Difference between revisions of "Lab 10 RS"
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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: | 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\ | + | <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> | Taking <math>R \ge 100\ k\Omega \Rightarrow I = \frac{12\ V}{100\ k\Omega} \le 0.12\ mA</math> |
Revision as of 06:47, 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
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 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 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
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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