Differential Pair — Worked Example

Discrete BJT diff-pair gain and CMRR

Given: $I_{\text{TAIL}} = 10\,\mu\text{A}$, $R_C = 10\,\text{k}\Omega$, $\beta = 200$, $V_A = 100\,\text{V}$ (Early voltage of tail source), $V_{CC} = 12\,\text{V}$, $V_T = 25\,\text{mV}$.

Find: $A_d$, $A_{cm}$, and CMRR.

1. Step 1 — Compute the transconductance $g_m$.
   $$g_m = \frac{I_{\text{TAIL}}}{2 V_T} = \frac{10\,\mu\text{A}}{2 \times 25\,\text{mV}} = 0.2\,\text{mA/V}$$

   Each transistor carries $I_{\text{TAIL}}/2 = 5\,\mu\text{A}$, so $g_m$ is modest at this low bias.

2. Step 2 — Differential voltage gain $A_d$.
   $$A_d = g_m \cdot R_C = 0.2\,\text{mA/V} \times 10\,\text{k}\Omega = 2\,\text{V/V}$$

   A gain of 2 V/V is modest for a single diff pair. Op-amps stack additional gain stages to reach much higher open-loop gain.

3. Step 3 — Tail source output resistance $r_{o,\text{tail}}$.
   $$r_{o,\text{tail}} = \frac{V_A}{I_{\text{TAIL}}} = \frac{100\,\text{V}}{10\,\mu\text{A}} = 10\,\text{M}\Omega$$

   The tail source looks like 10 MΩ to the emitter node. This high impedance is why common-mode signals get rejected.

4. Step 4 — Common-mode gain $A_{cm}$.
   $$A_{cm} = -\frac{R_C}{2\,r_{o,\text{tail}}} = -\frac{10\,\text{k}\Omega}{2 \times 10\,\text{M}\Omega} = -0.0005\,\text{V/V}$$

   Essentially zero: the common-mode signal barely makes it to the output.

5. Step 5 — CMRR.
   $$\text{CMRR} = \left|\frac{A_d}{A_{cm}}\right| = \left|\frac{2}{0.0005}\right| = 4000$$
   $$\text{CMRR}_{\text{dB}} = 20\log_{10}(4000) \approx 72\,\text{dB}$$

   72 dB is respectable for a discrete diff pair. IC op-amps with cascoded tail sources push well above 100 dB.