02 Low-Frequency Response & Cutoff Calculations

Here is your strictly PYQ-focused, comprehensive study note for the second major topic in the Frequency Response chapter: calculating the low-frequency response boundaries.

Topic 2: Low-Frequency Response & Cutoff Calculations

Exam-Focused Concept Note

This section is heavily tested via 10-to-15 mark numerical problems. The examiners will give you a circuit with multiple capacitors and ask you to find the overall lower cut-off frequency. You must also know the basic theory of why these specific capacitors affect the low-frequency region.

1. Core Concept: Why do these capacitors matter at low frequencies?

(Prerequisite Theory for PYQs)

In the mid-frequency range, the large network capacitors (the input coupling capacitor, the output coupling capacitor, and the emitter/source bypass capacitor) act as short circuits because their reactance is extremely small.
However, as the applied frequency drops into the low-frequency region, the capacitive reactance ( $X_{C} = \frac{1}{2 π f C}$ ) increases.
This increasing reactance takes an increasing share of the applied voltage, meaning less voltage reaches the amplifying transistor or the load. The specific frequency at which the output voltage drops to 70.7% (or -3 dB) of its maximum midband value is called the lower cut-off frequency,.

2. The “Highest is Boss” Rule (Crucial for Numericals)

In both BJT and FET circuits, you must calculate three separate cut-off frequencies (one for each capacitor).

The Rule: The highest of the three calculated low-cutoff frequencies will have the greatest impact and will essentially determine the overall lower cut-off frequency for the entire amplifier system,,.

3. BJT Low-Frequency Formulas (Memorize These)

For a standard Common-Emitter BJT amplifier with a voltage-divider bias and source resistance ( $R_{s}$ ), you must use these specific formulas to find the three break frequencies:

Capacitor	Purpose	Formula for Cut-off Frequency	Equivalent Resistance Formula
$C_{S}$	Input Coupling	$f_{L_{S}} = \frac{1}{2 π ( R _{s} + R _{i} ) C _{S}}$	$R_{i} = R_{1} ∥ R_{2} ∥ β r_{e}$
$C_{C}$	Output Coupling	$f_{L_{C}} = \frac{1}{2 π ( R _{o} + R _{L} ) C _{C}}$	$R_{o} = R_{C} ∥ r_{o} \approx R_{C}$
$C_{E}$	Emitter Bypass	$f_{L_{E}} = \frac{1}{2 π R _{e} C _{E}}$	$R_{e} = R_{E} ∥ (\frac{R _{s}^{'}}{β} + r_{e})$ (where $R_{s}^{'} = R_{s} ∥ R_{1} ∥ R_{2}$ )

4. JFET Low-Frequency Formulas (Memorize These)

For a standard JFET amplifier (like the one asked in your 2016 PYQ), the formulas are slightly different due to the high input impedance of FETs,:

Capacitor	Purpose	Formula for Cut-off Frequency	Equivalent Resistance Formula
$C_{G}$	Input Coupling	$f_{L_{G}} = \frac{1}{2 π ( R _{s i g} + R _{i} ) C _{G}}$	$R_{i} = R_{G}$
$C_{C}$	Output Coupling	$f_{L_{C}} = \frac{1}{2 π ( R _{o} + R _{L} ) C _{C}}$	$R_{o} = R_{D} ∥ r_{d} \approx R_{D}$
$C_{S}$	Source Bypass	$f_{L_{S}} = \frac{1}{2 π R _{e q} C _{S}}$	$R_{e q} = R_{S} ∥ \frac{1}{g _{m}}$

Exact PYQs to Master for this Topic:

Type 1: The Theoretical Explanation

Explain the frequency response of a common emitter amplifier and briefly discuss the impact of various capacitive element on transistor frequency response. (Asked in: 2023, 2020, 2017) (Answer Strategy: Explain that large coupling/bypass capacitors determine the low-frequency drop-off because their reactance increases at low frequencies, while small internal parasitic/wiring capacitors determine the high-frequency drop-off via the Miller effect. Draw the standard dome-shaped Bode plot,,,).

Type 2: The BJT Numerical (The 2022 Blueprint)

Determine the lower cutoff frequency for the network of Fig. 3(c) using following parameters: $C_{S} = 10 μ F, C_{E} = 20 μ F, C_{C} = 1 μ F, R_{S} = 1 k Ω, R_{1} = 40 k Ω, R_{2} = 10 k Ω, R_{E} = 2 k Ω, R_{C} = 4 k Ω, R_{L} = 2.2 k Ω, β = 100, r_{o} = \infty, V_{CC} = 20 V$ . (Asked in: 2022), (Note: This PYQ is an exact clone of Boylestad Example 9.11/9.12. You first find $r_{e}$ via DC analysis, then plug the values into the three BJT formulas above. $f_{L_{E}}$ will usually yield the highest frequency, typically around ~327 Hz, making it your final answer,,).

Type 3: The JFET Numerical (The 2016 Blueprint)

Determine the lower cut-off frequency for the network shown in Fig. 8(c) using the following parameters: $C_{G} = 0.01 μ F, C_{C} = 0.5 μ F, C_{S} = 2 μ F, R_{s i g} = 10 k Ω, R_{D} = 4.7 k Ω, R_{S} = 1 k Ω, R_{L} = 2.2 k Ω, I_{D SS} = 8 m A, V_{P} = - 4 V, r_{d} = \infty, V_{DD} = 20 V, R_{G} = 1 M Ω$ . (Asked in: 2016), (Note: This PYQ is an exact clone of Boylestad Example 9.13. You must first find $g_{m}$ using DC analysis, then calculate $f_{L_{G}}$ , $f_{L_{C}}$ , and $f_{L_{S}}$ using the JFET formulas above. The highest frequency will be your answer,).

Whenever you are ready, reply with “Topic 3”, and we will break down the final piece of this chapter: High-Frequency Response Calculations!

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