69.3. CS Amplifier Circuit Comparisons

We have here the CS amplifier circuit and we do have the corresponding bias resistors and all. So, we do have this resistor it is 9 kΩ, this resistance it is 3 kΩ. So, that gives us this voltage = 3 V. On the other hand, we have 25 kΩ here and we do have 95 kΩ.

In this section, we introduce the common source (CS) amplifier circuit. The circuit features two bias resistors: one with a resistance of 9 kΩ and the other with a resistance of 3 kΩ. The voltage developed across these resistors indicates how the circuit operates, leading to a total voltage of 3 V. This foundational information sets the stage for analyzing the performance of the CS amplifier.

Think of the resistors in this circuit like a water flow system where the resistors control how much water can flow through. The 9 kΩ and 3 kΩ resistors are like gates regulating water (electric current) flow, affecting how well the system (amplifier) performs.

In fact, intentionally, I have taken different value of this transconductance factor for transistor-1 and transistor-2. For transistor-1; we have 1 mA/V2 whereas, for this transistor; we have 4 mA/V2. Now again for this case it is I current of the 2 transistors should be equal.

The transconductance factor is used to measure how effectively a transistor can control the output current with respect to input voltage changes. Here, we intentionally assign different values to two transistors: 1 mA/V² for the first and 4 mA/V² for the second. Because of these differences, the currents through both transistors must be adjusted to be equal to achieve the desired output.

Imagine two runners in a race, where one can run faster (4 mA/V²) than the other (1 mA/V²). They need to adjust their strategies so that they finish at the same time (equal currents), highlighting the need to balance the performance of different components in a circuit.

To get the DC output voltage, I need to consider (1 + λV ) of the 2 transistors and if I equate them namely (1 + λV ) and to V and λ V . Since both the lambdas are equal. So, that gives us V = V.

In this chunk, we focus on calculating the operating point of the CS amplifier. To find the DC output voltage, we consider the relationship between the transistors using lambda (λ), which reflects the channel length modulation effect. By balancing the equations of voltages across the two transistors, we can determine that their voltages are equal, thus enabling us to calculate the operating point effectively.

Think of balancing a scale where each side has weights (voltages across transistors). By ensuring both sides are equal (balancing the voltage equations), we can see where the scale rests (operating point), which helps us maintain overall stability in our circuit.

So, the swing wise; if I say that +ve side we have output swing +ve side it is 5 V; 11 ‒ 6 = 5 V and then ‒ve side, we have 6 V ‒ 2 V. So, that is equal to 4 V.

This section discusses the voltage swing of the amplifier, which is the range in which the output voltage can vary without distortion. On the positive side, the maximum swing is calculated as 5 V, and on the negative side, it is 4 V. Understanding these limits is crucial for ensuring the amplifier operates within a linear range.

Imagine using a swing in a playground. The height of the swing can only go so high (positive swing) and not too low (negative swing). If you know your limits, you can enjoy your ride without falling off! Likewise, in our amplifier, knowing the voltage swings helps us maintain a smooth operation.

If you compare the common source amplifier having passive load and if its bias condition is equivalent... However, the 3 dB bandwidth got decreased and then input capacitance of course, it got increased. And this increase and this decrease, they are having the same factor mainly due to the change of this output resistance.

This chunk provides a comparative analysis of CS amplifiers with active versus passive loads. The active load design offers higher gain but results in a reduced bandwidth and increased input capacitance. This trade-off is essential for understanding which design approach to choose based on application requirements.

Imagine upgrading from basic bicycle brakes to sports car brakes. The sports car brakes (active load) can stop faster (higher gain) but may not hold up as well on long trips (reduced bandwidth). So, depending on your needs, you might prefer one over the other.

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