Difference between revisions of "Lab 10 RS"
Line 21: | Line 21: | ||
Determine the components needed in order to make the output ripple have a <math>\Delta V</math> less than 1 Volt. | 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> | ||
+ | |||
+ | Taking ADC from outlet equals <math>60\ Hz</math> my input pulse width is <math>\Delta t = 17\ ms</math> and using the say <math>C = 2.2\ uF</math> to make <math>\Delta\ V \less 1\ V</math> I need my current to be: | ||
+ | |||
+ | <math>I \less \frac{1\ V \cdot 2.2\ uF}{17\ mA} \less 0.129\ mA<math> | ||
+ | |||
The output ripple can be found by <math>\Delta V=\frac{I\Delta t}{C}</math> | The output ripple can be found by <math>\Delta V=\frac{I\Delta t}{C}</math> | ||
Line 35: | Line 42: | ||
<math>\Delta t = 17\ ms\ which\ corresponds\ to\ 60\ Hz</math> | <math>\Delta t = 17\ ms\ which\ corresponds\ to\ 60\ Hz</math> | ||
− | <math>V_{in} = | + | <math>V_{in} = 12\ V</math> |
Line 44: | Line 51: | ||
<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 + 1.2\ k\Omega =98.1\ k\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 + 1.2\ k\Omega =98.1\ k\Omega</math>. | ||
− | And the current becomes <math>I = \frac{ | + | And the current becomes <math>I = \frac{12\ V}{98.1\ k\Omega} = 0.122\ mA</math> |
− | So my output ripple becomes <math>\Delta V = \frac{0. | + | So my output ripple becomes <math>\Delta V = \frac{0.122\ mA \cdot 17\ ms}{2.2\ uF} = 0.9 V</math> |
Revision as of 06:33, 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 ADC from outlet equals
my input pulse width is and using the say to make I need my current to be:
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.
Go Back to All Lab Reports Forest_Electronic_Instrumentation_and_Measurement