79.3.1. Analysis of Current and Voltage Levels

So, 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 section, we are focusing on the differential amplifier using BJTs (Bipolar Junction Transistors). This amplifier is designed to boost the difference between two input signals while suppressing any common signals (known as common mode signals). In previous analyses, a resistor (labeled R) was divided into two equal parts. This division helps understand how signals behave in the amplifier—particularly, how differential signals (differences between inputs) and common mode signals (same signal at both inputs) propagate through the circuit.

Think of the differential amplifier like a person listening to two friends talking (the two input signals). If both friends are speaking at the same time about the same topic (common mode), it's hard to hear any individual opinions. But if one friend is giving a differing viewpoint (the differential signal), it's much easier to pick out those differences. Splitting the resistor is like giving each friend separate headphones, making it clearer to hear their distinct voices.

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 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.

This chunk discusses the key parameters of the BJTs used in the differential amplifier. The transistors, Q1 and Q2, have a current gain (β) that is not vital for this specific scenario, but it's noted for thoroughness. The base-emitter voltage is set at 0.6 V for both transistors, and they are characterized as identical devices. Additional specifications include an early voltage (which affects performance under varying collector voltages) and the power supply voltage set at 12 V. The load resistors R1 and R2 are both 5.2 kΩ, and the tail resistor (which sets the current in the circuit) is 1 kΩ.

You can think of this as setting up a team of athletes. Each athlete (transistor) has specific stats (parameters) that determine how they perform individually and as a team. Even if one athlete's strength (β) isn't crucial for the activity at hand, knowing fully about all of them helps in planning a strategy. The base-emitter voltage is like the starting position before a race, and the tail resistor determines the pace at which they start running.

So, to summarize the DC operating point, we do have 2.6 V is the base voltage and then at the emitter. So here also, it is 2.6 V and at the emitter we do have 2 V. Then, voltage here it is 6.8 V and here also it is 6.8 V and the collector current in both the transistors they are equal and they are 1 mA right.

In this section, we summarize the calculation of the DC operating point of the BJTs in the differential amplifier. The base voltage is determined to be 2.6 V, which is the voltage applied to the base terminals of both transistors. The emitter voltage (after accounting for the base-emitter voltage drop) is 2 V. At the collector terminals, the voltage drops to 6.8 V due to the current flowing through the load resistors. The collector current, which is the same for both transistors, is established at 1 mA. Thus, the operating point is used to ensure both transistors function in the active region.

Imagine adjusting the levels of water in two tanks connected by a pipe (the transistors in the amplifier). The base voltage is like the water level you set to start with, aiming for balanced flow through the pipe (the load resistors). The collector current is then like the speed at which water efficiently flows through the system, which you want to monitor to ensure everything runs smoothly.

In fact, we do have sufficient headroom. So, in case if you have say V_CE is at 0.3 V. So, this voltage it can come down as low as 2.3. So likewise, so we do have a swing here with respect to DC voltage it is 6.8 ‒ 2.3. So, that = 4.5. So, the negative side swing it is 4.5.

Here, the operating point's capabilities are analyzed in terms of voltage swing. The V_CE value (voltage from collector to emitter) can drop to 0.3 V, lower than which would push the transistors into saturation. The minimum collector voltage can, therefore, go down to 2.3 V, which allows for a voltage swing calculation where the DC output voltage (6.8 V) minus this minimum allows for a downward swing of 4.5 V. Such swings are important in ensuring the amplifier can handle varying input signals.

Think of the voltage swing as a swing set on a playground. The height at which the swing starts (the DC output voltage) must allow the swing to go low (down swing) without hitting the ground (saturation). The distance below the starting point that it can swing is the swing height. The swing needs enough room to move back and forth to be effective and enjoyable, just like the amplifier must process varying signals appropriately.

--