<?xml version="1.0" encoding="utf-8"?>
<document xmlns="http://cnx.rice.edu/cnxml" xmlns:m="http://www.w3.org/1998/Math/MathML" xmlns:md="http://cnx.rice.edu/mdml/0.4" xmlns:bib="http://bibtexml.sf.net/" xmlns:q="http://cnx.rice.edu/qml/1.0" id="m0053" module-id="" cnxml-version="0.6">
  <title>Dependent Sources</title>

  <metadata xmlns:md="http://cnx.rice.edu/mdml/0.4">
  <!-- WARNING! The 'metadata' section is read only. Do not edit below.
       Changes to the metadata section in the source will not be saved. -->
  <md:content-id>m0053</md:content-id>
  <md:title>Dependent Sources</md:title>
  <md:version>2.14</md:version>
  <md:created>2000/08/08</md:created>
  <md:revised>2009/06/04 12:20:42.028 GMT-5</md:revised>
  <md:authorlist>
    <md:author id="dhj">
        <md:firstname>Don</md:firstname>
        <md:surname>Johnson</md:surname>
        <md:fullname>Don Johnson</md:fullname>
        <md:email>dhj@rice.edu</md:email>
    </md:author>
  </md:authorlist>
  <md:maintainerlist>
    <md:maintainer id="dhj">
        <md:firstname>Don</md:firstname>
        <md:surname>Johnson</md:surname>
        <md:fullname>Don Johnson</md:fullname>
        <md:email>dhj@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="carolrb">
        <md:firstname>Carol</md:firstname>
        <md:surname>Bettoney</md:surname>
        <md:fullname>Carol Bettoney</md:fullname>
        <md:email>carolrb@alumni.rice.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  <md:license href="http://creativecommons.org/licenses/by/1.0"/>
  <md:licensorlist>
    <md:licensor id="dhj">
        <md:firstname>Don</md:firstname>
        <md:surname>Johnson</md:surname>
        <md:fullname>Don Johnson</md:fullname>
        <md:email>dhj@rice.edu</md:email>
    </md:licensor>
  </md:licensorlist>
  <md:subjectlist>
    <md:subject>Science and Technology</md:subject>
  </md:subjectlist>
  <md:abstract>Dependent sources.</md:abstract>
  <md:language>en</md:language>
  <!-- WARNING! The 'metadata' section is read only. Do not edit above.
       Changes to the metadata section in the source will not be saved. -->
</metadata>

<content>
    <para id="one"> A <term>dependent source</term> is either a
      voltage or current source whose value is proportional to some
      other voltage or current in the circuit. Thus, there are four
      different kinds of dependent sources; to describe an op-amp, we
      need a voltage-dependent voltage source.  However, the <link document="m1019" strength="2">standard circuit-theoretical model
      for a transistor</link> contains a current-dependent current
      source. Dependent sources do not serve as inputs to a circuit
      like independent sources. They are used to model
      <term>active circuits</term>: those containing electronic
      elements. The RLC circuits we have been considering so far are
      known as <term>passive circuits</term>.
    </para>

    <figure id="depsources">
      <title>dependent sources</title>
      <media id="id1165858351302" alt="">
        <image src="depsource.png" mime-type="image/png"/>
        <image src="depsource.eps" mime-type="application/postscript"/>
      </media>
      <caption>
	Of the four possible dependent sources, depicted is a
	voltage-dependent voltage source in the context of a generic
	circuit.
      </caption>
    </figure>

    <para id="two"> 
      <link target-id="opamp" strength="2"/>       
      shows the circuit symbol for the op-amp and its equivalent
      circuit in terms of a voltage-dependent voltage source.
    </para>

    <figure id="opamp">
      <title>op-amp</title> 
      <media id="id1165864341059" alt="">
        <image src="opamp10.png" mime-type="image/png"/>
        <image src="opamp10.eps" mime-type="application/postscript"/>
      </media>
      <caption>	
	The op-amp has four terminals to which connections can be
	made.  Inputs attach to nodes <emphasis>a</emphasis> and
	<emphasis>b</emphasis>, and the output is node
	<emphasis>c</emphasis>. As the circuit model on the right
	shows, the op-amp serves as an amplifier for the difference of
	the input node voltages.
      </caption>
    </figure>


    <para id="three">       
      Here, the output voltage equals an amplified version of the
      difference of node voltages appearing across its inputs. The
      dependent source model portrays how the op-amp works quite
      well. As in most active circuit schematics, the power supply is
      not shown, but must be present for the circuit model to be
      accurate. Most operational amplifiers require both positive and
      negative supply voltages for proper operation.
    </para>

    <para id="newone">
      Because dependent sources cannot be described as impedances, and
      because the dependent variable cannot "disappear" when you apply
      parallel/series combining rules, circuit simplifications such as
      current and voltage divider should not be applied in most
      cases. Analysis of circuits containing dependent sources
      essentially requires use of formal methods, like the 
      <link document="m0032" strength="3">node method</link>. 
      Using the node method for such circuits is not
      difficult, with node voltages defined across the source treated
      as if they were known (as with independent sources). Consider
      the circuit shown on the top in      
      <link target-id="fdop" strength="2"/>.
    </para>
    

    <figure id="fdop">
      <title>feedback op-amp</title>
      <media id="id1165863951770" alt="">
        <image src="opamp9.png" mime-type="image/png"/>
        <image src="opamp9.eps" mime-type="application/postscript"/>
      </media>
      <caption>
	The top circuit depicts an op-amp in a feedback amplifier
	configuration.  On the bottom is the equivalent circuit, and
	integrates the op-amp circuit model into the circuit.
      </caption>
    </figure>
    

    <para id="newtwo">Note that the op-amp is placed in the circuit "upside-down,"
      with its inverting input at the top and serving as the only
      input. As we explore op-amps in more detail in the next section,
      this configuration will appear again and again and its usefulness demonstrated. To determine how the output voltage is related
      to the input voltage, we apply the node method.
      Only two node voltages— 
      <m:math display="inline">
	<m:ci>v</m:ci>
      </m:math> 
      and  
      <m:math display="inline">
	<m:ci>
	  <m:msub>
	    <m:mi>v</m:mi>
	    <m:mi>out</m:mi>
	  </m:msub>
	</m:ci>
      </m:math>—need be defined; the remaining nodes are across
      sources or serve as the reference. The node equations are

      <equation id="eqone">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:divide/>
                <m:apply>
                  <m:minus/>
		  <m:ci>v</m:ci> 
		  <m:ci>
		    <m:msub>
		      <m:mi>v</m:mi>
		      <m:mi>in</m:mi>
		    </m:msub>
		  </m:ci>
                </m:apply>
                <m:ci>R</m:ci>
	      </m:apply>
	      <m:apply>
		<m:divide/>
                <m:ci>v</m:ci>
                <m:ci>
                  <m:msub>
                    <m:mi>R</m:mi>
                    <m:mi>in</m:mi>
                  </m:msub>
                </m:ci>
	      </m:apply>
	      <m:apply>
		<m:divide/>
                <m:apply>
                  <m:minus/>
		  <m:ci>v</m:ci> 
		  <m:ci>
		    <m:msub>
		      <m:mi>v</m:mi>
		      <m:mi>out</m:mi>
		    </m:msub>
		  </m:ci>
                </m:apply>
                <m:ci>
                  <m:msub>
                    <m:mi>R</m:mi>
                    <m:mi>F</m:mi>
                  </m:msub>
                </m:ci>
	      </m:apply>
	    </m:apply>
	    <m:cn>0</m:cn>
	  </m:apply>
	</m:math>
      </equation>

      <equation id="eqtwo">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:plus/>
	      <m:apply>
		<m:divide/>
                <m:apply>
                  <m:minus/>
		  <m:ci>
		    <m:msub>
		      <m:mi>v</m:mi>
		      <m:mi>out</m:mi>
		    </m:msub> 
		  </m:ci>
		  <m:apply>
		    <m:times/>
		    <m:apply>
		      <m:minus/>
		      <m:ci>G</m:ci>
		    </m:apply>
		    <m:ci>v</m:ci>
		  </m:apply>
                </m:apply>
                <m:ci>
                  <m:msub>
                    <m:mi>R</m:mi>
		    <m:mi>out</m:mi>
                  </m:msub>
                </m:ci>
	      </m:apply>
	      <m:apply>
		<m:divide/>
                <m:apply>
                  <m:minus/>
		  <m:ci>
		    <m:msub>
		      <m:mi>v</m:mi>
		      <m:mi>out</m:mi>
		    </m:msub>
		  </m:ci> 
		  <m:ci>v</m:ci>
                </m:apply>
                <m:ci>
                  <m:msub>
                    <m:mi>R</m:mi>
                    <m:mi>F</m:mi>
                  </m:msub>
                </m:ci>
	      </m:apply>
	      <m:apply>
		<m:divide/>
                <m:ci>
                  <m:msub>
                    <m:mi>v</m:mi>
                    <m:mi>out</m:mi>
                  </m:msub>
                </m:ci>
                <m:ci>
                  <m:msub>
                    <m:mi>R</m:mi>
                    <m:mi>L</m:mi>
                  </m:msub>
                </m:ci>
	      </m:apply>
	    </m:apply>
	    <m:cn>0</m:cn>
	  </m:apply>
	</m:math>
      </equation>

      Note that no special considerations were used in applying the
      node method to this dependent-source circuit. Solving these to
      learn how
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>v</m:mi>
	    <m:mi>out</m:mi>
	  </m:msub>
	</m:ci>
      </m:math> 
      relates to  
      <m:math>
	<m:ci>
	  <m:msub>
	    <m:mi>v</m:mi>
	    <m:mi>in</m:mi>
	  </m:msub>
	</m:ci>
      </m:math>
      yields

      <equation id="eqthree">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:times/>
	      <m:apply>
		<m:minus/>
                <m:apply>
                  <m:times/>
		  <m:apply>
		    <m:divide/>
		    <m:apply>
		      <m:times/>
		      <m:ci>
			<m:msub>
			  <m:mi>R</m:mi>
			  <m:mi>F</m:mi>
			</m:msub>
		      </m:ci>
		      <m:ci>
			<m:msub>
			  <m:mi>R</m:mi>
			  <m:mi>out</m:mi>
			</m:msub>
		      </m:ci>
		    </m:apply>
		    <m:apply>
		      <m:minus/>
		      <m:ci>
			<m:msub>
			  <m:mi>R</m:mi>
			  <m:mi>out</m:mi>
			</m:msub>
		      </m:ci> 
		      <m:apply>
			<m:times/>
			<m:ci>G</m:ci>
			<m:ci>
			  <m:msub>
			    <m:mi>R</m:mi>
			    <m:mi>F</m:mi>
			  </m:msub>
			</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		  <m:apply>
		    <m:plus/>
		    <m:apply>
		      <m:divide/>
		      <m:cn>1</m:cn>
		      <m:ci>
			<m:msub>
			  <m:mi>R</m:mi>
			  <m:mi>out</m:mi>
			</m:msub>
		      </m:ci>
		    </m:apply>
		    <m:apply>
		      <m:divide/>
		      <m:cn>1</m:cn>
		      <m:ci>
			<m:msub>
			  <m:mi>R</m:mi>
			  <m:mi>in</m:mi>
			</m:msub>
		      </m:ci>
		    </m:apply>
		    <m:apply>
		      <m:divide/>
		      <m:cn>1</m:cn>
		      <m:ci>
			<m:msub>
			  <m:mi>R</m:mi>
			  <m:mi>L</m:mi>
			</m:msub>
		      </m:ci>
		    </m:apply>
		  </m:apply>
		  <m:apply>
		    <m:plus/>
		    <m:apply>
		      <m:divide/>
		      <m:cn>1</m:cn>
		      <m:ci>R</m:ci>
		    </m:apply>
		    <m:apply>
		      <m:divide/>
		      <m:cn>1</m:cn>
		      <m:ci>
			<m:msub>
			  <m:mi>R</m:mi>
			  <m:mi>in</m:mi>
			</m:msub>
		      </m:ci>
		    </m:apply>
		    <m:apply>
		      <m:divide/>
		      <m:cn>1</m:cn>
		      <m:ci>
			<m:msub>
			  <m:mi>R</m:mi>
			  <m:mi>F</m:mi>
			</m:msub>
		      </m:ci>
		    </m:apply>
		  </m:apply>
                </m:apply>
                <m:apply>
                  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:ci>
		    <m:msub>
		      <m:mi>R</m:mi>
		      <m:mi>F</m:mi>
		    </m:msub>
		  </m:ci>
                </m:apply>
	      </m:apply>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>out</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	    <m:apply>
	      <m:times/>
	      <m:apply>
		<m:divide/>
                <m:cn>1</m:cn>
                <m:ci>R</m:ci>
	      </m:apply>
	      <m:ci>
		<m:msub>
		  <m:mi>v</m:mi>
		  <m:mi>in</m:mi>
		</m:msub>
	      </m:ci>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>

      This expression represents the general input-output relation for
      this circuit, known as the <term>standard feedback
      configuration</term>. Once we learn more about <link document="m0036" strength="3">op-amps</link>, in particular what
      its typical element values are, the expression will simplify
      greatly. Do note that the units check, and that the parameter
      <m:math>
	<m:ci>G</m:ci>
      </m:math> 
      of the dependent source is a dimensionless gain.
    </para>

  </content>
</document>
