88.3.2. Finding Currents at V = 11.4V

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 discuss that for certain transistors in a circuit, the current flowing through them must be equal for proper functioning. If the DC voltage at a certain point in the circuit (V_BE) is defined and results from the voltage drop across components, we can calculate the specific voltage present at any node. For instance, a supply voltage of 12V and a base-emitter drop of 0.6V gives us a voltage of 11.4V.

Think of a water pipeline where the pressure (voltage) must remain equal on both sides for water (current) to flow smoothly. If you have a pressure of 12 units at one end and a valve (like the base-emitter junction of a transistor) that requires a drop of 0.6 units to function correctly, the pressure after the valve will be 11.4 units, equalizing the flow on both sides.

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.

Here, the focus shifts to the concept of current mirroring. At the voltage of 11.4V, the expected current flowing through the circuit is estimated to be 2 mA. This is a common assumption in transistor circuits, where current mirrors facilitate the reproduction of this current in other similar components of the circuit.

Imagine a school where a teacher (current) is teaching a group of students (transistors). The teacher's ability to educate (current level) can be mirrored in different classrooms. If the teacher successfully educates 2 students (2 mA) in one classroom, similarly, other classrooms can replicate this effort if the environment (voltage) is conducive and identical conditions (equal V_BE) are provided.

If this voltage here it is say 11.4, then you can say that this current and this current they will be equal. In intuitively also you can say that if this voltage it is 11.4. So, that makes the current here and current here, they are equal and also the current flow in transistor-3 and transistor-4 they are equal.

This chunk emphasizes that if certain conditions (such as a fixed voltage of 11.4V) are met, equal currents will flow through respective transistors presuming they are identical in specifications (like β factors). Current mirroring relies on this principle, where two identical transistors (like transistor-3 and transistor-4) will share equal current under ideal conditions.

Think of two identical scales (transistors) side by side. If you place the same weight (current) on both and ensure they are calibrated identically (matched specifications), both will show equal readings as long as the environment remains steady (constant voltage supply).

In case if this current it is higher than the current flow through transistor-1 for 11.4 V DC, then the corresponding voltage here it would be higher +ve.

This section discusses how the early voltage and the currents affect the DC output voltage. If the required current flowing exceeds what is being mirrored through the transistor (transistor-1), then the output voltage may change and deviate from the expected value. The circuit's precision is influenced by how closely the actual performance of the devices matches theoretical predictions.

Recall the analogy of two water towers (transistors) regulating water flow. If one tower can handle more water flow than the other and tries to push more water than can be matched, the pressure in the pipelines (output voltage) can increase unexpectedly, leading to a need for adjustments to prevent overflow or system failure.

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