Industry-relevant training in Business, Technology, and Design to help professionals and graduates upskill for real-world careers.
Fun, engaging games to boost memory, math fluency, typing speed, and English skillsβperfect for learners of all ages.
Listen to a student-teacher conversation explaining the topic in a relatable way.
Signup and Enroll to the course for listening the Audio Lesson
Today, weβll discuss the common source amplifier. Can anyone explain what they think it does in an analog circuit?
It amplifies the input signal, converting it into a larger output signal.
Exactly! It's a type of voltage amplifier. The voltage gain can be represented by the formula A = gm * RD, where gm is the transconductance, and RD is the output resistance.
What factors affect the transconductance?
Great question! Transconductance (gm) is influenced by the drain current and the threshold voltage of the MOSFET. Remember the formula gm = 2Id/Vov, where Vov is the overdrive voltage. Letβs practice calculating this with an example.
Signup and Enroll to the course for listening the Audio Lesson
Letβs calculate the voltage gain of a common source amplifier. If we have RD = 3 kβ¦ and gm = 2 mA/V, how do we calculate it?
A = gm * RD, so it should be 2 mA/V * 3 k�
Yes! That gives us a gain of 6. Now, what about output resistance?
Is it just RD in this case?
Correct! The output resistance for a CS amplifier is primarily defined by RD, since we assume lambda (Ξ») to be very small. Let's add this to our calculations.
Signup and Enroll to the course for listening the Audio Lesson
Now, moving to the upper cutoff frequency. It's influenced by the load capacitance and output resistance. Can anyone remember the formula?
Is it fU = 1/(2Ο * RD * Cload)?
Exactly right! Let's apply that to a scenario where Cload = 100 pF and RD = 3 kβ¦. Whatβs the frequency?
Calculating gives us around 530 kHz?
Well done! This frequency indicates the amplifier's bandwidth limitations.
Signup and Enroll to the course for listening the Audio Lesson
Letβs talk about cascading. Why do we combine a common source (CS) with a common drain (CD) stage?
To enhance the bandwidth?
Yes! The benefits here include an increase in bandwidth while maintaining gain. The overall gain remains similar due to the CD stage having nearly a gain of 1, but how does this affect input resistance?
I guess it increases input resistance dramatically because of the feedback?
Precisely! The cascading helps combine benefits, which is vital in design for high-performance applications. Having the knowledge of both gain and frequency responses is crucial for our next exercises.
Signup and Enroll to the course for listening the Audio Lesson
To wrap up, let's apply what we've learned in a practical example. If we have a CS stage with a current of 2 mA and a cutoff frequency range mentioned earlier, what challenges might arise when we cascade it with another stage?
Maybe stability or distortion issues?
Excellent! Distortion and phase issues can arise if bandwidth is not addressed properly. Let's calculate for a CE configuration as well later today, to see how those impact our overall design!
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
The section focuses on common source amplifiers with MOSFETs, including calculations for output resistance, voltage gain, and upper cut-off frequency. It demonstrates the cascading effect of combining different amplifier stages to enhance performance.
In this section, Prof. Pradip Mandal explores multi-transistor amplifiers, specifically starting with the common source (CS) amplifier. The key parameters discussed include:
Throughout the section, numerical problems lead to significant learning by applying theoretical knowledge to practical situations, emphasizing why these configurations are essential in analog electronics and signal processing.
Dive deep into the subject with an immersive audiobook experience.
Signup and Enroll to the course for listening the Audio Book
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 Ξ» 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/VΒ²), 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).
Signup and Enroll to the course for listening the Audio Book
So, the voltage gain it was g R and so that becomes 2 m Γ 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 kβ¦. 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, 10β10 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 kβ¦, the result is a voltage gain of 6. The text goes on to discuss the output resistance defined by R (3 kβ¦) 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.
Signup and Enroll to the course for listening the Audio Book
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 informationβs 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 kβ¦.
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).
Signup and Enroll to the course for listening the Audio Book
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 k⦠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, itβs crucial to ensure adequate gate voltage to maintain proper current flow through the circuit and ensure the system works efficiently.
Signup and Enroll to the course for listening the Audio Book
So, the g of this transistor it is let me create some space here. So, g of transistor-2 it is we do have Γ (V β 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 doesnβt change the pressure but smoothens out the flow. This is how a common drain stage works β 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.
Signup and Enroll to the course for listening the Audio Book
Now, if I call it is fβ² it is having 2 candidates to define the upper cut off frequency; one is R . So, this is 2ΟR 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.
Signup and Enroll to the course for listening the Audio Book
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.
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Common Source Amplifier: An amplifier with high voltage gain.
Transconductance: Important parameter indicating current control efficiency.
Output Resistance: Resistance at the output of the amplifier that affects signal integrity.
Cascading: Technique to increase operating frequencies and input resistance.
See how the concepts apply in real-world scenarios to understand their practical implications.
A common source amplifier with a drain resistor of 3 k⦠and transconductance of 2 mA/V achieves a voltage gain of 6.
Cascading a common source amplifier with a common drain stage can provide a bandwidth increase to 4.24 MHz.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
CS stages can gain with ease, overhead voltage is the key, the current flow will never freeze.
Imagine a trainer (the amplifier) who boosts an athlete (the signal) by using their techniques (gain calculation) wisely to win the race (output performance)!
Remember: COT = Common Output Transconductance, outlining the connections in cascade configurations!
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Common Source Amplifier
Definition:
An amplifier configuration that provides high voltage gain, utilizing a MOSFET.
Term: Transconductance
Definition:
A measure of the effectiveness of a transistor as a controlled current source.
Term: Voltage Gain
Definition:
The ratio of output voltage to input voltage in an amplifier.
Term: Upper Cutoff Frequency
Definition:
The frequency at which the output power falls to half its peak value.
Term: Cascading
Definition:
The practice of connecting multiple stages of amplification to enhance performance.