79.1.1. Differential Amplifier: Analysis and Numerical Examples (Contd.)

So dear students, welcome back to our NPTEL online certification course on Analog Electronic Circuits. Myself Pradip Mandal from E and EC Department of IIT, Kharagpur. Today’s topic of discussion it is continuation of Differential Amplifier. In the previous lecture, we have completed analysis and today we will be talking about numerical examples.

The introduction sets the stage for today’s lecture, indicating it is a continuation of the previous discussions on the Differential Amplifier. It acknowledges that the analysis has been covered already, and the current focus will be on practical numerical examples to cement understanding.

You can think of this like learning to ride a bicycle. First, you learn the theory of balance and mechanics. Once you understand that, the next step is to practice by taking a few rides to apply what you learned.

So, the concepts we are planning to cover it is the following. As I said that, the analysis part it is done in the previous 2 lectures, and we are going to talk about numerical examples, and we do have primarily differential amplifier using BJT then we do have differential amplifier using MOSFET and then also we do have another example where we do have the differential amplifier, we do have both types of transistor MOSFET as well as BJT.

The session will cover numerical examples involving two different types of transistors: Bipolar Junction Transistors (BJTs) and Metal-Oxide-Semiconductor Field-Effect Transistors (MOSFETs). There will also be a mixed example using both transistor types, providing a comprehensive analysis of their functionalities.

Imagine learning different ways to cook a meal. First, you learn one recipe (BJT), then another (MOSFET). Finally, you might try a fusion dish that combines elements from both to create something new and exciting.

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.

This chunk introduces the differential amplifier circuit using BJTs, recalling that resistors can be split into two parts for analysis purposes. This splitting helps understand how signals function differently when we consider common mode versus differential mode inputs.

Think of it as splitting a stream into two smaller streams to see how water flows differently in each. By observing these separate flows, you can understand the overall movement more clearly.

So here how we do have the different device parameters namely, for BJT’s 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 VBE 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 section specifies the technical parameters of BJTs in the differential amplifier circuit, such as the base-emitter voltage (VBE), the early voltage, the supply voltage, and the load resistances. Understanding these parameters is crucial for analyzing the performance of the amplifier accurately.

Consider building a bridge: just as you need to know the weight limits, material strength, and dimensions to construct safely, you also need to know the electrical parameters to ensure the amplifier works well.

Then, 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. So, here we have picked up the value of this DC voltage well within its range, allowable range.

The DC voltage of 2.6V is crucial for ensuring that both transistors remain in their active regions for proper operation. This requires carefully picking a value that is high enough to avoid saturation yet low enough to maintain functionality.

Imagine a car engine: it needs a specific amount of fuel for optimal performance. Too little causes it to stall (similar to entering the cutoff region), while too much can flood it (saturation). Finding the right amount keeps it running smoothly.

So, to summarize the DC operating point, we do have 2.6 V as 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.

The DC operating point summary shows that both transistors have balanced voltages and currents, which is necessary for maintaining the symmetry of the differential amplifier circuit. The collector current is equal, ensuring optimal performance.

Think of balancing weights on a scale; if both sides are even, the scale remains stable, similar to how equal currents in the transistors help maintain the amplifier's balance.

So, now we obtain the output swing. So, we obtain operating point, we obtain the DC voltage; now, next thing is the small signal parameter of transistors.

After establishing the operating point, we analyze the output swing capacity of the amplifier. The swing refers to how much the amplifier's output can vary (above and below the DC operating point) while still functioning correctly.

Imagine a swing set: it can move back and forth within a certain range. If pushed too hard (or too close to its limits), it might hit the ground (saturation). Thus, knowing the swing range helps in safe and enjoyable play.

--