65.3. Numerical Examples

Welcome back after the short break. So, we are talking about the Cascode Amplifier using BJT sorry MOSFET. BJT part we already have completed now come here so, far we are talking about the passive load namely R it was 2 k now we are going to change this load to active kind of load, where our basic motivation is to for higher gain.

In this introduction, the speaker is emphasizing the transition from using BJTs to MOSFETs in cascode amplifiers. The focus is on changing from a passive load (which was 2 kΩ) to an active load aimed at achieving higher gain. This is significant because using active loads typically translates to better performance and higher amplification in circuits.

Imagine upgrading from a regular bicycle (passive load) to an electric bike (active load) to reach your destination faster (higher gain). Just as the electric bike provides better performance with less effort, an active load in an amplifier provides better gain without needing additional power sources.

So, we do have the cascode amplifier here with active load namely the I here it is 2 mA current and this R it is 5 MΩ. So, why did I take this 5 MΩ? It is whatever the parameter we have calculated small signal parameter based on that the equivalent resistance coming here it is 5 MΩ.

In this part, the speaker discusses the details of the cascode amplifier configuration, mentioning that the active load resistance (R) is set to 5 MΩ and the bias current (I) is 2 mA. The choice of 5 MΩ is based on calculations made from small signal parameters, which help optimize the performance of the amplifier by ensuring it can handle the current effectively while providing stability.

Think of designing a new road for traffic flow. The choice of road width (5 MΩ) is crucial for handling the expected traffic (current of 2 mA). If the road is too narrow, it may cause bottlenecks and slowdowns; similarly, the right resistance ensures the amplifier operates efficiently.

Now for that we need to know what will be the equivalent resistance coming here and this equivalent resistance it = (5 MΩ)/(2 mA * 50 kΩ). So, this gives us the current flowing though R namely i = half of this g v and hence we are getting the output voltage v out = ‒ R5 m1.

This segment focuses on calculating the equivalent resistance and how it impacts the voltage gain of the amplifier. The equivalent resistance is calculated using the values given, leading to the output voltage (v_out), which demonstrates how the product of resistance and current leads to amplified signals, a key concept in amplifier design.

Imagine filling a water tank (output voltage) through a pipe (resistance) using a pump (current). The broader the pipe and stronger the pump, the faster and higher you can fill the tank, just like the amplifier boosts the output signal based on the calculated resistance and current.

So, I should say this voltage gain A got increase from 4 magnitude wise to should say it got increase to 5000. In fact, this is I should say it is a big jump and as I said that for most transistor this cascode structure it is frequently use to enhance the gains.

Here, the comparison of voltage gain between using a passive load and an active load is highlighted. The gain jumped from 4 to 5000, indicating a significant improvement in performance due to the active load. This reinforces the concept that good design practices, like selecting suitable loads, can dramatically enhance circuit performance.

Think of it like upgrading from a small speaker (4) to a powerful sound system (5000). The improvement in the quality and loudness of sound represents how active loads improve amplification in electronic circuits.

To get the input capacitance C which is C + C(1 ‒ A) and what is the A? A = ‒ gv1 multiplied by r in parallel with the equivalent resistance coming from this circuit. This gives us the input capacitance value of C = 265 pF.

This part delves into calculating the input capacitance of the circuit. The formula involves both the capacitance of individual components and the effect of gain on overall capacitance. The final calculation results in an input capacitance of 265 pF, showcasing the relationship between gain and capacitance in amplifying circuits.

Imagine upgrading your phone. The larger battery (input capacitance) allows you to use your phone longer without charging (higher gain). However, if the battery is too large, it could make the phone bulkier (impact on circuit capacitance due to gain), thus affecting its usability.

The output resistance R in this circuit is of course, 5 MΩ and 5 M in parallel. So, that gives us 2.5 MΩ and with the C of say 100 pF the upper cutoff frequency may be quite a look.

This section reviews the output resistance of the circuit and its impact on bandwidth. By calculating the parallel resistance and considering the cutoff frequency with given capacitance, it emphasizes how design choices affect circuit performance in terms of frequency response.

Consider a water fountain. The size of the fountain (output resistance) and the size of the pipe (capacitance) dictate how high the water can shoot (bandwidth). A proper balance is necessary to achieve the desired aesthetic effect without wasting water.

In summary if I compared this cascode amplifier to boost to the gain and if I compare the performance of the standard common source amplifier... by the virtue of the cascode structure we can increase the gain by a big factor, but then the corresponding bandwidth it is getting affected.

The final takeaway is the trade-off that often exists between gain and bandwidth in amplifier designs. Cascode amplifiers provide substantial gain improvements, yet this also alters the bandwidth characteristics, a critical consideration for circuit designers aiming for optimal performance.

It's like a high-performance sports car that accelerates quickly (high gain) but may not handle corners as well (bandwidth trade-off). Engineers must balance speed and handling to achieve the best overall driving experience.

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