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class="gray">Diodes and rectification</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 20 minutes"><i class="fa fa-clock-o"></i> 20'</span> </ol> <article id="main-article"> <p><img alt="" src="../../em-induction/diode.jpg" style="float: left; width: 250px; height: 166px;">With most electricity generated to have alternating current, it is useful in some cases for this to be converted into direct current. This process is called rectification.</p> <p>At its heart is the diode, an electrical component that only conducts electricity in one direction.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Diode</p> </div> </div> <div class="panel-body"> <div> <p>A diode is an electronical component that conducts current in one direction (forward bias).</p> <p style="text-align: center;"><img alt="" src="../../em-induction/bias.png" style="width: 250px; height: 233px;"></p> <p>An ideal diode would have zero resistance in one direction and infinite) resistance in the other.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369634793"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Stages in rectification</p> </div> </div> <div class="panel-body"> <div> <h4>Half wave</h4> <p>The simplest form of rectification is the insertion of a diode in the series circuit. All current in the forward direction is transmitted; no current flows in reverse bias.</p> <p>Notice that the diode symbol acts as an arrow to show the direction in which conventional current will flow.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/half.png" style="width: 250px; height: 149px;"></p> <p>This circuit has the downside that current is only flowing for half of the time.</p> <h4>Full wave</h4> <p>Full wave rectification uses two diode pathways (in opposite directions) to enable the current to flow at all times. Each diode path runs through the same resistor (or desired electrical component).</p> <p>See if you can trace a finger around the following circuit diagram, first starting upwards from the power supply and then downwards.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/full.png" style="width: 250px; height: 206px;"></p> <p>While an improvement on half wave rectification, the output is still sinusoidal in magnitude.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633893"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Semiconductors</p> </div> </div> <div class="panel-body"> <div> <p>Most diodes are made from semiconducting materials such as silicon.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369634261"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Diode bias</p> </div> </div> <div class="panel-body"> <div> <p>The diode forward and reverse biases come from a PN junction connected to two electrical terminals.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633543"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Wheatstone bridge circuit</p> </div> </div> <div class="panel-body"> <div> <p>An understanding of rectification circuit diagrams can also be used to determine unknown values. The following circuit is known as a Wheatstone Bridge; the current directions are arbitrary.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/wheatstone.png" style="width: 250px; height: 192px;"></p> <p><span class="math-tex">\(R_1\)</span> and <span class="math-tex">\(R_3\)</span> are known fixed resistors. <span class="math-tex">\(R_2\)</span> is adjusted until <span class="math-tex">\(V_G\)</span> is zero, at which point there is no net current from D to B. <span class="math-tex">\(R_\text{x}\)</span> can be determined:</p> <p style="text-align: center;"><span class="math-tex">\(R_1+R_2=R_3+R_\text{x}\)</span></p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <div> <p><em>Use quizzes to practise application of theory.</em></p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#58a18e1c"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="58a18e1c"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Rectification <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-356-1137" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>The order of magnitude of the resistivity of silicon in Ωm is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>10<sup>-8</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>10<sup>-3</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>10<sup>12</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>10<sup>9</sup></span></label> </p></div><div class="q-explanation"><p>Silicon is a semiconductor. Its resistivity is the most moderate value, in between a conductor (10<sup>-8</sup>) and insulator (10<sup>9</sup>).</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph shows the <span class="math-tex">\(IV\)</span> characteristics for an LED.</p><p style="text-align: center;"><img alt="" height="195" src="../../ivdiode.png" width="209"></p><p><span class="math-tex">\(V_1\)</span> is approximately:</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>1 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>10 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 mV</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 kV</span></label> </p></div><div class="q-explanation"><p>The energy required to produce photons of visible light is in the order of 1 eV - therefore 1 V is needed to accelerate an electron across the diode.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Conduction in a P-type semiconductor is due to the movement of:</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>mainly holes but some electrons.</span></label> </p><p><label class="radio"> <input type="radio"> <span>only electrons.</span></label> </p><p><label class="radio"> <input type="radio"> <span>only holes.</span></label> </p><p><label class="radio"> <input type="radio"> <span>mainly electrons but some holes.</span></label> </p></div><div class="q-explanation"><p>Electrons are minority charge carriers.</p><p>One could argue that hole current is also due to electron movement but the correct answer is the best presented.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The image represents an LED and a 5 V AC source.</p><p style="text-align: center;"><img alt="" height="453" src="../../screenshot-2019-10-12-at-09.29.44.png" width="287"></p><p>Which graph represents the potential difference across the LED?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>A</span></label> </p><p><label class="radio"> <input type="radio"> <span>C</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>B</span></label> </p><p><label class="radio"> <input type="radio"> <span>D</span></label> </p></div><div class="q-explanation"><p>The LED requires about 1.5 V forward bias before it conducts.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The image shows an LED connected to a 5 V battery and a 1 kΩ resistor.</p><p style="text-align: center;"><img alt="" height="176" src="../../screenshot-2019-10-12-at-09.46.57.png" width="186"></p><p>The potential difference across the LED is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1.5 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>3.5 V</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>5 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>0 V</span></label> </p></div><div class="q-explanation"><p>The diode is in reverse bias so no current flows. The potential difference across the resistor <span class="math-tex">\(V_\text{R}=IR=0\text{ V}\)</span></p><p><span class="math-tex">\(\Rightarrow V_\text{LED}=5\text{ V}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The image shows an LED connected to a 5 V battery and a 1 kΩ resistor.</p><p style="text-align: center;"><img alt="" height="180" src="../../screenshot-2019-10-12-at-09.53.56.png" width="192"></p><p>The potential difference across the LED is:</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>1.5 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>5 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>0 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>3.5 V</span></label> </p></div><div class="q-explanation"><p>The diode is in forward bias. This means that the potential difference across it is that required to overcome the depletion layer - a value that you should be prepared to estimate.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The image shows two diodes connected to a 5 V battery and a 1 kΩ resistor.</p><p style="text-align: center;"><img alt="" height="199" src="../../screenshot-2019-10-12-at-10.05.26.png" width="180"></p><p>The potential difference across diode A is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1.5 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>2.5 V</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>5 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>0 V</span></label> </p></div><div class="q-explanation"><p>A is in reverse bias so has very large resistance; all potential is dropped across A.</p><p>The colour scheme in the simulation gives us a clue: g<span class="scayt-misspell-word" data-scayt-word="grey" data-wsc-lang="en_US">rey</span> is 0 V and green is 5 V.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The image shows a regular diode and an LED connected to a 5 V AC source and a 1 kΩ resistor.</p><p style="text-align: center;"><img alt="" height="468" src="../../screenshot-2019-10-12-at-10.34.29.png" width="291"></p><p>Which graph represents the potential difference across the resistor?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>B</span></label> </p><p><label class="radio"> <input type="radio"> <span>A</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>D</span></label> </p><p><label class="radio"> <input type="radio"> <span>C</span></label> </p></div><div class="q-explanation"><p>The LED requires a larger forward bias potential difference than a regular diode. Notice the asymmetry on the graph.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following circuits represents a full wave rectifier?</p><p style="text-align: center;"><img alt="" height="287" src="../../screenshot-2019-10-12-at-11.17.44.png" width="375"></p><p> </p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>A</span></label> </p><p><label class="radio"> <input type="radio"> <span>D</span></label> </p><p><label class="radio"> <input type="radio"> <span>B</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>C</span></label> </p></div><div class="q-explanation"><p>The current must pass through the resistor in the same direction for both halves of the cycle - so C is the correct response.</p><p>In A, notice that the two lower diodes are in the same direction.</p><p>In B, the resistor flow direction would change for each half of the cycle.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i> Prev </button> </div> <div class="pull-right pull-xs-none"> <button class="btn btn-success btn-xs-block text-xs-center btn-results" style="display: none"> <i class="fa fa-bar-chart"></i> Check Results </button> <button class="btn btn-default d-xs-none btn-next"> Next <i class="fa fa-arrow-right"></i> </button> <button class="btn btn-default btn-xs-block text-xs-center btn-close" data-dismiss="modal" style="display: none"> Close </button> </div> </div> </div> </div> </div></div> </div> </div> <div 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