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	<title>gaussmarkov: diy fx &#187; Capacitors</title>
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		<title>Capacitors: Math</title>
		<link>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-math/</link>
		<comments>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-math/#comments</comments>
		<pubDate>Wed, 04 Jul 2007 03:29:14 +0000</pubDate>
		<dc:creator>gaussmarkov</dc:creator>
				<category><![CDATA[Capacitors]]></category>

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		<description><![CDATA[For some time, I have been working on a description of some of the math behind equations that we use for passive filters.  I do this kind of writing to check my own understanding, and then I share it in the hope of finding my own mistakes and possibly helping others.
I have put everything [...]]]></description>
			<content:encoded><![CDATA[<p>For some time, I have been working on a description of some of the math behind equations that we use for passive filters.  I do this kind of writing to check my own understanding, and then I share it in the hope of finding my own mistakes and possibly helping others.<span id="more-60"></span></p>
<p>I have put everything into a single pdf document here: <a href="http://gaussmarkov.net/parts/capmath.pdf" title="Capacitor Mathematics" target="_blank">capmath.pdf</a>. This is not intended for everyone.  To read it, you will need to know enough calculus to differentiate and integrate the exponential function and the sine function and enough trigonometry to have seen the angle sum forumula for the sine function.  I chose not to use phasor notation.  I think that sine functions are more concrete and better for learning this material the first time.</p>
<p>In the spirit of keeping things real (pun intended), I included some calculations for the discharging filter caps of an amp with a resistor and some illustrations using the BMP tone stack.  Here&#8217;s an outline of the whole document:</p>
<p><span style="text-decoration: underline">Capacitor Mathematics</span></p>
<ul style="margin-top: 0pt; margin-bottom: 0pt">
<li>Summary</li>
<li>Capacitor Behaviour</li>
<li>Discharging and Charging
<ul style="margin-top: 0pt; margin-bottom: 0pt">
<li>Differential Equation Solution</li>
<li>Discharging</li>
</ul>
</li>
<li>Discharging a Capacitor with a Resistor</li>
<li>Alternating Current</li>
<li>RC Filters
<ul style="margin-top: 0pt; margin-bottom: 0pt">
<li>The Angle Sum Formula
<ul style="margin-top: 0pt; margin-bottom: 0pt">
<li>Phase</li>
<li>Summing Sine Waves</li>
</ul>
</li>
<li>Voltage of the Capacitor: Guess and Check</li>
<li>Current</li>
<li>Voltage of the Resistor</li>
</ul>
</li>
<li>High Pass Filter</li>
<li>Low Pass Filter</li>
<li>Cut-Off Frequency</li>
<li>A Final Note</li>
</ul>
<p>If you have any comments, please post them!</p>
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		<title>Capacitors 5: High Pass Filters</title>
		<link>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-5-high-pass-filters/</link>
		<comments>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-5-high-pass-filters/#comments</comments>
		<pubDate>Sat, 24 Feb 2007 00:24:36 +0000</pubDate>
		<dc:creator>gaussmarkov</dc:creator>
				<category><![CDATA[Capacitors]]></category>

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		<description><![CDATA[High Pass Filter
If the capacitor and resistor in the low pass filter change places, then the circuit becomes a high pass filter. Indeed the output of this high pass filter is just the output of the low pass filter subtracted from the input. So the frequencies that were attenuated most by one filter are the [...]]]></description>
			<content:encoded><![CDATA[<h2><a title="hpf" name="hpf"></a>High Pass Filter</h2>
<p>If the capacitor and resistor in the low pass filter change places, then the circuit becomes a high pass filter. Indeed the output of this high pass filter is just the output of the low pass filter subtracted from the input. So the frequencies that were attenuated most by one filter are the frequencies attenuated least by the other filter.   You could use this LTSpice setup to check this:<span id="more-59"></span></p>
<p style="text-align: center"><img src="/parts/capacitors/hpf_lpf.jpg" alt="[image: High and Low Pass Filters]" title="High and Low Pass Filters" /></p>
<p>One way to think about this is to use the rule that</p>
<blockquote><p><em>a series of capacitors and resistors has the same electronic properties no matter what order the components are connected</em>.</p></blockquote>
<p>This phenomenon is described for the simpler case of series of resistors <a href="/wordpress/parts/resistors/resistors-in-series/#order" class="std">here</a>. It is also true for any series of resistors and capacitors: viewed as a single multi-part component, a series of resistors and capacitors behaves the same no matter how the individual parts are ordered.</p>
<p>This rule can be inferred from the way resistors and capacitors conduct AC current and the basic laws of circuit analysis:</p>
<ul style="margin-left: 2em">
<li>the current is the same on both leads of a resistor or a capacitor,</li>
<li><a href="/wordpress/parts/resistors/resistors-in-parallel/#KCL" class="std">Kirchoff&#8217;s current law</a>,<br />
which implies that the same current will pass through all of the resistors and capacitors in a series, and</li>
<li><a href="/wordpress/parts/resistors/resistors-in-series/#KVL" class="std">Kirchoff&#8217;s voltage law</a>,<br />
which makes no distinction about order: the sum of the voltage differences across components must equal the voltage drop<br />
across the series. No matter what order you add a list of numbers, you always get the same sum.</li>
</ul>
<p>Now we apply these two conditions to a resistor and a capacitor in series. Given that the AC is the same through every component of the series, the voltage potential across the resistor is the same whether it sits at the beginning or the end of the series (using <a href="/wordpress/parts/resistors/resistors-limiting-current/#ohmslaw">Ohm&#8217;s law</a>). This is also true for the capacitor because its voltage potential must be the difference between the supply potential and the resistor&#8217;s potential (using <a href="/wordpress/parts/resistors/resistors-in-series/#KVL">Kirchoff&#8217;s voltage law</a>).</p>
<p>Therefore, the difference between the high and low pass filters is whether the output is the voltage potential across the resistor or across the capacitor. For the low pass filter, it is across the capacitor. For the high pass filter, it is the resistor.</p>
<p>Because the voltage <em>across</em> the capacitor is attenuating high frequencies more than low, the resistor is making up for this by simultaneously allowing the high frequencies to have a greater amplitude than the low. If the output voltage is the voltage across the resistor, we see that low frequencies are attenuated more than high frequencies. That is, we have a high pass filter.</p>
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		<title>Capacitors 4: Low Pass Filters</title>
		<link>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-4-low-pass-filters/</link>
		<comments>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-4-low-pass-filters/#comments</comments>
		<pubDate>Fri, 23 Feb 2007 23:54:36 +0000</pubDate>
		<dc:creator>gaussmarkov</dc:creator>
				<category><![CDATA[Capacitors]]></category>

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		<description><![CDATA[Resisting Voltage Change
Capacitors are often said &#8220;to oppose voltage change.&#8221;  When the current is finite, the change in the voltage across the plates of a capacitor generally lags behind the change in the voltage of the source. Also, the shape of voltage changes affects how the capacitor charges, the resulting shape of voltage across [...]]]></description>
			<content:encoded><![CDATA[<h2><a title="voltresist" name="voltresist"></a>Resisting Voltage Change</h2>
<p>Capacitors are often said &#8220;to oppose voltage change.&#8221;  When the current is finite, the change in the voltage across the plates of a capacitor generally lags behind the change in the voltage of the source. Also, the shape of voltage changes affects how the capacitor charges, the resulting shape of voltage across the plates of the capacitor, and the pattern of the current through the capacitor.<span id="more-58"></span></p>
<p>To illustrate a capacitor opposing voltage change, we will simulate a simple circuit with a resistor and a capacitor in series in LTSpice. As in a voltage divider, we will treat the junction between the resistor and capacitor as the output of the circuit. The voltage at this point is the voltage across the capacitor because the other lead of the capacitor is tied to ground. The resistor has a  <em>10K</em>  ohms resistance and the capacitor has a  <em>100µF</em> capacitance.  The resistor forces current to be finite.</p>
<p style="margin: 0pt auto; width: 500px"><img src="/parts/capacitors/cap_charge_schem.PNG" title="charging a capacitor schematic" alt="charging a capacitor schematic" align="middle" /></p>
<p>If we apply a rectangular pulse of positive voltage to a capacitor, then a capacitor charges and then discharges as shown in this figure. Its voltage does not change instantly with the supply voltage. Instead it rises or falls behind the supply. Note that the capacitor&#8217;s voltage never quite reaches the maximum voltage. This reflects the increasing difficulty of adding more electrons to one of its plates.</p>
<p align="center"><img src="/parts/capacitors/cap_charge_square.PNG" title="charging a capacitor schematic" alt="charging a capacitor schematic" align="middle" /></p>
<p>If we apply a triangle pulse, then the capacitor voltage is smoother. After the supply voltage has peaked, the capacitor continues to gain voltage. At the peak of the triangle, the capacitor voltage path switches from a convex shape to a concave shape, rather than changing direction. In other words, as long as the supply voltage is rising the capacitors voltage is rising faster and faster or <em>accelerating</em>. But when the supply voltage starts to fall, then the capacitor voltage starts to <em>decelerate</em>, at first slowing its rate of increase until the capacitor voltage eventually stops growing and starts to fall.  The capacitor voltage cannot rise above the supply voltage, so it peaks where the two voltages are equal.</p>
<p><img src="/parts/capacitors/cap_charge_triangle.PNG" alt="charging a capacitor schematic" /></p>
<p>There is one input AC shape that is not changed by a capacitor and that is the sine wave. Nevertheless, if we supply a sine wave pulse, then we get a <em>delayed</em> sine wave pulse. In this figure, we have a few cycles of the sine wave so that the capacitor voltage settles into its sine response. Because the capacitor resists voltage change, it also takes a few cycles to reach its steady-state response.</p>
<p><img src="/parts/capacitors/cap_charge_sine.PNG" alt="charging a capacitor schematic" /></p>
<p>All of these examples show the symptoms of a capacitor resisting a voltage change. First, the output voltage swing is smaller. Second, the output voltage changes lag behind the input voltage changes.</p>
<h2><a title="lpf" name="lpf"></a>Low Pass Filter</h2>
<p>Our simple circuit is actually a passive low pass filter. That is, if we compare the output for input AC with the same amplitude but different frequencies, the low frequency AC has a relatively higher amplitude at the output node. To illustrate how this occurs, we simulated the circuits shown in this figure in LTspice/SwitcherCAD III.</p>
<p><img src="/parts/capacitors/cap_charge_freq_schem.PNG" alt="charging a capacitor schematic" /></p>
<p>There are two identical circuits with AC voltage sources that differ only in the length of time a square wave has a positive voltage. First, compare the path of voltage across the two capacitors for a single pulse. Because the initial conditions are identical, the paths are identical until the shorter pulse turns off. After that point, one capacitor keeps charging while the other discharges. As a result, the shorter pulse reaches a lower peak in voltage than the longer pulse. This is why higher frequencies are have smaller amplitudes at the output: the capacitor has less time to charge over each cycle.</p>
<p><img src="/parts/capacitors/cap_charge_freq_1.PNG" alt="charging a capacitor schematic" /></p>
<p>When there is AC, this pattern is repeated. The next figure shows a 10-second portion after the pattern has been repeated for 65 seconds. The short pulse becomes the higher frequency signal, oscillating four times as fast as the long pulse. And its amplitude is about half the amplitude of the lower frequency signal. This phenomenon is the basic character of a low pass filter: low frequencies are attenuated less than high frequencies.</p>
<p><img src="/parts/capacitors/cap_charge_freq_2.PNG" alt="charging a capacitor schematic" /></p>
<p>The square input wave is useful for highlighting the differences in charge times for high and low frequency inputs. However, analyzing this low pass filter is more convenient when the source AC is a sine wave. In such a case, the output AC voltage is also a sine wave with the same frequency. The only differences between the input and output AC are amplitude and relative phase. There are no additional shape differences to consider. In the next figure, the previous calculations are repeated after replacing the square waves with sine waves having the same amplitudes and frequencies.</p>
<p><img src="/parts/capacitors/cap_charge_freq_3.PNG" alt="charging a capacitor schematic" /></p>
<p>Once again, the lower frequency input yields the output with the greater amplitude. Also, in both cases one can interpret the output as lagging behind the input. For example, look at the peaks of the sine waves. The output peaks appear to the right of the input peaks, indicating that the peaks occur later in time. And note that the peak of the lower frequency output is closer to the peak of the input (as a fraction of one cycle) than it is for the higher frequency case. The high frequency output is almost one quarter cycle behind its input. This is a general outcome for sine waves as well.</p>
<h2><a title="decoupling" name="decoupling"></a>Decoupling Capacitors</h2>
<p>You may want to jump to <a href="http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-5-high-pass-filters/">the high pass filter discussion</a>, but this is a good place to introduce the notion of a <em>decoupling capacitor</em> (as opposed to the <a href="/wordpress/parts/capacitors/capacitors-2-dc-and-ac/#coupling"><em>coupling</em> capacitor</a> described on the <a href="/wordpress/parts/capacitors/capacitors-2-dc-and-ac/">dc and ac</a> page). Decoupling capacitors appear in DC power supply filters to remove possible AC.  In a stompbox circuit, the power supply is supposed to be DC.  If there is any AC in the signal, it will appear in the output of the stompbox along with the guitar signal.  So care is taken to protect the guitar signal.</p>
<p>The simplest decoupling arrangement is to connect one lead of a capacitor to the power supply and the other lead to ground. Because the capacitor blocks DC, the desired DC power supply is unaffected.  But the capacitor transmits (some) AC to ground, effectively acting as a low pass filter.  You can think of DC as AC with a frequency of <em>0Hz</em>.  In the schematic for <a href="/wordpress/circuits/tubescreamer/">the TS808 clone</a> for example, a <em>100μF</em> polarized capacitor is used, so that the negative lead is connected to ground and the other lead to the power <em>rail</em>. Such capacitors are also commonly called <em>bypass</em> capacitors.</p>
<p>How do AC signals get into the power supply of a stompbox?  One common source is a 50Hz or 60Hz cycle that comes from transformers converting AC electricity imperfectly into DC electricity.  The bypass capacitor acts as part of a low pass filter to reduce this source of hum.  Now 60Hz is a pretty low frequency and this is why bypass capacitors have large values, like 100μF.  As I will explain later, the larger the capacitor value is the more AC signal a low pass filter attenuates.</p>
<p>There is another source of AC signals in a power supply and that is the operation of the stompbox ciruit itself. In a simple amplifier, for example, the current drawn by a transistor depends on the AC signal going through it.  The valleys in the signal correspond to moments when most current is drawn.  The internal resistance of the power source&#8211;this is especially an issue with old batteries&#8211;combines with cycles in current consumption to create cycles in the supply voltage.  This is <a href="/wordpress/parts/resistors/resistors-limiting-current/#ohmslaw">Ohm&#8217;s law</a> in action: voltage is proportional to current, where the factor of proportionality is constant resistance.</p>
<p>When this second source of AC in the power supply occurs, the various parts of the circuit that draw on the power supply are said to be <em>coupled</em>.  The current drawn by one part of the circuit causes the power supply seen by another part of the circuit to fluctuate.  Usually this is undesirable.  Look again at the schematic for <a href="/wordpress/circuits/tubescreamer/">the Tubescreamer clone</a> and you will see two transistors, Q1 (near the input) and Q2 (near the output), both with +9V power connections above them. Or return to the simpler <a href="/wordpress/tools/software/eagle/eagle/">Basic Fuzz Face</a> circuit, where there are two transistors fed by the same -9V power supply.  These are both cases where the two transistors are coupled.</p>
<p>A bypass capacitor in the power supply section <em>decouples</em> the two transistors by providing the additional momentary current needed to smooth out the voltage supplied.  The effect is not perfect.  Voltage does not become constant.  But the effect is sufficient in most cases to make the voltage supply effectively (or approximately) constant. So bypass capacitors are also called <em>decoupling</em> capacitors.  And this is a second way in which that capacitor attenuates AC in the power supply.</p>
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		<title>Capacitors 3: In Series and Parallel</title>
		<link>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-3-in-series-and-in-parallel/</link>
		<comments>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-3-in-series-and-in-parallel/#comments</comments>
		<pubDate>Fri, 23 Feb 2007 23:13:52 +0000</pubDate>
		<dc:creator>gaussmarkov</dc:creator>
				<category><![CDATA[Capacitors]]></category>

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		<description><![CDATA[The Capacitor &#8220;Law&#8221;
As described previously, capacitors are fundamentally different from resistors. They do not conduct DC current, but capacitors do transmit AC current because they can repeatedly accumulate and release a voltage potential like a rechargeable battery.
Charging and discharging are responses to changes in the voltage across the leads of a capacitor. There is a [...]]]></description>
			<content:encoded><![CDATA[<h2><a title="capacitor-law" name="capacitor-law"></a>The Capacitor &#8220;Law&#8221;</h2>
<p>As described previously, capacitors are fundamentally different from resistors. They do not conduct DC current, but capacitors do transmit AC current because they can repeatedly accumulate and release a voltage potential like a rechargeable battery.<span id="more-57"></span></p>
<p>Charging and discharging are responses to changes in the voltage across the leads of a capacitor. There is a simple equation that describes accurately the relationship between the current and the change in voltage for a capacitor with a capacitance of <em>C</em> farads:</p>
<table class="eqn">
<tr>
<td>I(t)   =   C ⋅</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
</tr>
</table>
<p>where <em>I(t)</em> is the current at time <em>t</em>, <em>V(t)</em> is the voltage at time <em>t</em>, and <em>dV(t)/dt</em> is the rate of change of voltage at time <em>t</em>.  I am using the calculus notation for a derivative to denote the rate of change of voltage.</p>
<p>This relationship plays the same role for analyzing capacitors that <a href="/wordpress/parts/resistors/resistors-limiting-current/" class="std">Ohm&#8217;s law</a> plays for resistors.  That&#8217;s why I called this section <em>The Capacitor &#8220;Law.&#8221;</em> Indeed, if we rearrange terms, this equation is similar to Ohm&#8217;s law:</p>
<table class="eqn">
<tr>
<td>
<table class="frac">
<tr>
<td class="numerator">dV(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
<td>=   I(t)⋅</td>
<td>
<table class="frac">
<tr>
<td class="numerator">1</td>
</tr>
<tr>
<td class="denominator">C</td>
</tr>
</table>
</td>
<td style="padding-left: 5em; padding-right: 5em">versus</td>
<td>V(t)   =   I(t) ⋅ R</td>
</tr>
</table>
<p>The differences between Ohm&#8217;s law and the capacitor law are that voltage <em>V(t)</em> has been replaced with the rate of change in voltage <em>dV(t)/dt</em> and the constant <em>R</em> has been replaced by the constant <em>1/C</em>. In what follows, voltage and rate of change in voltage can be treated analogously. The reciprocal notation for the constants of proportionality explains the differences between the formulas for combining capacitors and the formulas for combining resistors.</p>
<h2><a title="capsinparallel" name="capsinparallel"></a>Capacitors in Parallel</h2>
<p>The same argument applies to capacitors in parallel as resistors, so what is written here will be a briefer version of what I have written for <a href="/wordpress/parts/resistors/resistors-in-parallel/" class="std">resistors in parallel</a>.</p>
<p>The voltage across two parallel capacitors is the same for both capacitors. This applies to the change in voltage as well.  So we can also write  the capacitor law that applies to each capacitor as a function of a single voltage rate of change:</p>
<table class="eqn">
<tr>
<td>I<sub>1</sub>   =   C<sub>1</sub> ⋅</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>P</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
<td style="padding-left: 5em; padding-right: 5em">and</td>
<td>I<sub>2</sub>   =   C<sub>2</sub> ⋅</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>P</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
</tr>
</table>
<p>where  <em>C<sub>1</sub></em> is the capacitance of capacitor  <em>C1</em>, <em>C<sub>2</sub></em>  is the capacitance of capacitor  <em>C2</em>, and <em>V<sub>P</sub>(t)</em>  denotes the common voltage for the pair of capacitors.</p>
<p><a href="/wordpress/parts/resistors/resistors-in-parallel/#KCL" class="std"> Kirchoff&#8217;s Current Law</a> says that the currents flowing through each of the capacitors sum to the current flowing through them as a parallel pair:</p>
<table class="eqn">
<tr>
<td>I<sub>P</sub>  =  I<sub>1</sub> + I<sub>2</sub></td>
</tr>
</table>
<p>where <em>I<sub>P</sub></em> denotes the combined current, <em>I<sub>1</sub></em>  is the current through  <em>C1</em>, and <em>I<sub>2</sub></em>  is the current through  <em>C2</em>. If we combine these relationships  then we can figure out the effective capacitance of the capacitors. Substituting the capacitor law expressions into the current equation above gives</p>
<table class="eqn">
<tr>
<td>I<sub>P</sub>   =   C<sub>1</sub> ⋅</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>P</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
<td>+   C<sub>2</sub> ⋅</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>P</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
<td>=   (C<sub>1</sub> + C<sub>2</sub>) ⋅</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>P</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
</tr>
</table>
<p>This last equation tells us the combined capacitance of the parallel capacitors. It is the capacitor law for a capacitor with capacitance equal to the sum of the individual capacitances, <em>C<sub>1</sub></em> + <em>C<sub>2</sub></em>.</p>
<h2><a title="capsinseries" name="capsinseries"></a>Capacitors in Series</h2>
<p>Again, following <a href="http://gaussmarkov.net/wordpress/parts/resistors/resistors-in-series/">the argument for resistors</a>, Let <em>V<sub>S</sub></em> be the total voltage difference across two capacitors in series. <a href="/wordpress/parts/resistors/resistors-in-series/#KVL" class="std">Kirchoff&#8217;s Voltage Law</a> says that the voltage across the individual resistors,  <em>V<sub>1</sub></em>  and  <em>V<sub>2</sub></em>, sums to  <em>V<sub>S</sub></em>.  This also applies to rates of change:</p>
<table class="eqn">
<tr>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>S</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
<td>=</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>1</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
<td>+</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>2</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
</tr>
</table>
<p>At the same time, the currents through the capacitors are equal:</p>
<table class="eqn">
<tr>
<td>I<sub>S</sub>(t)   =   I<sub>1</sub>(t)   =   I<sub>2</sub>(t).</td>
</tr>
</table>
<p>Combining these equations with the capacitors law for each capacitor,</p>
<table class="eqn">
<tr>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>S</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
<td>=</td>
<td>
<table class="frac">
<tr>
<td class="numerator">I<sub>S</sub>(t)</td>
</tr>
<tr>
<td class="denominator">C<sub>1</sub></td>
</tr>
</table>
</td>
<td>+</td>
<td>
<table class="frac">
<tr>
<td class="numerator">I<sub>S</sub>(t)</td>
</tr>
<tr>
<td class="denominator">C<sub>2</sub></td>
</tr>
</table>
</td>
<td>=   I<sub>S</sub>(t) ⋅ (1/C<sub>1</sub> + 1/C<sub>2</sub>)</td>
</tr>
</table>
<p>or</p>
<table class="eqn">
<tr>
<td>I<sub>S</sub>(t)    =</td>
<td>
<table class="frac">
<tr>
<td class="numerator">C<sub>1</sub> C<sub>2</sub></td>
</tr>
<tr>
<td class="denominator">C<sub>1</sub> + C<sub>2</sub></td>
</tr>
</table>
</td>
<td>⋅</td>
<td>
<table class="frac">
<tr>
<td class="numerator">dV<sub>S</sub>(t)</td>
</tr>
<tr>
<td class="denominator">dt</td>
</tr>
</table>
</td>
</tr>
</table>
<p>we obtain the capacitor law for a capacitance of <em>C<sub>1</sub>⋅C<sub>2</sub>/(C<sub>1</sub> + C<sub>2</sub>)</em>, which we can interpret as the capacitance of two capacitors in series.</p>
<h2><a title="capsvsres" name="capsvsres"></a>Capacitors versus Resistors</h2>
<p>It appears as though the formulas for combining resistors are the reverse of the formulas for combining capacitors.  Resistors in series have a resistance<br />
equal to their sum whereas capacitors in series have a capacitance equal to their product divided by their sum</p>
<table class="eqn">
<tr>
<td>R<sub>1</sub>   +   R<sub>2</sub></td>
<td style="padding-left: 5em; padding-right: 5em">versus</td>
<td>
<table class="frac">
<tr>
<td class="numerator">C<sub>1</sub> C<sub>2</sub></td>
</tr>
<tr>
<td class="denominator">C<sub>1</sub> + C<sub>2</sub></td>
</tr>
</table>
</td>
</tr>
</table>
<p>while resistors in parallel have a combined resistance equal to their product divided by their sum whereas capacitors in parallel have a combined capacitance equal to their sum</p>
<table class="eqn">
<tr>
<td>
<table class="frac">
<tr>
<td class="numerator">R<sub>1</sub> R<sub>2</sub></td>
</tr>
<tr>
<td class="denominator">R<sub>1</sub> + R<sub>2</sub></td>
</tr>
</table>
</td>
<td style="padding-left: 5em; padding-right: 5em">versus</td>
<td>C<sub>1</sub>   +   C<sub>2</sub></td>
</tr>
</table>
<p>However, if we were to measure the value of a capacitor as the reciprocal of capacitance these calculations would be identical.  This is what is suggested by the comparison of Ohm&#8217;s law and the capacitor law above.  In words, the reciprocal capacitance of two capacitors in series is the sum of the individual reciprocal capacitances:</p>
<table class="eqn">
<tr>
<td>1/C<sub>1</sub>   +   1/C<sub>2</sub>   =</td>
<td>
<table class="frac">
<tr>
<td class="numerator">C<sub>1</sub> + C<sub>2</sub></td>
</tr>
<tr>
<td class="denominator">C<sub>1</sub> C<sub>2</sub></td>
</tr>
</table>
</td>
</tr>
</table>
<p>Similarly, the reciprocal capacitance of two capacitors in parallel is the product of the individual reciprocal capacitances divided by the sum of the individual reciprocal capacitances:</p>
<table class="eqn">
<tr>
<td>
<table class="frac">
<tr>
<td class="numerator">1/C<sub>1</sub>   1/C<sub>2</sub></td>
</tr>
<tr>
<td class="denominator">1/C<sub>1</sub>   +   1/C<sub>2</sub></td>
</tr>
</table>
</td>
<td>=</td>
<td>
<table class="frac">
<tr>
<td class="numerator">1</td>
</tr>
<tr>
<td class="denominator">C<sub>1</sub> + C<sub>2</sub></td>
</tr>
</table>
</td>
</tr>
</table>
]]></content:encoded>
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		<title>Capacitors 2: DC and AC</title>
		<link>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-2-dc-and-ac/</link>
		<comments>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-2-dc-and-ac/#comments</comments>
		<pubDate>Fri, 23 Feb 2007 22:27:48 +0000</pubDate>
		<dc:creator>gaussmarkov</dc:creator>
				<category><![CDATA[Capacitors]]></category>

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		<description><![CDATA[Preliminaries
First, let&#8217;s summarize some basic electronic simplifications:  electrons carry a negative charge and, therefore, repel each other.  Because they can move freely within a wire, electrons will distribute themselves uniformly throughout the wire so that there is no discernible polarity along the wire. On the other hand, if they are in the presence [...]]]></description>
			<content:encoded><![CDATA[<h2>Preliminaries</h2>
<p>First, let&#8217;s summarize some basic electronic simplifications:  electrons carry a negative charge and, therefore, repel each other.  Because they can move freely within a wire, electrons will distribute themselves uniformly throughout the wire so that there is no discernible polarity along the wire. On the other hand, <span id="more-56"></span>if they are in the presence of a voltage difference, all electrons will move some amount toward the positive pole because they are attracted to its positive charge. This is <em>direct current</em> (DC) or <em>continuous current</em>.</p>
<p>Second, here is a description of the basic internal structure of capacitors. Capacitors have two metal plates inside them that are separated by an insulating material called a <em>dielectric</em>. In an ideal capacitor, electrons cannot flow across the dielectric. For this reason, a capacitor does not conduct DC as wires or resistors do.  However, the plates are close enough that the electrons on one plate can be repelled by the presence of electrons on the other plate.  Alternatively, electrons can be attracted to one plate by an absence of electrons on the other plate.  Roughly speaking, the total number of electrons in a capacitor is constant so that every electron that enters a capacitor is balanced by an electron that exits.</p>
<h2><a title="analogy" name="analogy"></a>Battery Analogy</h2>
<p>For this reason, capacitors are like rechargeable batteries: by placing a voltage across a capacitor, some electrons gather on one plate and an equal number depart from the other. If the capacitor is then disconnected, it contains potential energy in the form of an electrostatic field across the plates, one with more electrons than the other.  Such a capacitor is said to be <em>charged</em>.</p>
<p>If the leads of a charged capacitor are shorted, then the electrons will flow off one plate on to the other to redistribute charge uniformly. If we connect a resistor (instead of a wire) across the leads, then the current of electrons will appear as a momentary voltage difference across the resistor (and capacitor) and the resistor will give off some heat.</p>
<p>In many capacitors, the two plates act symmetrically and the capacitor can be charged with relatively more electrons on either plate. <img class="floatright" style="margin: 10px;" title="Radial Polarized Capacitor" src="/parts/capacitors/Radial_Electrolytic_small.JPG" alt="Radial Polarized Capacitor" align="right" />The polarity of the voltage across such capacitors can be switched causing a corresponding accumulation of electrons on the opposite plate. <em>Polarized</em> capacitors must have the polarity of the voltage across them in only one direction.  In other words, electrons should be relatively abundant only on the plate of the lead marked with minus signs on the casing of the capacitor.</p>
<p>As a capacitor is charged, no electrons actually travel through the capacitor. But it is as though they do:  for every electron that goes into the capacitor through one lead another electron goes out through the opposite lead.  We could say that <em>current</em> flows through a capacitor even though <em>electrons</em> do not. But such current cannot flow forever: as electrons collect on a plate, it becomes more and more difficult to add additional electrons. If the voltage rating of a capacitor is exceeded and too many electrons appear on a plate, then the capacitor breaks down and there is a sudden short as electrons travel from one plate to the other. This can be dangerous.</p>
<h2><a title="AC" name="AC"></a>Alternating Current</h2>
<p><a title="AC" name="AC"></a>An <em>alternating current</em> (AC) is an electrical current with a cyclical magnitude and direction.  While DC cannot flow through an ideal capacitor, AC can because the accumulation of electrons on a plate can be reversed so that electrons accumulate on the opposite plate.  By repeating such reversals, we can alternate which plate is accumulating electrons and AC will flow &#8220;through&#8221; the capacitor.  The key is to keep changing the direction of the current.</p>
<p>An analogy with a ripple moving across the surface of a tank of water may be helpful. When the wave travels across the surface, the water does not flow. The surface merely rises and falls with each ripple as it passes. Alternating current through a capacitor is similar. The concentration of electrons rises and falls but electrons do not continuously flow in one direction.</p>
<p>Note that the polarity of voltage across a capacitor does not have to change for AC to flow. All that matters is the <em>difference</em> between two voltages:</p>
<ol style="margin-left: 2em;">
<li><a title="AC" name="AC"></a>the voltage across the plates within the capacitor and</li>
<li><a title="AC" name="AC"></a>the voltage across the leads of the capacitor caused by the source voltage.</li>
</ol>
<p><a title="AC" name="AC"></a>If there is a difference, then electrons in the capacitor will redistribute themselves towards matching the source voltage and current will flow. As a result, a polarized capacitor will also conduct AC even though the polarity of the voltage across its leads does not change. As long as the magnitude of the voltage is changing, current will flow through a capacitor.</p>
<p><a title="AC" name="AC"></a>It may be helpful to think of AC as similar to water moving back and forth in a garden hose.  The flow of the water repeatedly changes direction.  Now suppose that we hook up two hoses with a rubber diaphragm (an analogy to a capacitor) in between.  Water cannot flow from one hose into the other, but it can still move back and forth through the two hoses. In addition, if we add some extra water pressure from one end, this back and forth movement can still occur. It is the changing <em>magnitude</em> of pressure that matters not the changing <em>direction</em> of pressure.</p>
<p><a title="AC" name="AC"></a>If you would like to see some graphs of how a capacitor reacts to different kinds of AC, then jump to the <a href="http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-4-low-pass-filters/"> low pass filters</a> page. The next two sections on this page deal with a basic application of capacitors charging and transmitting AC that does not require discussing the details of the AC.</p>
<h2><a title="coupling" name="coupling"></a>Coupling Capacitors</h2>
<p>Because of their ability to conduct AC and inability to conduct DC, capacitors are often used to connect (or <em>couple</em>) stages of a circuit designed to conduct and alter AC. As such, they are called <em>coupling capacitors</em>. An amplification stage, for example, may place the center of an AC at a value like  <em>4.5V</em>,  half of a typical  <em>9V</em> DC voltage supply. This  <em>4.5V</em> is called a <em>DC offset</em>. To contain a DC offset within the amplification stage, capacitors typically appear as &#8220;sieves&#8221; for AC near the input and output of a stompbox circuit.</p>
<p style="text-align: center;"><img class="floatright" title="Basic Fuzz Face schematic" src="/images/bff-schem.png" alt="[image: Basic Fuzz Face schematic]" height="400" /></p>
<p>Look, for example, at the <a href="http://gaussmarkov.net/wordpress/tools/software/eagle/eagle/">Basic Fuzz Face</a> circuit. At the input, you will see capacitor  <em>C1</em> with a value of  <em>2.2uF</em>.  And at the output, sits capacitor  <em>C3</em> with a value of  <em>0.01uF</em>.</p>
<p>Using the schematic, one can see that there is a DC offset for the AC signal between the coupling capacitors by noting that the <em>-9V</em> power supply is just a couple of resistors away from the AC signal, which is vibrating on all of the connections &#8220;inside&#8221; of  <em>C1</em>,   <em>C2</em>,   <em>R2</em>,   <em>R3</em>, and the  <em>1K</em> GAIN pot at the bottom of the schematic.</p>
<p>An explanation of the DC offset involves describing how transistors can be used to make signal amplifiers. I do not have a companion explanation of transistors written yet, so I&#8217;ll just note that the amplification in this circuit works by creating a guitar signal that varies from  <em>0V</em> up to  <em>9V</em>, the range of voltages that one can produce with a <em>single supply voltage</em> like a battery. Because an audio signal tends to vary equally on either side of a resting point, it makes sense to place that point in the middle of the voltage range, at around <em>4.5V</em>.  So the coupling capacitors allow us to feed the circuit an input signal that varies around <em>0V</em> and to generate an output signal the also varies around <em>0V</em>.</p>
<h2><a title="leakage" name="leakage"></a>Capacitor Leakage</h2>
<p>Actual capacitors differ from the ideal just described, of course. Two important differences are that real capacitors</p>
<ol style="margin-left: 2em;">
<li> have some resistance and</li>
<li>conduct some DC, or <em>leak</em>.</li>
</ol>
<p>For many situations, these differences can be ignored. However, there is an exception for coupling capacitors that comes up right away in building stompboxes. The symptom is an annoying &#8220;pop&#8221; sound from the speaker of the guitar amp when switching the stompbox from by-pass into the path of the signal. As R.G. Keen explains in <a class="std" href="http://www.geofex.com/Article_Folders/box_pop.htm">Why does my stompbox pop when I hit the bypass switch?</a>, capacitor leakage can be the cause.</p>
<p>By-passing the stompbox circuit is often accomplished by switching the input and output connections to a wire or trace that runs directly between them. This arrangement is called <em>true by-pass</em> because no electronic components influence the signal of the guitar. When the circuit is by-passed, the outside lead of a coupling capacitor is not connected to anything; the lead is said to be <em>floating</em>. This would not be a problem if capacitors did not leak. The coupling capacitor would remain charged just as when it is connected.</p>
<p>But capacitors do leak and some of the charge maintained while the circuit is connected is lost while it is not. When the floating lead of the coupling capacitor is reconnected to the output, the capacitor faces a supply voltage again and must recharge to return to its operating voltage. As R.G. explains,</p>
<blockquote><p>&#8220;That sudden voltage difference and the charging current that brings the capacitors back to the working voltage is heard in the amplifier as a pop. If the effect has any gain, the input capacitor pop is further amplified by the gain of the pedal into a bigger pop.&#8221;</p></blockquote>
<p>To fix this, we need a <em>pull-down resistor</em>.  This is a resistor, typically <em>1M</em> or higher, connected on one side to the floating lead of the output coupling capacitor and connected to ground on the other side.  This high resistance connection keeps the voltage of the output plate of the capacitor at ground level where it is when the circuit is connected. The high resistance value ensures that when the circuit is connected, this pull-down resistor has little effect on the output.</p>
<p>If you look at the Basic Fuzz Face schematic above, you will see that the <em>LEVEL</em> potentiometer will serve as a pull-down resistor and another one is not needed for this circuit.  Even when the output is disconnected from the circuit on the wiper of the potentiometer, their is a 500K resistor connecting the floating lead of capacitor <em>C3</em> to ground.</p>
<p>Many stompbox builders prefer to put a pull-down resistor on the floating lead of the input coupling capacitor as well.</p>
<h2><a title="moreinfo" name="moreinfo"></a>Further Information</h2>
<p>R.G. Keen&#8217;s <a class="std" href="http://geofex.com/Article_Folders/How_It_Works/hiw.htm">How It Works &#8230;</a> is a beginner&#8217;s guide to understanding what electronics is about written by one of the masters of DIY stompboxes.  I don&#8217;t recall whether I first saw the battery analogy for capacitors in this article, but the analogy is there along with many other insights.  Another of his articles, <a class="std" href="http://www.geofex.com/Article_Folders/bypass/bypass.htm">The Ins and Outs of Effect Bypassing</a>, also contains closely related material.</p>
<p>Also see some of <a class="std" href="/wordpress/links/#electronicstheory">my electronics theory links</a>.</p>
]]></content:encoded>
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		<title>Capacitors 1: Description</title>
		<link>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-1-description/</link>
		<comments>http://gaussmarkov.net/wordpress/parts/capacitors/capacitors-1-description/#comments</comments>
		<pubDate>Fri, 23 Feb 2007 20:45:01 +0000</pubDate>
		<dc:creator>gaussmarkov</dc:creator>
				<category><![CDATA[Capacitors]]></category>

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		<description><![CDATA[Appearance

There are many kinds of capacitors and several popular ones are pictured here. Generally, capacitors have two leads. Some are axial leaded, like resistors, and others are radial leaded, with both leads at one end. Stompbox layouts seem to use radial leaded capacitors most often, but axial leaded are just as good. Unlike resistors, some [...]]]></description>
			<content:encoded><![CDATA[<h2><a title="appearance" name="appearance"></a>Appearance</h2>
<p><img src="/parts/capacitors/10uF_50V_Radial_Electrolytic_Capacitor.gif" alt="mylar capacitors" class="floatright" align="right" width="225" /></p>
<p>There are many kinds of capacitors and several popular ones are pictured here. Generally, capacitors have two leads. Some are <em>axial leaded</em>, like resistors, and others are <em>radial leaded</em>, with both leads at one end. Stompbox layouts <span id="more-55"></span>seem to use radial leaded capacitors most often, but axial leaded are just as good. Unlike resistors, some capacitors are <em>polarized</em>, with positive and negative leads: the voltage across such capacitors must agree with the polarity of the leads. Take care to orient polarized capacitors correctly in a circuit.</p>
<p style="clear: both"><img src="/parts/capacitors/mylar_caps.png" alt="mylar capacitors" class="floatright" align="right" width="300" /><br />
Generally, polarized capacitors have lead markings on the casing, like a colored band of minus (negative) signs. The radial electrolytic capacitor pictured above has a black casing with a gray band and you can just see one minus sign. Note also that the negative lead is shorter than the positive lead. On axial electrolytic capacitors, arrows often point toward the negative lead.  The blue capacitor shown here is an example.  When the arrows are not present, note that the aluminum can (housing) shows on the negative end while the positive end has a (black) seal insulated from the housing. Often, the indentation or groove around one end of the casing is on the positive end. You can also see some of these features on the blue capacitor below.</p>
<p><img src="/parts/capacitors/axial_electrolytic.gif" style="clear: both; background: white none repeat scroll 0% 50%; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial" alt="mylar capacitors" class="floatright" align="right" width="100" /></p>
<p><img src="/parts/capacitors/boxfilmcap.gif" style="margin-left: 0px" alt="mylar capacitors" class="floatright" align="right" width="54" /></p>
<p><img src="/parts/capacitors/ceramicdisccap.gif" alt="mylar capacitors" class="floatright" align="right" width="44" /></p>
<p>The other capacitors pictured are mylar film (the brown and green ones above), box film (yellow), and ceramic disc (light brown).</p>
<p>There are many types of capacitors because there are many ways to make them and each has its advantages.  There are accompanying disadvantages of course.  For stompboxes, important considerations are size and effects on an audio signal.  There is a lot of discussion about the latter, with many different views.  I offer my interpretation of good advice for beginners below under <a href="#values">Values</a>.</p>
<h2 style="clear: both"><a title="symbols" name="symbols"></a>Symbols</h2>
<p><img src="/parts/capacitors/cap_eagle_schem.png" alt="capacitor schem" class="floatleft" align="left" width="130" /></p>
<p>There are also a variety of schematic symbols for capacitors.  Two popular symbols are shown here, one for non-polarized and one for polarized capacitors. Often, the non-polarized symbol is made into a polarized symbol by adding a plus sign to show polarity.  The polarized symbol shown here does not always have its polarity marked, but the curved side is always the negative lead.  There is a nice table of other capacitor schematic symbols on <a href="http://en.wikipedia.org/wiki/Capacitor">the Capacitor page of Wikipedia</a>.</p>
<p>Capacitors are labelled with the capital letter <em>C</em> and a number, like <em>C2</em>.  In the Fuzz Face schematic described in <a href="/wordpress/tools/software/eagle/eagle/">Eagle 1: Description</a>, <em>C1</em> is a polarized capacitor with its positive lead on the input. <em>C2</em> is another polarized capacitor, positive lead going to ground. <em>C3</em> is <em>non-polarized</em>. This circuit has a <em>positive ground</em> (note that the voltage supply has a negative symbol) and this explains the orientation of the polarized capacitors. In a negative ground circuit, they would be reversed.</p>
<p style="text-align: center"><img src="/images/bff-schem.png" alt="[image: Basic Fuzz Face schematic]" title="Basic Fuzz Face schematic" style="margin-right: 5em" height="400" /></p>
<p><img src="/parts/capacitors/cap_eagle_layout.png" alt="capacitor board" class="floatright" align="right" width="130" /> The layout symbols for capacitors are ovals, boxes, circles, and the schematic symbol.</p>
<h2 style="clear: both"><a title="values" name="values"></a>Values</h2>
<p>Capacitor values are called <em>capacitance</em>, which is measured in <em>farad</em> units that are denoted by the capital letter <em>F</em>. In stompbox circuits, the largest capacitances are on the order of <em>10<sup>-6</sup></em> farads and their units are <em>microfarads</em>, denoted by <em>μF</em>. For typing convenience,</p>
<p style="text-align: center"><em><big><strong>μF</strong></big>       and       <big><strong>uF</strong></big>       and        <big><strong>mF</strong></big></em></p>
<p>are equivalent, where <em>u</em> has a similar appearance to <em>μ</em> and <em>m</em> is the Latin character that corresponds to <em>μ</em>. (The <em>mF</em> notation is awkward because <em>m</em> often denotes &#8220;milli&#8221; or &#8220;one-thousandth&#8221; as in <em>mA</em> for <em>milliamps</em> or <em>mm</em> for millimeter. Nevertheless you will see it occasionally, as in the Small Bear catalog.) The smallest capacitances one sees are <em>picofarads</em> or <em>10<sup>-12</sup></em> farads, denoted <em>pF</em>. In between, there are <em>nanofarads</em> (<em>10<sup>-9</sup></em> farads denoted <em>nF</em>). Many schematics avoid <em>nF</em>, writing <em>.01μF</em> instead of <em>10nF</em>. If there are no units for the capacitors on a stompbox schematic, one generally assumes that the units are <em>μF</em>.</p>
<p>Although the notation is usually reserved for resistors, one occasionally sees the decimal point in a capacitor value replaced with the capacitor&#8217;s magnitude. For example, 2.2nF is sometimes noted as 2n2.  Also the F (for farad) is also frequently dropped even when there is a decimal point:  2.2n instead of 2.2nF.</p>
<p>Tolerances are generally <em>±20%</em>, much less accurate than common resistors. As a result, capacitors generally come in fewer values than resistors, but the values are organized in the same way.  Capacitance values proportional to 10, 15, 22, 33, 47, and 68 are quite common.  See the <a href="/wordpress/parts/resistors/resistors-description/#values">Values section of Resistors 1: Description</a> for additional information or <a href="http://en.wikipedia.org/wiki/Capacitor_%28component%29#Standard_values">this Wikipedia entry</a>.</p>
<ul>
<li><strong>ceramic:</strong> often used for small capacitances in the <em>1pF</em> to <em>1000pF</em> range.</li>
<li><strong>polarized electrolytic:</strong> typically appear in power supply filters with values <em>10μF</em> and higher.</li>
<li><strong>film:</strong> come in various kinds and their values cover a large range, say <em>1,000pF</em> to several <em>μF</em>.</li>
<li><strong>mica:</strong> used for small capacitances like ceramic capacitors.</li>
<li><strong>tantalum:</strong> polarized and used in the signal path for their character.  For examples, see the schematic of <a href="/wordpress/circuits/tubescreamer" title="Tubescreamer">the tubescreamer clone</a> or the schematic of <a href="/wordpress/circuits/red-llama" title="Red Llama">the Red Llama clone</a>.</li>
</ul>
<h2><a title="ratings" name="ratings"></a>Ratings</h2>
<p>Capacitors are rated for the voltage potential across their leads. <em>16V</em> are typically used for <em>9V</em> stompbox circuits.  Move up to <em>25V</em> for an <em>18V</em> power supply.  Many builders just buy <em>25V</em> capacitors because they can be used in either case.  Some capacitor values are not available without moving up to higher voltage ratings like <em>50V</em>. There is no apparent problem with using such higher voltage ratings in stompbox circuits.</p>
<h2><a title="moreinfo" name="moreinfo"></a>Further Information</h2>
<p>The Wikipedia has (at least) two helpful pages for general reference: <a href="http://en.wikipedia.org/wiki/Capacitor">Capacitor</a> and <a href="http://en.wikipedia.org/wiki/Capacitor_%28component%29">Capacitor (component)</a>.</p>
<p>To learn more about types of capacitors and their audio properties, read <a href="http://www.muzique.com/cap_faq.htm">&#8220;the Cap FAQ on muzique.com&#8221;</a>. There is much discussion about the special character of certain capacitors and <a href="http://www.diystompboxes.com/smfforum/index.php">Aron&#8217;s stompbox forum</a> is a good place to get a sampling.</p>
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