59.1. Analog Electronic Circuits

So, in the next slide we will be talking about see common source amplifier again this numerical exercise we have seen before. So, this is prime and the main common source amplifier and sorry and then we have the information given here about the device namely which is 1 mA/V2, threshold voltage it is 1 V, supply voltage it is 12 V and so and so. And, we have seen that using this information we obtain V = 3 V and then I we GS DS obtain it was 2 mA and then corresponding small signal parameter g it was 2 mA/V. So, the corresponding voltage gain; voltage gain it was g into output resistance and you may ignore the r or other r we may consider this is very high assuming λ is very small.

This chunk introduces the common source amplifier, a fundamental component in analog electronics. The paragraph provides specific parameters of the device, such as its transconductance (1 mA/V²), threshold voltage (1 V), and supply voltage (12 V). Using these parameters, it calculates the gate-source voltage (3 V), the drain-source current (2 mA), and the small signal transconductance (g = 2 mA/V). The voltage gain is calculated using the output resistance and the transconductance. The text also notes simplifications made in the assumptions for the model, such as ignoring certain resistances.

Think of the common source amplifier like a water tap. The supply voltage represents the water pressure in the pipes (analogous to the supply voltage), while the gate-source voltage is like how far the tap is opened (which allows water flow). The current flowing through is the actual water flow, and the output resistance is similar to the size of the pipe after the tap; a larger pipe allows more flow (higher gain).

So, the voltage gain it was g R and so that becomes 2 m × R is 3 k; 3 k. So, the corresponding voltage gain it was only 6. So, whatever it is and then the output resistance for this case we see it is primarily defined by R and that is 3 kΩ. So, the upper cut off frequency for this case f it was into load capacitance of 100 pF. So, it was and then 3 k into this one 100 p; that means, 10‒10 yeah. And in fact, if you calculate it this gives us 530 kHz.

Here, the calculation of voltage gain is illustrated. Voltage gain (A_v) is derived as the product of transconductance (g) and the output resistance (R_D). With g at 2 mA/V and R_D at 3 kΩ, the result is a voltage gain of 6. The text goes on to discuss the output resistance defined by R (3 kΩ) and calculates the upper cut-off frequency (f_U) by multiplying output resistance with load capacitance (C_L = 100 pF), resulting in a frequency of 530 kHz.

In our water tap example, the current gain is like how strong the water jets out of the tap compared to the pressure at the source. The upper cut-off frequency is like the maximum rate at which you can open and close the tap before you cannot control the flow - if you open it too fast, the system cannot respond effectively, just like electronic signals can be lost if they are too high frequency for the circuit.

So, please recall or try to remember this information. In our next exercise where we will be cascading this CS stage by common drain stage. So, we do have in the next slide we do have that example here. So, all the information’s we are keeping it same we do have additional common drain stage coming out of the transistor M2 and the its bias circuit R and R it is given here it is 1.5 kΩ.

This part discusses the next step of integrating a common drain stage following the common source amplifier. The intention is to show how cascading these stages impacts performance, particularly in bandwidth. The setup retains the parameters from previous sections, such as the previous transistor M1 and now includes M2 with a common drain setup. The bias resistors are introduced here, which help maintain the operating point and stability of the circuit.

Imagine you're adding a second tap after the first one (common drain stage) to manage the same water flow more efficiently. The bias circuit works like a water filter, ensuring that the water goes through smoothly without clogging or losing pressure, allowing for better control of the overall water distribution system (signal processing in electronics).

And since there is no current flow even if you consider the 3 k resistance here, but still we can say that gate voltage it is 6 V. So, if I consider the loop here if I consider this loop and if I consider we do have 6 V coming here.

In this segment, the explanation focuses on understanding the operational loop of the transistor. Despite a 3 kΩ resistor, the setup indicates a 6 V gate voltage. The relation between gate voltage and source voltage gives critical insights into the transistor's operation, allowing calculations regarding the transistor's current flow and confirming whether it operates in saturation.

This is analogous to checking water levels in a connected system. Just like ensuring you have sufficient water input despite some resistance in the pipes, in electronics, it’s crucial to ensure adequate gate voltage to maintain proper current flow through the circuit and ensure the system works efficiently.

So, the g of this transistor it is let me create some space here. So, g of transistor-2 it is we do have × (V ‒ V ). So, here also we got g = 2 mA/V and then the output resistance and of course, since we do not have the equivalent whatever r we are having in BJT. And so the calculation here it is much simpler and you can approximate that the gain here A , it is very close to 1 and output resistance not A , A rather second stage gain.

This chunk explains how to calculate the transconductance (g) for the second transistor stage. The gain approximation shows that the common drain stage operates with gain close to 1, simplifying calculations for overall gain where it primarily reflects the first stage's performance rather than the second due to its buffering nature.

Think of adding a second water tap that doesn’t change the pressure but smoothens out the flow. This is how a common drain stage works – it maintains whatever the first stage provides while ensuring that the output is stable and ready for the next use, be it a tap or in this case, an electronic load.

Now, if I call it is f′ it is having 2 candidates to define the upper cut off frequency; one is R . So, this is 2πR and the corresponding load capacitance we do have C . So, if we connect the C here.

The segment discusses factors for calculating the upper cut-off frequency of the combined stages. It mentions two components: the resistance (R) and the load capacitance (C_L), which together define how quickly the circuit can respond to changing signals, determining the bandwidth.

Imagine the open pipe system again. The width of the pipe and the amount of water flowing through (represented by capacitive loading) determine how fast the tap can be opened and closed without losing pressure or causing a mess. Similarly, capacitors and resistors in circuits define how quickly signals can be processed before they diminish.

So, in this session what we have learned here it is the usefulness of the common collector and common drain and we have we have demonstrated through numerical examples. Basically, by considering CE and CC together and then CS and CD together to enhance the bandwidth upper cutoff frequency particularly getting increased by a factor of as is maybe a factor of 10 or more.

The conclusion reiterates the key points covered in the lecture, emphasizing the benefits of cascading different transistor stages to enhance performance. It highlights the numerical findings that suggest significant bandwidth improvements when combining various amplifier configurations.

This is analogous to building an efficient plumbing system where multiple pipes (amplifiers) work together to handle a larger flow of water (signals) than any single pipe alone could handle. Just as cascading pipelines allows for better management of water delivery and pressure, cascading amplifier stages enables improved signal handling in circuits.

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