52.2.5. Current Gain of Common Base 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. 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.

In this chunk, the discussion begins by replacing the ideal current source in the circuit with a signal current source 'i' that may have some finite conductance. The purpose here is to analyze how this signal current passes through to the emitter node of the transistor. The unloaded condition refers to the assumption that the output is shorted to AC ground, which is necessary for calculating the short-circuit output current in the circuit, focusing on how much current is delivered by the transistor when no load is applied.

Think of the current source as a water tap. When you open a tap (apply a signal current), water flows through the pipe (the circuit) into a bucket (the emitter node). In the unloaded condition, you can imagine the bucket being removed. The water (current) flows with no resistance, allowing us to assess how much comes out of the tap without any blockage or load.

For this current gain you may recall that the current gain is very close to 1. If I say that we are shorting this node here; the corresponding v = 0. And if you recall the small signal model here, we have this r we have g × v and v it is ‒ v and its ‒ sign can be taken out. So, the small signal model of this circuit can be written like this and then the output node it is shorted and we are observing the corresponding output current.

This chunk discusses the concept of current gain in the common base amplifier configuration. The current gain is denoted by 'A', and it is stated that this gain is very close to 1. When the node at the output is shorted (i.e., the collector is connected to ground), then the output voltage (v) becomes zero. Under this condition, the model simplifies as we look at how the input current produces an output current. The relationship we derive emphasizes that the current flowing out of the emitter is essentially the same as the current flowing into the collector, affirming that the amplifier is efficient with minimal loss.

Consider the concept like a funnel. When you pour a liquid into a narrow tube (the input current), most of it exits the funnel from the other side (the output current) if the tube is positioned correctly. The idea here is that if everything is properly arranged, nearly all the liquid you pour into the funnel comes out of the other end, just like in a common base amplifier, where nearly all current coming in is available at the output.

Now we can compare each of these terms and you may recall the value of g ; g is ℧ and r is 5.2 kΩ and R is 10 kΩ. So, we can say that 10 kΩ and then we have r is 100 kΩ. So, by considering all these practical values of different parameters naturally you may say this g will be dominating over in the denominator also this g will be dominating. So, we can directly say that this current gain is very close to 1.

Here, the values of several parameters (g, r, and R) are specified to derive the current gain mathematically. The small signal transconductance (g_m), intrinsic resistance (r_π), and the load resistance (R) are introduced to explain how to compute the current gain (A). In the end, it is established that due to the relative sizes of these parameters, the gain approaches 1. This result reinforces the notion that the common base amplifier passes nearly all the input current as output current.

Imagine you’re a conveyor belt at a factory. The speed at which goods (input current) travel along the belt almost determines how many goods come off the end into another area (output current). If the conveyor belt operates smoothly, nearly all goods sent in will come out from the end, reflecting the near-unity current gain found in the common base configuration.

If you want them to get better approximation probably you can retain this part and then you can find what will be the value instead of 1 to be more precise, it will be then ( ). In fact, this is α which means that if we have this i it entering at the emitter, the corresponding available current at the collector it will be α times of this transistor.

This chunk extends the discussion by suggesting that while the current gain is close to 1, it can be refined using the parameter alpha (α). Alpha represents the ratio of the collector current to the emitter current in a transistor. This shows that, while nearly all of the injected current is available at the output, there is a slight drop which can be quantified by this factor α. Therefore, the current at the collector is slightly less than that at the emitter, reflecting real-world transistor characteristics.

Think of a basketball player shooting hoops. Every shot taken (input current) results in a certain number of successful baskets (output current). However, not every shot will go in, and a few might miss, illustrating that although most attempts yield baskets (the current gain), a small number will fall short, akin to the influence of α on current gain.

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