Common-emitter amplifier

YOU WILL BE ABLE TO

Identify the common-emitter topology and its coupling capacitors.
Compute voltage gain with and without emitter bypass.
Compute input and output resistance.
Explain the gain-versus-stability trade the bypass capacitor makes.

Assemble the pieces: Lesson 2's bias network, Lesson 3's small-signal model, and CA-5's capacitors used as frequency-dependent switches. The result is the common-emitter amplifier — the workhorse single-transistor voltage amplifier, and the circuit where every idea in this unit finally does something.

The common-emitter amplifier: C_in and C_out pass signal while blocking DC, and C_E shorts the emitter resistor for signals only.

Three capacitors, three jobs, all of them CA-5's mathematics. $C_{in}$ couples the source to the base: at signal frequencies its impedance is negligible, so the signal passes; at DC its impedance is infinite, so it cannot disturb the bias. $C_{out}$ does the same at the output, delivering signal to the next stage while keeping the collector's $7$ V of DC to itself. $C_E$ is the interesting one: it shorts $R_E$ FOR SIGNALS ONLY. The emitter resistor stays in place for DC — bias stability preserved — but disappears at signal frequencies, which is exactly where it would otherwise destroy the gain.

With $R_E$ bypassed, the emitter is at signal ground, the whole input signal appears across $r_\pi$ , and the collector current $g_m v_{in}$ flows through $R_C$ . The gain follows immediately, and the minus sign is real: this stage INVERTS.

A_v = -g_m R_C \qquad\text{(emitter bypassed)}

A_v \approx -\frac{R_C}{R_E} \qquad\text{(unbypassed)}

voltage gain, both cases

The contrast between those two expressions is the whole design bargain. Bypassed, the gain is large but set by $g_m$ — which depends on bias current, and therefore on temperature and supply. Unbypassed, the gain is a ratio of two resistors: far smaller, but stable, predictable, and low-distortion, because the emitter resistor applies negative feedback to the signal as well as to the bias. Choosing between them is choosing between gain and linearity, and the general answer in modern design is to take the stable version and make up the gain elsewhere — a principle ME-4's op-amps take to its conclusion.

R_{in} = R_1 \parallel R_2 \parallel r_\pi

R_{out} = R_C

terminal resistances

WORKED EXAMPLE

The full amplifier, numerically

Continue Lesson 2's design: $R_1 = 47$ kΩ, $R_2 = 13$ kΩ, $R_C = 2.7$ kΩ, $R_E = 1$ kΩ, $I_C \approx 1.9$ mA, β = 100. From Lesson 3: $g_m \approx 73$ mS, $r_\pi \approx 1.37$ kΩ.
Gain, bypassed: $A_v = -g_m R_C = -0.073 \times 2700 \approx -197$ . Nearly 200×, inverting.
Gain, unbypassed: $A_v \approx -R_C/R_E = -2700/1000 = -2.7$ . The bypass capacitor is worth a factor of about 73 — which is $g_m R_E$ , not a coincidence.
Input resistance: $47\ \text{k} \parallel 13\ \text{k} = 10.2$ kΩ, then $\parallel 1.37\ \text{k} \approx 1.2$ kΩ. Modest — the base resistance dominates, and this is the common-emitter's main weakness as an input stage.
Output resistance: $R_C = 2.7$ kΩ.

A_v \approx -197

bypassed,

-2.7

unbypassed;

R_{in} \approx 1.2

kΩ,

R_{out} = 2.7

kΩ

WORKED EXAMPLE

Sizing a coupling capacitor

$C_{in}$ and $R_{in}$ form the high-pass filter of CA-6 L4: below its cutoff, signal is lost.
For audio down to $20$ Hz into $R_{in} \approx 1.2$ kΩ: $C = 1/(2\pi f_c R) = 1/(2\pi \times 20 \times 1200) \approx 6.6$ µF.
Choose $10$ µF, putting the corner at about $13$ Hz — comfortably below the band of interest, as filter design always should be.
$C_E$ must be larger still, because it works against the much smaller resistance at the emitter (roughly $1/g_m \approx 14\ \Omega$ in parallel with $R_E$ ) — typically $100$ µF for audio.

C_{in} = 10

µF;

C_E \approx 100

µF

PRACTICE

A common-emitter stage has $g_m = 40$ mS and $R_C = 3.3$ kΩ, with the emitter fully bypassed. Find the voltage gain.

The same stage has its bypass capacitor removed, with $R_E = 820\ \Omega$ . Find the approximate gain, and the factor by which it fell.

For a stage with $R_1 = 100$ kΩ, $R_2 = 22$ kΩ, and $r_\pi = 2.6$ kΩ, find the input resistance.

Explain why the bypassed gain $-g_m R_C$ is considered less desirable than the unbypassed $-R_C/R_E$ in most modern designs, despite being far larger.