Difference between revisions of "Lab 10 RS"

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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>
 
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 + 1.45\ M\Omega = 1.54\ M\Omega</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 + 1.45\ M\Omega = 1.54\ M\Omega</math>.
  
And the current becomes <math>I = \frac{24\ V}{98.1\ k\Omega} = 244\ uA</math>
+
And the current becomes <math>I = \frac{24\ V}{1.54\ M\Omega} = 15.5\ uA</math>
  
So my output ripple becomes <math>\Delta V = \frac{244\ uA \cdot 17\ ms}{2.2\ uF} = </math>
+
So my output ripple becomes <math>\Delta V = \frac{15.5\ uA \cdot 17\ ms}{2.2\ uF} = 0.119 V</math>
  
  

Revision as of 06:03, 8 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]

I have used the following components and input parameters:

[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]

and the following input parameters:


[math]\Delta t = 17\ ms\ which\ corresponds\ to\ 60\ Hz[/math]
[math]V_{in} = 24\ 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 + 1.45\ M\Omega = 1.54\ M\Omega[/math].

And the current becomes [math]I = \frac{24\ V}{1.54\ M\Omega} = 15.5\ uA[/math]

So my output ripple becomes [math]\Delta V = \frac{15.5\ uA \cdot 17\ ms}{2.2\ uF} = 0.119 V[/math]


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

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.

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


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