This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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....
View Full Document
- Fall '08
- 10%, #, VDS, VTN, Vgs Vtn, gm≈