Chapter 4 problems - Revision history

Fonggr: /* 4.20 */

2010-03-05T19:36:50Z

4.20

2010-03-05T19:32:53Z

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	:{\| class="wikitable" border="1" align="center"		:{\| class="wikitable" border="1" align="center"
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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		\| -.2\|\| 19 \|\| .8

2010-03-05T19:32:36Z

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← Older revision		Revision as of 12:32, 5 March 2010
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	! <math>v_{in}</math>!! <math>V_{CE}</math> V !! <math>i_C</math> mA		! <math>v_{in}</math>!! <math>V_{CE}</math> V !! <math>i_C</math> mA
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	\| -.2\|\| ~~.45~~ \|\| 2		\| -.2\|\| 19 \|\| .8
	\|-		\|-
	\| 0\|\| 16\|\| 2		\| 0\|\| 16\|\| 2
	\|-		\|-
	\| .2 \|\| ~~.62~~ \|\| 10		\| .2 \|\| 12 \|\| 4
	\|}		\|}

2010-03-05T19:31:05Z

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	! <math>v_{in}</math>!! <math>V_{BE}</math> !! <math>i_B</math>		! <math>v_{in}</math>!! <math>V_{BE} </math> V!! <math>i_B</math> uA
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	\| -.2\|\| .45 \|\| 2		\| -.2\|\| .45 \|\| 2
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	:*Writing a loop equation for the output: <math>-V_{CC}+R_CI_C+V_{CE}=0</math>		*Writing a loop equation for the output: <math>-V_{CC}+R_CI_C+V_{CE}=0</math>.
			:{\| class="wikitable" border="1" align="center"
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			! <math>v_{in}</math>!! <math>V_{CE}</math> V !! <math>i_C</math> mA
			\|-
			\| -.2\|\| .45 \|\| 2
			\|-
			\| 0\|\| 16\|\| 2
			\|-
			\| .2 \|\| .62 \|\| 10
			\|}

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2010-03-05T18:45:09Z

4.20

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	\| .2 \|\| .62 \|\| 10		\| .2 \|\| .62 \|\| 10
	\|}		\|}

			:*Writing a loop equation for the output: <math>-V_{CC}+R_CI_C+V_{CE}=0</math>

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2010-03-05T17:55:30Z

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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>
	:*<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.

			:{\| class="wikitable" border="1" align="center"
			\|+
			! <math>v_{in}</math>!! <math>V_{BE}</math> !! <math>i_B</math>
			\|-
			\| -.2\|\| .45 \|\| 2
			\|-
			\| 0\|\| .58 \|\| 5
			\|-
			\| .2 \|\| .62 \|\| 10
			\|}

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2010-03-05T17:42:02Z

4.20

← Older revision		Revision as of 10:42, 5 March 2010
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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>
	:*<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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2010-03-05T17:41:51Z

4.20

← Older revision		Revision as of 10:41, 5 March 2010
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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.		*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.

	===4.34===		===4.34===
	===4.45===		===4.45===
	===4.64===		===4.64===

2010-03-05T17:35:54Z

4.20

← Older revision		Revision as of 10:35, 5 March 2010
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	===4.20===		===4.20===
			*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.
			:*

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	===4.45===		===4.45===
	===4.64===		===4.64===

2010-03-03T22:17:52Z

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