neglect base currents.

  +VCC                 IO (from load)
    |                    |
    R                    |
    | IREF               |
    +------+             |
    |      |             |
    C1     |             C2
  Q1  B----+-------------B  Q2
    E1                   E2
    |                    |
    |                    RE
    |                    |
   GND                  GND

VBE1 - VBE2 = IO·RE
VT ln(IREF/IS) - VT ln(IO/IS) = IO·RE
RE = (VT/IO) · ln(IREF/IO)
   = (0.026 / 10×10⁻⁶) · ln(1 mA / 10 µA)
   = 2600 × ln(100)
   = 2600 × 4.6052
   = 11 973 Ω

Answer: RE ≈ 11.97 kΩ (about 12 kΩ). If VT = 25 mV is used, RE = 2500 × 4.6052 ≈ 11.51 kΩ. A simple mirror would need a resistor of about 1.4 MΩ (for VCC = 15 V) for the same 10 µA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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