48.1. Common Collector and Common Drain Amplifiers (Contd.): Numerical Examples (Part B)

Welcome back after the short break. So, we do have here it is the third example. So, it is you can see that essentially, we are adding one resistance at the emitter instead of having ideal bias current and also we do have the R at the base node. We can say the source resistance. Let you consider this R = 9.8 kΩ. So, let me change this one to 9.8 kΩ and let us try to analyze the circuit to find the operating point of the transistor.

In this part, we are setting up a circuit analysis example. A resistance of 9.8 kΩ is introduced at the emitter, changing how we analyze the transistor's operating point. The step involves replacing an ideal bias current with this resistor, affecting the current and eventually yielding a real-world application of the circuit.

Think of this circuit analysis like planning a trip. Instead of taking the fastest route (ideal conditions), you intentionally choose a route with obstacles (resistors) to learn how to navigate real-world traffic challenges.

So, to start with let me draw the DC loop here. So, we do have Vdd which is 6 V and then we do have the Rs, which is 100 k and then we do have the VBE(on) this diode drop and then we do have the resistance here RE. We can say the current flow here it is IB. And then of course, we do have the β times of this IB current it is coming from the collector. So, we can say that this is βIC.

Here, we are laying out the DC loop which helps in calculating the transistor's operating point. It shows how various components relate to each other. For instance, we have a power supply (Vdd), a resistor (Rs), and the base-emitter voltage drop (VBE(on)). Using these, we aim to find the current flowing through IB and relate it to the collector current IC using the transistor's current gain β.

Imagine you're preparing to bake a cake. You gather your ingredients (voltage and resistors) and measure them out carefully to ensure that the cake will rise properly (the circuit's operating point will be successful).

So, the current flowing through this resistor on the other hand it is (1 + β)IB. So, if I consider that this current it ≈ βIB and then if you analyze this circuit. Namely, if we consider the potential drop through this loop. So, what we can get it is in this loop. We do get 6 V here and then we do have 0.6 V here. So, 6 V equals to we do have IB × 100 k + 0.6 V.

This section calculates the current flow through a resistor in the circuit. We equate the voltage drops around the loop, remembering that the total voltage must equal the sum of the drops. By rearranging the equation, we derive a value for IB, which will help in calculating all other related currents in the circuit.

Consider a water pipe system. The water pressure (voltage) drops as it flows through different sections (resistors). Understanding how the pressure changes helps you to calculate the rate of water flow (current) throughout the system.

So, once we obtain this IB of say 5 µA, so, that gives us the IC = β × IB = 100 × 5 µA = 0.5 mA right. Then you can calculate what will be the voltage here. Similar to the previous case; the voltage here it will be 6 V here minus this drop, across this RE, which is 0.5 V. So, we can say the voltage here VSB = 6 - 0.5 = 5.5 V.

In this part, we calculate the collector current IC from the base current IB using the current gain β. We then use IC to find voltage drops across resistors. By subtracting the drop across RE from the source voltage, we find the output voltage at the transistor's emitter.

Imagine pouring water into a container with holes at the bottom (representing resistances). The amount of water collected (voltage at the emitter) depends on how fast you pour (current) and how much leaks out (voltage drop).

Now using this value of the quotient currents and voltages we can find the small signal parameter. And since the currents particularly, the collector current and base current they are the same. So, we can see that the gm value of the gm remains the same. Namely, A/V, and likewise ro = 5.2 k and then re = 100 k right.

In this section, we derive small-signal parameters needed for designing and evaluating amplifiers. The transconductance (gm) and output resistance (ro) are calculated, showing how they are derived from the previously computed currents and voltages in the circuit.

Think of small signal parameters like the measurements taken for a recipe's fine adjustments (like adding spice). They ensure that minor variations (small signals) still lead to a well-balanced final dish (operating amplifier performance).

So, we will see that, but for the time being let you consider voltage gain from precisely at the base terminal to the emitter terminal and even though this Rs is coming in parallel with ro because, this we do have (1 + β) getting multiplied. So, still and hence we can say that this is approximately 1.

This part discusses the voltage gain from the input (base terminal) to the output (emitter terminal). Each of the resistances affects this gain, but due to the specific arrangement of the circuit, the overall voltage gain is roughly equal to one, indicating that the output voltage is nearly equal to the input voltage.

Consider a tutor helping a student. The knowledge transferred from tutor to student (the input to output) might be almost equal because of effective communication, similar to how output voltage approximately equals input in this circuit.

So, the input resistance on the other hand it is as we have discussed before...

This section provides a deeper understanding of input and output resistance in the context of the amplifier. It highlights the significance of these resistances and how they affect the operation of the amplifier, impacting how much signal is successfully delivered or received.

Imagine a conversation where two people have to communicate important information. If one person is too quiet (high input resistance), the message might not get through effectively. Similarly, understanding these resistances helps determine the efficiency of our circuit's communication.

So, the conclusion here it is what we can say that even if you consider Rs and RE still we do have the overall performance of the circuit even with RE being very high.

This conclusion emphasizes that the overall performance of the circuit remains robust, even when introducing the resistors. It reassures that the desired outcomes can still be achieved, which is vital for successful circuit operation in real-world applications.

Much like a sports team succeeding despite individual injuries, the circuit can still perform well when robust components are thoughtfully chosen and integrated, ensuring collective success.

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