Find W/L for CMOS Circuit: Triode/Cutoff Modes

  • Thread starter jean28
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In summary, the question is asking how to find the value of W/L, which is already given as 2 in the exercise. The formula given in the hint is used to solve for W/L, but the resulting value is 0.08333 instead of the expected 2. It is noted that the circuit is to be analyzed in Triode and Cutoff modes as it is a CMOS circuit.
  • #1
jean28
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How to find W/L??

Homework Statement


The exercise already tells me that the value of W/L is 2. However, I can't seem to understand where that result came from. I'd like to know how I can get to that conclusion.

Here is the exercise with the graph of the inverter:
Inverter image:
http://i1226.photobucket.com/albums/ee410/jean28x/image2_zps9e7f8724.png

Exercise, part 1:
http://i1226.photobucket.com/albums/ee410/jean28x/image_zps136b69f7.png

Exercise, part 2:
http://i1226.photobucket.com/albums/ee410/jean28x/image_1_zps32b9a8ee.png

Homework Equations


Here is a picture of relevant formulas of MOSFETS:
http://i1226.photobucket.com/albums/ee410/jean28x/image_2_zpse3d8245e.png

There is also the formula in the hint that is given in the exercise:
http://i1226.photobucket.com/albums/ee410/jean28x/image_1_zps32b9a8ee.png

The Attempt at a Solution



So, using the formula given in the hint, I substitute the values (I already know the 48 KΩ part):

48k = 1/((125μ)(W/L)(2.5 - 0.5))

Solving for W/L, I get:

12 = 1 / (W/L)

W/L = 1 / 12 = 0.08333

However, the answer says that W/L is supposed to be 2. What am I doing wrong here?

Thanks a lot.P.S.

It might be worth noting that, since this is an exercise working with CMOS circuits, then the relevant state that it should be analyzed in is in Triode and Cutoff modes.
 
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  • #2


Here I will upload the pictures, just in case the links don't work.
 

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  • #3


Last one.
 

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1. What is the purpose of finding W/L for CMOS circuit in triode/cutoff modes?

The purpose of finding W/L (width/length) in CMOS circuits is to determine the optimal dimensions for the transistors in order to achieve the desired performance and functionality of the circuit. In triode/cutoff modes, the transistor operates in the linear region and the dimensions of W/L can greatly affect the current flowing through the transistor, thus impacting the overall performance of the circuit.

2. How is W/L calculated for CMOS circuit in triode/cutoff modes?

The W/L ratio is calculated by dividing the width of the transistor by its length. In triode/cutoff modes, the transistor operates as a resistor, so the W/L ratio can be adjusted to achieve the desired resistance value. The W/L ratio is typically expressed in terms of micrometers (µm) for width and micrometers (µm) or nanometers (nm) for length.

3. What are the factors that affect W/L in CMOS circuit design?

There are several factors that can affect the W/L ratio in CMOS circuit design. These include the desired resistance value, the technology node or fabrication process used, the supply voltage, and the load capacitance. Additionally, the transistor's geometry, such as channel length modulation and gate oxide thickness, can also affect the W/L ratio.

4. How does W/L impact the performance of a CMOS circuit in triode/cutoff modes?

The W/L ratio has a significant impact on the performance of a CMOS circuit in triode/cutoff modes. A smaller W/L ratio results in a smaller transistor, which can lead to a higher resistance and lower current flow. On the other hand, a larger W/L ratio can result in a larger transistor with lower resistance and higher current flow. The optimal W/L ratio is determined by the specific requirements and constraints of the circuit design.

5. Are there any limitations to consider when determining W/L for CMOS circuit in triode/cutoff modes?

Yes, there are some limitations to keep in mind when determining the W/L ratio for CMOS circuits in triode/cutoff modes. These include the minimum and maximum allowed values for W/L in a particular technology node, as well as design constraints such as power consumption, speed, and area. It is important to strike a balance between these limitations in order to achieve the best overall performance for the circuit.

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