Difference between revisions of "Lab 10 RS"
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The current through the circuit can be found as <math>I = \frac{V_{in}}{R_{tot}}</math> | The current through the circuit can be found as <math>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}} = 96.9\ k\Omega + \frac{98.7 \cdot 1.21}{98.7 + 1.21} \ k\Omega= (96.9 + 1.19)\ k\Omega = 98.1\ k\Omega</math>. | + | where <math> R_{tot} = R + \left|\frac{R_L \cdot \frac{1}{j\omega C}}{R_L + \frac{1}{j\omega C}} \right|= 96.9\ k\Omega + \frac{98.7 \cdot 1.21}{98.7 + 1.21} \ k\Omega= (96.9 + 1.19)\ k\Omega = 98.1\ k\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}{98.1\ k\Omega} = 244\ uA</math> |
Revision as of 05:43, 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
less than 1 Volt.The output ripple can be found by
I have used the following components and input parameters:
and the following input parameters:
The current through the circuit can be found as
where
.And the current becomes
So my output ripple becomes
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
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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