Single-stage amplifiers: CE, CS and follower configurations

Gain, input and output resistance of CE, CS, common-collector and common-drain stages, with degeneration and source loading.

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Why it matters

A sensor signal usually passes through at least one transistor stage before it reaches an ADC, and the choice of configuration decides how much gain you get, how much the stage loads the sensor, and how well it can drive the next stage or a cable. The three basic configurations (common emitter/source, common collector/drain, common base/gate) are the building blocks of every op-amp and instrumentation amplifier.

Key ideas

Naming. The configuration is named after the terminal that is common to input and output (at AC ground).

  • Common emitter (CE) / common source (CS): input at base/gate, output at collector/drain. Large inverting voltage gain, moderate input and output resistance.
  • Common collector (CC, emitter follower) / common drain (CD, source follower): input at base/gate, output at emitter/source. Voltage gain just below 1, non-inverting, high input resistance, low output resistance. Used as a buffer.
  • Common base (CB) / common gate (CG): input at emitter/source, output at collector/drain. Non-inverting gain similar in magnitude to CE, but very low input resistance (≈ 1/g_m). Used in cascodes and for wide bandwidth.

Small-signal method. Short the coupling and bypass capacitors, zero the DC supplies, replace the transistor by its hybrid-π model, then find A_v = v_o/v_i, the input resistance R_in and the output resistance R_out. The overall gain from a source with resistance R_sig is A_v·R_in/(R_in + R_sig): a stage with low R_in wastes much of the source voltage.

Emitter degeneration. An unbypassed emitter resistor R_E reflects into the base as (β + 1)·R_E, raising R_in, and reduces the gain to about −R_C/R_E when (β + 1)R_E ≫ r_π. The gain then depends on resistors rather than on β or I_C: this is local negative feedback, studied fully in the feedback topic. The MOSFET equivalent is the source resistor R_S, giving −g_m·R_D/(1 + g_m·R_S).

Followers. In an emitter follower the output resistance seen at the emitter is (r_π + R_sig)/(β + 1), i.e. the source resistance is divided by (β + 1). That is why a follower can drive a low-impedance load from a high-impedance source. The source follower has R_out ≈ 1/g_m, which is larger than a BJT follower's at the same current because MOSFET g_m is smaller.

BJT versus MOSFET. At equal bias current a BJT has higher g_m (I_C/V_T versus 2I_D/V_ov) and therefore more gain; a MOSFET has essentially infinite gate input resistance. Instrument front ends for high-impedance sensors use FET inputs for this reason.

Formulas

A_v = −g_m·(R_C ∥ R_L) — CE, R_E bypassed, r_o neglected; R_in = R_B ∥ r_π; R_out ≈ R_C.

A_v ≈ −β·(R_C ∥ R_L)/(r_π + (β + 1)·R_E) ≈ −(R_C ∥ R_L)/R_E — CE with unbypassed R_E; R_in = R_B ∥ (r_π + (β + 1)·R_E).

A_v = −g_m·(R_D ∥ R_L) — CS, source bypassed; with unbypassed R_S: A_v = −g_m·(R_D ∥ R_L)/(1 + g_m·R_S).

A_v = (β + 1)·R'_E / (r_π + (β + 1)·R'_E), with R'_E = R_E ∥ R_L — emitter follower.

R_in = r_π + (β + 1)·R'_E (in parallel with the bias resistors) — emitter follower.

R_out = R_E ∥ (r_π + R_sig)/(β + 1) — emitter follower; R_sig is the source resistance (Ω).

A_v = g_m·R'_S / (1 + g_m·R'_S), R_out = R_S ∥ (1/g_m) — source follower, R'_S = R_S ∥ R_L.

A_v = +g_m·(R_C ∥ R_L), R_in ≈ 1/g_m — common base/gate.

v_o/v_sig = A_v · R_in/(R_in + R_sig) — overall gain including source loading.

Worked examples

Example 1 (standard: emitter follower). β = 100, I_C = 2 mA, V_T = 26 mV, R_E = 1 kΩ, R_L = 1 kΩ, source resistance R_sig = 10 kΩ. Neglect the bias resistors and r_o. Find R_in, A_v, the overall gain and R_out.

  1. g_m = I_C/V_T = 2/26 = 76.9 mS; r_π = β/g_m = 1.30 kΩ.
  2. R'_E = 1 kΩ ∥ 1 kΩ = 500 Ω.
  3. R_in = r_π + (β + 1)R'_E = 1.30 kΩ + 101 × 0.5 kΩ = 51.8 kΩ.
  4. A_v = (β + 1)R'_E/R_in = 50.5/51.8 = 0.975.
  5. Overall: v_o/v_sig = A_v·R_in/(R_in + R_sig) = 0.975 × 51.8/61.8 = 0.817.
  6. R_out = R_E ∥ (r_π + R_sig)/(β + 1) = 1 kΩ ∥ (11.3 kΩ/101) = 1 kΩ ∥ 112 Ω = 101 Ω.

Answer: R_in ≈ 51.8 kΩ, A_v ≈ 0.975, v_o/v_sig ≈ 0.82, R_out ≈ 101 Ω. The 10 kΩ source now looks like about 100 Ω to the load.

Example 2 (GATE level: CE with source and load, then a source follower). (a) A CE stage has I_C = 1 mA, β = 100, V_T = 25 mV, R_C = 5 kΩ, R_L = 5 kΩ, bias network R_B = R1 ∥ R2 = 20 kΩ, emitter fully bypassed, R_sig = 1 kΩ. Find v_o/v_sig. (b) A source follower has g_m = 4 mS, R_S = 2 kΩ, R_L = 2 kΩ. Find A_v and R_out.

  1. (a) g_m = 1 mA/25 mV = 40 mS; r_π = 100/0.04 = 2.5 kΩ.
  2. R_in = R_B ∥ r_π = 20 ∥ 2.5 = 2.22 kΩ.
  3. A_v = −g_m(R_C ∥ R_L) = −0.04 × 2500 = −100.
  4. v_o/v_sig = A_v·R_in/(R_in + R_sig) = −100 × 2.22/3.22 = −69.0.
  5. (b) R'_S = 2 ∥ 2 = 1 kΩ; A_v = g_m·R'_S/(1 + g_m·R'_S) = 4/(1 + 4) = 0.80.
  6. R_out = R_S ∥ 1/g_m = 2 kΩ ∥ 250 Ω = 222 Ω.

Answer: (a) v_o/v_sig ≈ −69; (b) A_v = 0.80, R_out ≈ 222 Ω.

Common mistakes

  • Quoting the transistor-only gain when the question asks for gain from the source; include R_in/(R_in + R_sig).
  • Forgetting that the load R_L is in parallel with R_C (or R_E in a follower).
  • Using −R_C/R_E when R_E is bypassed, or −g_m·R_C when it is not.
  • Assuming a follower has gain exactly 1; with a light R'_E (or a MOSFET) it can be well below 1.
  • Forgetting the bias resistors R1 ∥ R2 when computing R_in of a CE or CC stage.
  • Thinking a source follower's output resistance is as low as an emitter follower's at the same current.

For GATE IN

Expect gain, input-resistance and output-resistance questions on CE, CS, CC and CD stages, often with emitter or source degeneration, and comparison questions (which configuration has the highest input resistance, lowest output resistance, non-inverting gain). Practise drawing the small-signal circuit and using the "(β + 1) reflection" rule in both directions.

Quick check

  1. Which configuration has a voltage gain slightly below 1 and no phase inversion?
  2. A CE stage has R_C = 6 kΩ and an unbypassed R_E = 300 Ω. Approximate A_v?
  3. A source follower has g_m = 10 mS and R_S → ∞. What is R_out?
  4. Why does a common-base stage have low input resistance?

Answers: 1. Common collector (emitter follower) or common drain (source follower). 2. About −20. 3. 100 Ω. 4. The signal enters at the emitter, where the resistance looking in is about 1/g_m (r_e).

Interview questions

All Analog Electronics interview questions

Try answering each one aloud before you open it.

  1. 1.What is a common-emitter (CE) amplifier and how does it work?Concept

    A common-emitter (CE) amplifier is a type of single-stage amplifier where the emitter terminal of a bipolar junction transistor (BJT) is common to both the input and output circuits. It works by using the transistor to amplify the input voltage signal. The input signal is applied to the base-emitter junction, and the amplified output is taken from the collector-emitter circuit. The CE amplifier provides high voltage gain and is widely used in audio and radio frequency applications.

  2. 2.Explain the operation of a common-source (CS) amplifier.Concept

    A common-source (CS) amplifier is a single-stage amplifier configuration using a field-effect transistor (FET) where the source terminal is common to both the input and output. The input signal is applied to the gate-source junction, and the output is taken from the drain-source circuit. The CS amplifier provides high voltage gain and is commonly used in analog circuits for amplification purposes. It is analogous to the common-emitter amplifier in BJTs.

  3. 3.What is a voltage follower and what are its key characteristics?Concept

    A voltage follower, also known as a buffer amplifier, is a configuration where the output voltage follows the input voltage. It has a gain of approximately one, meaning it does not amplify the voltage but provides high input impedance and low output impedance. This makes it ideal for impedance matching and isolating different stages of a circuit without affecting the signal voltage.

  4. 4.Why is a common-emitter stage used as the voltage-gain stage in audio preamplifiers?Application

    A CE stage gives large voltage gain (−g_m·R_C, often 50-200) together with current gain of about β, so it delivers the highest power gain of the single-BJT configurations. Its input resistance of a few kΩ is acceptable for microphones and line sources. It cannot drive a loudspeaker directly because its output resistance is about R_C, so it is followed by an emitter follower or a push-pull power stage.

  5. 5.What happens if the emitter resistor in a CE amplifier is bypassed with a capacitor?Application

    Bypassing the emitter resistor with a capacitor in a CE amplifier increases the AC gain of the amplifier. The capacitor provides a low impedance path for AC signals, effectively removing the emitter resistor from the AC signal path. This increases the voltage gain because the emitter resistor no longer reduces the AC gain, while the DC biasing remains unaffected.

  6. 6.How does the emitter bypass capacitor affect the frequency response of a CE amplifier?Application

    At mid and high frequencies the bypass capacitor shorts R_E and the full gain −g_m·R_C is obtained. At low frequencies its reactance becomes comparable with the resistance it sees, about R_E ∥ (r_e + R_sig'/(β + 1)), which is small, so the gain falls towards −R_C/R_E. Because that resistance is small, C_E usually sets the dominant lower cut-off frequency and must be much larger than the coupling capacitors.

  7. 7.Estimate the voltage gain of a CE amplifier with R_C = 4.7 kΩ, an unbypassed emitter resistor of 470 Ω and β = 100.Numerical

    With an unbypassed emitter resistor the exact gain is A_v = −β·R_C/(r_π + (β + 1)·R_E). When (β + 1)·R_E (here about 47.5 kΩ) is much larger than r_π, this reduces to A_v ≈ −R_C/R_E = −4700/470 = −10. The gain is then set by two resistors, nearly independent of β and bias current, which is the point of degeneration.

  8. 8.What is the effect of increasing the unbypassed source resistor in a CS amplifier on its gain?Application

    With an unbypassed source resistor the gain becomes A_v = −g_m·R_D/(1 + g_m·R_S). Increasing R_S therefore reduces the gain, and when g_m·R_S ≫ 1 the gain approaches −R_D/R_S. In return the gain becomes less sensitive to g_m, and linearity and bias stability improve, because R_S provides local negative feedback.

  9. 9.Explain why a voltage follower is used in impedance matching.Application

    A voltage follower is used in impedance matching because it has high input impedance and low output impedance. This allows it to connect a high-impedance source to a low-impedance load without significant signal loss. The voltage follower effectively isolates the source from the load, ensuring that the source can drive the load without being affected by its impedance.

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