Chapter 4 problems: Difference between revisions
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===4.20=== |
===4.20=== |
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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> |
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:*<math>V_{in}=-0.2, 0, 0.2</math> for the minimum, Q-point and maximum values. |
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:{| class="wikitable" border="1" align="center" |
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! <math>v_{in}</math> V!! <math>V_{BE} </math> V!! <math>i_B</math> uA |
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| -.2|| .45 || 2 |
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| 0|| .58 || 5 |
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| .2 || .62 || 10 |
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*Writing a loop equation for the output: <math>-V_{CC}+R_CI_C+V_{CE}=0</math>. |
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:{| class="wikitable" border="1" align="center" |
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! <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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===4.45=== |
===4.45=== |
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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
