HW7soln - PHYS3360/AEP3630 Electronic Circuits Spring 2011...

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PHYS3360/AEP3630 Electronic Circuits, Spring 2011, HW7 Solutions 1 MOSFET’s and Digital Circuits Solutions 1. a. In saturation, the current through Q3 is given by 𝐼 0 = 𝐼 𝐷𝑆3 = 𝐾 ( 𝑉 𝐺𝑆3 𝑉 𝑇 ) 2 so 𝑉 𝐺𝑆3 = 𝑉 𝑇 + 𝐼 0 𝐾 = 2.316 𝑉 Also, 𝑉 𝐺𝑆3 = 𝑉 𝐺3 – (- 𝑉 𝑆𝑆 ) and 𝑉 𝐺3 = 𝑅 3 𝑅 3 +𝑅 4 - 𝑉 𝑆𝑆 which yields 𝑅 3 = ( 𝑉 𝑆𝑆 𝑉 𝐺𝑆3 1) 𝑅 4 . Choosing 𝑅 4 =10kΩ gives 𝑅 3 = 11.59 kΩ. b. 𝐼 0 = 𝐼 𝐷1 + 𝐼 𝐷2 and 𝐼 𝐷1 = 𝐼 𝐷2 Then 𝐼 𝐷1 = 𝐼 𝐷2 = 𝐼 0 /2 = 5 mA so, respectively, 𝑉 𝐷𝐷 − 𝑉 𝐷1 , 2 = 𝐼 𝐷1 , 2 𝑅 𝐷1 , 2 which yields 𝑅 𝐷1 = 𝑅 𝐷2 = 500 Ω c. Small signal gain is calculated in the Lab Manual to be: 𝐺 = 𝑉 𝑜𝑢𝑡 𝑉 1 − 𝑉 2 = 1 2 𝑔 𝑚 𝑅 𝐷2 = 1 2 2 𝐾 𝐼 0 𝑅 𝐷2 = 11.18
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PHYS3360/AEP3630 Electronic Circuits, Spring 2011, HW7 Solutions
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