42.1. Analog Electronic Circuits

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, then 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.

In this introduction, the instructor sets the stage for discussing a numerical example related to an analog electronic circuit. They mention that they will be using a generalized equivalent circuit model and provide specific values for various components such as input resistance (1.3 kΩ), output resistance (3.3 kΩ), and source resistance (650 Ω). These parameters are crucial for understanding the performance of the circuit and will be used in subsequent calculations.

Think of this as preparing for a cooking recipe where you need to gather all the ingredients before you start cooking. Here, the values of resistors are like your ingredients, and knowing them allows you to understand the final dish—or in this case, the circuit's performance.

And then the load capacitance C 100 pF, C it is given here it is say 10 µF and then C L 1 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. And let you consider this voltage gain A or denoting this by A which is a 240 with a ‒ sign.

In this chunk, the instructor discusses the capacitances involved in the circuit. They mention the load capacitance of 100 picofarads (pF), another capacitance value of 10 microfarads (µF), and a few other values that contribute to the input capacitance. The mention of voltage gain, noted as -240, indicates that the amplifier inverts the input signal. The 'Miller effect' is briefly referenced as important for understanding how capacitances between nodes in a circuit can be perceived as larger than their actual values due to the voltage gain.

Imagine you are trying to capture sound in a noisy environment. The different capacitances are like the different microphones you might use. Some might pick up background noise more than others, just as some capacitances in the circuit affect the input signal differently based on how they interact with the circuit's voltage gain.

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.

The focus here shifts to assessing the frequency response of the amplifier based on the parameters discussed earlier. The instructor indicates that both lower and upper cutoff frequencies will be determined, along with the mid-frequency gain, which will help characterize the amplifier's ability to amplify signals at different frequencies without distortion.

Think of a musical instrument, like a guitar. Just as different strings produce different pitches, circuits respond differently to various frequencies of input signals. Understanding the cutoff frequencies helps us know what range of pitches (frequencies) the circuit can handle effectively.

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

In this segment, the calculation of input capacitance (C) involves a formula incorporating the voltage gain (-240), which reflects the influence of amplification on the effective capacitance in the circuit. Following this, the instructor prepares to calculate the first pole, essential for determining the lower cutoff frequency, a frequency below which the amplifier will start attenuating the input signal significantly.

Imagine trying to catch water in a funnel. The size of the hole at the bottom determines how quickly the water can flow out. Similarly, the lower cutoff frequency indicates the point where signals below it can no longer pass through efficiently, much like water slipping away through a small opening.

So, we do have the second pole it is given here 302 kHz. Now, you can also calculate the third pole p which is coming from R and then output resistance. So, p it is it is 3.3 k, so you have 3300 Ω and then output resistance it is when you have C = 100 pF and then the C it is almost coming as is, so we can see roughly 105 pF.

The instructor calculates the second pole that corresponds to the upper cutoff frequency (302 kHz) and mentions the third pole that involves the output resistance and a capacitance of 100 pF. This process of pole calculation is crucial as it helps define the frequency response limits of the amplifier, identifying where the amplification will reduce for higher frequencies.

Imagine you're at a concert, trying to hear the lead guitarist. If a loud drummer starts playing too close to you, their sound can drown out the guitar. Similarly, the upper cutoff frequency indicates the point where certain higher frequencies can no longer be amplified effectively, ensuring that only the 'lead guitarist' (desired frequencies) can be heard.

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 concluding chunk summarizes the overall analysis, reflecting on the high-frequency models of transistors and the influence of source resistance on the frequency response of common emitter and common source amplifiers. The highlights include understanding how these components affect gain and cutoff frequencies, which are essential for designing effective amplifiers.

Just as a chef learns various techniques to create the perfect dish, engineers need to understand different circuit parameters and their impact to design amplifiers that produce the best sound. Each component in a circuit adds its taste, influencing the performance of the final 'dish' or output.

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