Current mirror conditions and Widlar output resistance

Essential conditions for a current mirror

A current mirror is a circuit that copies (mirrors) a reference current IREF into an output branch. It works only if:

• 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

The Widlar current source is a simple mirror with a resistor RE in the emitter of the output transistor Q2. It gives a small output current (µA range) from a normal reference current.

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

DC relation (neglecting base currents):

VBE1 = VBE2 + IO·RE
VT ln(IREF/IS) = VT ln(IO/IS) + IO·RE
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.