# Difference between revisions of "Lab 10 RS"

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Taking <math>R \ge 100\ k\Omega \Rightarrow I = \frac{12\ V}{100\ k\Omega} \le 0.12\ 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 so my output ripple becomes less than 1 Volts. | + | that satisfy the condition above for current so my output ripple becomes less than 1 Volts. |

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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>\mbox{Zener}\ \mbox{Diode}\ 4.7\ V</math> | |

− | |||

− | |||

− | <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> | ||

+ | |||

+ | [[File:Tek00047.png | 800 px]] | ||

+ | |||

+ | |||

+ | 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> | ||

+ | |||

+ | |||

+ | |||

+ | |||

+ | <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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'''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.''' | '''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.''' | '''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. | ||

+ | |||

+ | [[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]] | ||

+ | |||

+ | |||

+ | 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

**Lab 10 Unregulated power supply**

Use a transformer for the experiment.

here is a description of the transformer.

File:TF EIM 241 transformer.pdf

# Half-Wave Rectifier Circuit

1.)Consider building circuit below.

**Determine the components needed in order to make the output ripple have a **
less than 1 Volt.

The output ripple can be found by

Taking AC signal from outlet equals to

my input pulse width is and using say I need my current to be:

Taking

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:

The current through the circuit can be found as

where

.

And the current becomes

So my output ripple becomes

As we can see from the sketch above my output voltage has ripple about , ripple time period is the same as input time , and the output DC voltage is about

# Full-Wave Rectifier Circuit

**Determine the components needed in order to make the above circuit's output ripple have a **
less than 0.5 Volt.

The output ripple in this case can be found by

Because we are now using

instead of than in previous case for half-wave rectifier theoretiacally we will be able to make 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:

Because now I need to replace by my output voltage now becomes two times less:

Now my DC output voltage becomes about

and ripple you can not even see for that scale.

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 , and the ripple time period now is two times less than input time period and is about .

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

Go Back to All Lab Reports Forest_Electronic_Instrumentation_and_Measurement