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> |
*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> |
:*<math>V_{in}=-0.2, 0, 0.2</math> for the minimum, Q-point and maximum values. |
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===4.34=== |
===4.34=== |
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Revision as of 09:42, 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.
