Benefits of Widlar source and output resistance

Benefits of the Widlar source over a simple mirror

In a simple mirror IO ≈ IREF = (VCC − VBE)/R, so a small current needs a huge resistor. The Widlar source adds RE in the emitter of Q2, giving IO·RE = VT ln(IREF/IO).

So the Widlar source gives small, stable bias currents with practical resistor values, a higher output resistance (better current source) and lower dependence on supply voltage. It is used in the input-stage biasing of the 741.

Output resistance of the Widlar current source

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

Small-signal model: Q1 is diode-connected, so its resistance is only about 1/gm1 and the base of Q2 is close to ac ground. Apply a test voltage vx at the collector of Q2 and let ix flow into it.

Emitter node: ix flows into RE ‖ rπ (rπ goes to the
grounded base), so
   ve  = ix (RE ‖ rπ) = ix R'E
Base is grounded:
   vbe = 0 - ve = -ix R'E
Current into collector:
   ix  = gm vbe + (vx - ve)/ro
       = -gm ix R'E + (vx - ix R'E)/ro
Multiply by ro and collect terms:
   vx  = ix ro + ix R'E (1 + gm ro)
   Ro  = vx/ix = ro + R'E (1 + gm ro)
Since gm ro >> 1:
   Ro ≈ ro (1 + gm R'E),   R'E = RE ‖ rπ

If RE << rπ, then Ro ≈ ro(1 + gm RE), where gm = IO/VT and ro = VA/IO. A simple mirror has Ro = ro only, so the Widlar source has an output resistance larger by the factor (1 + gm RE). Since gm·RE = IO·RE/VT = ln(IREF/IO), this factor is typically about 4 to 6 (for example, 5.6 when IREF/IO = 100). This is the result of the current-series negative feedback provided by the unbypassed RE: any rise in IO raises the drop across RE, lowers VBE2 and pulls IO back down.