Question

12. For the network of Fig 6.77, determine: (a) Va. (6) In, and Vas (c) Vp and Vs (a) Vrs 20 V 22 ka 910 kQ oss - 10 mA V=-35V Nong 110 1.1 k Figur

Answer 1

The current in the forward active region is

                                                             

ID = Ipss Ip = Ipss (1 - Ves)

Here

                                                

110k VG = 20 = 2.157 V 110k +9101.

Also

                                                   

Vs = ID * 1.1kΩ

So

                                                         

(-x-1) ssar = ar

Or

                                               

ID = 10mA (1- 2.157 – 1100 * ID -3.5

Let

                                                         

Ip = Ipo mA

Or

                                                

2.157 – 1.11 Do 0.11 po = (1+2 3.5

Or

                                                  

0.11 po = (5.657 – 1.11 Do 3.5

Or

                                                        

1.2251 po = 32 +1.21130 – 12.441 DO

Or

                                                        

1.2112 – 13.6651 do + 32 = 0

So

                                                            

Ipo = 8 or 3.3

So

                                                                    

Ip = 8mA or 3.3mA

For I_D = 8mA, V_GS = 2.157-1.1*8 = -6.643V < V_P. So it is not valid solution. So I_D = 3.3mA. So

                                              

VGSO = 2.157 – 3.3 * 1.1 = -1.473V

The drain voltage is

                                                      

Vp = 20 – 3.3 * 2.2 = 12.74V

The source voltage is

                                                              

Vs = 3.3 * 1.1 = 3.631

The Drain-source voltage is

                                            

Vpso = Vp - Vs = 12.74 – 3.63 = 9.11V