{"id":201315,"date":"2025-03-15T08:23:01","date_gmt":"2025-03-15T08:23:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201315"},"modified":"2025-03-15T08:23:09","modified_gmt":"2025-03-15T08:23:09","slug":"determine-the-common-mode-gain-of-an-op-amp-that-has-a-differential-voltage-gain-of-150x103-and-a-cmrr-of-90-db","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/15\/determine-the-common-mode-gain-of-an-op-amp-that-has-a-differential-voltage-gain-of-150x103-and-a-cmrr-of-90-db\/","title":{"rendered":"Determine the common-mode gain of an op amp that has a differential voltage gain of 150\u00d7103 and a CMRR of 90 dB"},"content":{"rendered":"\n<p>Determine the common-mode gain of an op amp that has a differential voltage gain of 150\u00d7103 and a CMRR of 90 dB.<\/p>\n\n\n\n<p>A differential amplifier has an open-loop voltage gain of 120 and a common input signal of 3.0 V to both terminals. An output signal of 24 mV results. Calculate the common-mode gain and the CMRR.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Common-Mode Gain and CMRR Calculation for the First Op-Amp<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Differential voltage gain (( A_d )) = 150,000 (or ( 150 \\times 10^3 ))<\/li>\n\n\n\n<li>CMRR = 90 dB<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Step 1: Convert CMRR from dB to a ratio<\/h4>\n\n\n\n<p>The CMRR (Common-Mode Rejection Ratio) in dB is related to the ratio of the differential gain (( A_d )) to the common-mode gain (( A_{cm} )):<\/p>\n\n\n\n<p>[<br>\\text{CMRR (dB)} = 20 \\log \\left( \\frac{A_d}{A_{cm}} \\right)<br>]<\/p>\n\n\n\n<p>Rearranging to solve for ( A_{cm} ):<\/p>\n\n\n\n<p>[<br>A_{cm} = \\frac{A_d}{10^{\\frac{\\text{CMRR (dB)}}{20}}}<br>]<\/p>\n\n\n\n<p>Substitute the given values:<\/p>\n\n\n\n<p>[<br>A_{cm} = \\frac{150 \\times 10^3}{10^{\\frac{90}{20}}}<br>]<\/p>\n\n\n\n<p>[<br>A_{cm} = \\frac{150 \\times 10^3}{10^{4.5}}<br>]<\/p>\n\n\n\n<p>[<br>A_{cm} = \\frac{150 \\times 10^3}{31622.8}<br>]<\/p>\n\n\n\n<p>[<br>A_{cm} \\approx 4.74 \\, \\text{mV\/V}<br>]<\/p>\n\n\n\n<p>Thus, the <strong>common-mode gain<\/strong> for the first op-amp is approximately <strong>4.74 mV\/V<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Common-Mode Gain and CMRR Calculation for the Differential Amplifier<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Open-loop voltage gain (( A_{OL} )) = 120<\/li>\n\n\n\n<li>Common-mode input voltage = 3.0 V<\/li>\n\n\n\n<li>Output signal = 24 mV<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Step 1: Calculate the common-mode gain<\/h4>\n\n\n\n<p>The common-mode gain (( A_{cm} )) can be calculated using the formula:<\/p>\n\n\n\n<p>[<br>A_{cm} = \\frac{V_{out}}{V_{in_cm}}<br>]<\/p>\n\n\n\n<p>Where ( V_{out} ) is the output signal, and ( V_{in_cm} ) is the common-mode input signal.<\/p>\n\n\n\n<p>Substitute the given values:<\/p>\n\n\n\n<p>[<br>A_{cm} = \\frac{24 \\times 10^{-3}}{3.0}<br>]<\/p>\n\n\n\n<p>[<br>A_{cm} = 0.008 \\, \\text{V\/V} \\, \\text{or} \\, 8 \\, \\text{mV\/V}<br>]<\/p>\n\n\n\n<p>Thus, the <strong>common-mode gain<\/strong> of the amplifier is <strong>8 mV\/V<\/strong>.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 2: Calculate the CMRR<\/h4>\n\n\n\n<p>The CMRR is calculated as:<\/p>\n\n\n\n<p>[<br>\\text{CMRR} = \\frac{A_d}{A_{cm}}<br>]<\/p>\n\n\n\n<p>Where ( A_d ) is the differential gain, and ( A_{cm} ) is the common-mode gain.<\/p>\n\n\n\n<p>Substitute the values:<\/p>\n\n\n\n<p>[<br>\\text{CMRR} = \\frac{120}{8 \\times 10^{-3}}<br>]<\/p>\n\n\n\n<p>[<br>\\text{CMRR} = 15,000<br>]<\/p>\n\n\n\n<p>To express the CMRR in dB:<\/p>\n\n\n\n<p>[<br>\\text{CMRR (dB)} = 20 \\log \\left( 15000 \\right)<br>]<\/p>\n\n\n\n<p>[<br>\\text{CMRR (dB)} \\approx 82.5 \\, \\text{dB}<br>]<\/p>\n\n\n\n<p>Thus, the <strong>CMRR<\/strong> of the amplifier is approximately <strong>82.5 dB<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>For the first op-amp, the <strong>common-mode gain<\/strong> is <strong>4.74 mV\/V<\/strong>.<\/li>\n\n\n\n<li>For the differential amplifier, the <strong>common-mode gain<\/strong> is <strong>8 mV\/V<\/strong>, and the <strong>CMRR<\/strong> is <strong>82.5 dB<\/strong>.<\/li>\n<\/ol>\n\n\n\n<p>These calculations show how an op-amp&#8217;s ability to reject common-mode signals is quantified, and how high CMRR is crucial for accurate differential amplification.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Determine the common-mode gain of an op amp that has a differential voltage gain of 150\u00d7103 and a CMRR of 90 dB. A differential amplifier has an open-loop voltage gain of 120 and a common input signal of 3.0 V to both terminals. An output signal of 24 mV results. Calculate the common-mode gain and [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-201315","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201315","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=201315"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201315\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201315"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201315"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201315"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}