42.2. Frequency Response Of CE/CS Amplifiers Considering High Frequency Models of BJT and MOSFET (Part C)

Welcome back after the short break. So, we are going to discuss about numerical example, and the circuit it is still that equivalent circuit we do have and what we are. So, we do have the generalized equivalent circuit, but then also we do have additional information namely the value of different components, R this input resistance is 1.3 k, R output resistance it is a 3.3 k and then let you consider source resistance 650 Ω that is also a typical value one possible value of typical signal source.

This chunk introduces the frequency response discussion and outlines the numerical example for analyzing CE (Common Emitter) and CS (Common Source) amplifiers. The essential circuit components are specified, such as input resistance (1.3 kΩ), output resistance (3.3 kΩ), and source resistance (650 Ω), which are crucial for frequency response calculations.

Think of an amplifier as a water pipe system. Just as water pressure and resistance in pipes affect water flow, the resistances and capacitances affect the signal flow in amplifiers.

And then the load capacitance C 100 pF, C it is given here it is say 10 µF and then C L which is one of the element contributing to input capacitance it is say 10 pF, C the Miller effected capacitance the capacitor which is breezing the input and output terminal of the circuit is 5 pF.

In this chunk, several capacitances relevant to the amplifier circuit are listed, including load capacitance (100 pF), an additional capacitance of 10 µF (likely bypass capacitance), input capacitance (10 pF), and the Miller effect capacitance (5 pF). Understanding these capacitances is essential as they all impact the frequency response of the amplifier.

Imagine using different sized balloons to hold water. Each balloon size changes how quickly water can flow. Similarly, different capacitances affect the speed of electrical signals in an amplifier circuit.

So, that means, actually this is ‒ and this is +. So, anyway with this information let we try to get the frequency response and particularly containing the mid frequency gain and then lower cutoff frequency and then upper cutoff frequency. So, how do we proceed?

This segment transitions into calculating the frequency response, including the mid-frequency gain, lower cutoff frequency, and upper cutoff frequency. The overall method is to analyze how these frequencies can be affected by the previously mentioned resistances and capacitances.

Consider tuning a musical instrument. To get the right sound (frequency), you need to adjust the tension of the strings (resistance) and their size (capacitance). The right adjustments provide the correct notes.

So, to start with let we calculate C and C = C (1 ‒ (‒240)), then + C . So, we do have here and then × 241 + 10. So, that gives us how much; we do have 1215, 1215 pF, yes.

Here, we begin calculating the input capacitance (C_in) based on the contributions from the Miller effect and additional components. This calculation gives a value of 1215 pF, which will influence the frequency response regarding pole frequencies.

It's like combining different weights to create an ideal balance for a scale. Each weight's contribution changes how the scale behaves, just like how capacitance change affects frequency response.

So, we do have the C is given here, with this C we can calculate the location of the second pole and let we calculate the first pole first. So, p which is defined by if I express this in the unit of Hz then we have to consider 2π. So, 2π× , so this = . R it is 650 and then R , R it is 1.3 k, so that is 1950 this resistance and then C it is 10 µF which means 10‒5.

In this step, the calculation of the first pole frequency (p1) is conducted using the resistances and capacitances of the circuit. By employing the formula involving 2π, the specific frequency values are determined, providing critical insights into the amplifier's operational limits.

Think of a seesaw; the position and weight of each child affect how balanced the seesaw is. In electronics, poles determine how signals behave based on the circuit’s configurations.

And if you do this calculation what you will be getting it is 459, in my calculation that is what I obtain 459 kHz.

The final pole calculation leads to finding the upper cutoff frequency, which is critical in defining the bandwidth of the amplifier. This value helps understand how quickly the amplifier can respond to frequency changes in the input signal.

Consider the limits of a sports car’s speed; just like a car's top speed, the cutoff frequency tells you how fast the amplifier can respond to high-frequency signals.

So, in summary what we have so far we have covered it is. So, in this module what we have covered here it is basically we have considered high frequency model of transistor and particularly in presence of source resistance R, what is its impact on the frequency response of common emitter and common source amplifiers.

This closing reflection points to the broader conclusions about frequency response in CE and CS amplifiers while emphasizing the effects of source resistance and how it shapes the amplifier’s performance. It wraps up the importance of understanding each component's contribution.

Consider a recipe; all ingredients (like resistors and capacitors) must be harmonized for the dish (the amplifier's operation) to turn out well. Each component has a role in achieving the desired result.

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