85.2.3. Common Mode and Differential Mode Improvement

this part and if we analyse this circuit it becomes like a common source amplifier, where the load it is diode connected. So, the circuit becomes like this. We do have g , we do have g connected to V and m1 m7 DD then this transistor half of the transistor we may call say M by 2. It is just a just a matter 3 of representation and at this point we are giving the small signal along with the DC.

In this part, we analyze a circuit resembling a common source amplifier. The active load is diode-connected, meaning it functions like a diode in the circuit. The transistors g, g connected to a voltage source (VDD) are introduced, with one transistor referred to as half of another one (M/2). We also introduce the concept of superimposing a small signal onto a direct current (DC) to analyze how it behaves in this circuit.

Think of a common source amplifier as a microphone input. When someone speaks into a microphone (the DC signal), there are also smaller background noises (the small signals) that influence the overall output. The microphone captures both, just like the circuit captures DC and AC signals.

Now, if you analyse this circuit or if you draw the small signal equivalent circuit, you can find that the voltage here v = ‒ g × . So, this into whatever input signal we are applying v . In fact, as I said the voltage here also it will be same.

Upon analyzing the small signal equivalent circuit, we derive that the output voltage (v) is proportional to the negative gain (−g) multiplied by the input signal (v_in_c). This tells us how the circuit responds to small variations in the input. Notably, the voltage across specific points in the circuit remains consistent when switching between common mode and differential mode operations.

Imagine tuning an audio equalizer where a small change in one frequency influences the overall sound output. Here, we denote how adjusting a small signal affects the output voltage, similar to how each small adjustment can significantly change our audio experience.

So, we may say that v under common mode operation is also = v and that gives us v = v common mode. And hence the corresponding common mode gain if I say A′′ which is defined by = ‒ g × .

We define that under common mode operation, the output voltage (v) is equal to the input voltage (v), which leads to the common mode output voltage equating to the common mode gain (A′′). This is formulated as the product of the negative gain and some component of the input signal.

Think of a seesaw: when both sides (input signals) are lifted equally (common mode), the center remains balanced—a representation of how the common mode gain functions, maintaining balance in the output despite equal inputs.

Now, again you can see here because the resistance now it is getting changed. So, this A it is much smaller than the previous common mode gain.

The change in circuit resistance indicates that the common mode gain (A′′) is reduced compared to previous configurations. As a result, using an active current mirror load results in better performance by diminishing the common mode gain, which is favorable for reducing noise in the output.

Consider a garden with different plants (signals) where some spread their roots farther, affecting others. By introducing a better barrier (active load), we can hinder those influences, improving the overall health of the garden (the signal output).

So, this is one consequence of having this active current mirror load and the gain at this point, gain at this point if I say that. So, that is becoming g ( ). In fact, you may say g or g both are same.

The active current mirror load allows for converting differential signals to a single-ended output without altering the gain significantly. When we express the differential input to single-ended output gain (g), we observe it remains consistent regardless of how it is categorized.

Think of a speaker producing stereo sound (differential) but you want to connect it to a single channel (single-ended). The output maintains richness while still staying true to the original sound—reflecting how this circuit preserves gain during conversion.

So, we can say that DC voltage here it is V ‒ V and hence this voltage can directly be used to bias PMOS transistor in the subsequent stage.

The voltage derived from the previous stages, calculated as the difference V – V, serves as a DC bias for the PMOS transistor in later amplifier stages. This highlights the utility of current mirrors in managing biasing effectively, ensuring that the transistor operates within its preferred range.

This situation is akin to a relay race where each runner (transistor) passes the baton (voltage) to the next. The successful transfer ensures that every runner stays energized and at their peak performance through each segment of the race.

To summarize what we have done it is we have started with a small signal model of current mirror particularly, under DC condition as well as whenever it is carrying the signal current for both BJT and MOSFET versions.

In the final summary, we encapsulate the key points discussed throughout the class. It includes the small signal model of current mirrors for both BJT and MOSFETs, their application as biasing elements in amplifiers, and the enhancement of common mode and differential mode gains using active loads.

It's like wrapping up a cooking class by summarizing how each ingredient (topic) combined to create a delicious dish (final understanding of current mirrors and their applications). Each component plays a role, just as each concept contributes to the overall comprehension of the circuit design.

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