Chapter 4 problems: Difference between revisions
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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> |
*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. |
:*<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>!! <math>V_{BE}</math> !! <math>i_B</math> |
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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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===4.34=== |
===4.34=== |
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Revision as of 10:55, 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.
-.2 .45 2 0 .58 5 .2 .62 10
