Difference between revisions of "Lab 10 RS"

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  <math>R_{scope} = 1\ M\Omega</math>
 
  <math>R_{scope} = 1\ M\Omega</math>
 
  <math>C = 2.2\ uF</math>
 
  <math>C = 2.2\ uF</math>
  <math>Zener\ Diode\ 4.7\ V</math>
+
  <math>\mbox{Zener}\ \mbox{Diode}\ 4.7\ V</math>
 
 
and the following input parameters:
 
 
 
<math>\Delta t = 17\ ms</math>
 
<math>V_{in} = 12\ V</math>
 
  
  
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So my output ripple becomes <math>\Delta V = \frac{0.122\ mA \cdot 17\ ms}{2.2\ uF} = 0.9 V</math>
 
So my output ripple becomes <math>\Delta V = \frac{0.122\ mA \cdot 17\ ms}{2.2\ uF} = 0.9 V</math>
  
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[[File:Tek00047.png | 800 px]]
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As we can see from the sketch above my output voltage has ripple about <math>\Delta V = 0.710\ V</math>, ripple time period is the same as input time <math>\Delta t = 16.6\ ms</math>, and the output DC voltage is about <math>V_{out} = 0.6\ V</math>
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<br><br><br><br><br><br><br><br><br><br>
 
=Full-Wave Rectifier Circuit=
 
=Full-Wave Rectifier Circuit=
 
[[File:TF_EIM_Lab10_FW_Rectifier.png| 500 px]]
 
[[File:TF_EIM_Lab10_FW_Rectifier.png| 500 px]]
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The output ripple in this case can be found by <math>\Delta V=\frac{I (\Delta t/2)}{C}</math>
 
The output ripple in this case can be found by <math>\Delta V=\frac{I (\Delta t/2)}{C}</math>
  
Because we are now using <math>(\Delta t/2)</math> instead of <math>(\Delta t)</math> than in previous case for half-wave rectifier theoretiacally we will be able to make <math>\Delta V</math> two times less then in previous case just by using exactly the same ellements and input parameters as before:
+
Because we are now using <math>(\Delta t/2)</math> instead of <math>(\Delta t)</math> than in previous case for half-wave rectifier theoretiacally we will be able to make <math>\Delta V</math> two times less then in previous case just by using exactly the same elements and input parameters as before. But here I realized by experiment that my Zener diode has low power limit and for this kind of circuit I need to use more powerful diode. Instead of 4.7 Zener Diode which has power limit 0.5 W I used rectifier diode 1N4001 which has power limit 50 W.
  
  
 
'''List the components below and show your instructor the output observed on the scope and sketch it below.'''
 
'''List the components below and show your instructor the output observed on the scope and sketch it below.'''
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I have used the following components:
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 +
<math>R = 96.9\ k\Omega</math>
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<math>R_L = 98.7\ k\Omega</math>
 +
<math>R_{scope} = 1\ M\Omega</math>
 +
<math>C = 2.2\ uF</math>
 +
<math>2\ \mbox{1N4001}\ \mbox{Diode}\ 50\ W</math>
 +
 +
 +
Because now I need to replace <math>\Delta t</math> by <math>(\Delta t)/2</math> my output voltage now becomes two times less:
 +
 +
<math>\Delta V = \frac{0.9\ V}{2} = 0.45\ V</math>
 +
 +
Now my DC output voltage becomes about <math>V = 5.2\ V</math> and ripple you can not even see for that scale.
 +
 +
[[File:Tek00053.png | 750 px]]
 +
 +
 +
To see ripple I used the AC mode of scope and they are plotted below. As we can see from this sketch my output voltage has ripple about <math>\Delta V = 68.4\ mV</math>, and the ripple time period now is two times less than input time period and is about <math>\Delta t = 8.4\ ms</math>.
 +
 +
[[File:Tek00054.png | 750 px]]
 +
 +
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Now I have almost perfect output DC current with small ripple.
 +
  
  
 
[https://wiki.iac.isu.edu/index.php/Electronics_RS Go Back to All Lab Reports] [[Forest_Electronic_Instrumentation_and_Measurement]]
 
[https://wiki.iac.isu.edu/index.php/Electronics_RS Go Back to All Lab Reports] [[Forest_Electronic_Instrumentation_and_Measurement]]

Latest revision as of 20:03, 11 March 2011

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Lab 10 Unregulated power supply


Use a transformer for the experiment.

here is a description of the transformer.

File:TF EIM 241 transformer.pdf

File:IN5230-B-T DataSheet.pdf

Half-Wave Rectifier Circuit

1.)Consider building circuit below.

TF EIM Lab10 HW Rectifier.png

Determine the components needed in order to make the output ripple have a [math]\Delta V[/math] less than 1 Volt.

The output ripple can be found by [math]\Delta V=\frac{I\Delta t}{C}[/math]

Taking AC signal from outlet equals to [math]60\ Hz[/math] my input pulse width is [math]\Delta t = \frac{1}{60\ sec} = 17\ ms[/math] and using say [math]C = 2.2\ uF[/math] I need my current to be:

[math]I \le \frac{1\ V \cdot 2.2\ uF}{17\ ms} \le 0.129\ mA[/math]

Taking [math]R \ge 100\ k\Omega \Rightarrow I = \frac{12\ V}{100\ k\Omega} \le 0.12\ mA[/math]

that satisfy the condition above for current so my output ripple becomes less than 1 Volts.



List the components below and show your instructor the output observed on the scope and sketch it below.

I have used the following components:

[math]R = 96.9\ k\Omega[/math]
[math]R_L = 98.7\ k\Omega[/math]
[math]R_{scope} = 1\ M\Omega[/math]
[math]C = 2.2\ uF[/math]
[math]\mbox{Zener}\ \mbox{Diode}\ 4.7\ V[/math]


The current through the circuit can be found as [math]I = \frac{V_{in}}{R_{tot}}[/math]

where [math] R_{tot} = R + \left|\frac{R_L \cdot \frac{1}{j\omega C}}{R_L + \frac{1}{j\omega C}} \right| = R + \sqrt{ \left(\frac{R_L}{1 + j\omega CR_L}\right)\left(\frac{R_L}{1 + j\omega CR_L}\right)^* } = R + \sqrt{ \left(\frac{R_L^2}{1 + (\omega CR_L)^2}\right) } [/math]

[math] = 96.9\ k\Omega + \sqrt{\frac{(98.7\ k\Omega)^2 }{1 + (2\pi\ 60\ sec^{-1})^2(2.2\ uF)^2 (98.7\ k\Omega)^2} }= 96.9\ k\Omega + 1.2\ k\Omega =98.1\ k\Omega[/math].

And the current becomes [math]I = \frac{12\ V}{98.1\ k\Omega} = 0.122\ mA[/math]

So my output ripple becomes [math]\Delta V = \frac{0.122\ mA \cdot 17\ ms}{2.2\ uF} = 0.9 V[/math]


Tek00047.png


As we can see from the sketch above my output voltage has ripple about [math]\Delta V = 0.710\ V[/math], ripple time period is the same as input time [math]\Delta t = 16.6\ ms[/math], and the output DC voltage is about [math]V_{out} = 0.6\ V[/math]













Full-Wave Rectifier Circuit

TF EIM Lab10 FW Rectifier.png


Determine the components needed in order to make the above circuit's output ripple have a [math]\Delta V[/math] less than 0.5 Volt.


The output ripple in this case can be found by [math]\Delta V=\frac{I (\Delta t/2)}{C}[/math]

Because we are now using [math](\Delta t/2)[/math] instead of [math](\Delta t)[/math] than in previous case for half-wave rectifier theoretiacally we will be able to make [math]\Delta V[/math] two times less then in previous case just by using exactly the same elements and input parameters as before. But here I realized by experiment that my Zener diode has low power limit and for this kind of circuit I need to use more powerful diode. Instead of 4.7 Zener Diode which has power limit 0.5 W I used rectifier diode 1N4001 which has power limit 50 W.


List the components below and show your instructor the output observed on the scope and sketch it below.

I have used the following components:

[math]R = 96.9\ k\Omega[/math]
[math]R_L = 98.7\ k\Omega[/math]
[math]R_{scope} = 1\ M\Omega[/math]
[math]C = 2.2\ uF[/math]
[math]2\ \mbox{1N4001}\ \mbox{Diode}\ 50\ W[/math]


Because now I need to replace [math]\Delta t[/math] by [math](\Delta t)/2[/math] my output voltage now becomes two times less:

[math]\Delta V = \frac{0.9\ V}{2} = 0.45\ V[/math]

Now my DC output voltage becomes about [math]V = 5.2\ V[/math] and ripple you can not even see for that scale.

Tek00053.png


To see ripple I used the AC mode of scope and they are plotted below. As we can see from this sketch my output voltage has ripple about [math]\Delta V = 68.4\ mV[/math], and the ripple time period now is two times less than input time period and is about [math]\Delta t = 8.4\ ms[/math].

Tek00054.png


Now I have almost perfect output DC current with small ripple.


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