In short

The common-emitter (CE) configuration is the workhorse amplifier circuit: the input signal is fed to the base, the output is taken across a collector resistor, and the emitter is grounded (or tied to a stable rail). A tiny AC voltage v_\text{in} at the base creates a tiny AC base current i_b = v_\text{in}/r_\text{in}, which is amplified by the transistor's current gain into a much larger AC collector current i_c = \beta\, i_b, which drops an AC voltage v_\text{out} = -i_c R_C across the collector resistor. The voltage gain is

\boxed{\;A_v = \frac{v_\text{out}}{v_\text{in}} \approx -\frac{\beta R_C}{r_\text{in}} \approx -\frac{R_C}{r_e}\;}

where r_e \approx 25\ \text{mV}/I_E is the small-signal emitter resistance at bias current I_E. A CE amplifier with R_C = 4.7\ \text{k}\Omega and I_E = 1 mA gives |A_v| \approx 190 — a 10 mV input signal comes out as a 1.9 V output swing, enough to drive a headphone.

The same transistor, driven hard in either direction, becomes a switch with only two states:

  • Cutoff — I_B = 0, I_C = 0, V_{CE} = V_{CC}. The transistor is "off" and acts like an open circuit between collector and emitter. Output is HIGH (equal to V_{CC}).
  • Saturation — I_B large enough that I_C is limited only by the external R_C, V_{CE} \approx 0.2 V. The transistor is "on" and acts like a closed switch. Output is LOW (≈ ground).

Between these two extremes is the active region where the transistor amplifies. A digital logic gate keeps the transistor out of the active region entirely — it flips between cutoff and saturation fast enough that the output is always either HIGH or LOW, never in between. This is how every NAND, NOR, and NOT gate inside every chip in every Indian phone actually works at the silicon level.

The input characteristic (I_B vs V_{BE} at fixed V_{CE}) looks like a forward-biased diode. The output characteristic (I_C vs V_{CE} at fixed I_B) shows three regions: a steep rise (saturation), a long nearly-flat plateau (active), and the cutoff line at the bottom. Pick a load line by drawing V_{CC} = V_{CE} + I_C R_C on the output graph; its intersection with the I_B curve fixes the operating point. That one picture explains both amplifier biasing and switching thresholds.

Plug a cheap electret microphone into a board with one BC547 transistor, a 10 kΩ resistor, a 4.7 kΩ collector resistor, a 9 V battery, and a small speaker at the output. Whisper into the microphone. A few millivolts of AC voltage ride on top of the steady bias at the base; the transistor scales them up by a factor of about 150; the speaker cone moves audibly, well above anything you could hear directly. One transistor, five passive components, a battery — you have built an amplifier. Swap the microphone for a push-button switch tied between 5 V and the base, and swap the speaker for a small relay coil: now the transistor is a switch. Press the button, the relay clicks, a 230 V AC lamp on the other side of the house turns on. Same silicon, same three legs; two completely different jobs.

This is the one component — one in a Bajaj portable radio, a million on a modern smartphone processor — that earns its keep by doing two things well at once. Most of electronics is figuring out which of those two jobs you want it to do today, and designing the circuit around the transistor so that it does that job reliably. This article takes the mental model built in the previous chapter — the BJT as a current-controlled current source with gain \beta — and applies it to the two classical jobs: analog amplification (the common-emitter amplifier) and digital switching (cutoff and saturation).

The common-emitter configuration

"Common-emitter" names which terminal is shared between the input side and the output side of the circuit. There are three possible choices — common-emitter (CE), common-base (CB), and common-collector (CC, also called the emitter follower) — and each has a different gain profile. The CE configuration is the most useful because it produces both voltage gain and current gain at the same time.

Common-emitter amplifier circuitAn npn transistor in common-emitter configuration. From the collector, a resistor R_C goes up to a positive supply V_CC. From the base, a biasing resistor R_B goes up to V_CC, and an input coupling capacitor C_in connects the base to an AC input v_in. The emitter is tied directly to ground. An output coupling capacitor C_out connects the collector to the output node v_out. Arrows show the quiescent currents I_B, I_C, I_E and the AC swings i_b, i_c, v_in, v_out. +V_CC R_B R_C B C E C_in ~ v_in C_out v_out I_B ↓ I_C ↓ I_E ↓
A common-emitter amplifier. The emitter sits on ground (common to both the input and the output side). The base is biased through $R_B$ to give a quiescent $I_B$ of about 10–100 µA; the input AC signal rides on top of this DC bias through the coupling capacitor $C_\text{in}$. The output is the AC voltage across $R_C$, taken through $C_\text{out}$ which strips off the DC.

Why the emitter is grounded — the two jobs of the input side

Two things must happen for the circuit to amplify:

  1. The base-emitter junction must be forward-biased (typically by about 0.7 V) so that a quiescent current I_B is flowing and the transistor is sitting in the active region — the linear regime where I_C = \beta I_B.
  2. A small AC signal v_\text{in} must be added to the base voltage so that it modulates I_B, which in turn modulates I_C.

Both jobs are performed on the same wire. The DC bias comes through R_B from V_{CC}; the AC signal comes through C_\text{in} from the input source. The capacitor blocks the DC component of the source (so the bias is not disturbed) and passes the AC component.

The emitter is tied directly to ground so that the base voltage swings relative to a fixed reference. If the emitter were floating or allowed to swing, changes in V_E would partially cancel changes in V_B and the effective input would be smaller than it looks. Grounding the emitter pins the reference.

The input and output characteristics

Every BJT is characterised by two families of I-V curves. They are drawn so that a single graph shows you everything you need to design a biasing or amplifier circuit.

Input characteristic: I_B vs V_{BE}

Sweep the base-emitter voltage V_{BE} from 0 to about 0.8 V while measuring the base current I_B. The result is almost identical to a forward-biased diode curve, because the base-emitter junction is a forward-biased p-n junction:

Input characteristic of an npn BJT in common-emitter configurationA graph of base current I_B in microamperes on the vertical axis against base-emitter voltage V_BE in volts on the horizontal axis. The curve stays near zero for V_BE below 0.5 volts, then rises sharply — an exponential diode-like curve — reaching about 100 microamperes at V_BE of 0.72 volts. V_BE (V) I_B (µA) 0.2 0.4 0.6 0.8 100 50 20 Q: V_BE=0.68V, I_B=40µA
The input characteristic of a typical npn BJT. Below about 0.5 V there is essentially no current; above the "knee" at 0.6–0.7 V the current rises exponentially. The Q-point (quiescent, or resting, operating point) sits on this curve wherever the DC biasing circuit places it.

The characteristic is barely affected by the collector voltage, so you can treat the input side as an ordinary diode — a good approximation for circuit design.

Output characteristic: I_C vs V_{CE} for fixed I_B

Now hold the base current I_B at some value (say 20 µA) and sweep the collector-emitter voltage V_{CE} while measuring I_C. The result is a curve with three distinct regions. Run the same experiment at several different I_B values; you get a family of curves:

Output characteristic of an npn BJT in common-emitter configurationA family of I_C vs V_CE curves for five values of base current I_B from 10 to 50 microamperes. Each curve rises steeply for small V_CE up to about 0.3 volts (the saturation region), then bends into a nearly flat plateau at I_C equal to beta times I_B (the active region). A load line V_CC equals V_CE plus I_C times R_C is drawn diagonally, intersecting the curves at five Q-points. V_CE (V) I_C (mA) 2 4 6 8 10 5 3 1.5 I_B=50µA I_B=40µA I_B=30µA I_B=20µA I_B=10µA load line: V_CE + I_C·R_C = V_CC Q SATURATION ACTIVE REGION CUTOFF (below the I_B=0 curve, I_C≈0)
The output characteristic for a BJT with $\beta = 100$, drawn with $R_C = 2\ \text{k}\Omega$ and $V_{CC} = 10$ V. The family of curves shows the collector current for fixed base currents (10, 20, 30, 40, 50 µA). The diagonal red dashed line is the **load line** — the set of operating points allowed by $V_{CC} = V_{CE} + I_C R_C$. The Q-point Q sits at the intersection of the load line with the $I_B = 30\ \mu\text{A}$ curve.

The three regions — and what they mean

Saturation (far left, small V_{CE}): Both the base-emitter and base-collector junctions are forward-biased. V_{CE} is about 0.2 V. The collector current is no longer controlled by I_B — it is limited by the external resistor R_C. Pushing more base current in does almost nothing to the collector current. The transistor is "on" like a closed switch.

Active region (middle plateau): Base-emitter is forward, base-collector is reverse. The collector current is I_C = \beta I_B and is almost independent of V_{CE}. This is where the transistor acts as a linear amplifier.

Cutoff (bottom): I_B = 0, so I_C \approx 0. The transistor is "off" like an open switch. V_{CE} rises all the way to V_{CC} because no current flows through R_C.

The load line is the one piece of external-circuit information that lets you read off the operating point. It comes from Kirchhoff's voltage law around the output loop: V_{CC} = V_{CE} + I_C R_C, rearranged to

I_C = \frac{V_{CC} - V_{CE}}{R_C}.

This is the equation of a straight line in the (V_{CE}, I_C) plane with slope -1/R_C. Every allowed operating point lies on this line. Where exactly? That depends on I_B — which is controlled by the input circuit.

Voltage gain of a common-emitter amplifier

Now the amplifier calculation. The key insight: small AC signals ride on top of the DC Q-point, and we can linearise the transistor's behaviour around Q.

Step 1. At the Q-point, small changes obey \Delta I_C = \beta\, \Delta I_B. A small AC base current i_b becomes an AC collector current

i_c = \beta\, i_b.

Why: in the active region the I_C–I_B relationship is linear with slope \beta. Differentiate I_C = \beta I_B and the AC version pops out.

Step 2. The input voltage v_\text{in} at the base translates into a base current through the transistor's small-signal input resistance r_\text{in}:

i_b = \frac{v_\text{in}}{r_\text{in}}, \qquad r_\text{in} \approx \beta\, r_e, \qquad r_e = \frac{V_T}{I_E} = \frac{kT/q}{I_E}.

Why: the base-emitter junction, linearised around its DC bias point, looks like a small resistor r_e in the emitter lead. Seen from the base, it looks \beta times larger, because for every 1 A of emitter current change, only 1/\beta A of base current change is needed. At room temperature V_T \approx 25.85 mV, so for I_E = 1 mA, r_e \approx 26\ \Omega — the BJT's emitter is a very small resistor at normal operating points.

Step 3. The AC collector current i_c flows through the collector resistor R_C. Since the top of R_C is tied to the fixed rail V_{CC} (which is AC-ground through the supply's bypass capacitors), the AC voltage swing at the collector is

v_\text{out} = -i_c R_C.

Why: when i_c rises, more current flows through R_C, so the voltage drop across R_C is larger, and the collector voltage (measured from ground) drops. The minus sign is the famous 180° phase inversion of the CE amplifier — an increase in input voltage produces a decrease in output voltage.

Step 4. Put the three steps together.

A_v = \frac{v_\text{out}}{v_\text{in}} = \frac{-i_c R_C}{v_\text{in}} = \frac{-\beta\, i_b\, R_C}{i_b\, r_\text{in}} = -\frac{\beta R_C}{r_\text{in}} = -\frac{R_C}{r_e}.

Why: cancel i_b, and use r_\text{in} = \beta r_e to eliminate \beta. The remarkable fact is that the voltage gain depends only on R_C and the Q-point current (via r_e = V_T/I_E) — not on the transistor's \beta. This is why CE amplifiers with real transistors, where \beta varies by a factor of 3–5 between individual units, still produce predictable voltage gains: the gain is set by the external resistor and the bias current.

The approximation A_v \approx -\beta R_C / R_B that appears in CBSE textbooks comes from assuming r_\text{in} \approx R_B — true only if the biasing resistor is much smaller than the transistor's intrinsic \beta r_e, which is uncommon in practice. The more accurate A_v = -R_C/r_e is the working formula.

Example 1: Voltage gain of a BC547 stage for an audio preamplifier

A BC547 with \beta = 200 is biased at I_C = 1 mA. The collector resistor is R_C = 4.7 kΩ, the supply is V_{CC} = 9 V. Compute the voltage gain. If a 10 mV peak-to-peak sine wave from an electret microphone is fed to the base (through a coupling capacitor), what is the output amplitude?

Step 1. Compute r_e at the bias point.

r_e = \frac{V_T}{I_E} = \frac{25.85\ \text{mV}}{1\ \text{mA}} = 25.85\ \Omega.

Why: at room temperature, V_T = kT/q \approx 25.85 mV. Since I_E \approx I_C for a BJT in the active region with large \beta, we use I_C = 1 mA.

Step 2. Compute the voltage gain.

A_v = -\frac{R_C}{r_e} = -\frac{4700}{25.85} = -182.

Step 3. Compute the output amplitude.

v_\text{out,pp} = |A_v|\, v_\text{in,pp} = 182 \times 10\ \text{mV} = 1.82\ \text{V peak-to-peak}.

Step 4. Check that the output does not clip. The Q-point sits at V_{CE,Q} = V_{CC} - I_C R_C = 9 - (0.001)(4700) = 4.3 V. The output can swing up to 9 V (into cutoff) and down to about 0.2 V (into saturation), so the maximum clean peak-to-peak swing is about 2 \times \min(9 - 4.3, 4.3 - 0.2) = 2 \times 4.1 = 8.2 V. An output of 1.82 V is well within this range — no clipping.

Result: |A_v| \approx 182, so a 10 mV input gives 1.82 V output (with a 180° phase inversion). Perfect for driving the input stage of a power amplifier or a pair of low-impedance headphones through another buffer stage.

Input and output waveforms of a common-emitter amplifier stageTwo sinusoidal waveforms plotted on the same time axis. The upper trace, labelled v_in, has a peak-to-peak amplitude of 10 millivolts and is in phase with the time axis. The lower trace, labelled v_out, is inverted — 180 degrees out of phase — and has a peak-to-peak amplitude of 1.82 volts. t t v_in (10 mV pp) v_out (1.82 V pp) 5 mV 0.91 V
The 10 mV input is amplified to 1.82 V and inverted (the peak of the output aligns with the trough of the input). The ratio of the amplitudes is the gain magnitude $|A_v| = 182$; the inversion is the sign.

What this shows: the transistor takes the whisper at the base and scales it up by the ratio of R_C to r_e. Nothing else in the circuit sets the gain — not the supply voltage, not the transistor's \beta, not the biasing resistor. Just R_C / r_e.

The transistor as a switch

Now take the same transistor and stop caring about small-signal behaviour. Drive the base hard enough that I_B is much larger than I_{C,\text{load}} / \beta — the transistor saturates, V_{CE} drops to about 0.2 V, and the collector current is locked at I_{C,\text{sat}} \approx V_{CC}/R_C. The transistor now behaves like a closed switch between collector and emitter. Drop the base drive to zero and the transistor goes into cutoff; I_C = 0 and V_{CE} = V_{CC}. The transistor is now an open switch.

Transistor as switch driving a relayAn npn transistor in switch configuration. A 5 V logic input at the left drives the base through a 10 kilohm resistor. The collector is connected through a relay coil to a 12 V rail. A freewheeling diode is drawn across the relay coil with its cathode at the positive rail. The emitter is grounded. Arrows show the two states: input HIGH, transistor saturated, relay energised, output LOW at 0.2 volts; input LOW, transistor cutoff, relay de-energised, output HIGH at 12 volts. +12 V relay coil (the load) D (freewheel) B C E R_B=10 kΩ V_in (5 V or 0 V)
A transistor switching a 12 V relay from a 5 V logic input. When $V_\text{in} = 5$ V, the base current through $R_B$ saturates the transistor, the full 12 V rail drives current through the relay coil, and the relay clicks on. When $V_\text{in} = 0$ V, the transistor is in cutoff, no current flows, the relay releases. The diode across the coil is a **freewheeling diode** — it catches the inductive kick when the coil is de-energised.

Sizing the base resistor for saturation

Suppose the relay coil has resistance 120 Ω and needs 100 mA to actuate, pulling the collector to near ground. The saturation collector current is I_{C,\text{sat}} = 12\ \text{V} / 120\ \Omega = 100\ \text{mA}. For the transistor to be solidly in saturation (not merely active), we want enough base current to overdrive the junction. A rule of thumb: provide base current equal to I_{C,\text{sat}}/\beta_\text{min} times a safety factor of about 5–10.

For a BC547 with \beta_\text{min} \approx 100:

I_{B,\text{min}} = \frac{I_{C,\text{sat}}}{\beta_\text{min}} = \frac{0.1}{100} = 1\ \text{mA}.

With a factor of 5 safety margin: I_B = 5 mA. The base resistor must provide this current from the 5 V logic input, with a drop of about 0.7 V across the base-emitter junction:

R_B = \frac{V_\text{in} - V_{BE}}{I_B} = \frac{5 - 0.7}{0.005} = 860\ \Omega.

A standard value of 1 kΩ works; in practice, 10 kΩ is the commonly recommended value when driving a small relay because the extra margin does not matter for low-current loads and the higher resistance reduces the load on the logic output. Anyone who has ever built an Arduino-driven relay module has met these numbers.

Why saturation is the desirable state of a digital switch

In saturation, V_{CE} \approx 0.2 V, so the power dissipated in the transistor is small: P = V_{CE} \cdot I_C = 0.2 \times 0.1 = 20 mW. In cutoff, I_C = 0, so P = 0. Between these two extremes (in the active region), V_{CE} can be several volts and I_C can be large, so the transistor dissipates far more power as heat. A digital switch therefore wants to transit between cutoff and saturation as quickly as possible — any time spent in the active region is wasted as heat.

This is also why modern CMOS logic (the technology inside every chip on your phone) is so much more efficient than old bipolar logic: in CMOS, both the pull-up and the pull-down transistor are fully off (cutoff) except during the brief switching transient, so the static power dissipation is nearly zero.

Common confusions

If you are here for the CBSE/JEE-level understanding of the CE amp and transistor switch, you can stop here. The rest is for readers who want the small-signal model, the distortion mechanisms, and the connection to integrated circuits.

The full hybrid-π small-signal model

The simplest model used above treats the BJT as a current-controlled current source (i_c = \beta i_b) with a small input resistance r_\text{in}. The more accurate model — called the hybrid-π model — adds two details:

  1. The collector current is also a function of V_{CE} (the Early effect). This appears as a finite output resistance r_o = V_A/I_C, where V_A \approx 50–100 V is the Early voltage. Typical r_o for I_C = 1 mA is 50–100 kΩ — usually much larger than R_C, so ignoring it is a good first approximation. When it matters (in high-gain op-amps and cascode stages), A_v = -g_m\, (R_C \parallel r_o).
  2. The transconductance g_m = I_C/V_T is the fundamental small-signal parameter. For I_C = 1 mA, g_m = 0.039 S (i.e., 39 mA/V). The voltage gain of a CE amp is really A_v = -g_m R_C — which is equivalent to -R_C/r_e since g_m = 1/r_e.

The hybrid-π model reappears in every integrated-circuit design class and is the starting point for the 741 op-amp, the LM358, the MOSFET extension, and essentially every transistor-level analog circuit designed since 1980.

Distortion — why the small-signal assumption breaks

The linear relationship i_c = g_m v_\text{in} is only approximate. The exact Ebers-Moll equation gives I_C = I_s (e^{V_{BE}/V_T} - 1), which is exponential. Expanding to second order:

I_C(V_{BE,Q} + v_\text{in}) = I_{C,Q}\left(1 + \frac{v_\text{in}}{V_T} + \frac{1}{2}\left(\frac{v_\text{in}}{V_T}\right)^2 + \dots\right).

The quadratic term produces a second harmonic at twice the input frequency. For a 10 mV input, the second-harmonic distortion is about v_\text{in}/(4 V_T) = 10/(4 \times 25.85) \approx 10\% — enormous for an audio amplifier. Practical CE amplifiers counter this with emitter degeneration: add a small resistor R_E in the emitter lead; the gain drops to A_v = -R_C/(r_e + R_E) but the distortion drops by the same factor, and the gain becomes nearly independent of the transistor.

Why the MOSFET replaced the BJT in digital logic

Modern chips are almost entirely MOSFETs, not BJTs. The reasons are all about power:

  • A MOSFET's gate is insulated — the input current is essentially zero, so driving one MOSFET from another dissipates no steady-state power. A BJT's base needs continuous current, which scales badly when you have a billion transistors on a chip.
  • CMOS logic (complementary n-channel and p-channel MOSFETs) has both transistors cut off between switching events, so the static power is nearly zero.
  • MOSFETs scale down better. A modern transistor has a channel length of about 3 nm — a BJT cannot be made that small because its action relies on minority-carrier diffusion across a thin base, which becomes quantum-mechanically badly behaved at those scales.

The BJT survives in analog circuits (where its higher transconductance per unit bias current matters), in discrete-component designs (where the BC547/2N3904 is cheaper than a MOSFET of equivalent specification), and inside integrated circuits as output drivers where high current capability matters.

Audio amplifier stages — a chain of CE amps

A typical discrete audio amplifier chain has three CE stages:

  1. Pre-amplifier: High-impedance, low-noise BJT in CE. Input from microphone or phono cartridge. Voltage gain ~50. R_C ~10 kΩ, I_C ~0.5 mA.
  2. Intermediate amplifier: Similar topology, higher bias current. Voltage gain ~30. R_C ~2 kΩ, I_C ~2 mA.
  3. Power amplifier: Complementary push-pull (a pair of npn and pnp in common-emitter). Voltage gain ~1, current gain ~50. I_C up to amperes.

Total voltage gain: 50 \times 30 \times 1 = 1500. Total current gain: up to 10^4. Input: microvolts. Output: volts into ohms. The same building block — one CE stage — scaled up three times. Every Ahuja or Bajaj or homemade stage amplifier uses a variant of this chain.

Where this leads next