Revision as of 05:26, 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 {math} R_{tot} = R + \frac{R_L \cdot \frac{1}{j\omega C}}{R_L + \frac{1}{j\omega C}}{/math}

So my output ripple becomes </math>\Delta V = \frac{}{}</math>

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.

Revision as of 05:26, 8 March 2011 (view source) Shaproma (talk \| contribs) (→‎Half-Wave Rectifier Circuit) ← Older edit		Revision as of 05:26, 8 March 2011 (view source) Shaproma (talk \| contribs) (→‎Half-Wave Rectifier Circuit) Newer edit →
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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}}{/math}	+	where {math} R_{tot} = R + \frac{R_L \cdot \frac{1}{j\omega C}}{R_L + \frac{1}{j\omega C}}{/math}