hw1_project1_solved

hw1_project1_solved - Lets assume we will use the value of...

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Question 2: Solution: There are 2 parasitic diodes, one between the bulk and the drain and the other between the bulk and the source a. Vb=0 Vs=0 so ID=0 Vb=0 Vd=5 so the diode is reverse biased. Hence ID=10nA b. Vb=1 and Vs=0 so the diode is Forward biased and hence I=10na(e^ 1/0.026 -1) Vb=1 and Vd=5 so the diode is reverse biased. Hence Id=10nA Question 3: Solution: V1=0.7V V2=-2.3V I1=2.3mA I2=.8mA I3=1.5mA
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Question 4: Solution Part a: Id=0uA for Vin=0V Vout=1.8V for vin =0 For Vin = 1.8V, transistor is in triode so the equation for the drain current is given by Also the current can be found using Now we have two equations and two unknowns hence Vout=3V, 0.75V The obvious choice for Vout is 0.75V Hence Solution Part b: There are three ways to compute VDsat for this device 1) Given on page 95 of Rabaey Hence Vdsat=0.242V Or using the equations on page 96 V Dsat =LEc=0.3 v Or V Dsat = L.v sat / μ n =0.5 volts. In all cases the answer is approximate.
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Unformatted text preview: Lets assume we will use the value of Vdsat=0.5 v Now there can be two different approaches: Approach1: Using equation We have Id = 800 uA and hence the value of VDS=1.8-R*I Dsat =1V This value of VDS>>V Dsat and hence due to velocity saturation effect the transistor has entered saturation. Approach2: Using equation We get the value of I Dsat as 300uA and the value of VDS as before as 1.5V . This shows that the transistor is in saturation which is again the effect of velocity saturation. Note: Both approaches get credit!!! Also note that this is a first order model for velocity saturation and thus different simplification give different answers, Hand analysis is usually used to approximate the region of operation and the general trend of the device and is then followed by accurate simulations....
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