{"id":58107,"date":"2024-04-16T01:13:41","date_gmt":"2024-04-16T01:13:41","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=58107"},"modified":"2024-04-16T01:13:41","modified_gmt":"2024-04-16T01:13:41","slug":"ic-a-1-a-ib-1-a","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/ic-a-1-a-ib-1-a\/","title":{"rendered":"IC = [a\/(1 &#8211; a )] IB + [. . . . . . . . \/(1 &#8211; a)]?"},"content":{"rendered":"<p>[amp_mcq option1=&#8221;ICBO&#8221; option2=&#8221;ICEO&#8221; option3=&#8221;IC&#8221; option4=&#8221;IE&#8221; correct=&#8221;option1&#8243;]<!--more--><\/p>\n<p>The correct answer is: $\\boxed{\\text{B}}$.<\/p>\n<p>The equation $I_C = \\frac{a}{1-a}I_B + \\frac{1}{1-a}I_{CEO}$ is the Ebers-Moll model for a BJT. It relates the collector current ($I_C$) to the base current ($I_B$) and the emitter-base junction reverse saturation current ($I_{CEO}$). The parameter $a$ is the common-emitter current gain.<\/p>\n<p>The first term on the right-hand side of the equation, $\\frac{a}{1-a}I_B$, is the base-emitter current multiplied by the current gain. This term represents the majority carriers that are injected into the base region from the emitter region. The second term on the right-hand side of the equation, $\\frac{1}{1-a}I_{CEO}$, is the emitter-base junction reverse saturation current divided by the current gain. This term represents the minority carriers that are injected into the base region from the emitter region due to the reverse bias across the emitter-base junction.<\/p>\n<p>The Ebers-Moll model is a useful tool for analyzing the operation of BJTs. It can be used to calculate the collector current for a given set of operating conditions. It can also be used to determine the current gain of a BJT.<\/p>\n<p>The other options are incorrect because they do not represent the correct equation for the Ebers-Moll model.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>[amp_mcq option1=&#8221;ICBO&#8221; option2=&#8221;ICEO&#8221; option3=&#8221;IC&#8221; option4=&#8221;IE&#8221; correct=&#8221;option1&#8243;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[970],"tags":[],"class_list":["post-58107","post","type-post","status-publish","format-standard","hentry","category-transistors","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>IC = [a\/(1 - a )] IB + [. . . . . . . . \/(1 - a)]?<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/ic-a-1-a-ib-1-a\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"IC = [a\/(1 - a )] IB + [. . . . . . . . \/(1 - a)]?\" \/>\n<meta property=\"og:description\" content=\"[amp_mcq option1=&#8221;ICBO&#8221; 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