Chapter 4 problems: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
Line 19: Line 19:
! <math>v_{in}</math>!! <math>V_{CE}</math> V !! <math>i_C</math> mA
! <math>v_{in}</math>!! <math>V_{CE}</math> V !! <math>i_C</math> mA
|-
|-
| -.2|| .45 || 2
| -.2|| 19 || .8
|-
|-
| 0|| 16|| 2
| 0|| 16|| 2
|-
|-
| .2 || .62 || 10
| .2 || 12 || 4
|}
|}



Revision as of 12:32, 5 March 2010

4.20

  • We want to establish a DC operating point for the BJT. Writing a loop equation gives us -VBB-Vin(t)+iBRB+vBE=0. To form a load line, we will let vin(t)=0 & iB=0. Thus vBE=VBB+vin(t) and iB(t)=(VBB+vin(t))/RB
  • for the minimum, Q-point and maximum values.
V uA
-.2 .45 2
0 .58 5
.2 .62 10
  • Writing a loop equation for the output: .
V mA
-.2 19 .8
0 16 2
.2 12 4

4.34

4.45

4.64