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Lab Circuit Analysis

week 05 — BJT common-emitter amplifier — transfer function & frequency response

Step-by-step analysis of a voltage-divider biased common-emitter BJT amplifier. We identify the DC operating point, build the AC small-signal model, derive the transfer function H(s) = Vo/Vs, find the poles, and sketch the Bode and Nyquist plots.

Part 01

Circuit Identification & Component Values

The circuit is a common-emitter (CE) BJT amplifier with voltage-divider biasing. Here are the components:

ComponentValueRole
Vs(t)10 mVpk, 5 kHz, 0°AC signal source
RS100 ΩSource resistance
CC10.1 μFInput coupling capacitor
RB1200 kΩUpper bias resistor
RB2100 kΩLower bias resistor
RC2 kΩCollector resistor
RE0.5 kΩEmitter resistor
CE220 μFEmitter bypass capacitor
CC20.1 μFOutput coupling capacitor
RL2 kΩLoad resistor
VCC5 VDC supply
BJT2N3904β = 200, fT = 300 MHz, Cμ = 4 pF
Node labels from the schematic: Node 7 = source input, Node 5 = after RS, Node 1 = base, Node 3 = collector, Node 4 = VCC rail, Node 2 = emitter, Node 0 = ground, Node 6 = output side of CC2.
Part 02

DC Operating Point (Q-Point)

At DC, all capacitors are open circuits. We need to find IC, VCE, and the transconductance gm for the small-signal model.

Step 1: Thevenin Base Voltage

The voltage divider RB1 and RB2 sets the base voltage:

VBB = VCC · RB2RB1 + RB2 = 5 · 100k200k + 100k = 1.667 V

Step 2: Thevenin Base Resistance

RBB = RB1 || RB2 = 200k · 100k200k + 100k = 66.67 kΩ

Step 3: Find IC (assuming β = 200)

Using the approximation IC ≈ (VBB − VBE) / (RE + RBB/β), with VBE ≈ 0.7 V:

IC ≈ VBB − VBERE + RBB/β = 1.667 − 0.7500 + 333.5 = 0.967833.5 = 1.160 mA

Step 4: VCE

VCE = VCC − IC(RC + RE) = 5 − 1.160m · 2.5k = 5 − 2.90 = 2.10 V
Check: VCE = 2.10 V > 0.2 V ⇒ the transistor is in active mode. The Q-point is valid.

Step 5: Small-Signal Parameters

gm = ICVT = 1.160 mA26 mV = 44.62 mA/V
rπ = βgm = 20044.62 mA/V = 4483 Ω ≈ 4.48 kΩ
Part 03

AC Small-Signal Equivalent Circuit

To build the AC model, we replace the BJT with its hybrid-π model, short VCC to ground, and keep all capacitors as impedances Z = 1/sC.

Rules for AC Model

Resulting AC Circuit

The AC equivalent has three capacitor-dependent paths:

Input Stage (Node 7 → Node 1)

Vs → RS → CC1 → (RB1 || RB2) || rπ ≈ 4.20 kΩ

The input coupling capacitor CC1 forms a high-pass filter with the total resistance it sees.

Emitter Stage (Node 2 → Ground)

RE in parallel with CE. At low frequencies, RE provides negative feedback (reducing gain). As frequency increases, CE bypasses RE, increasing the gain.

Output Stage (Node 3 → Node 6)

Collector node through CC2 to RL. Another high-pass filter.

Part 04

Transfer Function Derivation

We derive AV(s) = Vo(s) / Vs(s) as the product of the midband gain and frequency-dependent pole/zero terms. Define shorthand:

RBB = RB1 || RB2 = 66.67 kΩ    RC′ = RC || RL = 1 kΩ

Step 1: Midband Gain

At midband frequencies (coupling/bypass caps are shorts, parasitic caps are opens):

Rin = RBB || rπ = 66.67k · 4.48k66.67k + 4.48k ≈ 4.20 kΩ
Amid = −gm(RC || RL) · RBB || rπRS + RBB || rπ = −44.62 × 10−3 × 1000 × 4200100 + 4200 = −43.6 V/V  (≈ 32.8 dB)

Step 2: Low-Frequency Poles (High-Pass)

Input coupling (CC1):

ωL1 = 1CC1(RS + RBB || rπ) = 10.1μ · (100 + 4200) = 14.3 × 10−4 = 2326 rad/s  (fL1 ≈ 370 Hz)

Output coupling (CC2):

ωL2 = 1CC2(RC + RL) = 10.1μ · (2k + 2k) = 2500 rad/s  (fL2 ≈ 398 Hz)

Emitter bypass (CE): The resistance seen by CE is RE in parallel with the resistance looking into the emitter:

Rth = RS || RBB = 100 || 66.67k ≈ 100 Ω
ωL3 = 1CE · (RE || rπ + Rthβ + 1) = 1220μ · (500 || 22.8) = 1220μ · 21.8 = 208.5 rad/s  (fL3 ≈ 33 Hz)

Emitter bypass zero:

ωZ,CE = 1CE · RE = 1220μ · 500 = 9.09 rad/s  (fZ ≈ 1.4 Hz)

Step 3: High-Frequency Poles (Low-Pass, Miller Effect)

The BJT has internal capacitances Cπ and Cμ. Using fT = 300 MHz:

Cπ = gm2πfT − Cμ = 23.7 pF − 4 pF = 19.7 pF

Miller input capacitance:

Cin,Miller = Cπ + Cμ(1 + gmRC′) = 19.7 pF + 4 pF × (1 + 44.6) = 202.1 pF

Dominant input pole:

Rth,b = RS || RBB || rπ ≈ 97.7 Ω
ωH1 = 1Rth,b · Cin,Miller ⇒  fH1 ≈ 8.06 MHz

Output pole:

ωH2 = 1RC′ · Cμ = 11000 · 4 pF ⇒  fH2 ≈ 39.8 MHz

Step 4: Complete Transfer Function

Combining all stages:

AV(s) = Amid · ss + ωL1 · ss + ωL2 · s + ωZs + ωL3 · ωH1s + ωH1 · ωH2s + ωH2
Note: At very low frequencies (below ωZ), the emitter bypass is inactive and the gain reduces to gmRC′ / (1 + gmRE) ≈ 1.94 (5.8 dB). The bypass capacitor provides a gain boost of about 27 dB in the midband. The high-frequency response is limited by the Miller effect, with Cin,Miller ≈ 202 pF significantly exceeding the intrinsic Cπ.
Part 05

Poles & Zeros Summary
ParameterOriginFrequencyType
fL1CC1 input coupling370 HzHP pole
fL2CC2 output coupling398 HzHP pole
fL3CE emitter bypass33 HzHP pole
fZ,CECE emitter bypass1.4 HzZero
fH1Miller (Cπ + Cμ)8.06 MHzLP pole
fH2Output Cμ39.8 MHzLP pole
Key observations: The low-frequency −3 dB point is dominated by fL1 ≈ 370 Hz and fL2 ≈ 398 Hz (coupling capacitors). The high-frequency response is limited by the Miller effect at fH1 ≈ 8.06 MHz. In the midband, the phase is flat at −180°.
Part 06

Frequency Response & Nyquist

The Bode magnitude and phase plots show the frequency response of AV(s) from 10 Hz to 100 MHz. The Nyquist plot traces AV(jω) in the complex plane. Drag the frequency cursor to see the corresponding point on all three plots.

Circuit Parameters Interactive

0.10 μF
0.10 μF
220 μF
2.00 kΩ
2.00 kΩ
200
Amid =  |  fL1 =  |  fL2 =  |  fL3 =  |  fH1 =
10.0 kHz
|AV| =  |  ∠AV =  |  ∠ (0–360) =  |  Re =  |  Im =
Reading the plots: The midband gain is ≈ 32.8 dB. The coupling capacitor poles (fL1, fL2) cause the low-frequency roll-off. The Miller pole at fH1 dominates the high-frequency response. On the Nyquist plot, the locus starts at the origin (ω = 0), swings to the negative real axis at midband (phase ≈ −180°), and returns toward the origin at high frequencies.