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  <div class="livedemo-rendered livedemo-elem" style="height: 434px;"><h1>Chapter 2 - Diode Circuits</h1>
<h2>Section 2.1 - Rectifier Circuits</h2>
<ul>
<li>Diodes form the first stage of a <b>DC Power Supply</b><br><img src="http://i.imgur.com/9aNjedG.png" alt="alt text" title="DC Power Supply"></li>
</ul>
<ul>
<li><b>Rectification:</b> The process of converting an AC voltage into one that is limited to a single polarity<ul>
<li>Diodes are useful for converting AC voltage into a single polarity because of its non linear characteristics (that is, the current exists for one voltage polarity but is <em>essentially</em> zero for the opposite polarity).</li>
<li><u>Rectification</u> is either classified as a <b>Half-Wave</b> or <b>Full-Wave</b><ul>
<li><b>Half-Wave:</b> Considered "simpler"</li>
<li><b>Full-Wave:</b> Considered more efficient</li>
</ul>
</li>
</ul>
</li>
</ul>
<h3>Section 2.2.1 - Rectifier Circuits</h3>
<ul>
<li><p>For <b>Transformer Circuit</b> shown<br><img src="http://i.imgur.com/SbwkLXk.png" alt="alt text">  <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%5Cfrac%7Bv_I%7D%7Bv_S%7D%3D%5Cfrac%7BN_1%7D%7BN_2%7D" alt="\frac{v_I}{v_S}=\frac{N_1}{N_2}"><br>The <b>Transformer Turns ratio ( <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7B%20N_1%7D%7BN_2%7D" alt="{ N_1}{N_2}"> ) , </b> and the <b>input signal (<img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_I%7D" alt="{v_I}"> )</b> can be used to generate a particular <b>secondary voltage ( <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_S%7D" alt="{v_S}"> )</b>, which in turn, can used to generate a particular <b>output voltage ( <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_O%7D" alt="{v_O}"> )</b></p>
<ul>
<li>Assuming that the <b>Forward Resistance ( <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Br_f%7D" alt="{r_f}"> )</b> is equal to zero ( <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Br_f%3D0%7D" alt="{r_f=0}"> )<ul>
<li>If <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_S%20%3C%200%7D" alt="{v_S &lt; 0}"> or <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_S%20%3C%20v_%5Cgamma%7D" alt="{v_S &lt; v_\gamma}"><ul>
<li>The diode is <b>Reverse Biased</b> and <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7BI%3D0%7D" alt="{I=0}"> and <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_O%3D0%7D" alt="{v_O=0}"> </li>
<li>As long as <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=v_S%20%3C%20v_%5Cgamma" alt="v_S &lt; v_\gamma">, the  <em>diode is nonconducting</em>. Therefore, <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_O%3D0%7D" alt="{v_O=0}">  </li>
</ul>
</li>
<li>If <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_S%20%3E%20v_%5Cgamma%7D" alt="{v_S &gt; v_\gamma}"><ul>
<li>The diode becomes <b>Forward Biased</b> and we can write<br><img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bi_D%3D%5Cfrac%7Bv_S-V_%5Cgamma%7D%7BR%7D%7D" alt="{i_D=\frac{v_S-V_\gamma}{R}}"> and,<br><img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_O%3Di_DR%3Dv_S-V_%5Cgamma%7D" alt="{v_O=i_DR=v_S-V_\gamma}"></li>
</ul>
</li>
</ul>
</li>
</ul>
</li>
<li><p>For the same circuit, if the source is sinusoidal</p>
<ul>
<li><img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%7Bv_O%3Di_DR%3Dv_S-V_%5Cgamma%7D" alt="{v_O=i_DR=v_S-V_\gamma}"></li>
</ul>
</li>
</ul>
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