59.1.5. Multi-Transistor Amplifiers (Contd.):

Welcome back after the short break. So, we are talking about the CE-CC and now we will be moving to MOS counterpart. In the next slide we will be talking about see common source amplifier again this numerical exercise we have seen before. So, this is prime and the main common source amplifier.

This chunk introduces the topic of the common source amplifier, which is a crucial part of MOS transistor technology. The presenter reminds the audience of previous discussions on common emitter-common collector configurations before transitioning to discuss the common source amplifier, setting the foundation for a numerical example that follows. This implies that the attendees should have prior knowledge of amplifier configurations.

Think of the common source amplifier as a speaker system in your home. Just like you adjust the volume to amplify the sound, the common source amplifier adjusts and boosts electrical signals for clarity and strength in electronic circuits.

And, we have seen that using this information we obtain V = 3 V and then I we obtain it was 2 mA and then corresponding small signal parameter g it was 2 mA/V. So, the corresponding voltage gain; voltage gain it was g into output resistance and you may ignore the r or other r we may consider this is very high assuming λ is very small.

Here, we discuss crucial parameters obtained from the amplifier circuit, like the gate-source voltage (Vgs), drain-source current (Id), and transconductance (gm). The voltage gain of the amplifier is derived from multiplying gm by the output resistance. The mention of ignoring certain resistances simplifies the calculations under the assumption of ideal conditions, allowing focus on the main contributing factors to gain.

Imagine your amplifier setup as a water pipe. The voltage (Vgs) is the pressure pushing water through the pipe, the current (Id) represents the flow of water, and the transconductance (gm) describes how much more flow (output) you can get for a given increase in pressure (input).

So, the voltage gain it was g R and so that becomes 2 m × R is 3 k; 3 k. So, the corresponding voltage gain it was only 6.

In this chunk, the calculated voltage gain is obtained by multiplying transconductance (gm = 2 mA/V) with the output resistance (Rd = 3 kΩ), resulting in a voltage gain of 6. This indicates the strength of the amplifier, showing that the output signal is 6 times greater than the input signal.

Consider tuning a radio. When you turn the dial (input), the sound from the speaker (output) becomes 6 times louder. The radio circuit acts like the amplifier, enhancing the original sound signal significantly.

So, the corresponding output resistance for this case we see it is primarily defined by R and that is 3 kΩ. So, the upper cut off frequency for this case f it was into load capacitance of 100 pF.

This section emphasizes that the output resistance is primarily defined by the resistor, here noted as Rd = 3 kΩ. Additionally, it introduces the concept of the upper cutoff frequency (fU), which is dependent on this resistance and the load capacitance, measured in picofarads (pF). Understanding these parameters is crucial for determining the frequency response of the amplifier.

You can think of output resistance as the thickness of a garden hose. A thicker hose (higher resistance) allows water to flow without much back pressure. If the hose is too thin or has too many bends, it might restrict water flow (frequency response).

So, it was and then 3 k into this one 100 p; that means, 10‒10 yeah. And in fact, if you calculate it this gives us 530 kHz.

This segment covers the calculation of the upper cutoff frequency, which involves multiplying the output resistance by the load capacitance. The process leads to a calculated frequency of 530 kHz, indicating the highest frequency at which the amplifier can effectively operate before the gain begins to roll off.

Imagine the upper cutoff frequency like the maximum speed limit on a highway. Beyond this speed (frequency), the vehicle (signal) cannot maintain its performance due to interference or restrictions (circuit limitations).

So, the common source amplifier it is primarily it is having a gain of 6 and then upper cut off frequencies 530 kHz, we are not going to calculate the lower cut off frequency primarily because our intention here is to see the enhancement of the bandwidth by the use of common drain stage.

In this concluding statement about the common source amplifier, the overall gain of 6 and the upper cutoff frequency of 530 kHz are reiterated. The focus shifts towards cascading this stage with a common drain stage to enhance bandwidth, suggesting further exploration into advanced configurations for achieving better performance in electronic circuits.

Think of this amplifier setup as a concert where the band performs at a consistent volume (gain of 6). Now, to make the concert experience even better, they would consider different arrangements (cascading with common drain stage) to ensure everyone enjoys the music without distortion (enhanced bandwidth).

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