42.2.6. Upper and Lower Cutoff Frequencies in Common Source Amplifier

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, we establish that we will be working with a common source amplifier circuit. The focus will be on determining the frequency response, including the cutoff frequencies. The initial parameters are introduced: input resistance (1.3 kΩ), output resistance (3.3 kΩ), and source resistance (650 Ω), which you'll need to analyze the amplifier's performance.

Think of this stage as the blueprint of a building where the specific values (like resistance) are differentiated features such as the size of rooms, which determine how the building functions and flows overall.

First of all this resistance directly given there, but then need to calculate the input capacitance. 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.

We first need to calculate the input capacitance for the amplifier. Using the formula given, we can plug in the known values to find that the total input capacitance is 1215 pF. This capacitance is critical for defining how the amplifier responds to different frequency signals.

Imagine you are filling a balloon with air; the size and thickness of the balloon's material (capacitance) will determine how much air (signal) it can effectively hold before it bursts (causes distortion).

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. And if you calculate it, I have done this calculation for you. So, what you will get it is 8.16 Hz, right.

The lower cutoff frequency is determined by the characteristics of the input resistance and the input capacitance. Using the formula, we incorporate 650 Ω (source resistance) and the parallel combination of resistances to find that the lower cutoff frequency is around 8.16 Hz. This frequency indicates where the amplifier starts to lose sensitivity.

Consider a music speaker; if a sound is too low (below 8.16 Hz), it won’t be heard clearly. Just like with audio equipment, amplifiers have a point below which they cannot effectively 'amplify' signals.

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.

The upper cutoff frequency is calculated similarly by determining the second and third poles based on the output load capacitance and output resistance. For this setup, we find the upper cutoff frequency to be approximately 302 kHz, meaning signals above this frequency will begin to be attenuated by the amplifier.

Visualize it like a highway with a speed limit; if cars (signals) attempt to go faster than 302 kHz, they are more likely to be pulled over or slowed down, causing an inability to transmit effectively.

And to get the mid frequency gain, so mid frequency again it is of course, this multiplied by whatever the attenuation coming from these two elements. So, what is that attenuation? So, I should say mid frequency range equals to and R it is 2.3 this is 0.65 k.

The mid-frequency gain is calculated by taking into account the attenuation factors from the resistances in the circuit. It is defined as the voltage gain multiplied by the loss from these attenuations. Thus, it results in an overall gain of about 160. This gain represents how much input signal can be amplified effectively in the mid-frequency range.

Think of this as a conversation in a crowded room; your voice must cut through the noise (attenuation) to be heard clearly—how well you accomplish that is similar to the gain of the amplifier responding to signals.

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