Determining transistor sizes equal to a single inverter

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SUMMARY

The discussion focuses on determining the appropriate transistor sizes for a 2-input NAND gate to match the worst-case pull-down and pull-up performance of a single inverter with specific dimensions: Wn = 1 µm and Wp = 2 µm, while maintaining a transistor length of 0.25 µm. The key insight is that the equivalent width-to-length (W/L) ratio for the pull-down network (PDN) or pull-up network (PUN) of the NAND gate must equal the W/L ratio of the inverter under worst-case conditions. The participant initially struggled with the calculations but clarified that considering the worst-case scenario is essential for accurate sizing.

PREREQUISITES
  • Basic MOSFET knowledge
  • Understanding of W/L ratios in transistor sizing
  • Familiarity with digital logic gate performance metrics
  • Knowledge of worst-case analysis in circuit design
NEXT STEPS
  • Study MOSFET sizing techniques for digital circuits
  • Learn about worst-case analysis in CMOS design
  • Explore the impact of transistor dimensions on gate delay
  • Research design methodologies for NAND gates and their performance characteristics
USEFUL FOR

Electrical engineers, circuit designers, and students studying digital electronics who are interested in optimizing transistor sizing for logic gates.

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Homework Statement



Determine transistor sizes for (lets say a 2 input nand gate, my actual question is more complicated but i want to know the basic idea so i can do it myself) so that the worst case pull down and pull up is equivalent to a single inverter with Wn = 1 micro m and Wp= 2 Micro m
All transistor lengths are 0.25 micro meters


Homework Equations


Basic Mosfet knowledge should do i guess.


The Attempt at a Solution


i know the solutions, i just don't know how to arrive at it.
My guess was that the equivalent transistor W/L for the PDN or PUN should be the same as the W/L for the PDN or PUN of the inverter. But the numbers don't add up.
 
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Guys i think I've got it but I am not sure

the question said WORST CASE... i was basically missing that part.
so now that I've read it carefully I've found out (with working backwards on the solutions) that using the worst case scenario (slowest timing aka lowest equivalent W/L for PDN or PUN of the nand gate) your equivalent w/l of the worst case scenario should equal to the Wp or Wn given in the question.
 

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