60.1.1. CC followed by CE Amplifier

So, now, we do have another example, where we do have the CC followed by CE amplifier. And what we have here it is, the CC stage it is directly getting coupled to CE stage. And you can see here the in the CC stage, basically this part is the CC stage and normally we do have a current sink here for proper biasing; but here we assume that whatever the emitter current we do have out of Q1 that is entirely getting consumed to the base or base terminal of Q2.

This chunk introduces the concept of using a Common Collector (CC) amplifier followed by a Common Emitter (CE) amplifier. In this configuration, the CC stage is directly connected to the CE stage. Typically, a current sink is used for good biasing in circuits. However, in this setup, it's assumed that the current from the emitter of Q1 (the CC stage) flows completely into the base of Q2 (the CE stage), meaning that the extra biasing circuit isn't necessary.

Think of the CC amplifier as a helpful friend who supplies energy (current) directly to another friend (the CE amplifier) who needs it to perform a task. Instead of both needing to have their own power sources to function successfully, they support each other so that only one power source is needed effectively.

Now again here we assume that DC voltage of Q2, it is sufficient to feed the signal at the base of Q1; or you may say that, whatever the emitter current we do have out of Q1 that is good enough to make a bias of the Q2. So, in summary what will I like to say that, Q2 is biased by the base terminal of Q1. So, it is biased at its emitter by the base terminal of Q2. On the other hand Q2 is biased at its base terminal by emitter current of Q1.

In this part, it’s explained how the biasing works between the two stages. Q2's base is supplied with enough DC voltage from Q1's emitter current to operate correctly. Thus, Q1 helps to set the biasing for Q2 through its base terminal. This mutual biasing simplifies the circuit design, as the two transistors assist each other in maintaining their operating points.

Imagine a team of two people working together; one person's input (Q1's emitter current) supports and guides the other's task (Q2's operation). They rely on each other’s strengths to get the job done, resembling a close teamwork setup.

Now, here for comparison with our previous circuits, we are setting the value of this R1 such that the current flowing through this Q2 we are expecting that it will be say 1 mA. So, if the I_C of transistor-2 it is 1 mA and then we like to retain the same DC voltage at its emitter, we are maintaining R = 1.2 k and then now the required current here; since its β it is 100, so the required current here it is close to 10 µA. If I say that this 10 µA base current of transistor Q2 equals the emitter current of Q1.

This section discusses the current flow and the resistance needed in the circuit configuration. It calculates that in order to have a collector current (Ic) of 1 mA through Q2, certain resistances are required to maintain correct voltage levels. It further analyzes the base current required for Q2, which is determined using the beta (β) value of the transistor, establishing that the base current is significantly smaller. Hence, a very high resistance is necessary to support this current in the circuit.

Think of this as managing a line of dominoes. The first domino (emitter current) needs to be strong enough to tip the next one (base current of Q2) over. If the support (resistance) for that first domino isn't high enough, the chain reaction can't happen smoothly. You need to ensure they are aligned properly to maintain stability.

So, if I am having these two resistances are known, and then we can find what will be the input resistance of this circuit. The input resistance here is R, it is the parallel connection of R1 and whatever the resistance we do have. And what we have, the resistance here it is r_π1 coming in series with (1 + β)r_π2.

This chunk focuses on calculating the input resistance of the CC-CE amplifier configuration. It shows how to determine the overall input resistance as a combination of resistances from both stages, illustrating how to compute a more comprehensive resistance value that is essential for understanding circuit performance in actual applications.

Consider the input resistance like a water pipeline. The input resistance affects how easily water can flow through. If you have a pipe (resistance) that is too narrow or if there are too many intersections (parallel connections), it can impede water (signal) movement. The ideal scenario is to ensure every section of the pipeline facilitates a smooth flow.

So, C of the circuit it is we do have C of transistor-1. So, we do have C of transistor-1 in µ and this node it is AC ground. On the other hand, this C_π breezing the input base terminal and emitter terminal of Q1, its gain is very close to 1, so we may say that the effect of this C_π is negligible.

In this section, the impact of capacitance is analyzed. The capacitance of the first transistor is assessed, indicating how it interacts with the system; because the gain of this transistor is almost unity, its capacitance effect is considered minimal, thus not significantly affecting the overall performance of the circuit.

Imagine capacitance as a sponge soaking up water (signal). If the sponge is too small, it won’t affect how much water flows through (the signal). Since Q1’s gain is close to one, the sponge doesn't hold enough water to matter much, so we can ignore it and focus on bigger issues in the circuit.

I compare to CC-CE amplifier, there is another configuration something called Darlington pair and we will see that what the difference basic difference is. In fact, their configuration-wise they are very close; but there is a small difference, in the next slide we will be discussing about that.

This last chunk outlines how this CC-CE amplifier configuration can be compared to a Darlington pair configuration. Although both setups are similar, there are distinctions in how they operate, particularly regarding their gains and input resistances. This sets the stage for understanding why one might be preferred over the other depending on the application's requirements.

Visualize a sprinter (Darlington) who has another sprinter (CC-CE) in front of them. Both have great speed (gain) but the sprinter in the Darlington has a minor advantage in agility (input resistance) which may allow them to swiftly navigate corners (circuit complexity) during a race (active operation) making them slightly more efficient in certain situations.

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