Chapter 4 problems: Difference between revisions
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New page: ===4.20=== ===4.34=== ===4.45=== ===4.64=== |
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*We want to establish a DC operating point for the BJT. Writing a loop equation gives us -V<sub>BB</sub>-V<sub>in</sub>(t)+i<sub>B</sub>R<sub>B</sub>+v<sub>BE</sub>=0. To form a load line, we will let v<sub>in</sub>(t)=0 & i<sub>B</sub>=0. Thus v<sub>BE</sub>=V<sub>BB</sub>+v<sub>in</sub>(t) and i<sub>B</sub>(t)=(V<sub>BB</sub>+v<sub>in</sub>(t))/R<sub>B</sub> | |||
:*<math>V_{in}=-0.2, 0, 0.2</math> for the minimum, Q-point and maximum values. | |||
:{| class="wikitable" border="1" align="center" | |||
|+ | |||
! <math>v_{in}</math> V!! <math>V_{BE} </math> V!! <math>i_B</math> uA | |||
|- | |||
| -.2|| .45 || 2 | |||
|- | |||
| 0|| .58 || 5 | |||
|- | |||
| .2 || .62 || 10 | |||
|} | |||
*Writing a loop equation for the output: <math>-V_{CC}+R_CI_C+V_{CE}=0</math>. | |||
:{| class="wikitable" border="1" align="center" | |||
|+ | |||
! <math>v_{in}</math> V!! <math>V_{CE}</math> V !! <math>i_C</math> mA | |||
|- | |||
| -.2|| 19 || .8 | |||
|- | |||
| 0|| 16|| 2 | |||
|- | |||
| .2 || 12 || 4 | |||
|} | |||
*<math>A_v \cong (19-12)/.4=12.5</math> | |||
===4.34=== | ===4.34=== | ||
===4.45=== | ===4.45=== | ||
===4.64=== | ===4.64=== | ||
Latest revision as of 12:36, 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