52.3.5. Current Gain of Common Gate Amplifier

First of all on the stimulus part we need to replace current source. Signal current source is i and it may be having a finite conductance and this is conductance is . Now, this signal it is going here and again through this capacitor the signal is arriving to the emitter node.

In this section, the idea of analyzing the current gain of a common gate amplifier begins with replacing the signal source with a current source (i). This current source is connected to the amplifier's emitter node and may have a finite conductance, meaning it does not transmit current perfectly. This portion of the circuit is crucial as it sets the stage for understanding how the input current affects the output current.

Think of the current source as a water pipe supplying water (current) to a garden (amplifier). If the pipe has holes (finite conductance), it delivers slightly less water than intended. Understanding how much water is needed to make the plants grow (output current) helps us know how to adjust the pipe size or type.

Now, once we have this i we are feeding at the emitter node and then we like to see how much the current we will be getting here particularly in unloaded condition.

When analyzing the unloaded condition, it assumes that the output is short-circuited to the ground (AC ground). Under this condition, the current flowing through the output will be maximized since there’s no load to impede the flow. This simplification is crucial for understanding how ideal conditions would affect the current gain.

Imagine a garden hose (our current source) connected to a sprinkler (the amplifier). If the sprinkler is turned off (unloaded condition), all the water flows freely through the hose. Similarly, under unloaded conditions, the full current flows through the amplifier without resistance or obstruction, allowing us to measure the maximum output.

For this current gain you may recall that the current gain is very close to 1.

The current gain, often represented as A = Iout/Iin, is calculated under these ideal conditions. It can be approximated as very close to 1 because the output current closely matches the input current in unloaded conditions. This indicates that the amplifier can transfer current effectively, but this gain may not hold true under loaded conditions or in practical applications.

Consider the current gain like a direct pipeline transferring water. If you start with 100 liters of water (input), and after passing through the pipeline (amplifier), you still have 99 liters flowing out (output), the gain is effectively 1, showing minimal loss. However, if there were leaks (load), you'd see significantly less water at the end.

So; obviously, this α is very close to 1 depending on the value of the β though mathematically it is less than 1, but as I said it is very close to 1.

In discussing the current gain, the parameter α (or alpha) represents the ratio of the output current to the input current in a transistor. It indicates how effectively the device can amplify current. While it's mathematically represented as slightly less than 1, it implies strong current transfer capability. The approximation here indicates a roughly ideal scenario, simplifying calculations.

You can think of α like a water filtration system that, while losing a bit of water due to inefficiencies (less than 1), still delivers a nearly equal amount of clean water to the tap (the output) as it receives from the source (the input).

Now similar kind of analysis can be done for common gate amplifier. But I must say that for common gate circuit we can consider ideal bias.

The discussion wraps up the analysis by suggesting that similar evaluations can be carried out for a common gate amplifier, noting the variations in operation and the consideration of ideal biasing. Unlike common base configurations, common gate amplifiers may have unique characteristics, especially in terms of how they handle input and output signals.

Think about comparing two types of water pumps. Both move water effectively, but one works better under different conditions. The analysis suggests a need to understand these differences, much like considering the efficiency of different pumps based on their designs and operating environments. Each pump (or amplifier type) has its strengths and weaknesses.

--