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title: 'Lecture 22: Effect of Emitter Bypass Capacitor and Coupling Capacitor on...'
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  ## DI02011011: Electronics Circuit and Application (ECA)
  Frequency Response of Transistor Amplifier
---

# DI02011011: Electronics Circuit and Application (ECA)
## Lecture 22: Effect of Emitter Bypass Capacitor and Coupling Capacitor on...
### Frequency Response of Transistor Amplifier

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Welcome to Lecture 22. In this lecture, we examine the quantitative mechanisms by which external coupling and bypass capacitors restrict the low-frequency performance of BJT amplifiers. We will derive the equivalent resistance seen by each capacitor and establish how to calculate the lower 3-dB cutoff frequency.
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# Lecture Agenda & Learning Roadmap

- Low-Frequency Limitations and High-Pass RC Filter Action in Amplifiers
- Input Coupling Capacitor ($C_{C1}$) Network & Cutoff Frequency ($f_{L1}$) Derivation
- Output Coupling Capacitor ($C_{C2}$) Network & Cutoff Frequency ($f_{L2}$) Derivation
- Emitter Bypass Capacitor ($C_E$) AC Impedance & Cutoff Frequency ($f_{LE}$) Derivation
- Pole-Zero Analysis of the Emitter Bypass Impedance Network $Z_E(s)$
- Dominant Pole Approximation & Low-Frequency Asymptotic Bode Plotting
- Detailed Step-by-Step Numerical Calculation of $f_{L1}, f_{L2}, f_{LE}$, and Overall $f_L$
- Low-Frequency Phase Lead Response and Pulse Distortion (Sag/Tilt)
- Engineering Design Guidelines for Low-Frequency Capacitor Selection
- Design Calculation Example for Target Lower Cutoff Frequency $f_L = 50\text{ Hz}$
- Comprehensive Summary and Design Rules

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This roadmap details our study of low-frequency response. We will systematically analyze each capacitor's equivalent network, compute individual poles, solve a comprehensive numerical example, and discuss design criteria.
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# Low-Frequency Signal Attenuation Mechanisms in BJT Amplifiers

- Capacitive Reactance Frequency Dependency: AC reactance $X_C(f) = \frac{1}{2\pi f C}$ approaches infinity as frequency $f \to 0\text{ Hz}$, acting as a open circuit to DC and attenuating low-frequency AC signals.
- High-Pass Filter Topology: Coupling capacitors $C_{C1}$ and $C_{C2}$ are in series with the signal path, forming high-pass RC filter sections with circuit resistances.
- Bypass Capacitor Degeneration: Emitter bypass capacitor $C_E$ is in parallel with emitter resistor $R_E$. At low frequencies where $X_{C E}$ is large, negative feedback across $R_E$ reduces AC voltage gain.
- Midband Transition: As frequency increases above $f_L$, $X_C \to 0$, capacitors act as AC short circuits, and amplifier gain stabilizes at midband value $A_M$.

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At low frequencies, coupling and bypass capacitors no longer act as short circuits. As f decreases, XC rises, creating voltage divider losses across coupling capacitors and introducing unbypassed emitter resistance RE that dramatically cuts gain.
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# Input Coupling Capacitor ($C_{C1}$) Network & Cutoff Frequency ($f_{L1}$)

- Equivalent Input Resistance ($R_{in,C1}$): Seen looking into the nodes connected to $C_{C1}$: $R_{in,C1} = R_S + R_{in,stage} = R_S + (R_1 \parallel R_2 \parallel R_{in,base})$, where $R_{in,base} = r_\pi + (1+\beta_0)R_E$ if unbypassed at DC.
- Transfer Function Contribution: $T_1(s) = \frac{V_{b}(s)}{V_s(s)} = \left(\frac{R_{in,stage}}{R_S + R_{in,stage}}\right) \frac{s}{s + \omega_{L1}}$, where $\omega_{L1} = \frac{1}{C_{C1} R_{in,C1}}$.
- Lower Cutoff Pole Frequency ($f_{L1}$): $f_{L1} = \frac{1}{2\pi C_{C1} R_{in,C1}} = \frac{1}{2\pi C_{C1} \left[ R_S + (R_1 \parallel R_2 \parallel R_{in,base}) \right]}$.
- Low-Frequency Roll-Off & Phase Shift: Below $f_{L1}$, voltage transfer drops at $+20\text{ dB/decade}$ with frequency decrease, and introduces a phase lead of $+45^\circ$ at $f = f_{L1}$.

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To find fL1, we turn off independent sources and calculate the Thevenin equivalent resistance connected across CC1. Notice that the signal source resistance RS adds directly to the stage input resistance Rin,stage.
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# Output Coupling Capacitor ($C_{C2}$) Network & Cutoff Frequency ($f_{L2}$)

- Equivalent Output Resistance ($R_{out,C2}$): Seen looking from the terminals of $C_{C2}$: $R_{out,C2} = R_{out,stage} + R_L = R_C + R_L$ (assuming transistor output resistance $r_o \gg R_C$).
- Output Transfer Function Expression: $T_2(s) = \frac{V_o(s)}{V_c(s)} = \left(\frac{R_L}{R_C + R_L}\right) \frac{s}{s + \omega_{L2}}$, where $\omega_{L2} = \frac{1}{C_{C2} (R_C + R_L)}$.
- Output Cutoff Pole Frequency ($f_{L2}$): $f_{L2} = \frac{1}{2\pi C_{C2} (R_C + R_L)}$.
- Loading Effect Impact: Larger load resistance $R_L$ increases $R_{out,C2}$, which lowers the required capacitance $C_{C2}$ for a given low-frequency cutoff target $f_{L2}$.

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The output coupling capacitor CC2 sees the collector resistor RC in series with the load resistor RL. Calculating fL2 is straightforward because the collector node impedance is simply RC plus RL.
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# Emitter Bypass Capacitor ($C_E$) AC Impedance & Cutoff Frequency ($f_{LE}$)

- Equivalent Resistance Seen by $C_E$ ($R_{eq,E}$): Seen looking into the emitter terminal: $R_{eq,E} = R_E \parallel R_{emitter,in} = R_E \parallel \left[ r_e + \frac{R_S \parallel R_1 \parallel R_2}{\beta_0 + 1} \right]$, where small-signal emitter resistance $r_e = \frac{V_T}{I_C} = \frac{r_\pi}{\beta_0 + 1}$.
- Emitter Pole Frequency Formula ($f_{LE}$): $f_{LE} = \frac{1}{2\pi C_E R_{eq,E}} = \frac{1}{2\pi C_E \left( R_E \parallel \left[ r_e + \frac{R_S \parallel R_1 \parallel R_2}{\beta_0 + 1} \right] \right)}$.
- Small Equivalent Resistance Implication: Because $r_e$ is small (typically $10 - 30\ \Omega$), $R_{eq,E}$ is very small ($15 - 50\ \Omega$). Thus, $f_{LE}$ is much higher than $f_{L1}$ and $f_{L2}$ for equal capacitor values.
- Dominant Pole Role: To keep $f_{LE}$ down near audio frequencies, $C_E$ must be significantly larger ($47 - 220\ \mu\text{F}$) than $C_{C1}$ or $C_{C2}$ ($1 - 10\ \mu\text{F}$).

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Pay close attention to Req,E! Because the base circuit resistance is divided by (beta + 1) when reflected into the emitter, Req,E is very small—often under 30 ohms. A tiny resistance requires a huge capacitance CE to achieve a low cutoff frequency.
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# Pole-Zero Derivation of the Emitter Bypass Impedance Network $Z_E(s)$

- Impedance Expression for $R_E \parallel C_E$: $Z_E(s) = R_E \parallel \frac{1}{s C_E} = \frac{R_E}{1 + s R_E C_E}$.
- Unbypassed Gain Equation: Incorporating $Z_E(s)$ into CE amplifier voltage gain: $A_v(s) = -\frac{\beta_0 (R_C \parallel R_L)}{r_\pi + (\beta_0 + 1) Z_E(s)} = -g_m (R_C \parallel R_L) \frac{1 + s R_E C_E}{1 + s R_E C_E + g_m R_E}$.
- Zero Frequency Location ($\omega_z$): The numerator zero occurs at $\omega_z = \frac{1}{R_E C_E} \implies f_z = \frac{1}{2\pi R_E C_E}$.
- Pole Frequency Location ($\omega_p$): The denominator pole occurs at $\omega_p = \frac{1 + g_m R_E}{R_E C_E} = \frac{1}{C_E R_{eq,E}} \implies f_{LE} = \frac{1}{2\pi C_E R_{eq,E}}$.
- Gain Transition: At frequencies below $f_z$, gain drops to unbypassed value $A_{v,unbypassed} = -\frac{R_C \parallel R_L}{R_E}$. Above $f_{LE}$, gain rises to full midband bypassed gain $A_M = -g_m (R_C \parallel R_L)$.

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This pole-zero derivation reveals the complete behavior of the bypass network. The zero fz occurs at lower frequency 1/(2*pi*RE*CE), setting the minimum unbypassed gain floor, while the pole fLE sets the transition to full midband gain.
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# Dominant Pole Approximation & Low-Frequency Asymptotic Response

- Overall Low-Frequency Transfer Function: $A_L(s) = A_M \cdot \left(\frac{s}{s + \omega_{L1}}\right) \left(\frac{s}{s + \omega_{L2}}\right) \left(\frac{s + \omega_z}{s + \omega_{LE}}\right)$.
- Dominant Pole Condition: If one pole frequency is significantly higher than all other low-frequency poles (e.g., $f_{LE} \ge 4 f_{L1}$ and $f_{LE} \ge 4 f_{L2}$), $f_{LE}$ dominates the lower 3-dB cutoff: $f_L \approx f_{LE}$.
- Non-Dominant Pole Combination Formula: When poles are closely spaced, overall lower cutoff $f_L$ is estimated using short-circuit sum or root-square approximation: $f_L \approx \sqrt{f_{L1}^2 + f_{L2}^2 + f_{LE}^2}$ or $f_L \approx f_{L1} + f_{L2} + f_{LE}$.
- Asymptotic Bode Slope Progression: Gain magnitude slope increases from $+20\text{ dB/decade}$ below $f_{LE}$, to $+40\text{ dB/decade}$ below $f_{L1}$, to $+60\text{ dB/decade}$ below $f_{L2}$.

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When analyzing multiple high-pass cutoffs, the largest pole frequency determines where gain first drops by 3 dB as frequency decreases from midband. This highest low-frequency pole is called the dominant low-frequency pole.
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# Numerical Example 1: Determining $f_{L1}, f_{L2}, f_{LE}$ and Overall $f_L$

- Circuit Parameters: $V_{CC}=12\text{ V}, R_1=40\text{ k}\Omega, R_2=10\text{ k}\Omega, R_C=4\text{ k}\Omega, R_E=2\text{ k}\Omega, R_S=600\ \Omega, R_L=10\text{ k}\Omega, \beta_0=100, I_C=1\text{ mA}$. Capacitors: $C_{C1}=10\ \mu\text{F}, C_{C2}=4.7\ \mu\text{F}, C_E=47\ \mu\text{F}$. Small-signal parameters: $V_T=26\text{ mV} \implies r_e = 26\ \Omega, g_m = 38.46\text{ mS}, r_\pi = 2.6\text{ k}\Omega$.
- Step 1 - Calculate $f_{L1}$: $R_{B} = 40\text{ k} \parallel 10\text{ k} = 8\text{ k}\Omega$. $R_{in,stage} = R_B \parallel r_\pi = 8\text{ k} \parallel 2.6\text{ k} = 1.962\text{ k}\Omega$. $R_{in,C1} = 600 + 1962 = 2.562\text{ k}\Omega$. $f_{L1} = \frac{1}{2\pi \times 10\mu\text{F} \times 2562} = 6.21\text{ Hz}$.
- Step 2 - Calculate $f_{L2}$: $R_{out,C2} = R_C + R_L = 4\text{ k} + 10\text{ k} = 14\text{ k}\Omega$. $f_{L2} = \frac{1}{2\pi \times 4.7\mu\text{F} \times 14000} = 2.42\text{ Hz}$.
- Step 3 - Calculate $f_{LE}$: $R_{base,eq} = R_S \parallel R_1 \parallel R_2 = 600 \parallel 8000 = 558\ \Omega$. $R_{emitter,in} = r_e + \frac{558}{101} = 26 + 5.52 = 31.52\ \Omega$. $R_{eq,E} = 2000 \parallel 31.52 = 31.03\ \Omega$. $f_{LE} = \frac{1}{2\pi \times 47\mu\text{F} \times 31.03} = 109.13\text{ Hz}$.
- Step 4 - Overall $f_L$: Dominant pole $f_{LE} = 109.13\text{ Hz}$. Root-square estimate $f_L = \sqrt{6.21^2 + 2.42^2 + 109.13^2} = 109.33\text{ Hz}$.

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Examine the numbers carefully! fLE is 109.13 Hz, whereas fL1 is 6.21 Hz and fL2 is 2.42 Hz. fLE is over 17 times larger than fL1, proving beyond doubt that the emitter bypass capacitor CE completely dominates the low-frequency response.
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# Low-Frequency Phase Lead & Signal Distortion

- Total Low-Frequency Phase Lead Equation: $\theta(f) = +\arctan\left(\frac{f_{L1}}{f}\right) + \arctan\left(\frac{f_{L2}}{f}\right) + \arctan\left(\frac{f_{LE}}{f}\right) - \arctan\left(\frac{f_z}{f}\right)$.
- Midband vs Low-Frequency Phase: At midband ($f \gg f_L$), $\theta(f) \approx 0^\circ$ relative to the $180^\circ$ inversion of Common-Emitter configuration.
- Pulse Response Distortion (Square Wave Sag/Tilt): Passing a pulse of width $T_W$ through high-pass cutoff $f_L$ causes exponential rooftop decay.
- Percentage Sag Formula: Fractional voltage droop $S = \frac{\Delta V}{V_{peak}} \approx \pi \frac{f_L}{f_{pulse}} \times 100\%$. Keeping sag under $5\%$ requires $f_L \le 0.016 f_{pulse}$.

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High-pass cutoff networks do not merely attenuate AC amplitude; they shift phase lead. In digital and audio pulse applications, low-frequency phase lead causes rooftop tilt or sag in square waves, degrading signal fidelity.
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# Capacitor Selection Strategies for Target Lower Cutoff Frequency

- Design Budgeting Strategy: When designing for a target lower 3-dB cutoff $f_L$, allocate contributions across capacitors to minimize physical component sizes and costs.
- Optimal Pole Budget Rule of Thumb: Set the emitter bypass pole $f_{LE} = 0.8 f_L$, and input/output coupling poles $f_{L1} = 0.1 f_L$, $f_{L2} = 0.1 f_L$.
- Minimizing Bypass Capacitor Size: Since $C_E = \frac{1}{2\pi f_{LE} R_{eq,E}}$, allocating $80\%$ of $f_L$ to $f_{LE}$ maximizes $f_{LE}$, reducing required physical capacitance $C_E$.
- Practical Considerations: Electrolytic capacitors used for $C_E$ have wide tolerances ($\pm 20\%$) and Equivalent Series Resistance (ESR) that must be factored into high-precision designs.

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When designing an amplifier for a specific fL target, allocating 80% of the frequency budget to fLE allows the engineer to select the smallest possible value for CE, saving PCB area and component cost.
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# Numerical Example 2: Designing Capacitors for a Specified $f_L = 50\text{ Hz}$

- Design Requirement: Size capacitors $C_{C1}, C_{C2}, C_E$ for a CE stage with $R_{in,C1} = 3\text{ k}\Omega, R_{out,C2} = 12\text{ k}\Omega, R_{eq,E} = 25\ \Omega$ to achieve a lower 3-dB cutoff $f_L = 50\text{ Hz}$.
- Step 1 - Allocate Pole Budgets: Target $f_{LE} = 0.8 \times 50\text{ Hz} = 40\text{ Hz}$, $f_{L1} = 0.1 \times 50\text{ Hz} = 5\text{ Hz}$, $f_{L2} = 0.1 \times 50\text{ Hz} = 5\text{ Hz}$.
- Step 2 - Calculate Input Coupling Capacitor $C_{C1}$: $C_{C1} = \frac{1}{2\pi \times 5\text{ Hz} \times 3000\ \Omega} = 10.61\ \mu\text{F} \implies \text{Select standard } 15\ \mu\text{F}$.
- Step 3 - Calculate Output Coupling Capacitor $C_{C2}$: $C_{C2} = \frac{1}{2\pi \times 5\text{ Hz} \times 12000\ \Omega} = 2.65\ \mu\text{F} \implies \text{Select standard } 3.3\ \mu\text{F}$.
- Step 4 - Calculate Emitter Bypass Capacitor $C_E$: $C_E = \frac{1}{2\pi \times 40\text{ Hz} \times 25\ \Omega} = 159.15\ \mu\text{F} \implies \text{Select standard } 220\ \mu\text{F}$.

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Here we see practical component selection in action. Notice how after calculating theoretical values, we round up to standard commercial capacitor values to guarantee meeting or exceeding the 50 Hz specification.
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# Summary of Bypass and Coupling Capacitor Effects on Frequency Response

- Coupling Capacitors ($C_{C1}, C_{C2}$): Determine high-pass cutoffs $f_{L1} = \frac{1}{2\pi C_{C1} R_{in,C1}}$ and $f_{L2} = \frac{1}{2\pi C_{C2} R_{out,C2}}$.
- Bypass Capacitor ($C_E$): Determines cutoff $f_{LE} = \frac{1}{2\pi C_E R_{eq,E}}$. Requires largest capacitance because $R_{eq,E} = R_E \parallel \left[ r_e + \frac{R_{base}}{\beta_0+1} \right]$ is very small.
- Pole-Zero Behavior: Bypass network creates a zero $f_z = \frac{1}{2\pi R_E C_E}$ and pole $f_{LE}$. Gain transitions from unbypassed value below $f_z$ to bypassed midband value above $f_{LE}$.
- Dominant Pole Approximation: Lower 3-dB cutoff $f_L \approx \sqrt{f_{L1}^2 + f_{L2}^2 + f_{LE}^2} \approx f_{LE}$ when $f_{LE}$ is dominant.

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To wrap up Lecture 22, remember that low-frequency response is entirely dictated by external coupling and bypass capacitors. Emitter bypass capacitor CE is almost always the dominant factor due to the small equivalent resistance seen at the emitter node.
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