64.1.3. Indian Institute of Technology, Kharagpur

Before I go to the next topic, I must see here that this calculation of the C I did a small mistake. It should be 135 because C = C + in in π 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 begins by correcting an error in a previous calculation of the capacitance (C). Initially, it was calculated incorrectly, but the correct formula incorporates additional components. When recalculating, the individual capacitances of 10 µF and 1025 pF are combined, resulting in a total capacitance of 1035 pF. This correction is significant for subsequent calculations involving circuit behavior.

Think of calculating the total weight of a mixture. If you forget to include one ingredient's weight (the 10 µF capacitor in this case), the total will be off, impacting recipes (or calculations) that rely on accurate totals.

The advantage of cascode amplifier with respect to standard CE amplifier is that in cases if you are retaining this passive element for both cases, then gain wise we do not get much advantage. But then we have seen that in case if we have the input resistance and the source resistance together forming a significantly low-frequency pole due to this large value of the C.

The professor explains that while the gains of a cascode amplifier compared to a standard common-emitter (CE) amplifier might not be impressive when passive components are involved, the true benefit lies in its ability to create a low-frequency pole in the frequency response of the circuit. This allows the cascode amplifier to better handle bandwidth and improve performance, especially when faced with significant source resistances.

Consider a multi-tiered cake's effectiveness for a birthday party. While each tier might seem similar in size (gain), the base layer (cascode design) offers better stability and aesthetic appeal (signal fidelity) when it bears most of the weight (source resistance).

To demonstrate the capability of the cascode amplifier to increase the gain first of all let us consider a different situation instead of having this R if you put some active circuit there.

The next step involves changing the scenario from using a passive resistor (R) to employing an active circuit, such as a higher resistance (10 MΩ) current source. This change is critical because the higher resistance will better showcase the cascode amplifier's high gain capabilities by eliminating some limitations present in previous setups.

Imagine a student preparing for an exam. Initially studying alone (passive resistance) might yield average results, but when they engage a tutor (active circuit), their understanding and grades significantly improve, thus enhancing their capability.

If I say that the current here it is 2 mA supported by this bias current, then naturally the output voltage will be 12 V.

In this section, we’re establishing the relationship between current, voltage, and resistance in the circuit. If the current flowing through a defined circuit is 2 mA, the resulting voltage will depend on how it’s shared across the circuit components. If one of these components has negligible resistance, the maximum voltage achieved can be expected to align closely to the supply voltage, here quoted as 12 V.

Thinking of a water pump system where the pump's ability to push water (current) determines how much pressure (voltage) will be available at the outlet. If there are no restrictions, you can expect maximum water pressure at exit points.

So, we can say that for all practical purposes both the devices are in good condition and hence we can move to the small signal equivalent circuit.

Here, the lecture indicates that the operating conditions of the transistors (devices) are stable, allowing for simplification of their analysis through the small signal equivalent circuit technique. This technique is used to predict circuit behavior under small deviations in voltage and current, which is essential for linear applications.

Consider tuning a musical instrument. When it's well-tuned (good condition), you can play soft notes effectively (equivalent small signal behavior). If it’s out of tune, it will distort the sound (non-linear behavior), making proper tuning essential for quality performance.

Now if I try to see what is the voltage it is coming here in the small signal equivalent circuit.

In this procedure, the professor outlines how to calculate output voltage using the small signal equivalent circuit. It's emphasized that understanding the interdependencies between circuit elements, like load resistances and transconductance, is key to determining how the output voltage will behave.

Think of a factory assembly line where each worker's efficiency contributes to the output. If one worker (circuit element) improves their speed (transconductance), it positively affects the total quantity of items produced (output voltage).

If I consider the total resistance here and the ideal current here it is 2 mA, this is also 2 mA. So, then this voltage will be 6 V.

The lecture highlights how the total impedance in the circuit affects the voltage levels, specifically stating that with the given conditions, an expected voltage of 6 V can be realized. It emphasizes the significance of impedance on voltage division and cutoff frequency adjustments in AC signals through this understanding.

Think of a traffic system where the total number of cars (impedance) determines how smoothly traffic flows (voltage levels) to a certain area. High congestion (high impedance) could limit the throughput, similar to how cutoff frequencies limit an AC signal's behavior.

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