Current mirror requirements, Widlar resistance, best source

Basic requirements of current mirror circuits

• 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.

Output resistance of the Widlar current source

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

DC design relation: IO·RE = VT ln(IREF/IO).

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.

Best current source

Among the common sources, the Wilson current source (and the cascode mirror) is usually the best:

• Its output resistance is about β·ro/2, much higher than ro (simple mirror) and ro(1 + gm RE) (Widlar).
• Its β error is very small: IO = IREF[1 − 2/(β² + 2β + 2)], compared with IREF/(1 + 2/β) in the simple mirror.

Between the two derived here, the Widlar source is better than the simple mirror because it gives a higher output resistance and small currents with small resistors.