64.1.1. Prof. Pradip Mandal

Before I going to the next topic I must see here that this calculation of the C I did a small mistake hear, it should be 135 because C = C + C(1 + 204) and so, here we do have 5 and also here we do have 10. So, I miss this 10 µ part. So, 10 + 1025. So, that gives us 1035 pF capacitance.

The professor points out an error in the calculation of the capacitance (C). Initially, the capacitance was incorrectly calculated. The formula C = C + C(1 + 204) was applied, where two values (5 and 10 µ) were noted, but there was a missed component, which changed the final value. After correction, the final capacitance is calculated to be 1035 pF, which is an important component in understanding the circuit design.

Imagine baking a cake where you mistakenly measure the flour—if you only add partial flour, the recipe will not turn out right. Similarly, just like adding the correct amount of flour matters for a good cake, accurately calculating components like capacitance is crucial for a successful circuit.

So, far what we have discussed that the advantage of cascode amplifier with respect to standard CE amplifier and namely what you have seen is that in case if you are retaining this passive element for both the cases, then gain wise we do not get much advantage.

The professor discusses the advantages of a cascode amplifier compared to a standard Common Emitter (CE) amplifier. While both configurations use passive elements, the cascode amplifier shows a notable advantage in extending bandwidth and increasing gain, especially when high source resistance is involved, even if the passive elements used are similar.

Think of a water fountain that uses a pump (CE amplifier) versus one that uses an additional height to push the water higher (cascode amplifier). The second fountain may not have a more powerful pump, but the height helps achieve greater water flow, similar to how the cascode amplifier manages to enhance performance despite similar component specs.

So, to demonstrate the capability of the cascode amplifier to increase the gain first of all let we consider a different situation instead of having this R we put some active circuit there...

The professor suggests using a different R value, replacing a passive element with an active circuit to show how a cascode amplifier can gain significantly. For instance, when using a 10 MΩ load instead of 2.8 kΩ, this alteration leads to an increase in gain due to better handling of current, demonstrating how the circuit can be optimized for performance.

Imagine upgrading from a small battery (passive element) to a high-capacity power source (active circuit). Just like a more powerful source can run heavier machinery better, modifying the cascode amplifier this way allows it to boost gain effectively.

So, I it is supporting this 2 mA of current. So, then you may say that why do you have this resistance at all...

In this section, the professor explains how biasing supports a quiescent current of 2 mA, which is significant for maintaining consistent operation of the cascode amplifier. Even if certain resistances seem to be unnecessary, they play an essential role during circuit implementation, especially concerning the nature of current sources, which can experience variations in real applications.

Consider a well-regulated train system (the biasing current) ensuring that trains always have enough power to reach their destinations despite possible fluctuations in passenger numbers (variations in current loads). This ensures smooth, uninterrupted service, just like how biasing stabilizes the amplifier's function.

If I consider practical circuit and let me use the analysis slight analysis here...

The professor explains the effects of practical circuit parameters and resistance on output voltage, emphasizing how these resistances interact. When calculating, evaluating each component's impact on the total output is essential. For example, a particular resistance setup will yield different values that affect the gain and the circuit's behavior overall.

Imagine you have different lengths of water pipes, and each pipe (resistor) can either allow more or less flow (current) of water based on its diameter (value). Balancing these out will determine how much water ultimately reaches your garden (output voltage)—if one is too small or too large, it changes the flow entirely.

So, in summary if I consider this is 10 MΩ and the equivalent resistance of this part it is 10 MΩ and the ideal current here it is 2 mA...

In this summary, the professor consolidates the results of previous calculations regarding output voltage and operational conditions in the cascode amplifier. If everything is consistent with theoretical values, it verifies circuit functionality, ultimately confirming whether both transistors operate within the desired parameters.

Think of a recipe’s final checklist before cooking. If all ingredients align correctly, you're set to cook a great meal (functioning circuit). But if there’s a miss here or there, it could lead to a culinary disaster (circuit failure). Hence, confirmation steps are vital.

Now in before we go into the small signal equivalent circuit, I like to recall that value of g and g both are then r and m1 m2 o1...

This portion mentions transitioning to small signal analysis, which is crucial in understanding the circuit's behavior under varying inputs. It highlights parameters relevant for this analysis—specifically, the transconductance values and the output characteristics that will dictate the circuit's efficiency and gain.

Consider tuning a musical instrument—small adjustments (small signals) can drastically affect the overall sound quality. An understanding of these fine details in circuits leads to better performance, just as fine-tuning musical notes leads to harmony.

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