33.2. Prof. Pradip Mandal

Welcome back after the short break and we are about to start the small signal equivalent circuit for the Common Source Amplifier. The first thing is that we are making the DC bias to be 0. The capacitor we have sorted here and the capacitor we have sorted here and then DC current in this voltage dependent current source we made it 0; living behind the small signal current i which is a linear function of the v_ds, gs here.

In this segment, we are discussing the small signal equivalent circuit. When analyzing circuits like the Common Source Amplifier, we start by eliminating the DC components, which simplifies our calculations. We set the DC bias to zero and ignore any DC currents. This allows us to focus solely on the small signal behavior, represented by 'i', which is dependent on the voltage ('v_ds') across the transistor's drain and source terminals. The signals produced can be considered linear, simplifying our analysis for understanding amplification.

Think of a crowded room where two people are having a conversation. Initially, the noise of the crowd (DC components) makes it hard to hear what they are saying. If we focus exclusively on their conversation (the small signal), we would start filtering out the crowd noise, making it easier to understand their communication.

The output voltage is –R_D × i and that is given as – R_D × g_m × v_gs. The voltage gain A is defined as v_out / v_in = – R_D × g_m.

After establishing the small signal conditions, we move to voltage gain calculations. The output voltage is derived from the product of the drain resistance (R_D) and the small signal current (i), which can be expressed in terms of the transconductance (g_m) and the gate-source voltage (v_gs). The voltage gain 'A' indicates how much the input signal is amplified, calculated as the ratio of output voltage to input voltage, highlighting the influence of R_D and g_m in achieving the desired amplification.

Imagine you are amplifying a faint sound using a microphone and speaker system. The microphone captures your voice (input signal), and the speaker (acting like the output) increases the sound level significantly. The amount the sound is amplified depends on the quality of the microphone (analogous to g_m) and the settings of the speaker volume (similar to R_D).

If you observe this circuit from outside, we will be doing this as we said that we can connect a signal source and then we need to make this part equals to 0. This makes this current zero, resulting in an expression derived as v_out = R_D × i.

In analyzing output resistance, we consider how to connect a signal source while ensuring that unnecessary currents are set to zero. This simplification allows us to derive the expression for output voltage directly related to the output resistance. Understanding the output resistance helps deduce how the amplifier will respond under various loads, which is crucial in circuit design.

Imagine trying to use a water hose. If you want to measure how fast the water is coming out (output resistance), you turn off the faucet (set the current to zero) to gauge how much pressure is inside the hose without extra water flow influencing your reading. This gives you a clear understanding of the strength of the water pressure (output voltage) just waiting to come out.

If we see what is the corresponding resistance at the input port since the circuit here is open, we can assume that this is equal to zero. If we stimulate this port, we will observe the corresponding current i_s flowing through R1 and R2.

Next, we analyze the input resistance of the amplifier. If we consider the amplifier's circuitry without any input signal (open circuit), the input current can be assumed to be effectively zero. Therefore, the input resistance can be evaluated based on the parallel connection of resistances at the input port, which is pivotal in determining how the amplifier will respond to incoming signals.

Think of a dry sponge that has not absorbed any water. If you place it under a dripping faucet (incoming signal), it won't take in the water immediately due to its resistance to flow. The sponge's condition without additional water indicates its input resistance—meaning how receptive it is to the incoming water (signal).

The common source amplifier can either map into a voltage amplifier or a transconductance amplifier. The input remains the same voltage, but at the output side, it can be current if it is a transconductance amplifier.

Finally, we differentiate between two types of amplifiers that can be constructed from the common source amplifier: voltage amplifiers and transconductance amplifiers. In the voltage amplifier configuration, both the input and output are voltage signals. In contrast, the transconductance amplifier converts the voltage input into a current output. Understanding the differences between these configurations helps in selecting appropriate applications for the amplifier.

Imagine a light dimmer switch (voltage amplifier) that adjusts the brightness of a light without changing its type of illumination. Conversely, think of a water pump (transconductance amplifier) that takes in electrical energy and converts it into water flow, acting distinctly from the original input type. Both serve desirable functions based on the context they are used in.

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