88.3.4. Voltage Variation Due to Additional Current

As I said that the current flow current flow here and here should be equal and if you see it carefully the DC voltage here it is defined by this V ‒ V drop. So, I should say the voltage here it is 12 V. So, the voltage here it is 12 ‒ 0.6 so, that is 11.4 V.

In this chunk, we are discussing the requirement for balance in current flow through different transistors (specifically through the current mirror). The text indicates that the DC voltage can be determined using the drop across the transistors. Given the specified values (12V and 0.6V), we find the voltage at the node is 11.4V. This balance in current flow is crucial for the proper operation of amplifiers using current mirrors.

Think of a water flow system where you have two pipes (transistors) connected to a reservoir (the voltage supply). For the system to work efficiently, the water flow (current) through both pipes must be equal. If one pipe is clogged (higher resistance), it will not allow the same amount of water to flow, changing the pressure (voltage) recorded downstream.

Now, with this 11.4 V here we can say that whatever the current we do have. So, that may be 2 m which is of course, 1 approximation that we are assuming ( ) = 1 please stop here.

This segment explains that at the node voltage of 11.4V, we can determine the collector current (2 mA) flowing through the transistors, allowing for the assumption that they are operating similarly. This mirroring behavior is essential as it demonstrates how the current in one transistor reflects that in another, facilitating consistent performance in circuits.

Imagine two identical water fountains. If the gauge shows that one fountain is pouring out 2 liters of water per minute (2 mA), we can assume the second fountain, which is designed the same way, will pour out the same amount of water if it's condition remains unchanged. The idea of mirroring current is just like this; both systems should respond the same way under the same conditions.

So, that gives us the DC output voltage V = 11.4 V right. But of course, if the values of early voltage or in case we cannot ignore say this base current and then of course, the corresponding the current here and here there will be a mismatch and the DC voltage here it will deviate from here.

In this part, the text conveys that the calculated output voltage is 11.4V under ideal conditions. However, it warns that practical scenarios (such as variations in early voltage or base current effects) can lead to mismatches in current flow, resulting in a deviation from this voltage. This introduces the importance of considering real-world imperfections in circuit design.

Consider a light fixture that is supposed to shine at 11.4 lumens (similar to our voltage). If the bulbs you use differ slightly in power (like mismatched early voltage) or if the wiring has weak spots (analogous to base current), the actual brightness may vary, and not all bulbs will shine evenly. Understanding these imperfections is key to achieving desired outcomes.

In fact, this is the non-ideality factor is . And then the corresponding DC voltage it is given here yeah. So, that is how you can approach.

This final segment discusses how imperfections in current matching can affect the DC voltage at the output. Specifically, the text hints at a factor of non-ideality that complicates the calculation of DC output voltage. This reinforces the importance of precise component values in designing robust electronic circuits.

Imagine baking a cake where the ingredients are not mixed evenly (non-ideality). If the flour is unevenly distributed, parts of the cake may be denser or fluffier than others, affecting the overall taste and appearance. In electronics, if the currents through the components are not matched precisely, the overall performance can be negatively impacted, leading to inconsistencies.

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