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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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This note was uploaded on 02/18/2010 for the course ECE 255 taught by Professor Staff during the Fall '08 term at Purdue University.
- Fall '08