Common-Collector & Common-Drain Follower — Worked Example

Buffering a CE pre-amp into 16 Ω headphones

Scenario: A CE pre-amp (Rᴄ = 2 kΩ, Iᴄ = 2 mA, β = 150) must drive a pair of 16 Ω headphones. First we calculate what happens without a buffer, then we add a CC emitter-follower and compare.

Parameters: $V_{CC} = 12\,\text{V}$, $I_C = 2\,\text{mA}$, $\beta = 150$, $R_C = 2\,\text{k}\Omega$, $R_L = 16\,\Omega$, $R_{E2} = 100\,\Omega$.

1. Spec — drive 16 Ω headphones from a 2 kΩ CE pre-amp.
   $$g_m = \frac{I_C}{V_T} = \frac{2\,\text{mA}}{25\,\text{mV}} = 80\,\text{mA/V}, \quad r_{\pi} = \frac{\beta}{g_m} = \frac{150}{0.08} = 1{,}875\,\Omega$$
   The problem: Rᴄ = 2 kΩ is 125× larger than Rʟ = 16 Ω. Direct coupling will collapse the gain.
2. Without a follower — direct drive into 16 Ω.
   $$A_{v,\text{direct}} = -g_m(R_C \| R_L) = -0.08 \times \frac{2000 \times 16}{2000 + 16}$$
   $$= -0.08 \times 15.9 \approx \mathbf{-1.3\,\text{V/V}} \quad \text{(collapsed!)}$$
   The headphone load is so heavy it essentially short-circuits the collector — gain disappears.
3. Add a CC follower — compute R_in of the CC stage.With $R_{E2} = 100\,\Omega$ and $R_L = 16\,\Omega$:
   $$R_{E2} \| R_L = \frac{100 \times 16}{100 + 16} = 13.8\,\Omega$$
   $$R_{in,CC} = r_{\pi} + (\beta+1)(R_{E2} \| R_L) = 1{,}875 + 151 \times 13.8 \approx \mathbf{3{,}960\,\Omega}$$
   The CC follower presents nearly 4 kΩ to the CE collector — a manageable load.
4. Pre-amp gain with R_in_CC as the effective load.
   $$R_C \| R_{in,CC} = \frac{2{,}000 \times 3{,}960}{2{,}000 + 3{,}960} = 1{,}328\,\Omega$$
   $$A_{v,CE} = -g_m(R_C \| R_{in,CC}) = -0.08 \times 1{,}328 \approx \mathbf{-106\,\text{V/V}}$$
   The CE gain is largely restored because $R_{in,CC}$ is much larger than $R_L$.
5. CC stage gain and total system gain.
   $$A_{v,CC} = \frac{g_m(R_{E2} \| R_L)}{1 + g_m(R_{E2} \| R_L)} = \frac{0.08 \times 13.8}{1 + 0.08 \times 13.8} = \frac{1.10}{2.10} \approx 0.52$$
   $$A_{v,\text{total}} = A_{v,CE} \times A_{v,CC} = -106 \times 0.52 \approx \mathbf{-55\,\text{V/V}}$$
   Note: with a larger $R_{E2}$ (e.g. 1 kΩ), $A_{v,CC}$ approaches 1 but $R_{in,CC}$ grows further — the designer trades off gain per stage vs. impedance ratio.