Lab 10 RS: Difference between revisions
| Line 30: | Line 30: | ||
<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>\Delta t = 17\ ms | |||
and the following input parameters: | |||
<math>\Delta t = 17\ ms\ which\ corresponds\ to\ 60\ Hz</math> | |||
<math>V_{in} = 24\ V</math> | <math>V_{in} = 24\ V</math> | ||
Revision as of 05:26, 8 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 [math]\displaystyle{ \Delta V }[/math] less than 1 Volt.
The output ripple can be found by [math]\displaystyle{ \Delta V=\frac{I\Delta t}{C} }[/math]
I have used the following components and input parameters:
[math]\displaystyle{ R = 96.9\ k\Omega }[/math] [math]\displaystyle{ R_L = 98.7\ k\Omega }[/math] [math]\displaystyle{ R_{scope} = 1\ M\Omega }[/math] [math]\displaystyle{ C = 2.2\ uF }[/math]
and the following input parameters:
[math]\displaystyle{ \Delta t = 17\ ms\ which\ corresponds\ to\ 60\ Hz }[/math] [math]\displaystyle{ V_{in} = 24\ V }[/math]
The current through the circuit can be found as [math]\displaystyle{ I = \frac{V_{in}}{R_{tot}} }[/math]
where {math}R_{tot} = R + \frac{R_L \cdot \frac{1}{j\omega C}}{R_L + \frac{1}{j\omega C}}{/math}
So my output ripple becomes </math>\Delta V = \frac{}{}</math>
List the components below and show your instructor the output observed on the scope and sketch it below.
Full-Wave Rectifier Circuit
Determine the components needed in order to make the above circuit's output ripple have a [math]\displaystyle{ \Delta V }[/math] less than 0.5 Volt.
List the components below and show your instructor the output observed on the scope and sketch it below.
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