41.2.5. Transfer Function Derivation

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_s, and then signal coupling capacitor C_1, and then if I consider this is the main amplifier where we do have the input resistance represented by this R_1. 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_2. And then C_3 and C_4, they are representing you know either C_π, C_µ, C_gd based on whether the circuit it is CE or CS amplifier.

This chunk provides an overview of the CE (Common Emitter) and CS (Common Source) amplifier configuration. The introduction sets the stage for discussing the generalized model of these amplifiers, focusing on key components such as the input signal source, source resistance (R_s), coupling capacitor (C_1), input resistance (R_1), a voltage-dependent voltage source, and output resistance (R_2). Additionally, it mentions other capacitors (C_3 and C_4) that play roles depending on the circuit type (CE or CS). Understanding these components is crucial before delving into the specifics of the transfer function derivation.

Think of the amplifier as a water pipe system. The input signals are like water entering a network of pipes (components) with certain resistance (source resistance R_s). The coupling capacitor (C_1) acts like a valve that controls the flow before it enters the main pipeline (amplifier). The water's flow is influenced by how wide or narrow the pipes (R_1, R_2) are and how they branch out (C_3, C_4). This analogy helps illustrate how these electrical components impact the overall signal flow in the amplifier.

This particular capacitor 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_4 is C_4(1 ‒ A), where A is the voltage gain A_o. And if the polarity is opposite, then we can consider that C_4 is just C_4 multiplied by (1 + |A|). So, the net input capacitance C_in is given by C_3 + C_4(1 ‒ A), while the net output capacitance is C_L + C_4.

This section discusses how capacitors are treated in the amplifier model, particularly during the analysis of frequency response. The input and output capacitances are derived based on the presence of feedback (the voltage gain A). Depending on the circuit configuration (positive or negative feedback), the values of the capacitors can change, which affects the overall input and output capacitances. Summing these capacitances is important for determining how they influence the amplifier's frequency response.

Imagine adjusting the size of water tanks connected to our earlier water pipe analogy. The input port capacitance (C_in) acts like a reservoir that stores water before it flows into the main pipeline. Depending on how full the tank is (related to gain A), it may allow more or less water to flow through. The output capacitance is like a collection tank at the end of the pipe, which collects the flow from the pipeline. Understanding how these tanks interact gives us a clear picture of how efficiently the system can deliver water (or signals).

Our first task is to find the frequency response from this point to this point, namely in Laplace domain we can see, and then we can find what is the corresponding transfer function we are getting. We do have this circuit, we do have R_in, and then C, and then R_in in series with R for the frequency response.

This portion aims to derive the frequency response of the amplifier circuit using the Laplace domain. The frequency response helps understand how the amplifier behaves with different signal frequencies, allowing us to derive the transfer function. The circuit is examined considering its resistances (R_in, etc.) and capacitors to ascertain how they interact in the frequency domain.

Returning to our water analogy, this step is like testing how the water system responds when there's a gradual increase in demand. We are observing how quickly and efficiently the system (amplifier) responds to changing water flow rates (frequencies). Analyzing the frequency response tells us at what points the system might struggle (like blockages) and how we can optimize it for different materials or needs.

So, we can get the transfer function by considering the impedance which is R_s in series with R in parallel with C. Simplifying this further, we arrive at the numerator and denominator expressions for the transfer function.

This part focuses explicitly on calculating the transfer function of the amplifier circuit. By using the series and parallel combinations of resistors and capacitors, we derive formulas to express the system's behavior in response to various frequencies. This mathematical manipulation is critical for analyzing how each component contributes to the overall transfer function.

Consider this calculation similar to mixing different chemicals in precise amounts to achieve a specific reaction. The ratios of these chemicals (R and C values) represent how changes in the input affect the final output. Understanding how to balance these ratios helps us predict how the amplifier will act under various conditions.

The transfer function here which is written here you can see that it is having a 0 here, due to this s term, and also since it is a second order polynomial in the denominator, so we are expecting two poles...

This section explains the significance of poles and zeros in the transfer function derived earlier. The presence of a zero indicates a frequency at which the output goes to zero, while poles represent frequencies where the output can become infinitely large. Understanding their locations facilitates deeper insights into the amplifier's performance across different frequencies.

Think of poles as points where there are blockages along a river - if the water flows too fast (high frequency), it can flood the banks (increase output), but if it’s slow and hits these blockages (poles), it may barely trickle through. The zeros, conversely, are points where the river runs dry (no output); knowing where these affect the flow enables us to manage the overall river system effectively.

So, we do have the first pole here p_1 given by that expression and then we do have the second pole p_2 coming from the other factor,...

This chunk delves into the expected behavior of the frequency response based on the derived poles. It highlights how the interaction between different components leads to specific frequency limitations, effectively determining how the amplifier behaves at various frequencies. Understanding these behaviors is key to applying the amplifier in practical scenarios.

Imagine the behavior of crowds at a festival. At first, everyone is loosely gathered, representing how the signal flows without much interruption (low frequency). If a bottleneck occurs (like poles), attendees (signals) can either get stalled or redirected, affecting the entire flow (amplifier gain). Knowing where these bottlenecks exist allows managers to improve the experience of festival-goers, just as understanding frequency responses ensures optimal amplifier performance.

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