EE 435 Homework 4 Solutions Spring 2021 Problem 1 NMOS
Transcript of EE 435 Homework 4 Solutions Spring 2021 Problem 1 NMOS
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EE 435 Homework 4 Solutions Spring 2021
Problem 1
NMOS
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PMOS
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Problem 2
Part A
The current through M1 is ๐
๐๐ท๐ทโ๐๐๐โ
1
2= 83.33๐๐ด. So we can find ๐1 as follows:
๐1 =2๐ผ๐ฟ
๐๐ถ๐๐ฅ๐๐ธ๐ต2 =
2 โ 83.3๐ โ 2๐
112.8๐ โ 0.22= 73.84๐๐
Part B We know that ๐๐ธ๐ต1 = ๐๐บ๐ โ ๐๐๐ = 0.2๐. Quiescently, ๐๐บ = 0๐. So:
๐๐บ โ ๐๐ โ ๐๐๐ = โ๐๐ โ 0.79๐ = 0.2 โ ๐๐ = โ0.99๐
Part C Create an expression for ๐๐๐๐ divided by ๐๐ผ๐:
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๐๐1๐๐ผ๐ = ๐๐๐๐ (๐ ๐ถ1 +1
๐ 1)
๐ด(๐ ) =๐๐1
๐ ๐ถ1 + 1/๐ 1
Part D Start by finding the DC gain:
๐ด0 = ๐๐๐ 1 =2 โ 83.33๐๐ด
0.2โ 50๐ฮฉ = 41.66510 = 32.39๐๐ต
Now find the corner frequency:
๐ ๐ถ1 +1
๐ 0= 0 โ ๐ = โ
1
๐ 1๐ถ1= 500๐ ๐๐๐/๐ ๐๐
Part E As found in Part D, the 3dB bandwidth is 500๐ ๐๐๐/๐ ๐๐.
Part F Increasing the power increases the gain-bandwidth, but not the bandwidth. The bandwidth is only dependent on ๐ 1 and ๐ถ1.