From a faint signal to a loud one
Hold a guitar pickup or a dynamic microphone and what comes out is almost nothing — a wobble of a few millivolts riding on top of silence. A loudspeaker, by contrast, wants whole volts swinging back and forth to push air. The amplifier is the lever that connects the two: a tiny input controls a much larger reservoir of power sitting in the supply rail, the way a light touch on a faucet handle controls a heavy gush of water. The transistor is that faucet, and in rungs 3 and 4 you learned its two essential personalities — how to bias it (set the faucet half-open so it can swing both ways) and how to model its small-signal behaviour (how much the flow changes for a tiny twist of the handle).
This rung fuses those two ideas into one circuit. A bipolar junction transistor arranged so the emitter is the shared, grounded terminal gives us the common-emitter amplifier; the same trick with a MOSFET and a grounded source gives the common-source amplifier. They are the workhorses of discrete analog electronics — the first real circuit most engineers build that actually does something useful with a transistor rather than just switching an LED.
The common-emitter, drawn and biased
Here is the classic single-stage common-emitter amplifier. Don't be intimidated by the parts count — every component has exactly one job, and we'll name them all. The four resistors set the bias; the three capacitors handle the signal.
+Vcc (e.g. +12 V)
|
+---------+---------+
| |
[R1] [Rc] collector load
| |
| Cc1 +-----||----o Vout
Vin o--+------||----+ | Cc2
| | collector
[R2] |/ C base current sets
| base -->| BJT (NPN)
| |\ E
| | emitter
+------------+----+----+
| |
[Re] [Ce] bypass cap
| |
GND GND- R1 and R2 form a voltage divider that pins the base at a steady DC voltage — typically around a quarter of Vcc. This sets the operating point, the bias you learned to design in rung 3.
- Re sits below the emitter and provides negative feedback for DC: if the transistor heats up and tries to conduct more, the rising emitter current lifts the emitter voltage, which shrinks the base–emitter drive and pulls the current back. It makes the bias rock-solid against temperature and part spread.
- Rc is the collector load. The collector current flows through it, and the signal voltage appears across it — this resistor is where small-signal current becomes output voltage. Its value is half the gain story.
- Cc1, Cc2 are coupling capacitors. They pass the AC signal but block DC, so the source and the next stage can't disturb the carefully set bias point — and the bias can't push a damaging DC offset into them.
- Ce is the bypass capacitor across Re. At signal frequencies it acts like a short circuit, so it gives the AC signal a clean path to ground around Re — restoring full gain — while leaving Re's DC stabilising job intact. It's the resolution of a genuine conflict, which we'll unpack below.
Where the gain comes from: gm and Rc
Now collapse the circuit into its small-signal model. For signals, the DC supply is just a fixed voltage — an AC ground — so Vcc and the top of Rc connect to the same imaginary ground for the wiggle. The transistor becomes a single magic component: a current source whose output current equals transconductance gm times the input voltage. That one parameter, gm, is the heart of every amplifier.
The logic is two short steps. The input voltage v_in develops a small-signal collector current i_c = gm · v_in. That current is forced through Rc, and (because the top of Rc is AC ground) it produces an output voltage v_out = −i_c · Rc. The minus sign is real: when the input rises, the transistor conducts harder, more current flows down through Rc, and the collector node is pulled down. The common-emitter inverts.
Voltage gain of a common-emitter stage (Ce bypassing Re):
Av = v_out / v_in = -gm * Rc
For a BJT, gm = Ic / VT, with VT ~ 26 mV at room temperature.
Worked example:
Ic = 1 mA -> gm = 1 mA / 26 mV = 38.5 mA/V (= 38.5 mS)
Rc = 4.7 k
Av = -38.5 mA/V * 4.7 k = -181 (about 45 dB, inverting)
If Re is NOT bypassed (no Ce), the gain drops to roughly:
Av ~ -Rc / Re (degeneration: trades gain for linearity)The common-source MOSFET tells the same story with a different gm. There, gm = √(2 · k · Ic) in the square-law region (k bundles the device width, length, and oxide), so gm grows only with the square root of current, not linearly. A MOSFET at the same current typically has a smaller gm than a BJT — one reason BJTs still win where raw gain-per-milliamp matters, and MOSFETs win where you need that near-infinite input resistance.
Reading the load line: gain versus headroom
The load line is the single most illuminating picture in analog design. Plot the transistor's output current (vertical) against its collector–emitter voltage (horizontal). The transistor's own physics gives a family of curves; the external resistor Rc imposes one straight line across them — KVL written as a graph: Vcc = Ic·Rc + Vce. The amplifier lives where the bias picks a point on that line, and the signal makes the operating point slide up and down it.
Ic
^
|\
| \ <- load line, slope = -1/Rc
| \
| \ . Q (quiescent / bias point, ~half of Vcc)
| \ /|
| X | signal swings the point up & down the line
| / \|
| / \
+--+-----+------------> Vce
0 Vsat Vq Vcc
|<-------headroom------->|
sat region cutoff (Ic=0)Now the central trade-off becomes visible. Make Rc large to get more gain (Av = −gm·Rc) and the load line tilts shallow — but a large Rc with the same current means a big DC drop across it, leaving little Vce for the transistor and a short distance to the saturation cliff. The output can only swing a small amount before it clips flat against the rail. Gain and headroom pull against each other. You cannot have a huge gain and a huge undistorted swing from one stage on a fixed supply; you choose where to spend your volts.
Impedances, capacitors, and clean supply rails
A stage doesn't live alone — something feeds it and it feeds something else. Its input impedance is what the source sees, and its output impedance is what the next stage sees. For the common-emitter, the input impedance is the bias divider R1‖R2 in parallel with the transistor's own base resistance rπ (a few kΩ) — moderate, a few kilohms. The output impedance is essentially Rc itself — looking back into the collector you see a current source (very high resistance) shunted by Rc. The common-source is gentler on the input: a MOSFET's gate is an insulator, so its input impedance is set almost entirely by the bias divider and is enormous, which is exactly why MOSFET front-ends suit high-impedance sensors.
This is why the coupling and bypass capacitors matter so much. Each capacitor with the resistance it sees forms a high-pass corner f = 1/(2π·R·C); below that frequency, gain falls off. Choose Cc1, Cc2, and especially Ce large enough that all their corners sit well below your lowest signal frequency. The bypass capacitor Ce is the touchy one — it sees a small resistance (roughly 1/gm in parallel with Re), so it usually needs to be the largest, often tens or hundreds of microfarads for audio. If Ce is too small, your bass disappears.
Distinguish the three roles cleanly, because beginners blur them: a coupling cap is in series with the signal and blocks DC between stages; a bypass cap is in parallel with Re to restore AC gain; a decoupling cap is in parallel with the supply to absorb current transients. Same component, three jobs — telling them apart is a real mark of fluency.
Putting it together — a simulated stage
Words become belief once you watch the numbers move. Here is a SPICE deck for the 1 mA, Rc = 4.7 kΩ stage we computed above. Run it and you should measure an inverting gain of roughly 180, an output that clips when the input grows too large, and bass roll-off if you shrink Ce.
* Common-emitter amplifier, Vcc = 12 V, Ic ~ 1 mA Vcc vcc 0 12 Vin in 0 AC 1 SIN(0 5m 1k) ; 5 mV, 1 kHz test tone R1 vcc b 68k R2 b 0 12k Rc vcc c 4.7k Re e 0 1k Cc1 in b 10u Cc2 c out 10u Ce e 0 100u ; bypass: try 1u to hear bass vanish Cdec vcc 0 0.1u ; decoupling cap on the rail RL out 0 100k ; light load Q1 c b e QNPN .model QNPN NPN (BF=200 IS=1e-15) .op ; check Vce ~ 6 V at the Q-point .ac dec 20 10 1meg ; gain vs frequency .tran 10u 5m ; watch the waveform clip if Vin grows .end
To turn this into a common-source, swap Q1 for an n-channel MOSFET (M1 d g s s, with a .model NMOS), rename collector→drain and emitter→source, and re-tune the divider so the gate sits above threshold. Everything else — Rd as the load, a source-degeneration resistor with a bypass cap, coupling caps, the same load-line and headroom reasoning — carries straight over. Build one of each and you've built the foundation of nearly all discrete analog electronics.