89.3.1. Summary of Numerical Examples on Current Mirror

In the numerical examples, we will be reestablishing the same thing, to start with we do have to find the DC currents through all the transistors.

This chunk introduces the concept of current mirrors within the context of example problems. The basic task is to determine the direct current (DC) flowing through each transistor in the circuit. Understanding the flow of current in transistors is essential, as it helps in analyzing the performance and behavior of current mirrors.

Imagine a water system where pipes (transistors) carry water (current). To ensure that the right amount of water flows to each pipe, you need to calculate how much is needed at each point. In a circuit, similar calculations ensure each transistor receives the correct amount of current, allowing the circuit to function properly.

Again the value of this resistor it is 11.4 k, so that makes this current = 1 mA. So, the current flow here it is also 1 mA...

This section provides specific values that illustrate how to calculate the DC currents through different components in the current mirror circuit. Initially, a resistor is given a value of 11.4 kOhm, which leads to a current of 1 mA. By applying the principles of circuit analysis, we see how this current splits into half (0.5 mA each) across two paths involving more transistors.

Consider two garden hoses splitting water from a single source. If the total flow is 1 mL of water per second and the hoses are identical, each one will carry 0.5 mL per second. This is similar to how the current splits between the transistors in the circuit. Understanding this split helps to ensure that each component operates correctly.

So, since this is transistor-7 it is diode connected. So, the voltage here it is decided by the 12 V ‒ 0.6 V so the voltage here it is 11.4 V...

The voltage across a diode-connected transistor is crucial for determining the output states within the circuit. By referencing a 12 V supply and the diode forward voltage drop of 0.6 V, a constant voltage of 11.4 V is established at this point. This voltage contributes to the stability required for the current mirror to function effectively.

Think of this scenario like a water tower maintaining pressure by a drop in its spout. The pressure at the bottom (11.4 V) remains consistent regardless of the total amount of water (12 V) in the tank due to the outlet's resistance. This ensures every connected pipe receives a steady flow, just as transistors do in the current mirror design.

So, yeah so now our next calculation it is we need to find differential mode gain and common mode gain...

This part explains how to calculate the differential mode gain and common mode gain, which are key performance metrics for amplifiers. The differential mode gain measures how well the circuit amplifies the difference in input signals, while the common mode gain focuses on signals that are common to both inputs. Understanding these concepts is vital for assessing amplifier performance in practical scenarios.

Imagine you are measuring the height of two plants. If you want to know which plant is taller (differential gain), you compare their heights. But if you want to see if both plants reach a certain mark on a wall (common mode gain), you look at the height relative to that mark. Similarly, these gains in circuits help compare how well signals are amplified.

Now, coming to the conclusion of this entire session it is what we have talked about basically we started with numerical examples of simple current mirror...

In concluding the numerical examples section, the main takeaway is a summary of various current mirror implementations, calculations of quiescent currents, and an analysis of both differential and common mode gains. Competence in these calculations signifies a solid understanding of how current mirrors function within electronic circuits.

Think of wrapping up a project presentation where you summarize your findings on how best to build a model. You outline what you did step by step, similar to retelling the calculations of the current mirrors in your circuit design, ensuring that your audience understands the importance of each part of your process.

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