Lab 6 RS - Revision history

Shaproma: /* What happens if R is doubled and C is halved?( 5 pnts.) */

2011-02-24T17:07:16Z

What happens if R is doubled and C is halved?( 5 pnts.)

Shaproma: /* What happens if R is doubled and C is halved?( 5 pnts.) */

2011-02-24T17:06:45Z

What happens if R is doubled and C is halved?( 5 pnts.)

Shaproma: /* Rise time \tau measurements */

2011-02-14T18:41:48Z

Rise time \tau measurements

Shaproma: /* Rise time \tau^{\prime} measurements */

2011-02-14T18:38:53Z

Rise time \tau^{\prime} measurements

Shaproma: /* Rise time \tau^{\prime} measurements */

2011-02-14T18:38:29Z

Rise time \tau^{\prime} measurements

Shaproma: /* Questions */

2011-02-14T18:27:35Z

Questions

Shaproma: /* Compare the measurement of \tau^{\prime} to what you expected based on your measured values of C_1, R_1 and R.( 15 pnts.) */

2011-02-14T18:26:19Z

Compare the measurement of \tau^{\prime} to what you expected based on your measured values of C_1, R_1 and R.( 15 pnts.)

Shaproma: /* Compare the measurement of \tau^{\prime} to what you expected based on your measured values of C_1, R_1 and R.( 15 pnts.) */

2011-02-14T18:22:37Z

Compare the measurement of \tau^{\prime} to what you expected based on your measured values of C_1, R_1 and R.( 15 pnts.)

Shaproma: /* Pulse Sharpener */

2011-02-14T18:21:26Z

Pulse Sharpener

Shaproma: /* Compare the measurement of \tau^{\prime} to what you expected based on your measured values of C_1, R_1 and R.( 15 pnts.) */

2011-02-14T18:20:25Z

Compare the measurement of \tau^{\prime} to what you expected based on your measured values of C_1, R_1 and R.( 15 pnts.)

@@ Line 86: / Line 86: @@
 ===What happens if R is doubled and C is halved?( 5 pnts.)===
-  The times constant <math>\tau_{RC} = \frac{1}{(R/2)(2C)} = \frac{1}{RC}</math> becomes unchangeable so nothing is really happened
+  The times constant <math>\tau_{RC} = (R/2)(2C) = RC</math> becomes unchangeable so nothing is really happened
 =Pulse Sharpener=

@@ Line 52: / Line 52: @@
 ===What happens if R is doubled and C is halved?( 5 pnts.)===
-  The times constant <math>\tau_{RC} = \frac{1}{(R/2)(2C)} = \frac{1}{RC}</math> becomes unchangeable so nothing is really happened
+  The times constant <math>\tau_{RC} =(R/2)(2C) = RC</math> becomes unchangeable so nothing is really happened
 =Integrator=

@@ Line 138: / Line 138: @@
 :At time <math>t = \tau \rightarrow V_{in} = 6 \left ( 1 - e^{-1}\right ) = 3.79\ V</math>
-:From the plot above my measurement of rise time is:
+To define the rise time I have used different definition. It is time when <math>t = \tau</math> under the exponent function so the exponent becomes <math>e^{-1}</math>
   <math>\tau = (52 \pm 0.2)\ us</math>

@@ Line 152: / Line 152: @@
 :At time <math>t = \tau^{\prime} \rightarrow V_0^{\prime} = 3.04 \left ( 1 - e^{-1}\right ) = 1.92\ V</math>
-So to define the rise time I have used different definition. It is time when <math>t = \tau</math> under the exponent function so the exponent becomes <math>e^{-1}</math>
+To define the rise time I have used different definition. It is time when <math>t = \tau</math> under the exponent function so the exponent becomes <math>e^{-1}</math>
 From the plot above my measurement of rise time is:

@@ Line 152: / Line 152: @@
 :At time <math>t = \tau^{\prime} \rightarrow V_0^{\prime} = 3.04 \left ( 1 - e^{-1}\right ) = 1.92\ V</math>
-:From the plot above my measurement of rise time is:
+So to define the rise time I have used different definition. It is time when <math>t = \tau</math> under the exponent function so the exponent becomes <math>e^{-1}</math>
   <math>\tau^{\prime} = (26.0 \pm 2.0)\ us</math>

← Older revision		Revision as of 18:27, 14 February 2011
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	2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)		2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)
		+
		+	<math>V_{out} = \frac{R_1}{R+R_1}V_0 (1- e^{-t/\tau_{out}})</math>
		+
		+	So the output voltage is always less then input one by the factor <math>\frac{R_1}{R+R_1}</math>

← Older revision		Revision as of 18:26, 14 February 2011
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−	And in order to this result be true the <math>\tau_{in}</math> should be equal to <math>R\ C</math>. Then the term <math>\left ( \frac{1}{\tau_{in}} - \frac{1}{CR}\right )</math> in equation above will be zero and the solution becomes <math>V_{out} = \frac{R\ R_1}{R+R_1}V_0 (1- e^{-t/\tau_{out}})</math>	+	And in order to this result be true the <math>\tau_{in}</math> should be equal to <math>R\ C</math>. Then the term <math>\left ( \frac{1}{\tau_{in}} - \frac{1}{CR}\right )</math> in equation above will be zero and the solution becomes <math>V_{out} = \frac{R_1}{R+R_1}V_0 (1- e^{-t/\tau_{out}})</math>
	where <math>\tau_{out} = C \frac{RR_1}{R+R_1} </math> that gives us the result above about rising time of output voltage.		where <math>\tau_{out} = C \frac{RR_1}{R+R_1} </math> that gives us the result above about rising time of output voltage.

← Older revision		Revision as of 18:22, 14 February 2011
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−	In real situation the term <math>\left ( \frac{1}{\tau_{in}} - \frac{1}{CR}\right )</math> is not equal to zero and we can not really use the formula <math> \tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right )C_1</math>. That can be seen from my measurements of rising time of input pulse for two cases: 1) with capacitor shorted the oscillator but without any additional circuit in output <math>\tau = 64\ us<math> and 2) with real circuit <math>\tau = 30.9\ us<math>. In this case the theoretical values of expected error is difficult to handle. We need somehow to solve this equation exactly that could be done numerically or we need somehow to estimate the error of not including this term in equation.	+	In real situation the term <math>\left ( \frac{1}{\tau_{in}} - \frac{1}{CR}\right )</math> is not equal to zero and we can not really use the formula <math> \tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right )C_1</math>. That can be seen from my measurements of rising time of input pulse for two cases: 1) with capacitor shorted the oscillator but without any additional circuit in output <math>\tau = 68\ us</math> and 2) with real circuit <math>\tau = 52.0\ us</math>. In this case the theoretical values of expected error is difficult to handle. We need somehow to solve this equation exactly that could be done numerically or we need somehow to estimate the error of not including this term in equation.

	==Questions==		==Questions==

← Older revision		Revision as of 18:21, 14 February 2011
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−	~~Taking <math>R = 49.9\ \Omega ,</math> <math>R_1 = 51.2\ \Omega</math> and <math>C_1 = 1.225\ \mu F\ \Rightarrow \tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right )C_1 = 31.0\ us</math>~~

	==Make measurements of the rise time <math>\tau</math> and <math>\tau^{\prime}</math>. The rise time is defined as the time it take the pulse to go from 10% of its max value to 90% of its max value.( 5 pnts.)==		==Make measurements of the rise time <math>\tau</math> and <math>\tau^{\prime}</math>. The rise time is defined as the time it take the pulse to go from 10% of its max value to 90% of its max value.( 5 pnts.)==

← Older revision		Revision as of 18:20, 14 February 2011
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−	In real situation the term <math>\left ( \frac{1}{\tau_{in}} - \frac{1}{CR}\right )</math> is not equal to zero and we can not really use the formula <math> \tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right )C_1</math>. In this case the theoretical values of expected error is difficult to handle. We need somehow to solve this equation exactly that could be done numerically or we need somehow to estimate the error of not including this term in equation.	+	In real situation the term <math>\left ( \frac{1}{\tau_{in}} - \frac{1}{CR}\right )</math> is not equal to zero and we can not really use the formula <math> \tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right )C_1</math>. That can be seen from my measurements of rising time of input pulse for two cases: 1) with capacitor shorted the oscillator but without any additional circuit in output <math>\tau = 64\ us<math> and 2) with real circuit <math>\tau = 30.9\ us<math>. In this case the theoretical values of expected error is difficult to handle. We need somehow to solve this equation exactly that could be done numerically or we need somehow to estimate the error of not including this term in equation.

	==Questions==		==Questions==