59.3. Common Drain Stage Example

So, in the next slide we do have that example here. So, all the information’s are we are keeping it same we do have additional common drain stage coming out of the transistor M2 and the its bias circuit R and R it is given here it is 1.5 kΩ. Note that its biasing it is done directly from the DC voltage available at the drain of transistor-1.

In this section, we are discussing how a common drain stage is set up. This configuration includes a transistor, M2, which is the second transistor in the circuit. We start by noting that the additional components involved here are a bias circuit with potentially resistors mapped as R4 and R3, which is given as 1.5kΩ. This means that the biasing, or how we establish a stable operating point for our transistor, is done using the direct current (DC) voltage available from the previous transistor, which is M1. This is crucial because it allows M2 to operate consistently based on the conditions set by M1.

Think of the biasing process like setting the temperature of a heater based on outdoor conditions. Just as the data from outside decides how warm to make the inside of a home, the voltage from transistor M1 helps determine how M2 will behave in the circuit.

The DC voltage coming here if you see the current flow of here it is 2 mA and R it is 3 k. So, drop across this 3 k it is 6 V. So, we can say that we do have a DC voltage of 6 V coming to the gate of transistor-2.

Here, we are calculating the source voltage and gate voltage in the common drain configuration. The current in the circuit is specified as 2 mA. There is a resistor R3 of 3 kΩ connected, and hence, we calculate the voltage drop across it by using Ohm's law (V = IR). With the known current, we find the voltage drop, which results in a DC voltage of 6 V at the gate of transistor-2.

It's akin to how water pressure is measured in a pipe. The water flow (the current) and the resistance in the pipe (the 3 kΩ resistor) gives the pressure drop along the pipe, which is analogous to the voltage drop we calculate here.

How do you find the corresponding current here I_DS? So, there is a method first of all this I_DS flowing through this R creating a drop which is defining the source voltage here and then at the gate we do have 6 V.

To determine the drain-source current (I_DS) in this stage, we notice how this current flows through the resistor R4, creating a voltage drop that sets the source voltage. Since we know both the gate voltage (6 V) and how current flows in relation to this setup, we can apply similar principles from earlier calculations to establish the current level through transistor-2.

Imagine a scenario of filling a tank with water. The inlet (gate) provides water pressure (voltage) to fill up to a certain level. The current flowing through the drain can be likened to the amount of water flowing out once the tank reaches this specific level.

Now, since this equation it is coming primarily in terms of V_GS - V_th let me consider this is x. So, we can say that this is x = 5 - R4(1.5 k)(1m).

In this chunk, we are manipulating the equation to solve for the gate-source voltage (V_GS) concerning the threshold voltage (V_th). We designate the equation variables and solve for x, which in this case simplifies down to a quadratic equation. Ultimately, our solution reveals two potential values for x, but only one solution is practical, leading us to deduce that V_GS = V_th + 2V.

Think of balancing your bank account. You subtract (threshold voltage) from your total balance (V_GS) to see how much you can actually spend. Here, we are effectively determining how much voltage we can utilize after accounting for the threshold voltage needed to operate the transistor.

So, the g of this transistor it is × (V_GS - V_th). So, here also we got g = 2 mA/V and then the output resistance and of course, since we do not have the equivalent whatever r we are having in BJT.

In this segment, we calculate the transconductance (g) and output resistance of the transistor in the common drain stage. The transconductance is derived from the earlier voltage findings, which gives us a value of 2 mA/V. Since the structure of MOS transistors differs from BJTs, this output resistance can be considered much simpler, allowing us to assume the output gain is nearly one.

Consider a tuning fork or a spring: the sensitivity to changes (transconductance) will define how well it resonates (amplifies). In this analogy, the output resistance allows us to understand how 'quiet' or stable the output signal will be when affected by these changes.

Now, if I call it is f′ it is having 2 candidates to define the upper cut off frequency; one is R_O2 and the corresponding load capacitance we do have C_L.

Here we explore the upper cutoff frequencies resulting from our circuit's design. In any electronic circuit, the cutoff frequency is critical to determining how capable the circuit is at amplifying various frequencies. We consider two potential contributors to the cutoff frequency: the load capacitance and the resistance from the circuit. By comparing calculations and ensuring we capture the lower limit, we can efficiently optimize bandwidth.

This is similar to how a sound system has different frequencies it can play. For example, a subwoofer handles low frequencies while tweeters handle high ones. Just as the system needs both types of speakers to produce a full range of sound, we take into account multiple sources of cutoff frequency to ensure our circuit can operate over a broad spectrum of signals.

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