19.2. Numerical Example

In this numerical example if you see we have the same basic circuit, the main thing is at this bias we do not have any resistance. So, directly the voltage you are applying both the bias as well as the signal to the base. So, that makes this circuit simple and then we have say supply voltage say 10 V and let you consider this V such that the current here it is say 10 µA.

This chunk introduces a numerical example that simplifies the circuit by eliminating resistances that might complicate the analysis. The circuit operates under a supply voltage of 10V and we have a current of 10µA flowing through it. Not having resistors means that the signals can be applied directly to the base, allowing for easier calculations.

Imagine simplifying a recipe by not using any complex cooking tools. You just mix the basic ingredients directly, which makes the process faster and more straightforward. Similarly, in this circuit, not using resistors simplifies our calculations.

So, I am giving this V in term such that the I it is 10 µA and then say β of the transistor it is say 200. And so that gives us the quiescent current here it is 2 mA.

The voltage applied corresponds to a base current of 10 µA, and with a transistor current gain (beta, β) of 200, we can calculate the quiescent current in the circuit, which ends up being 2 mA. The quiescent current is important as it represents the stable operating point of the transistor when no input signal is present.

Think of this like setting a steady pace in a race. Just like a racer maintains a certain speed to achieve optimal performance, in the circuit, the transistor must maintain a steady quiescent current (2 mA) to function effectively.

Now, to get a good linearity range here namely to get good linearity range here we like to keep the operating point will away from this point as well as whatever the saturation limit or active region limit.

In order to maintain linearity in the circuit's response, the operating point should be positioned away from both saturation limits and the cutoff region. This ensures that the transistor operates efficiently within its active region, allowing for reliable amplification of input signals without distortion.

Consider an athlete who must stay within a certain running lane. If they stray too far to the left or right (saturation limits), they could lose balance and slow down. Similarly, the transistor must maintain its operating point in an optimal range to function properly.

So, suppose this information’s are given. So, what I have done here it is. In fact, there may be two ways of framing the numerical problem, either we give this information and then directly we can ask draw the small signal equivalent circuit, find the values of the small signal parameters and then find the small signal voltage gain.

The next step would involve calculating the necessary resistances, ensuring they align with the defined operating conditions. There are various approaches to this problem depending on what values are initially provided, including drawing the small signal equivalent circuit and calculating small signal parameters, like the voltage gain.

It's like planning a road trip. Depending on whether you start with a map or a GPS, the steps you take may differ, but ultimately you're trying to reach the same destination, which in this case is understanding the behavior of the circuit and calculating its voltage gain.

Let me consider this is something different may be 2.2 kΩ which is a practical value, 2.5 normally we do not get. So, probably you can try out find the operating point namely the I is given, collector current is also very straightforward.

In practical scenarios, we might choose resistors that are more commonly available. Therefore, exploring a resistor value like 2.2 kΩ offers a good learning moment as it illustrates that real-world choices can affect calculations. Subsequently, determining the current from the transistor is straightforward, facilitating the final computations.

Think of it as choosing ingredients for a dish. Sometimes you have to adjust your recipe based on what ingredients are available in your pantry, as you would when selecting resistor values for your circuit based on what is commonly found in stores.

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