{"id":233784,"date":"2025-06-13T11:58:53","date_gmt":"2025-06-13T11:58:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233784"},"modified":"2025-06-13T11:58:55","modified_gmt":"2025-06-13T11:58:55","slug":"using-the-formula-of-transconductance-of-fet-in-ac-mode","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/using-the-formula-of-transconductance-of-fet-in-ac-mode\/","title":{"rendered":"Using the formula of transconductance of FET in AC mode"},"content":{"rendered":"\n

a) Using the formula of transconductance of FET in AC mode, prove that: i) for n-channel D-MOSFET, gm = 2\u00e2\u02c6\u0161(KID) ii) for n-channel E-MOSFET, Rp = 1.5k b) For the circuit shown below, find the value of Rs required to set the drain current at a quarter of loss. Loss = 12mA, Vp = 3V For the same value of Rs, find VGVs, VoVos, and VGs. R = 1MO<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

Correct Answer<\/strong><\/h3>\n\n\n\n

Part a) Proofs:<\/strong>
The proofs for the transconductance formulas are detailed in the explanation below by differentiating the standard current equations for D-MOSFETs and E-MOSFETs.<\/p>\n\n\n\n

Part b) Circuit Analysis:<\/strong><\/p>\n\n\n\n

    \n
  • The required value of the source resistor is\u00a0Rs = 500 \u03a9<\/strong>.<\/li>\n\n\n\n
  • The DC voltage values are:\n
      \n
    • VGS = -1.5 V<\/strong><\/li>\n\n\n\n
    • VG = 0 V<\/strong><\/li>\n\n\n\n
    • VS = 1.5 V<\/strong><\/li>\n\n\n\n
    • VD = 10.5 V<\/strong><\/li>\n\n\n\n
    • VDS = 9.0 V<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Detailed Explanation<\/strong><\/h3>\n\n\n\n

      This problem involves two parts: first, deriving the AC transconductance (g_m) formulas for different types of MOSFETs, and second, performing a DC analysis of a JFET\/D-MOSFET self-bias circuit.<\/p>\n\n\n\n

      Part a) Transconductance Proofs<\/strong><\/h4>\n\n\n\n

      i) For n-channel D-MOSFET:<\/strong><\/p>\n\n\n\n

      The transconductance, g_m, is defined as the change in drain current (I_D) with respect to the change in gate-to-source voltage (V_GS), which is mathematically expressed as the partial derivative g_m = \u2202I_D \/ \u2202V_GS.<\/p>\n\n\n\n

      The drain current for a D-MOSFET (or JFET) in the saturation region is given by the Shockley equation:
      I_D = I_DSS * (1 – V_GS \/ V_p)\u00b2<\/p>\n\n\n\n

      Differentiating I_D with respect to V_GS using the chain rule yields:
      g_m = I_DSS * 2 * (1 – V_GS \/ V_p) * (-1 \/ V_p)
      g_m = (-2 * I_DSS \/ V_p) * (1 – V_GS \/ V_p)<\/p>\n\n\n\n

      The maximum transconductance, g_mo, occurs at V_GS = 0. Substituting V_GS = 0 into the equation gives:
      g_mo = -2 * I_DSS \/ V_p<\/p>\n\n\n\n

      By substituting g_mo back into the general g_m equation, we get:
      g_m = g_mo * (1 – V_GS \/ V_p)<\/p>\n\n\n\n

      From the original Shockley equation, we can also express (1 – V_GS \/ V_p) as \u221a(I_D \/ I_DSS). Substituting this relationship into the equation for g_m proves the required formula:
      g_m = g_mo * \u221a(I_D \/ I_DSS)<\/p>\n\n\n\n

      ii) For n-channel E-MOSFET:<\/strong><\/p>\n\n\n\n

      The drain current for an E-MOSFET in saturation is given by:
      I_D = K * (V_GS – V_th)\u00b2, where K is a device constant and V_th is the threshold voltage.<\/p>\n\n\n\n

      Differentiating I_D with respect to V_GS:
      g_m = \u2202I_D \/ \u2202V_GS = K * 2 * (V_GS – V_th)<\/p>\n\n\n\n

      From the E-MOSFET current equation, we can write (V_GS – V_th) = \u221a(I_D \/ K). Substituting this into the g_m equation:
      g_m = 2 * K * \u221a(I_D \/ K)
      g_m = 2 * \u221a(K\u00b2 * I_D \/ K)
      g_m = 2 * \u221a(K * I_D)<\/p>\n\n\n\n

      Part b) Circuit DC Analysis<\/strong><\/h4>\n\n\n\n

      The circuit shown is a self-bias configuration for an n-channel JFET\/D-MOSFET. The provided parameters are I_DSS = 12 mA, V_p = -3 V (for an n-channel device), and R_D = 1.5 k\u03a9.<\/p>\n\n\n\n

      1. Find the value of Rs:<\/strong><\/p>\n\n\n\n

      The problem requires setting the drain current I_D to a quarter of I_DSS.
      I_D = I_DSS \/ 4 = 12 mA \/ 4 = 3 mA<\/p>\n\n\n\n

      To find the V_GS required for this I_D, we use the Shockley equation:
      I_D = I_DSS * (1 – V_GS \/ V_p)\u00b2
      3 mA = 12 mA * (1 – V_GS \/ (-3 V))\u00b2
      1\/4 = (1 + V_GS \/ 3)\u00b2
      Taking the square root of both sides: \u00b11\/2 = 1 + V_GS \/ 3. For operation, V_GS must be between V_p (-3V) and 0. This condition is met for the negative root.
      1\/2 = 1 + V_GS \/ 3 => V_GS \/ 3 = -1\/2 => V_GS = -1.5 V<\/p>\n\n\n\n

      In a self-bias circuit, the gate current I_G is approximately zero. Therefore, there is no voltage drop across R_G, and the gate voltage V_G is 0 V. The source voltage is V_S = I_D * R_s.
      The gate-to-source voltage is V_GS = V_G – V_S = 0 – I_D * R_s = -I_D * R_s.
      We can now solve for R_s:
      -1.5 V = -(3 mA) * R_s
      R_s = 1.5 V \/ 3 mA = 0.5 k\u03a9 or 500 \u03a9<\/strong>.<\/p>\n\n\n\n

      2. Find VG, VS, VD, VDS, and VGS:<\/strong><\/p>\n\n\n\n

      Using the calculated value of R_s = 500 \u03a9 and the operating point I_D = 3 mA:<\/p>\n\n\n\n

        \n
      • VGS<\/strong>: Already calculated to establish the operating point.\u00a0VGS = -1.5 V<\/strong>.<\/li>\n\n\n\n
      • VG<\/strong>: As established, with no gate current,\u00a0VG = 0 V<\/strong>.<\/li>\n\n\n\n
      • VS<\/strong>:\u00a0V_S = I_D * R_s = 3 mA * 500 \u03a9 = 1.5 V. (Check:\u00a0V_GS = V_G – V_S = 0 V – 1.5 V = -1.5 V. Correct.)<\/li>\n\n\n\n
      • VD<\/strong>: The drain voltage is calculated from the supply voltage and the drop across\u00a0R_D.
        V_D = V_DD – I_D * R_D = 15 V – (3 mA * 1.5 k\u03a9) = 15 V – 4.5 V =\u00a010.5 V<\/strong>.<\/li>\n\n\n\n
      • VDS<\/strong>: The drain-to-source voltage is the difference between the drain and source voltages.
        V_DS = V_D – V_S = 10.5 V – 1.5 V =\u00a09.0 V<\/strong>.<\/li>\n<\/ul>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        a) Using the formula of transconductance of FET in AC mode, prove that: i) for n-channel D-MOSFET, gm = 2\u00e2\u02c6\u0161(KID) ii) for n-channel E-MOSFET, Rp = 1.5k b) For the circuit shown below, find the value of Rs required to set the drain current at a quarter of loss. Loss = 12mA, Vp = 3V […]<\/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-233784","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233784","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=233784"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233784\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233784"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233784"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233784"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}