CSE 463M and CSE 563M students:
1) Solve problems (10 points each)
3.1, 3.2, 3.3, 3.6, 3.10, 3.12.
from Chapter 3 from the textbook. Please show all your work and be neat and circle your answers.
2) Cadence exercise (20 points):
Repeat Transistor I-V characteristics exercise from the Cadence wiki. Print out and submit
Transistor schematics and I-V graph just like it is done in the Cadence wiki. To access Cadence you need to login to a Linux Machine at https://linuxlab.engr.wustl.edu
Once you login to the Linux Machine start running Cadence exercise.
CSE 563M students only:
In addition to the problem set above, please solve the following problem from the textbook:
3.9 (10 points).
3.1
+ We know the following:
𝑡𝑜𝑥 = 1.6 𝑛𝑚 = 1.6 ⋅ 10−7𝑐𝑚
Φ𝐺𝐶 = −1.04 𝑉
𝑁𝐴 = 2.8 ⋅ 1018𝑐𝑚−3
𝑄𝑜𝑥 = 𝑞4 ⋅ 1010𝐶/𝑐𝑚2 = 1.6 ⋅ 10−19 ⋅ 4 ⋅ 1010𝐶/𝑐𝑚2 = 6.4 ⋅ 10−9𝐶/𝑐𝑚2
𝜀𝑜𝑥 = 3.97𝜀0 = 3.97 ⋅ 8.85 ⋅ 10−14 𝐹/𝑐𝑚 (1 𝐹𝑎𝑟𝑎𝑑 = 1 𝑉 𝐶 )
𝐹 𝐶
𝜀𝑠𝑖 = 11.7𝜀0 = 11.7 ⋅ 8.85 ⋅ 10−14 (1 𝐹𝑎𝑟𝑎𝑑 = 1 )
𝑐𝑚 𝑉
𝑛𝑖 = 1.45 ⋅ 1010𝑐𝑚−3
𝑘 = 1.38 ⋅ 10−23𝐽/𝐾
𝑞 = 1.6 ⋅ 10−19𝐶
+ To get the threshold voltage, solve the equation below:
𝑄𝐵0 𝑄𝑜𝑥
𝑉𝑇0 = Φ𝐺𝐶 − 2𝜙𝐹(𝑠𝑢𝑏) − −
𝐶𝑜𝑥 𝐶𝑜𝑥
+ First, 𝜙𝐹(𝑠𝑢𝑏):
𝑘𝑇 𝑛𝑖
𝜙𝐹(𝑠𝑢𝑏) = ln ( )
𝑞 𝑁𝐴
𝑘𝑇 1.38 ⋅ 10−23𝐽/𝐾 ⋅ 300𝐾
= −19𝐶 = 0.026 V
𝑞 1.6 ⋅ 10
18𝑐𝑚−3
𝑁𝐴 2.8 ⋅ 10
𝑘𝑇 𝑛𝑖
𝜙𝐹(𝑠𝑢𝑏) = ln ( ) = 0.026𝑉 ⋅ −19 = −0.496𝑉
𝑞 𝑁𝐴
+ Second, 𝑄𝐵0:
𝑄𝐵0
= − √2 ⋅ (1.6 ⋅ 10−19𝐶) ⋅ (2.8 ⋅ 1018𝑐𝑚−3) ⋅ (11.7 ⋅ 8.85 ⋅ 10−14 𝐹/𝑐𝑚) ⋅ | − 2(−0.496𝑉)|
= − 9.593 ⋅ 10−7𝐶/𝑐𝑚2
+ Third, 𝐶𝑜𝑥:
𝐶𝑜𝑥 = 𝜀𝑜𝑥 = 3.97 ⋅ 8.85 ⋅ 10−14 𝐹/𝑐𝑚 −6𝐹/𝑐𝑚2
= 2.195 ⋅ 10
𝑡𝑜𝑥 1.6 ⋅ 10−7𝑐𝑚
+ Finally, plug in all values above into equation:
𝑄𝐵0 𝑄𝑜𝑥
𝑉𝑇0 = Φ𝐺𝐶 − 2𝜙𝐹(𝑠𝑢𝑏) − −
𝐶𝑜𝑥 𝐶𝑜𝑥
= 0.386𝑉
3.1.b.
+ The amount of channel implant:
𝑞𝑁1
∆𝑉 = 0.6 − 𝑉𝑇0 = 0.6 − 0.386 = 0.214 =
𝐶𝑜𝑥
𝑁𝐼 = 𝑞 𝑜𝑥 = 0.2141.⋅62⋅.19510−19⋅ 10−6 = 2.953 ⋅ 1012𝑐𝑚−2 ∆𝑉 ⋅ 𝐶
+ The type is p-type since threshold voltage should be increased.
3.2
+ Constant values:
𝐹 𝐶
𝜀𝑠𝑖 = 11.7𝜀0 = 11.7 ⋅ 8.85 ⋅ 10−14 (1 𝐹𝑎𝑟𝑎𝑑 = 1 )
𝑐𝑚 𝑉
𝑛𝑖 = 1.45 ⋅ 1010𝑐𝑚−3
𝑞 = 1.6 ⋅ 10−19𝐶
+ The expression for the junction capacitance:
𝐶𝑗(𝑉) = 𝐴
+ for 𝜙0:
⋅ 𝑁𝐷 2 ⋅ 1020 ⋅ 2 ⋅ 10
𝜙0 𝑞 𝑛𝑖2 (1.45 ⋅ 1010)2
+ for 𝐴:
𝐴 = 0.2 ⋅ 0.4 + 2(0.2 ⋅ 0.032) + 2(0.4 ⋅ 0.032) = 1.184 ⋅ 10−9
+ Plug-in values:
𝐶𝑗
3.3.
They are not identical.
The electric channel length and the mask channel length have a relationship below:
𝐿 = 𝐿𝑀 − 2𝐿𝐷,𝑤ℎ𝑒𝑟𝑒 𝐿𝐷 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑙𝑒𝑛𝑔𝑡ℎ.
3.6
+ we know the following:
𝑁𝐷 = 2 ⋅ 1020𝑐𝑚−3
𝑁𝐴 = 2 ⋅ 1020𝑐𝑚−3
𝑋𝑗 = 32 𝑛𝑚
𝐿𝐷 = 10 𝑛𝑚
𝑡𝑜𝑥 = 1.6 𝑛𝑚
𝑉𝑇0 = 0.53 𝑉
𝐶ℎ𝑎𝑛𝑛𝑒𝑙 𝑠𝑡𝑜𝑝 𝑑𝑜𝑝𝑖𝑛𝑔 = 16.0 𝑋 (𝑝 𝑡𝑦𝑝𝑒 𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑡𝑒 𝑑𝑜𝑝𝑖𝑛𝑔)
𝐹 𝐶
𝜀𝑠𝑖 = 11.7𝜀0 = 11.7 ⋅ 8.85 ⋅ 10−14 (1 𝐹𝑎𝑟𝑎𝑑 = 1 )
𝑐𝑚 𝑉
𝑞 = 1.6 ⋅ 10−19𝐶
+ Drain capacitance equation:
𝐶𝑑𝑟𝑎𝑖𝑛 = 𝐴 ⋅ 𝐶𝑗0 ⋅ 𝐾𝑒𝑞 + 𝑃 ⋅ 𝐶𝑗𝑠𝑤𝐾𝑒𝑞(𝑠𝑤)
+ for 𝜙0:
⋅ 𝑁𝐷 2 ⋅ 1020 ⋅ 2 ⋅ 10 𝜙0 𝑞 𝑛𝑖2 (1.45 ⋅ 1010)2
+ for 𝜙0𝑠𝑤:
𝑘𝑇 𝑁𝐴(𝑠𝑤) ⋅ 𝑁𝐷 16 ⋅ 2 ⋅ 1020 ⋅ 2 ⋅ 1020 𝜙0𝑠𝑤 = 𝑞 ln ( 𝑛 2 ) = 0.026ln ( (1.45 ⋅ 1010)2 ) = 1.28
𝑖
+ for 𝐶𝑗0,𝐶𝑗𝑠𝑤:
𝐶𝑗0 =
0
𝐹/𝑐𝑚2
𝜀𝑠𝑖 ⋅ 𝑞 𝑁𝐴(𝑠𝑤) ⋅ 𝑁𝐷 1
𝐶𝑗0𝑠𝑤 = ( ) ⋅
0
𝐹/𝑐𝑚2
𝐶𝑗𝑠𝑤 = 𝑋𝑗𝐶𝑗0𝑠𝑤 = 32 ⋅ 10−9 ⋅ 3.49 ⋅ 10−6 = 0.11 ⋅ 10−12 𝐹/𝑐𝑚
+ for A:
𝐴 = 𝑌 ⋅ 𝑊 = 6 ⋅ 10 = 60 𝜇𝑚2=60 ⋅ 10−8𝑐𝑚2
+ for P:
𝑃 = 2(𝑌 + 𝑊) = 2(6 + 10) = 32𝜇𝑚 = 32 ⋅ 10−4 𝑐𝑚
+ for 𝐾𝑒𝑞,𝐾𝑒𝑞(𝑠𝑤):
𝐾𝑒𝑞
𝐾𝑒𝑞(𝑠𝑤) = − (√𝜙0𝑠𝑤 − 𝑉2 − √𝜙0𝑠𝑤 − 𝑉1) = 0.768
𝑉2 − 𝑉1
+ Now, we can plugin values into Drain capacitance equation:
𝐶𝑑𝑟𝑎𝑖𝑛 = 𝐴 ⋅ 𝐶𝑗0 ⋅ 𝐾𝑒𝑞 + 𝑃 ⋅ 𝐶𝑗𝑠𝑤 ⋅ 𝐾𝑒𝑞(𝑠𝑤)
= 60 ⋅ 10−8 ⋅ 2.61 ⋅ 10−6 ⋅ 0.759 + 32 ⋅ 10−4 ⋅ 0.11 ⋅ 10−12 ⋅ 0.768
= 1.188 𝑝𝐹
3.10
+ we know the following:
𝑘′ = 168 𝜇𝐴/𝑉2
𝑉𝑇0 = 0.48 𝑉
𝛾 = 0.52 𝑉1/2
|2𝜙𝐹| = 1.01 𝑉
+ Gate voltage is high and the Vx is expected to low. The load is in saturation and the driver is in linear region.
+ for 𝑉𝑇:
𝑉𝑇
𝑉𝑇
+ From KCL:
𝐼𝐷 = 𝐼𝐷,𝐷𝑟𝑖𝑣𝑒𝑟 = 𝐼𝐷,𝐿𝑜𝑎𝑑
1 ′ 𝑊 (1 − 𝑉𝑥 − 𝑉𝑇,𝐿(𝑉𝑥))2 = 1 𝑘′ 𝑊 (2(1 − 𝑉𝑇0)𝑉𝑥 − 𝑉𝑥2)
𝑘
2 𝐿 2 𝐿
(1 − 𝑉𝑥 − 𝑉𝑇,𝐿(𝑉𝑥))2 = (2(1 − 0.48)𝑉𝑥 − 𝑉𝑥2)
(1 − 𝑉𝑥 − 𝑉𝑇,𝐿(𝑉𝑥))2 = 1.04𝑉𝑥 − 𝑉𝑥2
+ To get Vx, plug in 𝑉𝑇 into KCL equation:
(1 − 𝑉𝑥 − 𝑉𝑇,𝐿(𝑉𝑥))2 = 1.04𝑉𝑥 − 𝑉𝑥2
𝑉𝑥2
𝑉𝑥 = 0.1357
+ To get Id, plug in values into KCL equation:
𝐼𝐷 = 𝐼𝐷,𝐷𝑟𝑖𝑣𝑒𝑟 = 𝐼𝐷,𝐿𝑜𝑎𝑑 1 𝑘′ 𝜇𝐴
3.12
+ For 𝜆:
Choose row2 and row4 from the table.
𝐼𝐷1 − 𝐼𝐷2
𝜆 =
𝑉𝐷𝑆1𝐼𝐷2 − 𝑉𝐷𝑆2𝐼𝐷1
𝜆
+ For 𝑉𝑇0:
Choose row1 and row2 from the table.
𝐼𝐷1 (𝑉𝐺𝑆1 − 𝑉𝑇0)2
= 2
𝐼𝐷2 (𝑉𝐺𝑆2 − 𝑉𝑇0)
𝑉𝐺𝑆1
𝑉𝑇
𝑉𝑇 𝑉
+ For 𝑘:
Choose row2 from the table.
𝐼𝐷𝑆𝐴𝑇 = 𝑘 2(1 + 𝜆𝑉𝐷𝑆)
(𝑉𝐺𝑆 − 𝑉𝑇)
2
𝑘
𝑘 = 1.08 𝑚𝐴/𝑉2
2𝐼𝐷𝑆𝐴𝑇/ ((𝑉𝐺𝑆 − 𝑉𝑇)2(1 + 𝜆𝑉𝐷𝑆)) = 𝑘
+ For 𝛾:
Choose row3 from the table.
𝑉𝑇(𝑉𝐵𝑆 = −0.3) = 0.55 𝑉
𝑉𝑇
0.
𝛾 = 0.52 𝑉1/2
3.9(For 563M)
+ We know the following:
𝑡𝑜𝑥 = 1.6 𝑛𝑚 = 1.6 ⋅ 10−7𝑐𝑚
Φ𝐺𝐶 = −1.04 𝑉
𝑁𝐷,𝑠𝑢𝑏 = 2 ⋅ 1016𝑐𝑚−3
𝑁𝐷,𝑝𝑜𝑙𝑦 = 1020𝑐𝑚−3
𝑄𝑜𝑥 = 𝑞4 ⋅ 1010𝐶/𝑐𝑚2 = 1.6 ⋅ 10−19 ⋅ 4 ⋅ 1010𝐶/𝑐𝑚2 = 6.4 ⋅ 10−9𝐶/𝑐𝑚2
𝜀𝑜𝑥 = 3.97𝜀0 = 3.97 ⋅ 8.85 ⋅ 10−14 𝐹/𝑐𝑚 (1 𝐹𝑎𝑟𝑎𝑑 = 1 𝑉 𝐶 )
𝐹 𝐶
𝜀𝑠𝑖 = 11.7𝜀0 = 11.7 ⋅ 8.85 ⋅ 10−14 (1 𝐹𝑎𝑟𝑎𝑑 = 1 )
𝑐𝑚 𝑉
𝑛𝑖 = 1.45 ⋅ 1010𝑐𝑚−3
𝑘 = 1.38 ⋅ 10−23𝐽/𝐾
𝑞 = 1.6 ⋅ 10−19𝐶
𝑘𝑇 1.38 ⋅ 10−23𝐽/𝐾 ⋅ 300𝐾
= −19𝐶 = 0.026 V 𝑞 1.6 ⋅ 10
+ To get the Φ𝐺𝐶:
Φ𝐺𝐶 = 𝜙𝐹(𝑠𝑢𝑏) − 𝜙𝐹(𝑔𝑎𝑡𝑒)
𝜙𝐹(𝑠𝑢𝑏) = 𝑘𝑇 ln (𝑁𝐷,𝑠𝑢𝑏) = 0.026⋅ ln ( 2 ⋅ 1016 𝑉
𝑞 𝑛𝑖 1.
𝑘𝑇 𝑁𝐷,𝑝𝑜𝑙𝑦 1020 𝜙𝐹(𝑔𝑎𝑡𝑒) = ln ( ) = 0.026 ⋅ ln ( 𝑉
𝑞 𝑛𝑖 1.
Φ𝐺𝐶 = 0.367 − 0.589 = −0.221 𝑉
+ Second, 𝑄𝐵0:
𝑄𝐵0
= √2 ⋅ (1.6 ⋅ 10−19𝐶) ⋅ (2 ⋅ 1016𝑐𝑚−3) ⋅ (11.7 ⋅ 8.85 ⋅ 10−14 𝐹/𝑐𝑚) ⋅ |2 ⋅ (0.374𝑉)|
= 6.974 ⋅ 10−8𝐶/𝑐𝑚2
+ Third, 𝐶𝑜𝑥:
𝐶𝑜𝑥 = 𝜀 𝑡𝑜𝑥𝑜𝑥 = 3.97 ⋅18..685⋅ 10⋅ 10−7−𝑐𝑚14 𝐹/𝑐𝑚 = 2.195 ⋅ 10−6𝐹/𝑐𝑚2
+ Finally, plug in all values above into equation:
𝑄𝐵0 𝑄𝑜𝑥
𝑉𝑇0 = Φ𝐺𝐶 − 2𝜙𝐹(𝑠𝑢𝑏) − −
𝐶𝑜𝑥 𝐶𝑜𝑥
= −0.9896 𝑉
