{"id":218813,"date":"2025-05-24T13:32:25","date_gmt":"2025-05-24T13:32:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218813"},"modified":"2025-05-24T13:32:28","modified_gmt":"2025-05-24T13:32:28","slug":"a-diode-has-reverse-saturation-current","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/a-diode-has-reverse-saturation-current\/","title":{"rendered":"A diode has reverse saturation current"},"content":{"rendered":"\n

A diode has reverse saturation current Is \u0002 \u0003 10 18 Aand nonideality factor \b \u0002 1.05. If diode has current of 70\u0006A, then diode voltage is(A) 0.63 V (B) 0.87 V(C) 0.54 V (D) 0.93 V21. An ideal pn junction diode is operating in theforward bais region. The change in diode voltage, thatwill cause a factor of 9 increase in current, is(A) 83 mV (B) 59 mV(C) 43 mV (D) 31 mV<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 1:<\/strong><\/h3>\n\n\n\n

Given:<\/strong><\/p>\n\n\n\n

    \n
  • Reverse saturation current, IS=10\u221218\u2009AI_S = 10^{-18} \\, \\text{A}<\/li>\n\n\n\n
  • Non-ideality factor, n=1.05n = 1.05<\/li>\n\n\n\n
  • Current, I=70\u2009\u03bcA=70\u00d710\u22126\u2009AI = 70 \\, \\mu\\text{A} = 70 \\times 10^{-6} \\, \\text{A}<\/li>\n\n\n\n
  • Temperature, T=300\u2009KT = 300 \\, \\text{K} (assumed standard)<\/li>\n\n\n\n
  • Thermal voltage, VT=kTq\u224825.85\u2009mVV_T = \\frac{kT}{q} \\approx 25.85 \\, \\text{mV}<\/li>\n<\/ul>\n\n\n\n

    Using the diode equation: I=IS(eVnVT\u22121)I = I_S \\left( e^{\\frac{V}{nV_T}} – 1 \\right)<\/p>\n\n\n\n

    Since I\u226bISI \\gg I_S, we can approximate: I\u2248ISeVnVTI \\approx I_S e^{\\frac{V}{nV_T}}<\/p>\n\n\n\n

    Solving for VV: V=nVTln\u2061(IIS)V = nV_T \\ln\\left( \\frac{I}{I_S} \\right)<\/p>\n\n\n\n

    Substitute values: V=1.05\u00d725.85\u00d7ln\u2061(70\u00d710\u2212610\u221218)=27.14\u00d7ln\u2061(7\u00d71013)V = 1.05 \\times 25.85 \\times \\ln\\left( \\frac{70 \\times 10^{-6}}{10^{-18}} \\right) = 27.14 \\times \\ln(7 \\times 10^{13})<\/p>\n\n\n\n

    Break it down: ln\u2061(7\u00d71013)=ln\u2061(7)+ln\u2061(1013)\u22481.9459+13ln\u2061(10)=1.9459+13\u00d72.3026=31.8847\\ln(7 \\times 10^{13}) = \\ln(7) + \\ln(10^{13}) \\approx 1.9459 + 13 \\ln(10) = 1.9459 + 13 \\times 2.3026 = 31.8847 V\u224827.14\u00d731.8847\u2248865\u2009mV=0.87\u2009V\u21d2Answer: (B)V \\approx 27.14 \\times 31.8847 \\approx 865 \\, \\text{mV} = \\boxed{0.87 \\, \\text{V}} \\Rightarrow \\text{Answer: (B)}<\/p>\n\n\n\n

    \n\n\n\n

    Question 2:<\/strong><\/h3>\n\n\n\n

    Given:<\/strong> Ideal diode. The current increases by a factor of 9.
    We want to find the voltage change \u0394V\\Delta V.<\/p>\n\n\n\n

    From the ideal diode equation (n = 1), ignoring ISI_S again: I2I1=e\u0394VVT\u21d2\u0394V=VTln\u2061(I2I1)\\frac{I_2}{I_1} = e^{\\frac{\\Delta V}{V_T}} \\Rightarrow \\Delta V = V_T \\ln\\left( \\frac{I_2}{I_1} \\right)<\/p>\n\n\n\n

    Given: I2I1=9,VT=25.85\u2009mV\u21d2\u0394V=25.85\u00d7ln\u2061(9)\u224825.85\u00d72.1972\u224856.8\u2009mV\\frac{I_2}{I_1} = 9, \\quad V_T = 25.85 \\, \\text{mV} \\Rightarrow \\Delta V = 25.85 \\times \\ln(9) \\approx 25.85 \\times 2.1972 \\approx 56.8 \\, \\text{mV}<\/p>\n\n\n\n

    Closest option: (B) 59 mV<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    \u2705 Final Answers:<\/h3>\n\n\n\n
      \n
    1. (B) 0.87 V<\/strong><\/li>\n\n\n\n
    2. (B) 59 mV<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      Explanation (Summary):<\/h3>\n\n\n\n

      Diodes follow an exponential I-V relationship given by the Shockley diode equation. For forward-biased conditions, if the current is much larger than the reverse saturation current, we can ignore the \u201c\u20131\u201d term for simplicity. This leads to a simple logarithmic relationship between current and voltage. In the first question, we used this to find the voltage corresponding to a given forward current, factoring in non-ideality. In the second, the ideal case simplifies analysis further, and we used the thermal voltage to calculate the voltage increment that leads to a 9x current increase. The logarithmic and exponential nature of diode characteristics is the key to solving such problems accurately.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      A diode has reverse saturation current Is \u0002 \u0003 10 18 Aand nonideality factor \b \u0002 1.05. If diode has current of 70\u0006A, then diode voltage is(A) 0.63 V (B) 0.87 V(C) 0.54 V (D) 0.93 V21. An ideal pn junction diode is operating in theforward bais region. The change in diode voltage, thatwill cause […]<\/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-218813","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218813","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=218813"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218813\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218813"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218813"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218813"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}