{"id":1435,"date":"2018-04-20T10:56:56","date_gmt":"2018-04-20T10:56:56","guid":{"rendered":"https:\/\/www.tina.com\/resources\/?page_id=1435"},"modified":"2018-06-21T11:05:57","modified_gmt":"2018-06-21T11:05:57","slug":"8-inverting-amplifier","status":"publish","type":"page","link":"https:\/\/www.tina.com\/resources\/home\/practical-operational-amplifiers\/8-inverting-amplifier\/","title":{"rendered":"8. Inverting Amplifier"},"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; 8. Inverting Amplifier<\/div>\n<div class=\"previous-page\"><a href=\"https:\/\/www.tina.com\/resources\/home\/7-non-inverting-amplifier\/\"><img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/leftarrow5.png\" \/>PREVIOUS- 7. Non-inverting amplifier<\/a><\/div>\n<div class=\"next-page\"><a href=\"https:\/\/www.tina.com\/resources\/practical-operational-amplifiers\/9-differential-summing\/\">NEXT- 9. Differential summing<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><br \/>\nFigure 36(a) illustrates an inverting amplifier. Figure 36(b) shows the equivalent circuit using the op-amp model developed earlier in this chapter.<\/p>\n<div id=\"attachment_1444\" style=\"width: 533px\" class=\"wp-caption alignnone\"><a href=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-36.png\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1444\" class=\"wp-image-1444 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-36.png\" alt=\"Inverting amplifier\" width=\"523\" height=\"205\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-36.png 523w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-36-300x118.png 300w\" sizes=\"auto, (max-width: 523px) 100vw, 523px\" \/><\/a><p id=\"caption-attachment-1444\" class=\"wp-caption-text\">Figure 36 &#8211; Inverting amplifier<\/p><\/div>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">8.1 Input and Output Resistance<\/span><\/b><\/span><\/em><\/h6>\n<p>Figure 36(b) is reduced to Figure 37(a) if we let,\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1447\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-37equation.png\" alt=\"\" width=\"245\" height=\"28\" \/><\/p>\n<div id=\"attachment_1449\" style=\"width: 535px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1449\" class=\"wp-image-1449 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-37.png\" alt=\"Simplified inverting amplifier model\" width=\"525\" height=\"210\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-37.png 525w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-37-300x120.png 300w\" sizes=\"auto, (max-width: 525px) 100vw, 525px\" \/><p id=\"caption-attachment-1449\" class=\"wp-caption-text\">Figure 37 \u2013 Simplified inverting amplifier model<\/p><\/div>\n<p>It is reasonable to assume that these inequalities apply because, were they not true, the output would load the input and the gain would be reduced.<\/p>\n<p>A voltage-divider relationship can be used to yield<\/p>\n<div id=\"attachment_1451\" style=\"width: 104px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1451\" class=\"wp-image-1451 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-71.png\" alt=\"\" width=\"94\" height=\"42\" \/><p id=\"caption-attachment-1451\" class=\"wp-caption-text\">(71)<\/p><\/div>\n<p>and a loop equation yields<\/p>\n<div id=\"attachment_1452\" style=\"width: 310px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1452\" class=\"wp-image-1452 size-medium\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-72-300x41.png\" alt=\"\" width=\"300\" height=\"41\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-72-300x41.png 300w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-72.png 333w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><p id=\"caption-attachment-1452\" class=\"wp-caption-text\">(72)<\/p><\/div>\n<p>The input resistance, <em>R<sub>in<\/sub><\/em>, is obtained from Figure 37(b), where we have replaced the dependent source with an equivalent resistance. The value of this resistor is <em>v<sub>&#8211;<\/sub><\/em>\/<em>i&#8221;<\/em> which is found from Equation (72) . For large <em>G <\/em>(i.e., <img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1455\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-Equation37-1.png\" alt=\"\" width=\"68\" height=\"30\" \/>), the rightmost resistance in Figure 37(b) is approximately zero, and\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1456\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-Equation37-2.png\" alt=\"\" width=\"59\" height=\"26\" \/>.<\/p>\n<p>The output resistance of the inverting amplifier is the same as that of the non-inverting amplifier. Thus,<\/p>\n<div id=\"attachment_1457\" style=\"width: 146px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1457\" class=\"wp-image-1457 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-73.png\" alt=\"\" width=\"136\" height=\"46\" \/><p id=\"caption-attachment-1457\" class=\"wp-caption-text\">(73)<\/p><\/div>\n<p>&nbsp;<\/p>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">8.2 Voltage Gain<\/span><\/b><\/span><\/em><\/h6>\n<p>We use the equivalent circuits of Figure 36(b) and Figure 37(a) to determine the voltage gain. The inverting input gain, <em>A<sub>&#8211; <\/sub><\/em>= <em>v<sub>out<\/sub><\/em>\/<em>v<sub>in<\/sub><\/em>, is obtained from the circuit of Figure 37(a) by again making the same assumptions that we made in finding the output resistance.<\/p>\n<p>These assumptions reduce the circuit to that shown in Figure 38(a), where we have changed the voltage source in series with a resistance to a current source in parallel with a resistance. The resistors can then be combined to yield the circuit of Figure 38(b). Finally, the current source is converted back to the voltage source to yield the simplified circuit of Figure 38(c).<\/p>\n<p>The loop equation for this circuit is given by<\/p>\n<div id=\"attachment_1461\" style=\"width: 310px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1461\" class=\"wp-image-1461 size-medium\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-74-300x42.png\" alt=\"\" width=\"300\" height=\"42\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-74-300x42.png 300w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-74.png 325w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><p id=\"caption-attachment-1461\" class=\"wp-caption-text\">(74)<\/p><\/div>\n<p>&nbsp;<\/p>\n<p>Since <em>v<sub>out<\/sub><\/em> = <em>G<sub>o<\/sub>v<sub>d<\/sub><\/em>, the inverting voltage gain is<\/p>\n<div id=\"attachment_1463\" style=\"width: 310px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1463\" class=\"wp-image-1463 size-medium\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-75-300x44.png\" alt=\"\" width=\"300\" height=\"44\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-75-300x44.png 300w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-75.png 316w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><p id=\"caption-attachment-1463\" class=\"wp-caption-text\">(75)<\/p><\/div>\n<div id=\"attachment_1467\" style=\"width: 505px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1467\" class=\"wp-image-1467 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-38abc.png\" alt=\"\" width=\"495\" height=\"381\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-38abc.png 495w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-38abc-300x231.png 300w\" sizes=\"auto, (max-width: 495px) 100vw, 495px\" \/><p id=\"caption-attachment-1467\" class=\"wp-caption-text\">Figure 38 (parts a, b, c)- Inverting input gain<\/p><\/div>\n<p>We can verify this result relative to the gain of the ideal op-amp by making the approximations: <em>R<sub>A<\/sub><\/em> &lt;&lt; 2<em>R<sub>cm<\/sub><\/em> and <em>G<\/em> &gt;&gt; 1. Then<\/p>\n<div id=\"attachment_1469\" style=\"width: 153px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1469\" class=\"wp-image-1469 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-76.png\" alt=\"\" width=\"143\" height=\"91\" \/><p id=\"caption-attachment-1469\" class=\"wp-caption-text\">(76)<\/p><\/div>\n<p>This is the same as the result found earlier for the simplified model.<\/p>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">8.3 Multiple-Input Amplifiers<\/span><\/b><\/span><\/em><\/h6>\n<div id=\"attachment_1486\" style=\"width: 231px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1486\" class=\"wp-image-1486 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-ampFigure9-39.png\" alt=\"\" width=\"221\" height=\"258\" \/><p id=\"caption-attachment-1486\" class=\"wp-caption-text\">(39)<\/p><\/div>\n<p>If the voltages <em>v<sub>a<\/sub><\/em>, <em>v<sub>b<\/sub><\/em>, &#8230; , <em>v<sub>m<\/sub><\/em> are applied to the summing junction (inverting input to op-amp) through resistors <em>R<sub>a<\/sub><\/em>, <em>R<sub>b<\/sub><\/em>, &#8230;, <em>R<sub>m<\/sub><\/em>, respectively, as shown in Figure 39, the output voltage is<\/p>\n<div id=\"attachment_1473\" style=\"width: 266px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1473\" class=\"wp-image-1473 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-77.png\" alt=\"\" width=\"256\" height=\"46\" \/><p id=\"caption-attachment-1473\" class=\"wp-caption-text\">(77)<\/p><\/div>\n<p>To achieve bias balance, we choose<\/p>\n<div id=\"attachment_1475\" style=\"width: 152px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1475\" class=\"wp-image-1475 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-78.png\" alt=\"\" width=\"142\" height=\"32\" \/><p id=\"caption-attachment-1475\" class=\"wp-caption-text\">(78)<\/p><\/div>\n<p>Let us define<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1477 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-79.png\" alt=\"\" width=\"121\" height=\"32\" \/><\/p>\n<p><span style=\"font-size: 14px;\">(79)<\/span><\/p>\n<div class=\"mceTemp\">\n<p>The output resistance is then<\/p>\n<\/div>\n<div id=\"attachment_1479\" style=\"width: 140px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1479\" class=\"wp-image-1479 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-80.png\" alt=\"\" width=\"130\" height=\"45\" \/><p id=\"caption-attachment-1479\" class=\"wp-caption-text\">(80)<\/p><\/div>\n<p>Suppose now that only two inputs are used. The output voltage is then<\/p>\n<div id=\"attachment_1481\" style=\"width: 131px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1481\" class=\"wp-image-1481 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-81.png\" alt=\"\" width=\"121\" height=\"42\" \/><p id=\"caption-attachment-1481\" class=\"wp-caption-text\">(81)<\/p><\/div>\n<p>The input resistance at <em>v<sub>a<\/sub><\/em> is approximately equal to <em>R<sub>a<\/sub><\/em>, and the input resistance at <em>v<sub>b<\/sub><\/em> is approximately <em>R<sub>b<\/sub><\/em>. We can make this circuit a unity-gain two-input summer with an output voltage of<\/p>\n<div id=\"attachment_1483\" style=\"width: 124px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1483\" class=\"wp-image-1483 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-82.png\" alt=\"\" width=\"114\" height=\"25\" \/><p id=\"caption-attachment-1483\" class=\"wp-caption-text\">(82)<\/p><\/div>\n<p>by setting <em>R<sub>F<\/sub><\/em> = <em>R<sub>a<\/sub><\/em> = <em>R<sub>b<\/sub><\/em>. The resistance from the non-inverting input terminal to ground is chosen to achieve bias balance. Thus, <em>R<sub>1<\/sub><\/em> = <em>R<sub>F<\/sub><\/em>\/3, and we have<\/p>\n<div id=\"attachment_1485\" style=\"width: 146px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1485\" class=\"wp-image-1485 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-83.png\" alt=\"\" width=\"136\" height=\"45\" \/><p id=\"caption-attachment-1485\" class=\"wp-caption-text\">(83)<\/p><\/div>\n<p>An equal-gain (i.e., not unity) two-input summer is obtained by setting\u00a0\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1492\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-84equation-1.png\" alt=\"\" width=\"79\" height=\"21\" \/>\u00a0and <img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1490\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-84equation2.png\" alt=\"\" width=\"96\" height=\"28\" \/>. In this case, the output voltage is<\/p>\n<div id=\"attachment_1494\" style=\"width: 140px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1494\" class=\"wp-image-1494 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-84.png\" alt=\"\" width=\"130\" height=\"44\" \/><p id=\"caption-attachment-1494\" class=\"wp-caption-text\">(84)<\/p><\/div>\n<p>The input resistance is approximately <em>R<\/em>. Since <em>R<sub>A<\/sub><\/em> = <em>R<\/em>\/2,<\/p>\n<div id=\"attachment_1497\" style=\"width: 114px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1497\" class=\"wp-image-1497 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-85.png\" alt=\"\" width=\"104\" height=\"42\" \/><p id=\"caption-attachment-1497\" class=\"wp-caption-text\">(85)<\/p><\/div>\n<p>If <em>m<\/em> inputs are summed through equal resistors (say <em>R<\/em>), the output voltage is<\/p>\n<div id=\"attachment_1499\" style=\"width: 108px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1499\" class=\"wp-image-1499 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-86.png\" alt=\"\" width=\"98\" height=\"46\" \/><p id=\"caption-attachment-1499\" class=\"wp-caption-text\">(86)<\/p><\/div>\n<p>For this equal-gain multiple-input inverting summer, the input resistance to each input is approximately <em>R<\/em>. Since <em>R<sub>A <\/sub><\/em>= <em>R<\/em>\/<em>m<\/em>,<\/p>\n<div id=\"attachment_1501\" style=\"width: 124px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1501\" class=\"wp-image-1501 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-87.png\" alt=\"\" width=\"114\" height=\"42\" \/><p id=\"caption-attachment-1501\" class=\"wp-caption-text\">(87)<\/p><\/div>\n<p>and<\/p>\n<div id=\"attachment_1503\" style=\"width: 89px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1503\" class=\"wp-image-1503 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-88.png\" alt=\"\" width=\"79\" height=\"44\" \/><p id=\"caption-attachment-1503\" class=\"wp-caption-text\">(88)<\/p><\/div>\n<p>The output resistance is<\/p>\n<div id=\"attachment_1506\" style=\"width: 143px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1506\" class=\"wp-image-1506 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/04\/Practical-op-amp9-89.png\" alt=\"\" width=\"133\" height=\"44\" \/><p id=\"caption-attachment-1506\" class=\"wp-caption-text\">(89)<\/p><\/div>\n<h6><u>Example<\/u><\/h6>\n<p>Design and analyze a three-input inverting amplifier using a 741 op-amp where<\/p>\n<p><a href=\"https:\/\/www.tina.com\/resources\/home\/practical-operational-amplifiers\/8-inverting-amplifier\/practicalamplifier-example9-9equation\/#main\" rel=\"attachment wp-att-3222\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-3222 alignnone\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/06\/PracticalAmplifier-Example9-9Equation.png\" alt=\"\" width=\"124\" height=\"22\" \/><\/a><\/p>\n<p>and the input resistance is <em>R<sub>min<\/sub><\/em> = 8 k\u03a9.<\/p>\n<p><strong><em>Solution:<\/em><\/strong> We use the design method of Chapter &#8220;Ideal Operational Amplifiers&#8221; to find <em>X<\/em> = 0, <em>Y<\/em> = 9, <em>Z<\/em> = -10.<\/p>\n<p>Then<\/p>\n<p><a href=\"https:\/\/www.tina.com\/resources\/home\/practical-operational-amplifiers\/8-inverting-amplifier\/practicalamplifier-example9-9equation2-7\/#main\" rel=\"attachment wp-att-3224\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-3224 alignnone\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/06\/PracticalAmplifier-Example9-9Equation2-7.png\" alt=\"\" width=\"190\" height=\"249\" \/><\/a><\/p>\n<p>The gain multiplier of the amplifier is 1+<em>R<sub>F<\/sub><\/em>\/<em>R<sub>A <\/sub><\/em>= 10. We find the input resistance as follows:<\/p>\n<p><a href=\"https:\/\/www.tina.com\/resources\/home\/practical-operational-amplifiers\/8-inverting-amplifier\/practicalamplifier-example9-9equation8-10\/#main\" rel=\"attachment wp-att-3226\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-3226 alignnone\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/06\/PracticalAmplifier-Example9-9Equation8-10.png\" alt=\"\" width=\"192\" height=\"131\" \/><\/a><\/p>\n<p>The output resistance is approximately 75(10)\/10<sup>5<\/sup> = 7.5 m\u03a9. To achieve bias balance, we set<\/p>\n<p><a href=\"https:\/\/www.tina.com\/resources\/home\/practical-operational-amplifiers\/8-inverting-amplifier\/practicalamplifier-example9-9equation8-11\/#main\" rel=\"attachment wp-att-3228\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-3228 alignnone\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/06\/PracticalAmplifier-Example9-9Equation8-11.png\" alt=\"\" width=\"192\" height=\"31\" \/><\/a><\/p>\n<div class=\"previous-page\"><a href=\"https:\/\/www.tina.com\/resources\/home\/7-non-inverting-amplifier\/\"><img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/leftarrow5.png\" \/>PREVIOUS- 7. Non-inverting amplifier<\/a><\/div>\n<div class=\"next-page\"><a href=\"https:\/\/www.tina.com\/resources\/practical-operational-amplifiers\/9-differential-summing\/\">NEXT- 9. Differential summing<img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/rightarrow5.png\" \/><\/a><\/div>\n<div class=\"clear\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>CURRENT &#8211; 8. Inverting Amplifier PREVIOUS- 7. Non-inverting amplifier NEXT- 9. Differential summing Figure 36(a) illustrates an inverting amplifier. Figure 36(b) shows the equivalent circuit using the op-amp model developed earlier in this chapter. 8.1 Input and Output Resistance Figure 36(b) is reduced to Figure 37(a) if we let,\u00a0 It is reasonable to assume that [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":917,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"yst_prominent_words":[480,485,477,483,484,325,70,172,168,479,267,248,482,171,238,478,476,486,462,481],"class_list":["post-1435","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/1435","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=1435"}],"version-history":[{"count":62,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/1435\/revisions"}],"predecessor-version":[{"id":3416,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/1435\/revisions\/3416"}],"up":[{"embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/917"}],"wp:attachment":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/media?parent=1435"}],"wp:term":[{"taxonomy":"yst_prominent_words","embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/yst_prominent_words?post=1435"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}