{"id":655,"date":"2018-04-12T12:50:30","date_gmt":"2018-04-12T12:50:30","guid":{"rendered":"https:\/\/www.tina.com\/resources\/?page_id=655"},"modified":"2018-07-04T22:11:36","modified_gmt":"2018-07-04T22:11:36","slug":"7-other-op-amp-applications","status":"publish","type":"page","link":"https:\/\/www.tina.com\/resources\/home\/ideal-operational-amplifiers-2\/7-other-op-amp-applications\/","title":{"rendered":"7. Other Op-amp Applications"},"content":{"rendered":"<div class=\"top-navigation\">\n<div class=\"current-page\"><img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/people-location-30.png\" \/>CURRENT &#8211; 7. Other op-amp applications<\/div>\n<div class=\"previous-page\"><a href=\"https:\/\/www.tina.com\/resources\/home\/6-design-of-op-amp-circuits\/\"><img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/leftarrow5.png\" \/>PREVIOUS- 6. Design of op-amp circuits<\/a><\/div>\n<div class=\"next-page\"><a href=\"https:\/\/www.tina.com\/resources\/1-ideal-op-amps\/\">NEXT- 1. Ideal op-amps<img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/rightarrow5.png\" \/><\/a><\/div>\n<div class=\"clear\"><\/div>\n<\/div>\n<p><script>\nfunction setStyle(){ \nheight = document.getElementsByClassName('entry-header')[0].clientHeight;\ndocument.getElementsByClassName('entry-header')[0].style.marginTop = height +\"px\";\ndocument.getElementsByClassName('top-navigation')[0].style.top = Math.round(-(height-20)*2) +\"px\";\n}\nsetStyle();\nwindow.onresize = function(event) {\nsetStyle();\n};\n<\/script><\/p>\n<p align=\"justify\"><strong><em><span style=\"font-size: 22px;\">Other op-amp applications<\/span><\/em><\/strong><\/p>\n<p>We have seen that the op-amp can be used as an amplifier, or as a means of combining a number of inputs in a linear manner. We now investigate several additional important applications of this versatile linear IC.<\/p>\n<h6><em> <span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">7.1 Negative Impedance Circuit<\/span><\/b><\/span><\/em><\/h6>\n<div style=\"clear: both;\">\n<div id=\"attachment_659\" style=\"width: 267px\" class=\"wp-caption alignnone\"><a href=\"https:\/\/www.youtube.com\/watch?v=JfeUb-gbMBo&amp;list=PLScrHAmLYSQC8NWGyG6B8UtK9Ju2BYBvW&amp;index=1&amp;t=1s\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-659\" class=\"wp-image-659 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/OpampFigure2-17.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"257\" height=\"211\" \/><\/a><p id=\"caption-attachment-659\" class=\"wp-caption-text\">Figure 17 Negative Impedance Circuit<\/p><\/div>\n<p>The circuit shown in Figure (17) produces a negative input resistance (impedance in the general case).<\/p>\n<p>This circuit can be used to cancel an unwanted positive resistance. Many oscillator applications depend on a negative resistance op-amp circuit. The input resistance, <em>R<sub>in<\/sub><\/em>, is the ratio of input voltage to current.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-657\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Op-amp2-43.png\" alt=\"\" width=\"85\" height=\"45\" \/><br \/>\n(43)<\/p>\n<p>A voltage divider relationship is used to derive the expression for <em>v<sub>&#8211; <\/sub><\/em>since the current into the op-amp is zero.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-661\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-44.png\" alt=\"\" width=\"102\" height=\"44\" \/><br \/>\n(44)<\/p>\n<p>We now let <em>v<sub>+<\/sub> = v<\/em><sub>&#8211;<\/sub> and solve for <em>v<sub>out<\/sub><\/em> in terms of <em>v<sub>in<\/sub><\/em>, which yields,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-677 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-45.png\" alt=\"\" width=\"106\" height=\"48\" \/><br \/>\n(45)<\/p>\n<p>Since the input impedance to the <em>v<sub>+<\/sub><\/em> terminal is infinite, the current in <em>R<\/em> is equal to <em>i<sub>in<\/sub><\/em> and can be found as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-678\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-46.png\" alt=\"\" width=\"138\" height=\"44\" \/><br \/>\n(46)<\/p>\n<p>The input resistance, <em>R<sub>in<\/sub><\/em>, is then given by<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-679\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-47.png\" alt=\"\" width=\"106\" height=\"44\" \/><br \/>\n(47)<\/p>\n<p>Equation (47) shows that the circuit of Figure (17) develops a negative resistance. If <em>R<\/em> is replaced by an impedance, <em>Z<\/em>, the circuit develops a negative impedance.<\/p>\n<div>\n<p><em><strong><u>APPLICATION<\/u><\/strong><\/em><\/p>\n<p>Analyze the following circuit online with the TINACloud circuit simulator by clicking the link below.<\/p>\n<p><a href=\"https:\/\/www.tinacloud.com\/tinademo\/tina.php?c=5afeb2f75f394:899866&amp;amp;act;act=&amp;amp;noad&amp;amp;inframe=https%3A%2F%2Fwww.tina.com\" target=\"_blank\" rel=\"noopener\"><em><strong><span style=\"color: #466ce0;\"><em><strong>1- Negative Impedance Circuit Simulation <\/strong><\/em><\/span><\/strong><\/em><\/a><\/p>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">7.2 Dependent-Current Generator<\/span><\/b><\/span><\/em><\/h6>\n<div>A dependent-current generator produces a load current which is proportional to an applied voltage, <em>v<sub>in<\/sub><\/em>, and is independent of the load resistance. It can be designed using a slight modification of the negative-impedance circuit. The circuit is shown in Figure 18 (a).<\/div>\n<div id=\"attachment_680\" style=\"width: 545px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-680\" class=\"wp-image-680 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-18.png\" alt=\"\" width=\"535\" height=\"196\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-18.png 535w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-18-300x110.png 300w\" sizes=\"auto, (max-width: 535px) 100vw, 535px\" \/><p id=\"caption-attachment-680\" class=\"wp-caption-text\">Figure 18 \u2013 Dependent current generator<\/p><\/div>\n<p>Suppose we let <em>R<sub>F<\/sub> = R<sub>A<\/sub><\/em>. Equation (47)then indicates that the input resistance to the op-amp circuit (enclosed in the dashed box) is <em>-R<\/em>. The input circuit can then be simplified as shown in Figure 18(b). We wish to calculate <em>i<sub>load<\/sub><\/em>, the current in <em>R<sub>load<\/sub><\/em>. Although the resistance is negative, the normal Kirchhoff&#8217;s laws still apply since nothing in their derivations assumes positive resistors. The input current, <em>i<sub>in<\/sub><\/em>, is then found by combining the resistances into a single resistor, <em>R<sub>in<\/sub><\/em>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-681 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-48.png\" alt=\"\" width=\"309\" height=\"48\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-48.png 309w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-48-300x48.png 300w\" sizes=\"auto, (max-width: 309px) 100vw, 309px\" \/><br \/>\n(48)<\/p>\n<p>We then apply a current-divider ratio to the current split between <em>R<sub>load<\/sub> and -R to <\/em>obtain<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-682 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-49.png\" alt=\"\" width=\"281\" height=\"46\" \/><br \/>\n(49)<\/p>\n<p>Thus the effect of the addition of the op-amp circuit is to make the current in the load proportional to the input voltage. It does not depend upon the value of the load resistance, <em>R<sub>load<\/sub><\/em>. The current is therefore independent of changes in the load resistance. The op-amp circuit effectively cancels out the load resistance. Since the current is independent of the load but depends only upon the input voltage, we call this a <em>current generator<\/em> (or voltage-to-current converter).<\/p>\n<p>Among the many applications of this circuit is a <em>dc<\/em> regulated voltage source. If we let <em>v<sub>in<\/sub> = E<\/em> (a constant), the current through <em>R<sub>load<\/sub><\/em> is constant independent of variations of <em>R<sub>load<\/sub><\/em>.<\/p>\n<p><em><strong><u>APPLICATION<\/u><\/strong><\/em><\/p>\n<p>Analyze the following circuit online with the TINACloud circuit simulator by clicking the link below.<\/p>\n<p><a href=\"https:\/\/www.tinacloud.com\/tinademo\/tina.php?c=5b05471d18dd9:231049&amp;amp;act;act=&amp;amp;noad&amp;amp;inframe=https%3A%2F%2Fwww.tina.com\" target=\"_blank\" rel=\"noopener\"><em><strong><span style=\"color: #466ce0;\"><em><strong>2- Dependent Current Generator Circuit Simulation <\/strong><\/em><\/span><\/strong><\/em><\/a><\/p>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">7.3 Current-to-Voltage Converter<\/span><\/b><\/span><\/em><\/h6>\n<div><\/div>\n<div id=\"attachment_684\" style=\"width: 277px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-684\" class=\"wp-image-684 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-19.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"267\" height=\"192\" \/><p id=\"caption-attachment-684\" class=\"wp-caption-text\">Figure 19 &#8211; Current -to-Voltage converter<\/p><\/div>\n<p>The circuit of Figure (19)\u00a0 produces an output voltage that is proportional to the input current (this can also be viewed as a <em>unity-gain inverting amplifier<\/em>). We analyze this circuit in using the properties of ideal op-amps.\u00a0 We solve for the voltages at the input terminals to find<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-683\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-50.png\" alt=\"\" width=\"136\" height=\"45\" \/><br \/>\n(50)<\/p>\n<p>Hence, the output voltage, <em>v<sub>out<\/sub> = -i<sub>in<\/sub>R<\/em>, is proportional to the input current, <em>i<sub>in<\/sub><\/em>.<\/p>\n<p><em><strong><u>APPLICATION<\/u><\/strong><\/em><\/p>\n<p>Analyze the following circuit online with the TINACloud circuit simulator by clicking the link below.<\/p>\n<p><a href=\"https:\/\/www.tinacloud.com\/tinademo\/tina.php?c=5afec31e3a7b0:683283&amp;amp;act;act=&amp;amp;noad&amp;amp;inframe=https%3A%2F%2Fwww.tina.com\" target=\"_blank\" rel=\"noopener\"><em><strong><span style=\"color: #466ce0;\"><em><strong>3- Current to Voltage Converter Circuit Simulation <\/strong><\/em><\/span><\/strong><\/em><\/a><\/p>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">7.4 Voltage-to-Current Converter<\/span><\/b><\/span><\/em><\/h6>\n<div><\/div>\n<div id=\"attachment_686\" style=\"width: 265px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-686\" class=\"wp-image-686 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-20.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"255\" height=\"177\" \/><p id=\"caption-attachment-686\" class=\"wp-caption-text\">Figure 20 \u2013 Voltage to current converter<\/p><\/div>\n<p>The circuit of Figure (20), is a voltage-to-current converter. We analyze this circuit as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-687\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-51.png\" alt=\"\" width=\"92\" height=\"45\" \/><br \/>\n(51)<\/p>\n<p>From Equation (51) we find,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-688\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-52.png\" alt=\"\" width=\"64\" height=\"46\" \/><br \/>\n(52)<\/p>\n<p>Therefore, the load current is independent of the load resistor, <em>R<sub>load<\/sub><\/em>, and is proportional to the applied voltage, <em>v<sub>in<\/sub><\/em>. This circuit develops a voltage-controlled current source. However, a practical shortcoming of this circuit is that neither end of the load resistor can be grounded.<\/p>\n<div>As an alternative, the circuit shown in Figure (21) provides a voltage-to-current converter with one end of the load resistance grounded.<\/div>\n<div id=\"attachment_689\" style=\"width: 300px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-689\" class=\"wp-image-689 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-21.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"290\" height=\"216\" \/><p id=\"caption-attachment-689\" class=\"wp-caption-text\">Figure 21 \u2013 Voltage-to-current converter<\/p><\/div>\n<p>We analyze this circuit by writing node equations as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-690\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-53.png\" alt=\"\" width=\"236\" height=\"86\" \/><br \/>\n(53)<\/p>\n<p>The last equality uses the fact that <em>v<sub>+<\/sub> = v<sub>&#8211;<\/sub><\/em>. There are five unknowns in these equations (<em>v<sub>+<\/sub><\/em>, <em>v<sub>in<\/sub><\/em>, <em>v<sub>out<\/sub><\/em>, <em>v<\/em>, and <em>i<sub>load<\/sub><\/em>). We eliminate <em>v<sub>+<\/sub><\/em> and <em>v<sub>out<\/sub><\/em> to obtain,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-691\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-54.png\" alt=\"\" width=\"105\" height=\"44\" \/><br \/>\n(54)<\/p>\n<p>The load current, <em>i<sub>load<\/sub><\/em>, is independent of the load, <em>R<sub>load<\/sub><\/em>, and is only a function of the voltage difference, (<em>v<sub>in<\/sub> &#8211; v<\/em>).<\/p>\n<p><em><strong><u>APPLICATION<\/u><\/strong><\/em><\/p>\n<p>Analyze the following circuit online with the TINACloud circuit simulator by clicking the link below.<\/p>\n<p><a href=\"https:\/\/www.tinacloud.com\/tinademo\/tina.php?c=5b065cfe233af:742756&amp;amp;act;act=&amp;amp;noad&amp;amp;inframe=https%3A%2F%2Fwww.tina.com\" target=\"_blank\" rel=\"noopener\"><em><strong><span style=\"color: #466ce0;\"><em><strong>4-Voltage to Current Converter Circuit Simulation <\/strong><\/em><\/span><\/strong><\/em><\/a><\/p>\n<h6><em> <span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">7.5 Inverting Amplifier with Generalized Impedances<\/span><\/b><\/span><\/em><\/h6>\n<div><\/div>\n<div id=\"attachment_692\" style=\"width: 544px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-692\" class=\"wp-image-692 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-22.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"534\" height=\"242\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-22.png 534w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-22-300x136.png 300w\" sizes=\"auto, (max-width: 534px) 100vw, 534px\" \/><p id=\"caption-attachment-692\" class=\"wp-caption-text\">Figure 22 \u2013 Use of generalized impedance in place of resistance<\/p><\/div>\n<p>The relationship of Equation (17) is easily extended to include non-resistive components if <em>R<sub>j<\/sub><\/em> is replaced by an impedance, <em>Z<sub>j<\/sub><\/em>, and <em>R<sub>F<\/sub><\/em> is replaced by <em>Z<sub>F<\/sub><\/em>. For a single input, as shown in Figure 22 (a), the output reduces to<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-693\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-55.png\" alt=\"\" width=\"86\" height=\"44\" \/><br \/>\n(55)<\/p>\n<p>Since we are dealing in the frequency domain, we use uppercase letters for the voltages and currents, thus representing the <em>complex amplitudes<\/em>.<\/p>\n<p>One useful circuit based on Equation (55) is the <em>Miller integrator<\/em>, as shown in Figure 22(b). In this application, the feedback component is a capacitor, <em>C<\/em>, and the input component is a resistor, <em>R<\/em>, so<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-694\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-56.png\" alt=\"\" width=\"57\" height=\"40\" \/><br \/>\n(56)<\/p>\n<p>In Equation (56) , <em>s<\/em>\u00a0 is the Laplace transform operator. For sinusoidal signals,\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-695\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2sinosodialsignal.png\" alt=\"\" width=\"47\" height=\"28\" \/> . When we substitute these impedances into Equation (55) , we obtain<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-696\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-57.png\" alt=\"\" width=\"92\" height=\"44\" \/><br \/>\n(57)<\/p>\n<p>In the complex frequency domain, <em>1\/s<\/em> corresponds to integration in the time domain. This is an <em>inverting integrator<\/em> because the expression contains a negative sign. Hence the output voltage is<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-697\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-58.png\" alt=\"\" width=\"204\" height=\"46\" \/><br \/>\n(58)<\/p>\n<p>where <em>v<sub>out<\/sub><\/em>(0) is the initial condition. The value of <em>v<sub>out<\/sub><\/em> is developed as the voltage across the capacitor, <em>C<\/em>, at time <em>t = <\/em>0. The switch is closed to charge the capacitor to the voltage <em>v<sub>out<\/sub><\/em>(0) and then at <em>t = <\/em>0 the switch is open. We use electronic switches, which we discuss more fully in Chapter 16. In the event that the initial condition is zero, the switch is still used to reset the integrator to zero output voltage at time <em>t = <\/em>0.<\/p>\n<div id=\"attachment_841\" style=\"width: 246px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-841\" class=\"wp-image-841 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-23.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"236\" height=\"185\" \/><p id=\"caption-attachment-841\" class=\"wp-caption-text\">Figure 23 &#8211; Example of an inverting differentiator<\/p><\/div>\n<p>If the feedback element is a resistor, and the input element is a capacitor, as shown in Figure (23), the input-output relationship becomes<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-698\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-59.png\" alt=\"\" width=\"85\" height=\"44\" \/><br \/>\n(59)<\/p>\n<p>In the time domain, this becomes<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-699\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-60.png\" alt=\"\" width=\"112\" height=\"40\" \/><br \/>\n(60)<br \/>\n<em><strong><u>APPLICATION<\/u><\/strong><\/em><\/p>\n<p>Analyze the following circuit online with the TINACloud circuit simulator by clicking the link below.<\/p>\n<p><a href=\"https:\/\/www.tinacloud.com\/tinademo\/tina.php?c=5b07ff91d9cfb:453821&amp;amp;act;act=&amp;amp;noad&amp;amp;inframe=https%3A%2F%2Fwww.tina.com\" target=\"_blank\" rel=\"noopener\"><em><strong><span style=\"color: #466ce0;\"><em><strong>5- Example of an inverting differentiator Circuit Simulation <\/strong><\/em><\/span><\/strong><\/em><\/a><\/p>\n<p>The circuit is operating as an <em>inverting differentiator<\/em>. Note that the input capacitor, <em>Z<sub>a<\/sub> = 1\/sC<\/em>, does not provide a path for <em>dc<\/em>. This does not affect the result since the derivative of a constant is zero. For simplicity, let\u2019s use a sinusoidal input signal. Rearranging Equation (59) and substituting the numeric values for this circuit, we obtain<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-700\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-61.png\" alt=\"\" width=\"258\" height=\"93\" \/><br \/>\n(61)<\/p>\n<p>The input voltage is inverted (180\u00b0 shift) by this circuit and then scaled and shifted again (90\u00b0 by the <em>j<\/em>-operator) by the value of <em>RCs <\/em>where\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-837\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-RC2.png\" alt=\"\" width=\"158\" height=\"21\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-RC2.png 158w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-RC2-150x21.png 150w\" sizes=\"auto, (max-width: 158px) 100vw, 158px\" \/>.<\/p>\n<p>Results of the simulation are shown in Figure (24).<\/p>\n<div id=\"attachment_847\" style=\"width: 583px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-847\" class=\"wp-image-847 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-24.png\" alt=\"\" width=\"573\" height=\"230\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-24.png 573w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-24-300x120.png 300w\" sizes=\"auto, (max-width: 573px) 100vw, 573px\" \/><p id=\"caption-attachment-847\" class=\"wp-caption-text\">Figure 24 \u2013 Simulation results for inverting differentiator<\/p><\/div>\n<p>The input waveform peaks at 0.5 volts. The output voltage has a net shift (delay) of 90 degrees and the output voltage peaks at approximately 0.314 volts. This is in good agreement with the result of Equation (61) .<\/p>\n<p>We may also use the waveforms to show that this circuit performs the task of an inverting differentiator. We will confirm that the output waveform represents the slope of the input signal times a constant. The constant is the voltage gain of the circuit. The greatest rate of change of the input voltage waveform occurs at its zero-crossing. This corresponds with the time that the output waveform reaches its maximum (or minimum). Picking a representative point, say at time0.5 ms, and using graphical techniques, we compute the slope of the input voltage waveform as<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-850\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-62-300x35.png\" alt=\"\" width=\"300\" height=\"35\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-62-300x35.png 300w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-62.png 340w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\n(62)<\/p>\n<p>Scaling this rate of change (i.e.,<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-851\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2Scaling.png\" alt=\"\" width=\"32\" height=\"40\" \/> ) by the circuit voltage gain according to Equation (60)we expect the peak output voltage to be<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-853\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-63.png\" alt=\"\" width=\"262\" height=\"90\" \/><br \/>\n(63)<\/p>\n<h6><em> <span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">7.6 Analog Computer Applications<\/span><\/b><\/span><\/em><\/h6>\n<p>In this section we present the use of interconnected op-amp circuits, such as summers and integrators, to form an analog computer which is used to solve differential equations. Many physical systems are described by linear differential equations, and the system can therefore be analyzed with the aid of an analog computer.<\/p>\n<div id=\"attachment_860\" style=\"width: 292px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-860\" class=\"wp-image-860 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-25.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"282\" height=\"147\" \/><p id=\"caption-attachment-860\" class=\"wp-caption-text\">Figure 25 &#8211; Analog computer application<\/p><\/div>\n<p>Let us solve for the current, i(t), in the circuit of Figure 25. The input voltage is the driving function and the initial conditions are zero. We write the differential equation for the circuit as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-865\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-64.png\" alt=\"\" width=\"173\" height=\"46\" \/><br \/>\n(64)<\/p>\n<p>Now solving for di\/dt, we obtain<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-867\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-65.png\" alt=\"\" width=\"176\" height=\"46\" \/><br \/>\n(65)<\/p>\n<p>We know that for t &gt; 0,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-869\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-66.png\" alt=\"\" width=\"153\" height=\"46\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-66.png 153w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-66-150x46.png 150w\" sizes=\"auto, (max-width: 153px) 100vw, 153px\" \/><br \/>\n(66)<\/p>\n<p>From Equation (65) we see that -di\/dt is formed by summing three terms, which are found on Figure 26 at the input to the first integrating amplifier.<\/p>\n<div id=\"attachment_873\" style=\"width: 545px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-873\" class=\"wp-image-873 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-26.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"535\" height=\"225\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-26.png 535w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-26-300x126.png 300w\" sizes=\"auto, (max-width: 535px) 100vw, 535px\" \/><p id=\"caption-attachment-873\" class=\"wp-caption-text\">Figure 26 &#8211; Analog computer solution for Figure 25<\/p><\/div>\n<p>The three terms are found as follows:<\/p>\n<p>1. The driving function, -v(t)\/L, is formed by passing v(t) through an inverting summer (Summer) with gain, 1\/L.<br \/>\n2. Ri\/L is formed by taking the output of the first integrating amplifier (Integrator 1) and adding it at the amplifier input to the output of the summing amplifier (Summer).<br \/>\n3. The term<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-876\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-67.png\" alt=\"\" width=\"86\" height=\"46\" \/><br \/>\n(67)<br \/>\nis the output of the second integrator (Integrator 2). Since the sign must be changed, we sum it with the unity gain inverting summer (Summer).<br \/>\nThe output of the first integrator is +i, as seen from Equation (66). The constants in the differential equation are established by proper selection of the resistors and capacitors of the analog computer. Zero initial conditions are accomplished by switches across the capacitors, as shown in Figure 22(b).<\/p>\n<h6><em> <span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">7.7 Non-Inverting Miller Integrator<\/span><\/b><\/span><\/em><\/h6>\n<div id=\"attachment_879\" style=\"width: 289px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-879\" class=\"wp-image-879 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opampFigure2-27.png\" alt=\"other op-amp applications, circuit simulation, circuit simulator, circuit design\" width=\"279\" height=\"211\" \/><p id=\"caption-attachment-879\" class=\"wp-caption-text\">Figure 27 \u2013 Non-inverting integrator<\/p><\/div>\n<p>We use a modification of the dependent current generator of the previous section to develop a non-inverting integrator. The circuit is configured as shown in Figure 27.<br \/>\nThis is similar to the circuit of Figure 21, but the load resistance has been replaced by a capacitance. We now find the current, Iload. The inverting voltage, V-, is found from the voltage division between Vo and V- as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-888\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-68.png\" alt=\"\" width=\"173\" height=\"80\" \/><br \/>\n(68)<\/p>\n<p>Since V+ = V-, we solve and find<br \/>\nIL = Vin \/R. Note that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-889\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-69.png\" alt=\"\" width=\"168\" height=\"40\" \/><br \/>\n(69)<\/p>\n<p>where s is the Laplace transform operator. The Vout \/Vin function is then<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-891\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-70.png\" alt=\"\" width=\"82\" height=\"44\" \/><br \/>\n(70)<\/p>\n<p>Thus, in the time domain we have<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-893\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/opamp2-71.png\" alt=\"\" width=\"141\" height=\"46\" \/><br \/>\n(71)<\/p>\n<p>The circuit is therefore a non-inverting integrator.<\/p>\n<p><em><strong><u>APPLICATION<\/u><\/strong><\/em><\/p>\n<p>Analyze the following circuit online with the TINACloud circuit simulator by clicking the link below.<\/p>\n<p><a href=\"https:\/\/www.tinacloud.com\/tinademo\/tina.php?c=5b06a89d14673:968481&amp;amp;act;act=&amp;amp;noad&amp;amp;inframe=https%3A%2F%2Fwww.tina.com\" target=\"_blank\" rel=\"noopener\"><em><strong><span style=\"color: #466ce0;\"><em><strong>6-Non-inverting integrator Circuit Simulation <\/strong><\/em><\/span><\/strong><\/em><\/a><\/p>\n<div><\/div>\n<p>&nbsp;<\/p>\n<h2><strong>SUMMARY<\/strong><\/h2>\n<p>The operational amplifier is a very useful building block for electronic systems. The real amplifier operates almost as an ideal amplifier with very high gain and almost infinite input impedance. For this reason, we can treat it in the same way we treat circuit components. That is, we are able to incorporate the amplifier into useful configurations prior to studying the internal operation and the electronic characteristics. By recognizing the terminal characteristics, we are able to configure amplifiers and other useful circuits.<br \/>\nThis chapter began with an analysis of the ideal operational amplifier, and with development of equivalent circuit models using dependent sources. The dependent sources we studied early in this chapter form the building blocks of equivalent circuits for many of the electronic devices we study in this text.<br \/>\nWe then explored the external connections needed to make the op-amp into an inverting amplifier, a non-inverting amplifier, and a multiple input amplifier. We developed a convenient design technique eliminating the need for solving large systems of simultaneous equations.<br \/>\nFinally, we saw how the op-amp could be used to build a variety of more complex circuits, including circuits which are equivalent to negative impedances (which can be used to cancel the effects of positive impedances), integrators and differentiators.<\/p>\n<div class=\"previous-page\"><a href=\"https:\/\/www.tina.com\/resources\/home\/6-design-of-op-amp-circuits\/\"><img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/leftarrow5.png\" \/> <span style=\"font-size: 20px;\">PREVIOUS- 6. DESIGN OF OP-AMP CIRCUITS<\/span><\/a><\/div>\n<div class=\"next-page\"><a style=\"font-size: 20px;\" href=\"https:\/\/www.tina.com\/resources\/1-ideal-op-amps\/\">NEXT- 1. Ideal op-amps <img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/rightarrow5.png\" \/><\/a><\/div>\n<div class=\"clear\"><\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>CURRENT &#8211; 7. Other op-amp applications PREVIOUS- 6. Design of op-amp circuits NEXT- 1. Ideal op-amps Other op-amp applications We have seen that the op-amp can be used as an amplifier, or as a means of combining a number of inputs in a linear manner. We now investigate several additional important applications of this versatile [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1357,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"yst_prominent_words":[216,325,575,578,582,70,326,172,249,332,580,576,238,327,581,577,329,579,574,32],"class_list":["post-655","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/655","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/comments?post=655"}],"version-history":[{"count":201,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/655\/revisions"}],"predecessor-version":[{"id":3596,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/655\/revisions\/3596"}],"up":[{"embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/1357"}],"wp:attachment":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/media?parent=655"}],"wp:term":[{"taxonomy":"yst_prominent_words","embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/yst_prominent_words?post=655"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}