03 High-Frequency Response & Cutoff Calculations

Here is your strictly PYQ-focused, comprehensive study note for the third topic in the Frequency Response chapter.

Topic 3: High-Frequency Response & Cutoff Calculations

Exam-Focused Concept Note

This section tests your ability to apply the Miller Effect in a mathematical scenario. While the low-frequency response is governed by large physical capacitors, the high-frequency limits are dictated by invisible, microscopic capacitances inside the device itself.

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1. Core Concept: Why do high frequencies drop off?

(Prerequisite Theory)

• At high frequencies, the large coupling and bypass capacitors ($C_S,C_C,C_E$) that hindered low frequencies now act as perfect short circuits and are completely ignored.
• Instead, the small internal parasitic capacitors (like the base-emitter capacitance $C_{be}$ or gate-source capacitance $C_{gs}$) and the wiring capacitances ($C_{Wi},C_{Wo}$) become the problem.
• As frequency increases, the reactance of these tiny capacitors decreases ($X_C=\frac{1}{2\pi fC}$). They begin to act as shorting paths to ground, siphoning the signal away from the load and severely reducing the voltage gain.
• The “Lowest is Boss” Rule: You will calculate the high-frequency cutoff at the input ($f_{Hi}$) and at the output ($f_{Ho}$). The lowest of these calculated frequencies will dominate and establish the overall upper cutoff frequency for the amplifier.

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2. The High-Frequency Formulas to Memorize

Whether the exam asks for a BJT or a JFET, the calculation structure is identical. You simply use the specific capacitive parameters for that device.

Step 1: Calculate the Miller Capacitances Because the amplifier is inverting, the feedback capacitor ($C_f$) between the input and output is magnified by the Miller effect.

• Input Miller: $C_{Mi}=(1-A_v)C_f$
• Output Miller: $C_{Mo}=\left(1-\frac{1}{A_v}\right)C_f$ (Note: For a BJT, $C_f$ is $C_{bc}$. For a JFET, $C_f$ is $C_{gd}$. Because $A_v$ is a negative number, the input equation becomes an addition: $1+|A_v|$).

Step 2: Calculate Total Input and Output Capacitances

Parameter	BJT Amplifier Formulas	JFET Amplifier Formulas
Total Input Capacitance ($C_i$)	$C_i=C_{Wi}+C_{be}+C_{Mi}$	$C_i=C_{Wi}+C_{gs}+C_{Mi}$
Total Output Capacitance ($C_o$)	$C_o=C_{Wo}+C_{ce}+C_{Mo}$	$C_o=C_{Wo}+C_{ds}+C_{Mo}$

Step 3: Calculate the High-Cutoff Frequencies To find the frequencies, you must combine the total capacitances with the Thévenin equivalent resistances seen at the input ($R_{Thi}$) and output ($R_{Tho}$) terminals.

Parameter	BJT Resistance	JFET Resistance	The Cutoff Formula
Input ($f_{Hi}$)	$R_{Thi}=R_s\parallel R_1\parallel R_2\parallel R_i$	$R_{Thi}=R_{sig}\parallel R_G$	$f_{Hi}=\frac{1}{2\pi R_{Thi}{C}_i}$
Output ($f_{Ho}$)	$R_{Tho}=R_C\parallel R_L\parallel r_o$	$R_{Tho}=R_D\parallel R_L\parallel r_d$	$f_{Ho}=\frac{1}{2\pi R_{Tho}{C}_o}$

(Note: The exam will typically provide the resistance values or give you the circuit diagram to calculate them quickly).

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Exact PYQ to Master for this Topic:

The Guaranteed High-Frequency Numerical (The JFET 2022 Blueprint)

• The Question: “Determine the high-cutoff frequencies for the network shown in Fig. 6(c) using the following parameters: $A_v=-3$, $r_d=\infty$, $C_{gd}=2pF$, $C_{gs}=4pF$, $C_{ds}=0.5pF$, $C_{Wi}=5pF$, $C_{Wo}=6pF$.” (Asked in 2022).

How to solve this exact PYQ (Based on Boylestad Example 9.15): This is a direct clone of an example from your textbook. Here is the step-by-step mathematical solution you should practice:

1. Find the Input Miller Capacitance: $C_{Mi}=(1-A_v)C_{gd}=(1-(-3))(2pF)=(4)(2pF)=8pF$.
2. Find Total Input Capacitance: $C_i=C_{Wi}+C_{gs}+C_{Mi}=5pF+4pF+8pF=17pF$.
3. Find the Output Miller Capacitance: $C_{Mo}=(1-\frac{1}{A_v})C_{gd}=(1-\frac{1}{-3})(2pF)=(1+0.333)(2pF)=2.67pF$.
4. Find Total Output Capacitance: $C_o=C_{Wo}+C_{ds}+C_{Mo}=6pF+0.5pF+2.67pF=9.17pF$.
5. Calculate the Frequencies: Use the provided $R_{Thi}$ and $R_{Tho}$ from the exam’s circuit diagram and plug them into the formulas: $f_{Hi}=\frac{1}{2\pi R_{Thi}(17pF)}$ and $f_{Ho}=\frac{1}{2\pi R_{Tho}(9.17pF)}$. (In the textbook version of this problem, $R_{Thi}$ was $9.9k\Omega$ and $R_{Tho}$ was $1.5k\Omega$, resulting in $f_{Hi}=945.67\text{ kHz}$ and $f_{Ho}=11.57\text{ MHz}$. Because $f_{Hi}$ is the lowest, it dictates the overall high-frequency cutoff).

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Whenever you are ready, reply with “Topic 4”, and we will quickly knock out the final conceptual piece of this chapter: Multistage Bandwidth Effects!