41.1.5. Frequency Response of CE/CS Amplifiers Considering High Frequency Models of BJT and MOSFET (Part B)

Yeah. So, welcome after the break. So, we are talking about the, in fact, what we got it is the generalized model of CE and CS amplifier here. What it is having here it is the input signal source, having the source resistance of R, and then signal coupling capacitor C, and then if I consider this is the main amplifier where we do have the input resistance represented by this R. And then we do have voltage dependent voltage source, which means that this is the core of the amplifier, then we do have the output resistance R. And then C and C, they are representing you know either C, C or C and C based on whether the circuit it is CE amplifier or CS amplifier.

Here, we are introduced to the generalized model of CE (Common Emitter) and CS (Common Source) amplifiers. The model includes essential components such as the input signal source, source resistance, coupling capacitors, input resistance, output resistance, and voltage-dependent voltage sources. These variables are crucial for understanding how the amplifier operates under different conditions. The presence of capacitors C and C showcases the adaptability of the model for both transistor types (BJT and MOSFET), effectively demonstrating how the design can change based on specific circuit requirements.

Think of the analog amplifier as a water distribution system. The input signal source is like a water main, source resistance acts like a narrowing pipe that limits flow, while the coupling capacitors ensure water only flows when the pressure (voltage) is appropriate. Just as different pipes can carry water differently based on their diameter and material, different transistor types influence the performance of the amplifier.

So, this particular this capacitor it can be converted into two equivalent capacitance; one is for the input port, the other one is for the output port. And then, the input port part coming out of the C it is what we said is that C (1 ‒ A ) or in this case A is equal to A. In fact, if you see here we are putting a ‒ sign here assuming that the polarity of the voltage dependent voltage source, here it is +ve.

In this section, we discuss the role of capacitors in the circuit. They're transformed into equivalent capacitances for both the input and output ports. This is vital for analyzing how signals pass through the amplifier. The term C(1 - A) highlights that the input capacitance is affected by the gain of the amplifier (A). The negative sign indicates an inverted relationship based on the voltage source polarity, which is an essential concept in understanding signal processing within amplifiers.

Imagine a dam that holds water (signal) and releases it downstream. The capacitors serve as adjustable gates in this dam. The input capacitance changes based on the amount of water being let through (gain). If more water is released too quickly (high gain), it may affect the pressure in the remaining pipes, akin to how signal inversion impacts the amplifier's performance.

So, in summary what we have it is at this node we do have the C and then at this node we do have the net C. Now, to get the frequency response of this circuit namely starting from this point till the primary output what we have it is we do have one network here and then we do have of course, the main amplifier starting from this point to this point and then of course, at this point we do have the C.

Here, the focus is on understanding the total effective load capacitance at a specific node in the circuit. This capacitance affects the circuit's frequency response, as capacitive loads impact how quickly the circuit can respond to changes in input signals. By establishing connections between network nodes and the main amplifier, the analysis of how the complete system behaves at different frequencies can begin.

Consider a car hitting a speed bump (the load capacitance) on a road. The car’s ability to respond (frequency response) varies based on how heavy the load is (capacitance) and how many speed bumps (nodes) it encounters along the way. A lighter load or fewer bumps means a quicker response time.

So, our first task is to find the frequency response from this point to this point, namely maybe in Laplace domain we can see, and then we can find what is the corresponding transfer function we are getting.

The analysis begins with determining the frequency response, which indicates how the circuit behaves at different frequencies. Utilizing the Laplace domain allows for easier manipulation of the differential equations that represent the circuit. The objective is to derive the transfer function, a mathematical representation that describes the output signal in relation to the input signal, vital for predicting circuit performance.

Imagine tuning a radio to different frequencies to find your favorite station. Each frequency you tune is similar to the analysis of the circuit's response. The transfer function acts like the clarity of the signal at each frequency; it tells you how well the circuit (radio) can amplify and transmit the signal (music) you want.

Now, if you see here to get the frequency response of this circuit namely in Laplace domain, we can get the transfer function by considering this impedance which is R ⫽ and then R in series with + R ⫽.

This step involves deriving the transfer function by evaluating the impedance of the circuit. By calculating the total impedance from resistances in series and parallel, we can identify how the circuit responds to a sinusoidal input. This step is crucial for analyzing stability and gain at various frequencies.

Think of a water park slide (impedance). When you add more slide sections (resistances), the overall speed (response) changes. Calculating total slide length (impedance) at different angles (input frequencies) determines how quickly someone can reach the bottom of the slide (circuit output).

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