BJT as an amplifier (common emitter) — Hardware notes

WHAT is a common-emitter amplifier?

WHY the active region? Only there does $I_C=\beta I_B$ hold cleanly. In cutoff there is no current to modulate; in saturation the collector "runs out of room" and gain collapses. We want to sit at a Q-point (quiescent point) in the middle so the AC signal can swing both up and down without clipping.

HOW it works — deriving the gain from scratch

Step 1 — DC bias (the operating point)

Apply Kirchhoff's Voltage Law (KVL) around the output loop: $V_{CC} = I_C R_C + V_{CE}$

Why this step? The supply voltage must be shared between the drop across $R_C$ and the voltage left across the transistor. This is the load line.

To pick $I_C$ , KVL on the base loop with a fixed base voltage $V_B$ : $V_{BE} = 0.7\text{ V},\qquad I_B=\frac{V_B-0.7}{R_B},\qquad I_C=\beta I_B$

Step 2 — The transconductance (the heart)

The collector current depends exponentially on $V_{BE}$ (Ebers–Moll): $I_C = I_S\,e^{V_{BE}/V_T},\qquad V_T=\frac{kT}{q}\approx 26\text{ mV at room temp}$

Differentiate to find how much $I_C$ changes for a small change in $V_{BE}$ : $g_m \equiv \frac{dI_C}{dV_{BE}} = \frac{I_S}{V_T}e^{V_{BE}/V_T} = \frac{I_C}{V_T}$

Step 3 — Small-signal voltage gain

Output voltage (AC part). KVL: $v_{out}=V_{CC}-I_C R_C$ . Only the changing part matters for a signal, and $V_{CC}$ is constant, so: $v_{out} = -R_C\,i_c = -R_C\,(g_m\,v_{be}) = -g_m R_C\,v_{in}$

Why the minus sign (deep reason): More base voltage → more $I_C$ → bigger drop across $R_C$ → the collector node is pulled down toward ground. Rising input ⇒ falling output.

Figure — BJT as an amplifier (common emitter)

Emitter degeneration (the practical CE)

If you add $R_E$ in the emitter, gain becomes stabilized and less dependent on the wobbly $g_m$ : $A_v \approx -\frac{R_C}{R_E}\quad(\text{when }g_m R_E\gg1)$ Why: the emitter resistor provides negative feedback — a rise in $I_C$ raises $V_E$ , which reduces $V_{BE}$ , opposing the rise. You trade gain for stability and linearity.

Worked examples

Common mistakes

Active recall

Recall Feynman: explain to a 12-year-old

Imagine a garden tap where a tiny twist of your finger opens a huge flow of water. The finger-twist is the small base signal; the big water flow is the collector current. To turn that flow into a "voltage picture," we push the water through a narrow pipe ( $R_C$ ) and measure the pressure. Push more water → more pressure drop → the far end sags down. So when you push the input up, the output goes down — that's why the amplifier flips the signal. And because a small twist controls a big flow, the picture at the output is a giant copy of your tiny finger-wiggle.

Flashcards

What configuration shares the emitter between input and output?

Common-emitter

Formula for transconductance

g_m

g_m=I_C/V_T\approx I_C/26\text{mV}

Small-signal CE voltage gain?

A_v=-g_mR_C=-I_CR_C/V_T

Why is there a minus sign in CE gain?

I_C

→ larger drop across

R_C

→ collector node pulled down → output inverted (180°)

What is

V_T

at room temperature?

\approx 26\text{ mV}

(

=kT/q

)

CE gain with emitter degeneration (strong feedback)?

A_v\approx -R_C/R_E

Why add

R_E

Negative feedback → stabilizes gain vs

\beta

/temperature, improves linearity (trades gain for stability)

Which region must the BJT operate in to amplify?

Active region (

V_{BE}\approx0.7

V forward, base–collector reverse)

Load-line KVL for CE output loop?

V_{CC}=I_CR_C+V_{CE}

Where should the Q-point sit for max symmetric swing?

Near mid-supply (

V_{CE}\approx V_{CC}/2

)

Difference between

\beta

and

A_v

\beta

= DC current gain of device;

A_v

= small-signal voltage gain

=-g_mR_C

Connections

BJT operating regions (cutoff, active, saturation)
Ebers–Moll model
Small-signal hybrid-pi model
Load line analysis
Negative feedback in amplifiers
FET as an amplifier (common source) — the field-effect analogue
Thermal voltage $V_T=kT/q$

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, common-emitter amplifier ka core idea bilkul simple hai: base pe ek chhota sa signal do, aur collector me uska bada copy nikalta hai. BJT ek current-controlled current source hai — $I_C=\beta I_B$ . Lekin voltage gain chahiye, isliye hum collector me ek resistor $R_C$ lagate hain. Jab current badhti hai to $R_C$ pe voltage drop badhta hai, aur collector node niche gir jaata hai. Isiliye output ulta (inverted) hota hai — input upar to output niche. Yeh minus sign ka asli reason hai.

Gain ka formula yaad rakho: $A_v=-g_mR_C$ , jahan $g_m=I_C/V_T$ aur $V_T\approx26$ mV. Yeh 26 mV bahut chhota hai, isiliye gain bahut bada aata hai (Example me ~190x). Ek common galti: log sochte hain gain $\beta$ hota hai — nahi! $\beta$ current gain hai device ka, voltage gain alag cheez hai jo bias current aur $R_C$ pe depend karta hai.

Bias point (Q-point) ka dhyan rakho — transistor ko active region me mid-supply ke aas-paas baithao, taaki signal upar-niche dono taraf swing kar sake bina clip hue. Agar edge pe biased hoga to ek side clip ho jaayega, sound/signal distort. Aur agar gain ko stable banana ho (temperature/beta se independent), to emitter me $R_E$ daal do — tab gain approx $-R_C/R_E$ ban jaata hai, thoda kam par bahut reliable. Yeh negative feedback ka jaadu hai: gain thoda kam, par bharosa zyada.