79.4. Extended Analysis with Split Resistors

We do have differential amplifier realized by BJT. So, this is the circuit we have discussed before and you may recall that in our most of our analysis we used to split this resistor R into two identical elements in parallel. And the intention of that was to get more insight of the circuits particularly, to see how the differential signal and common mode signal they are getting propagated from primary input port to the primary output port.

In this chunk, the focus is on using split resistors in the analysis of a differential amplifier circuit, particularly using BJTs (Bipolar Junction Transistors). The principle behind splitting the resistor is to simplify the understanding of how signals pass through the circuit. Differential amplifiers need to differentiate between signals applied to their inputs. By splitting the resistor into two parallel parts, it becomes easier to analyze how different types of signals (differential and common mode) behave in the circuit.

Think of the circuit as a traffic system. The split resistors are like traffic lights that help manage the flow of cars (signals) on two different roads (inputs). By separating the two paths, we can see how each type of car interacts with the traffic lights, which helps us understand the overall flow of traffic in that area.

So here how we do have the different device parameters namely, for BJTs we do have β. In this case, this β may not be having much of use, but for the sake of completeness we are keeping the parameter. And then we do have the V_BE(on) of both the transistors = 0.6. In fact, we are considering Q1 and Q2, they are identical and then we also have the early voltage of the 2 transistors = 100 V. And then we do have the supply voltage = 12 V and then the loads R1 and R2, both are equal; and they = 5.2 kΩ and the tail resistor it is 1 kΩ.

In this section, several important parameters and specifications relevant to the BJTs used in the differential amplifier are presented. The beta (β) of a transistor indicates how much the output current is amplified compared to the input current. Although β may not play a significant role here, it's vital to have a complete set of parameters for analysis. The V_BE(on) represents the base-emitter voltage required to turn the transistor on, which is essential for ensuring they operate in the active region. The overall supply voltage, tail resistor, and load values are also established, ensuring that the circuit is properly set up for analysis.

Imagine you're setting up a team for a project: each team member has a specific role (parameters like β), they need certain tools (V_BE(on)), and you have a goal to achieve (supply voltage and loads). Ensuring everyone knows their role and what resources they have will help you successfully complete the project.

Now to start with, we do have this DC voltage given to us which is 2.6. In fact, this DC voltage should be sufficiently high, so that Q1 and Q2 should be in active region. And on the other hand this DC voltage should not be too high otherwise, Q1 and Q2 may enter into saturation region.

The operating point of a transistor in a differential amplifier is crucial for its performance. Here, a DC voltage of 2.6 V is selected, striking a balance to keep both transistors (Q1 and Q2) in their active region. The active region is where transistors amplify signals correctly, while staying away from saturation, where the transistor stops functioning properly for amplification. Understanding how to find this point is fundamental for designing effective differential amplifiers.

Consider the operating point like the temperature setting on an oven. If it’s too low, the food won't cook (transistors won't amplify correctly), but if it’s too high, the food burns (transistors saturate). Finding the right temperature ensures the food cooks just right, similar to how the operating point makes sure the transistors amplify signals optimally.

So, we assume that of course, this is the emitter current 1 mA. So, we assume that the base current is very small. So, we can say that the collector current of transistor 1 as well as transistor 2 both of them we can well approximate by 1 mA.

In this segment, we learn about how to calculate the collector currents for the BJTs in a differential amplifier. It’s stated that the emitter current is 1 mA and, because the base current is small in comparison, both transistors are approximated to have a collector current of about 1 mA as well. This approximation allows for simpler calculations during analysis and helps in assessing circuit performance.

This can be likened to two workers in a factory. If you estimate that one worker (transistor) can handle a certain amount of product (current) efficiently, and find that they are both working at similar capacity, you can simplify management (calculations) instead of treating each worker as if they require unique oversight.

So, we obtain the small signal parameters of both the transistors. Next thing is we need to find the small signal gain namely, a differential mode gain and common mode gain.

After determining the DC operating point and collector currents, the next step is analyzing how the differential amplifier responds to small signals. The small signal parameters provide critical information about how the circuit will react to tiny fluctuations in input voltage. The differential mode gain indicates how well it amplifies the difference between the inputs, while the common mode gain shows how much unwanted signals (common mode) are amplified.

This is similar to evaluating how a speaker (amplifier) responds to the sound of different instruments (signals). We want to see how well it amplifies the variance between two instruments versus the general background noise (common mode gain). This comparison helps determine the speaker's effectiveness.

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