TF EIMLab14 Writeup - Revision history

Foretony at 23:15, 15 March 2018

2018-03-15T23:15:45Z

2018-03-15T23:15:07Z

2018-03-15T23:14:35Z

2011-04-01T01:58:32Z

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2011-03-30T17:58:13Z

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2011-03-30T17:56:57Z

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2011-03-30T17:51:10Z

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2011-03-30T17:49:53Z

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2011-03-30T17:38:17Z

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2011-03-30T17:32:47Z

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@@ Line 19: / Line 19: @@
 so <math>V_{CC} < 40</math>
-To set this collector current you ned to choose V_{CC} and R_C to give 0.5 mA and be in the middle of the load line.
+To set this collector current you ned to choose <math>V_{CC}</math> and <math>R_C</math> to give 0.5 mA and be in the middle of the load line.
-From previous lab we should try V_{CE} = 1 Volt
+From previous lab we should try <math>V_{CE} = 1 Volt</math>
 Loop theorem \Rightarrow

@@ Line 48: / Line 48: @@
 <math>R_E = 200.0 \Omega</math>
-R_C = 2k\Omega
+<math>R_C = 2k\Omega</math>
-R_1 = 2k\Omega
+<math>R_1 = 2k\Omega</math>
-R_2 = 15 k\Omega
+<math>R_2 = 15 k\Omega</math>

@@ Line 44: / Line 44: @@
 :e. <math>V_{BE} \approx 0.6 V</math>
 :f. <math>I_D \approx 10 I_B < 1</math> mA
 .)Draw a load line using the <math>I_{C}</math> -vs- <math>I_{CE}</math> from the previous lab 13.  Record the value of <math>h_{FE}</math> or <math>\beta</math>.

@@ Line 36: / Line 36: @@
 Unknowns are<math> V_{CC}</math> and <math>R_C</math>
-:<math>R_C = \frac{V_{CC}}{2 I_C} - R_E = \frac{V_{CC}}}{2 0.5 mA} - 200 \Omega</math>
+:<math>R_C = \frac{V_{CC}}{2 I_C} - R_E = \frac{V_{CC}}{2 0.5 mA} - 200 \Omega</math>
 :e. <math>V_{BE} \approx 0.6 V</math>

@@ Line 29: / Line 29: @@
 :<math>V_{CE} = V_{CC}/2</math>
-:<math>\RIghtarrow  V_{CC}/2 = I_C R_C + I_E R_E</math>
+:<math>\Rightarrow  V_{CC}/2 = I_C R_C + I_E R_E</math>
 : Since <math>I_C \approx I_E</math>
-: <math>V_{CC} = 2I_E (R_C+R_E)</math>
+: <math>V_{CC} = 2I_C (R_C+R_E)</math>

← Older revision		Revision as of 17:38, 30 March 2011
Line 13:		Line 13:
	To set this collector current you ned to choose V_{CC} and R_C to give 0.5 mA and be in the middle of the load line.		To set this collector current you ned to choose V_{CC} and R_C to give 0.5 mA and be in the middle of the load line.

		+
		+	From previous lab we should try V_{CE} = 1 Volt
		+
		+	If V_{CC} = 10 Volts then R = \frac{10}{0.0005} = 200 k \Omega

	:c. <math>V_{CE} \approx V_{CC}/2 > 2</math> V		:c. <math>V_{CE} \approx V_{CC}/2 > 2</math> V

@@ Line 10: / Line 10: @@
 :a. Try <math>R_E \approx 220 \Omega</math>
 :b. <math>I_C > 0.5</math> mA DC with no input signal
 :c. <math>V_{CE} \approx V_{CC}/2 > 2</math> V
 :d. <math>V_{CC} < V_{CE}(max)</math> to prevent burnout