Sol HW3 ECE 255 Prof Ye

# Sol HW3 ECE 255 Prof Ye - 4-Prob 4.19 Using Vtn and Kn from...

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ECE 255 Prof. P. Ye Solutions to HW #     3    1.-  Prob. 4.3. Using the standard vales of the named constants and plugging in the value of N B  in the  equations given in the problem. It is calculated that Cd =10.5x10 -9  F/cm 2 . 2.- Prob. 4.6 Using the well-known relationship v n = - μ n E   it is obtained that: a) v n =-1x10 cm/s   and   b) v n =-1.6x10 cm/s    3.- Prob. 4.18 From the graph, for Vgs=4 in the saturation region, the current is approximately 395  μ A.  Therefore: 395  μ A =(1/2)K n (4- Vtn) 2  . By the same reasoning, for Vgs=3:  140  μ =(1/2)K n (3- Vtn) 2 . Dividing the two previous equations:    395/140 = (4- Vtn) 2  /(3- Vtn) 2 . Therefore Vtn=1.5V. Substituting Vtn back in either of the two equations and solving for Kn one gets  Kn=125 μ A/V 2 . Since it is a NMOS transistor and Vtn >0, necessarily, it must be an enhancement type  NMOS. The ratio W/L will not be required for this problem.

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Unformatted text preview: 4.-Prob. 4.19 Using Vtn and Kn from the previous problem: Id sat =(1/2)K n (Vgs- Vtn) 2 and Vd sat =Vgs-Vtn (using Vgs=3.5V and Vgs=4.5V). In order to add the curves corresponding to Vgs=3.5V and Vgs=4.5V in the linear region it must be used Id trio =Kn[(Vgs-Vtn)Vds - Vds/2]Vds ( this equation must be used from Vds =0V to Vds= Vd sat ). For saturation region (Vds > Vd sat ) is simply Id sat =(1/2)K n (Vgs- Vtn) 2 . In the end the graph should look something like this: 5.-Prob. 4.28 Using equation 4.20 and plugging in the values obtained in 4.18 and 4.19: gm 316 ≈ μ S (for Vgs=4) and gm 200 ≈ μ S (for Vgs=3). Approximations using ∆Id/∆Vgs (instead of dId/dVgs) are accepted if results are within a 10% error from the values shown above....
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Sol HW3 ECE 255 Prof Ye - 4-Prob 4.19 Using Vtn and Kn from...

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