Difference between revisions of "Lab 10 RS"

From New IAC Wiki
Jump to navigation Jump to search
Line 37: Line 37:
 
'''List the components below and show your instructor the output observed on the scope and sketch it below.'''
 
'''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:
+
I have used the following components:
  
 
  <math>R = 96.9\ k\Omega</math>
 
  <math>R = 96.9\ k\Omega</math>

Revision as of 06:49, 8 March 2011

Go Back to All Lab Reports


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]

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\ 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]

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:

[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} = 12\ 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 + 1.2\ k\Omega =98.1\ k\Omega[/math].

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{0.122\ mA \cdot 17\ ms}{2.2\ uF} = 0.9 V[/math]

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.


Go Back to All Lab Reports Forest_Electronic_Instrumentation_and_Measurement