Chapter 4 problems: Difference between revisions
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! <math>v_{in}</math>!! <math>V_{BE} </math> V!! <math>i_B</math> uA |
! <math>v_{in}</math> V!! <math>V_{BE} </math> V!! <math>i_B</math> uA |
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| -.2|| .45 || 2 |
| -.2|| .45 || 2 |
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! <math>v_{in}</math>!! <math>V_{CE}</math> V !! <math>i_C</math> mA |
! <math>v_{in}</math> V!! <math>V_{CE}</math> V !! <math>i_C</math> mA |
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| -.2|| 19 || .8 |
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| 0|| 16|| 2 |
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| .2 || 12 || 4 |
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*<math>A_v \cong (19-12)/.4=12.5</math> |
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===4.34=== |
===4.34=== |
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Latest revision as of 11:36, 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 V uA -.2 .45 2 0 .58 5 .2 .62 10
- Writing a loop equation for the output: .
V V mA -.2 19 .8 0 16 2 .2 12 4
