93.1. Feedback System (Part-D)

So, yeah dear students, so welcome back after the short break and before the break we are talking about the change of input resistance of the different configuration. And whereas, we are going to talk about change in output resistance to start with let we consider it is a voltage amplifier and we want to see the change due to feedback connection.

In this introduction, we are shifting focus from input resistance to output resistance in feedback systems. The output resistance is vital for understanding how different amplifier configurations react to feedback. We will begin by examining a voltage amplifier's behavior under feedback.

Think of an amplifier as a water faucet. When you turn the faucet (the amplifier) on, it allows water flow (signal) to come out. The output resistance can be compared to how much pressure is lost in the pipes when water is flowing out – we want to see how feedback changes that.

And since it is a voltage amplifier as we have discussed, the circuit is given here. The configuration here it is referred to as shunt series feedback or voltage series Feedback circuit.

This chunk describes the setup of a voltage amplifier. The shunt series feedback configuration indicates how feedback is applied: a shunt connection at the input and a series connection at the output. This configuration affects the input and output resistances of the amplifier.

Imagine a classroom where the teacher (the amplifier) listens and responds to students’ questions (feedback). If there are many students (high input resistance), it’s easier for one student to be heard (lower output resistance).

To start with let we consider the feedback network it is ideal namely its input resistance, it is infinite and the output resistance as it is producing voltage, output resistance it is 0.

In an ideal feedback scenario, the input resistance is considered infinite (meaning it does not affect the circuit) and the output resistance is zero (the output can provide unlimited current). These assumptions simplify calculations for output resistance.

Think of an ideal feedback network as a magical wall that doesn't absorb energy, allowing all that gets input through it to flow freely and without resistance, similar to water flowing out of a valve into an open pipe.

However, for the forward amplifier we are considering finite input resistance and finite output resistance. We do have the voltage dependent voltage source A_v, v it is appearing at the input of v_in in the forward amplifier.

In reality, amplifiers have finite resistances. We introduce the dependent voltage source, which represents how the output voltage (Av) is influenced by the input voltage (vin). This relationship is essential for calculating actual output resistance.

It's like a special type of water pump that increases its output depending on how much you push the handle (input). The stronger your push (input voltage), the more water (output voltage) comes out.

Now, to know the output resistance what we can do? We can stimulate this output port by say voltage source v and then we observe the corresponding current say i and then the port impedance or port resistance it is defined by R_out.

To find output resistance, we connect a voltage source at the output and measure how much current flows in response. The ratio of voltage (v) to current (i) gives the output resistance (R_out).

Imagine you connect a hose to your faucet and measure how much water flows when you apply different pressures. The speed at which water flows gives you an idea of the output resistance of the faucet.

So, while we are doing this exercise we have to keep the input port appropriate. So, that it will support the feedback connection, namely we in this case the signal here it is voltage and so, we do have the ideal signal voltage source connected. However, we have to keep its magnitude, signal magnitude should be 0.

For accurate measurement, we must ensure that the input signal is effectively shorted (set to 0). This allows us to isolate the feedback effects without introducing additional variables during output resistance calculations.

Think of testing the resistance of a hose by clamping its end while measuring how much water pressure is applied. This ensures that the pressure reading truly reflects the hose's characteristics, without creating added complications.

Now, that is the condition and let we derive the expression of this R_out_f in terms of R_out and probably A and β.

The derivation to express output resistance involves variables such as the original output resistance (R_out) and feedback factors like A (gain) and β (feedback factor), which determine how feedback influences the output resistance.

It's like tweaking a recipe based on feedback. If everyone says it’s too salty (high A), you might decide to adjust the salt level (β adjustment) to improve the final dish (output resistance).

So let me clear yeah. So, if you see if I consider this is short, then if I consider KCL in this mixer what we are getting it is v, input voltage of the amplifier it is ‒ v. But then v it is same as βv...

When we account for non-ideal elements, like finite source resistance, we realize that each component affects the overall output resistance. This complicates calculations because now we need to consider how much feedback is actually influencing the output.

Imagine adjusting the water flow through multiple taps (resistances) in different rooms. Each adjustment changes how strong the pressure is at the end of each tap (output resistance).

In that case, we will get the output resistance under this condition, in presence of this R_s, where β′ it is given...

Here we wrap up the findings for ideal conditions and transition into considering more complex, realistic situations where not all components behave optimally. This signals a shift in focus to analyzing how these non-ideal conditions impact output resistance.

This is like preparing for an emergency by considering possible failures – how does each room (component) react when one tap (part) is not functioning as expected?

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