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.

Half-Wave Rectifier Circuit

1.)Consider building circuit below.

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

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 .

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.

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.

@@ Line 40: / Line 40: @@
 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>