88.2.1. Using Current Mirror

We are using current mirror and transistor-1; it is the amplifying device then we are assuming that Q and Q they are identical and also we are assuming that whatever this Q and Q are also identical.

Current mirrors are essential components in analog circuit design. They are used to create precise current sources or sinks by mirroring a reference current. In this setup, we have transistor-1 functioning as an amplifier, and we assume identical transistors Q1 and Q2. This assumption allows us to create consistent and predictable behavior in the circuit since matched transistors will exhibit similar electrical characteristics, such as current gain (β).

Think of a current mirror like a set of identical twins; if one twin (transistor) performs a task, the other twin (the second transistor) can perform the same task in almost the same way. This similarity ensures that the results are consistent, just like how identical twins often behave similarly.

To get the I current of transistor-1 and collector current or transistor-4 equal, we want the current flow through transistor-2 should be equal to current flow through transistor-1.

In this circuit, the aim is to ensure that the collector current of transistor-1 (Ic1) is equal to the collector current of transistor-4 (Ic4). To achieve this, we must ensure that the current flowing through transistor-2 (I2) equals the current flowing through transistor-1 (I1). This condition is necessary for proper functioning of the current mirror, where the mirrored current should exactly replicate the reference current, allowing for effective amplification.

Imagine a relay race where the first runner completes their lap and passes the baton exactly to the next runner. Just like in the race, it is crucial that the baton (current) is handed over correctly; if it isn’t, the overall performance will suffer. Here, if transistor-2 does not accurately mirror the current from transistor-1, the output will not meet design expectations.

So, the value of this resistance bias resistance, based biased resistance R1 should be identical to this transistor the resistor R2.

To properly bias the transistors and ensure the correct operation of the amplifier, the bias resistance (R1) must match the corresponding resistance (R2) connected to the transistors. This match helps in maintaining the same base current in both transistors, allowing them to operate under similar conditions. The calculation involves the transistor characteristics and desired operating points, ensuring a proper balance in the circuit.

Consider this like maintaining equal weights on each side of a balanced scale. If one side is heavier (the bias resistance is not matched), the scale will tip, leading to an incorrect reading (mismatched currents). To keep the scale balanced, you need equal weights on both sides.

So, to get the value of this R1 to get 20 µA, the R1 should be = 570 kΩ. So, the value of this resistance as well as this resistance they are = 570 kΩ.

Now, based on the calculations, to obtain a collector current of 20 microamperes (µA), we determine that the bias resistance R1 should equal 570 kΩ. By knowing the desired collector current, we can derive the necessary resistance values using the transistor parameters like β (current gain) to ensure that transistors are optimally functioning.

Think of it like cooking a recipe—if you want to make a certain amount of soup (20 µA current), you need to use the right quantity of each ingredient (resistance values like 570 kΩ). If you use too much or too little, the final dish (output current) won’t turn out as desired.

So, we are assuming both the devices are in active region. So, the output resistance R_out = r_o1 + r_o4.

When analyzing the amplifier, we need to understand its output resistance, which is determined by the intrinsic resistances of the individual transistors (r_o1 and r_o4). By adding these resistances together, we obtain the total output resistance (R_out). Small signal analysis is crucial for understanding how the amplifier will respond to varying input signals.

Imagine the output resistance as the thickness of a water pipe. The thicker the pipe (higher output resistance), the less pressure is lost when water flows through it. Similarly, in the circuit, higher output resistance allows for better preservation of current throughout the load.

The gain of the amplifier of course, the voltage gain of this amplifier it is g_m * R_out with a ‒ sign.

The voltage gain of the amplifier is calculated using the transconductance (g_m) multiplied by the output resistance (R_out). The negative sign indicates that there is a phase inversion, meaning that the output signal is opposite in phase to the input signal. This calculation shows how effectively the amplifier boosts the input signal.

Think of the amplifier like a karaoke speaker. When you sing (input), the speaker outputs music at a much louder volume, but if you listen—your voice comes out delayed and slightly altered (phase inverted). The speaker amplifies your voice, just like the circuit amplifies the input signal.

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