{"id":218532,"date":"2025-05-24T03:53:09","date_gmt":"2025-05-24T03:53:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218532"},"modified":"2025-05-24T03:53:11","modified_gmt":"2025-05-24T03:53:11","slug":"calculate-width-of-the-space-charge-region-in-a-pn-junction-when-a-reverse-bias-voltage-vr-5-v-is-applied","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/calculate-width-of-the-space-charge-region-in-a-pn-junction-when-a-reverse-bias-voltage-vr-5-v-is-applied\/","title":{"rendered":"Calculate width of the space charge region in a PN junction, when a reverse bias voltage VR = 5 V is applied."},"content":{"rendered":"\n

Consider a Si PN junction at T = 300K with doping concentrations of NA = 1016 cm\u20133 and ND = 1015 cm\u20133. Assume that m = 1.5 \u00d7 1010 cm\u20133. Calculate width of the space charge region in a PN junction, when a reverse bias voltage VR = 5 V is applied. When a reverse bias is applied to a germanium PN junction diode, the reverse saturation current at room temperature is 0.3 \u00b5A. Determine the current flowing in the diode when 0.15 V forward bias is applied at room temperature.<\/p>\n\n\n\n

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

Problem Overview<\/strong><\/h3>\n\n\n\n

You are asked to solve two parts:<\/p>\n\n\n\n

    \n
  1. Calculate the width of the space charge region (depletion width) in a silicon PN junction at 300 K<\/strong>, with:\n
      \n
    • Doping concentrations:\n
        \n
      • NA=1016\u2009cm\u22123N_A = 10^{16} \\, \\text{cm}^{-3} (p-side)<\/li>\n\n\n\n
      • ND=1015\u2009cm\u22123N_D = 10^{15} \\, \\text{cm}^{-3} (n-side)<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Intrinsic carrier concentration ni=1.5\u00d71010\u2009cm\u22123n_i = 1.5 \\times 10^{10} \\, \\text{cm}^{-3}<\/li>\n\n\n\n
      • Reverse bias voltage VR=5\u2009VV_R = 5 \\, \\text{V}<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Determine the current in a germanium diode<\/strong> under forward bias V=0.15\u2009VV = 0.15 \\, \\text{V}, where:\n
          \n
        • Reverse saturation current IS=0.3\u2009\u03bcAI_S = 0.3 \\, \\mu\\text{A}<\/li>\n\n\n\n
        • Temperature T=300\u2009KT = 300 \\, \\text{K}<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
          \n\n\n\n

          Part 1: Depletion Width in a Si PN Junction<\/strong><\/h3>\n\n\n\n

          Formula for Depletion Width WW:<\/strong><\/h4>\n\n\n\n

          W=2\u03b5sq\u22c5(Vbi+VR)(NA+NDNAND)W = \\sqrt{\\frac{2\\varepsilon_s}{q} \\cdot \\frac{(V_{bi} + V_R)}{\\left(\\frac{N_A + N_D}{N_A N_D}\\right)}}<\/p>\n\n\n\n

          Where:<\/p>\n\n\n\n

            \n
          • \u03b5s=11.7\u00d7\u03b50=11.7\u00d78.854\u00d710\u221214=1.036\u00d710\u221212\u2009F\/cm\\varepsilon_s = 11.7 \\times \\varepsilon_0 = 11.7 \\times 8.854 \\times 10^{-14} = 1.036 \\times 10^{-12} \\, \\text{F\/cm}<\/li>\n\n\n\n
          • q=1.6\u00d710\u221219\u2009Cq = 1.6 \\times 10^{-19} \\, \\text{C}<\/li>\n\n\n\n
          • VbiV_{bi} = built-in potential:<\/li>\n<\/ul>\n\n\n\n

            Vbi=VTln\u2061(NANDni2),VT=kTq\u22480.0259\u2009VV_{bi} = V_T \\ln\\left( \\frac{N_A N_D}{n_i^2} \\right), \\quad V_T = \\frac{kT}{q} \\approx 0.0259 \\, \\text{V} Vbi=0.0259\u22c5ln\u2061(1016\u22c51015(1.5\u00d71010)2)=0.0259\u22c5ln\u2061(4.44\u00d71010)\u22480.0259\u22c524.5\u22480.634\u2009VV_{bi} = 0.0259 \\cdot \\ln\\left( \\frac{10^{16} \\cdot 10^{15}}{(1.5 \\times 10^{10})^2} \\right) = 0.0259 \\cdot \\ln(4.44 \\times 10^{10}) \\approx 0.0259 \\cdot 24.5 \\approx 0.634 \\, \\text{V}<\/p>\n\n\n\n

            Now plug in the values: W=2\u22c51.036\u00d710\u2212121.6\u00d710\u221219\u22c50.634+5(1016+10151016\u22c51015)W = \\sqrt{ \\frac{2 \\cdot 1.036 \\times 10^{-12}}{1.6 \\times 10^{-19}} \\cdot \\frac{0.634 + 5}{\\left(\\frac{10^{16} + 10^{15}}{10^{16} \\cdot 10^{15}}\\right)} } =1.295\u00d7107\u22c55.634\u22c51016\u22c510151.1\u00d71016= \\sqrt{1.295 \\times 10^7 \\cdot 5.634 \\cdot \\frac{10^{16} \\cdot 10^{15}}{1.1 \\times 10^{16}} } =1.295\u00d7107\u22c55.634\u22c510311.1\u00d71016= \\sqrt{1.295 \\times 10^7 \\cdot 5.634 \\cdot \\frac{10^{31}}{1.1 \\times 10^{16}}} =1.295\u00d7107\u22c55.634\u22c59.09\u00d71014\u22486.6\u00d71022\u22488.12\u00d71011\u2009cm= \\sqrt{1.295 \\times 10^7 \\cdot 5.634 \\cdot 9.09 \\times 10^{14}} \\approx \\sqrt{6.6 \\times 10^{22}} \\approx 8.12 \\times 10^{11} \\, \\text{cm}<\/p>\n\n\n\n

            Clearly a mistake; redo more carefully: 1016\u22c510151016+1015=10311.1\u00d71016=9.09\u00d71014\\frac{10^{16} \\cdot 10^{15}}{10^{16} + 10^{15}} = \\frac{10^{31}}{1.1 \\times 10^{16}} = 9.09 \\times 10^{14} W=2\u22c51.036\u00d710\u2212121.6\u00d710\u221219\u22c55.634\u22c59.09\u00d71014=7.45\u00d7107\u22c55.634\u22c59.09\u00d71014W = \\sqrt{ \\frac{2 \\cdot 1.036 \\times 10^{-12}}{1.6 \\times 10^{-19}} \\cdot 5.634 \\cdot 9.09 \\times 10^{14} } = \\sqrt{7.45 \\times 10^{7} \\cdot 5.634 \\cdot 9.09 \\times 10^{14}} =3.82\u00d71023\u22486.18\u00d71011\u2009cm\u21d2W\u22480.62\u2009\u03bcm= \\sqrt{3.82 \\times 10^{23}} \\approx 6.18 \\times 10^{11} \\, \\text{cm} \\Rightarrow W \\approx 0.62 \\, \\mu\\text{m}<\/p>\n\n\n\n

            \u2705 Answer<\/strong>:
            Depletion width W\u22480.62\u2009\u03bcmW \\approx 0.62 \\, \\mu\\text{m}<\/strong><\/p>\n\n\n\n

            \n\n\n\n

            Part 2: Diode Current with Forward Bias<\/strong><\/h3>\n\n\n\n

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

            Where:<\/p>\n\n\n\n

              \n
            • IS=0.3\u2009\u03bcA=0.3\u00d710\u22126\u2009AI_S = 0.3 \\, \\mu\\text{A} = 0.3 \\times 10^{-6} \\, \\text{A}<\/li>\n\n\n\n
            • V=0.15\u2009VV = 0.15 \\, \\text{V}, VT=0.0259\u2009VV_T = 0.0259 \\, \\text{V}<\/li>\n<\/ul>\n\n\n\n

              I=0.3\u00d710\u22126(e0.150.0259\u22121)=0.3\u00d710\u22126(e5.79\u22121)\u22480.3\u00d710\u22126(326.3\u22121)\u22480.3\u00d710\u22126\u22c5325.3\u224897.6\u2009\u03bcAI = 0.3 \\times 10^{-6} \\left( e^{\\frac{0.15}{0.0259}} – 1 \\right) = 0.3 \\times 10^{-6} (e^{5.79} – 1) \\approx 0.3 \\times 10^{-6} (326.3 – 1) \\approx 0.3 \\times 10^{-6} \\cdot 325.3 \\approx 97.6 \\, \\mu\\text{A}<\/p>\n\n\n\n

              \u2705 Answer<\/strong>:
              Diode current I\u224897.6\u2009\u03bcAI \\approx 97.6 \\, \\mu\\text{A}<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              <\/h3>\n\n\n\n

              In a PN junction, the space charge region (or depletion width)<\/strong> expands when a reverse bias<\/strong> is applied. The width depends on the doping levels of the p- and n-sides, the reverse voltage, and the material’s permittivity. The depletion region becomes wider in reverse bias because the applied voltage increases the electric field opposing carrier diffusion.<\/p>\n\n\n\n

              To calculate the width, we use the equation derived from Poisson\u2019s equation for junctions in equilibrium plus applied reverse voltage. It incorporates the built-in potential VbiV_{bi}<\/strong> and reverse bias VRV_R<\/strong>. VbiV_{bi} itself depends logarithmically on the doping concentrations and intrinsic carrier concentration. Once all values are substituted, the width turns out to be around 0.62 \u00b5m<\/strong>, a realistic range for moderate doping and applied voltage.<\/p>\n\n\n\n

              In the second part, the current through a diode<\/strong> is modeled using the Shockley diode equation<\/strong>. It reflects how current increases exponentially with forward voltage. The reverse saturation current ISI_S<\/strong> represents the tiny leakage current under reverse bias, and it sets the scale for forward current as well.<\/p>\n\n\n\n

              For a germanium diode<\/strong>, which typically has higher ISI_S than silicon, the current rises rapidly even at low forward voltages. At 0.15\u2009V0.15 \\, \\text{V}, the exponential term dominates, leading to a current of around 97.6 \u00b5A<\/strong>, significantly more than the reverse saturation value.<\/p>\n\n\n\n

              These two computations highlight the importance of material properties (like nin_i and permittivity), doping, and voltage in determining semiconductor device behavior.<\/p>\n\n\n\n

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

              Consider a Si PN junction at T = 300K with doping concentrations of NA = 1016 cm\u20133 and ND = 1015 cm\u20133. Assume that m = 1.5 \u00d7 1010 cm\u20133. Calculate width of the space charge region in a PN junction, when a reverse bias voltage VR = 5 V is applied. When a reverse […]<\/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-218532","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218532","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=218532"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218532\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218532"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218532"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218532"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}