{"id":47462,"date":"2025-07-02T12:19:49","date_gmt":"2025-07-02T12:19:49","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=47462"},"modified":"2025-07-02T12:19:51","modified_gmt":"2025-07-02T12:19:51","slug":"now-we-replace-the-resistor-in-the-circuit-of-problem-1-with-a-diode-connected-pmos-device-as-shown","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/now-we-replace-the-resistor-in-the-circuit-of-problem-1-with-a-diode-connected-pmos-device-as-shown\/","title":{"rendered":"Now we replace the resistor in the circuit of problem 1 with a diode-connected PMOS device as shown"},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">Now we replace the resistor in the circuit of problem 1 with a diode-connected PMOS device as shown<\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"583\" height=\"443\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-121.png\" alt=\"\" class=\"wp-image-47464\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-121.png 583w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/image-121-300x228.png 300w\" sizes=\"auto, (max-width: 583px) 100vw, 583px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here is the correct answer.<\/p>\n\n\n\n<p><strong>To solve this problem, we must assume typical process parameters for the MOSFETs, as they are not provided. Let&#8217;s assume the following values for a standard CMOS process, ignoring channel-length modulation (\u03bb=0):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Electron mobility parameter, k&#8217;n = \u03bcnCox = 100 \u00b5A\/V\u00b2<\/li>\n\n\n\n<li>Hole mobility parameter, k&#8217;p = \u03bcpCox = 50 \u00b5A\/V\u00b2<\/li>\n\n\n\n<li>NMOS threshold voltage, Vtn = 0.7 V<\/li>\n\n\n\n<li>PMOS threshold voltage, Vtp = -0.7 V<\/li>\n<\/ul>\n\n\n\n<p>All transistors in the reference leg carry I_REF = 25 \u00b5A and are assumed to be in saturation.<\/p>\n\n\n\n<p><strong>a. What W\/L ratio is needed for M6 in order to set I_REF = 25 \u00b5A?<\/strong><\/p>\n\n\n\n<p>The total voltage drop across the reference leg is V_DD &#8211; V_SS = 5V &#8211; (-5V) = 10V. This voltage is distributed across the three series-connected transistors M4, M6, and M1. Using Kirchhoff&#8217;s Voltage Law:<br>V_DD &#8211; V_SS = V_SG4 + V_SG6 + V_GS1 = 10V<\/p>\n\n\n\n<p>First, we find the gate-source voltages for M1 and M4 using the saturation current equation I_D = \u00bd * k&#8217; * (W\/L) * (V_GS &#8211; V_t)\u00b2:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For M1 (NMOS): 25\u00b5A = \u00bd * (100 \u00b5A\/V\u00b2) * (5\/1) * (V_GS1 &#8211; 0.7V)\u00b2, which gives V_GS1 \u2248 1.016 V.<\/li>\n\n\n\n<li>For M4 (PMOS): 25\u00b5A = \u00bd * (50 \u00b5A\/V\u00b2) * (5\/1) * (V_SG4 &#8211; 0.7V)\u00b2, which gives V_SG4 \u2248 1.147 V.<\/li>\n<\/ul>\n\n\n\n<p>Now, we find V_SG6 from the KVL equation:<br>V_SG6 = 10V &#8211; V_GS1 &#8211; V_SG4 = 10V &#8211; 1.016V &#8211; 1.147V = 7.837 V.<\/p>\n\n\n\n<p>Finally, we calculate the W\/L ratio for the pFET M6:<br>25\u00b5A = \u00bd * (50 \u00b5A\/V\u00b2) * (W\/L)6 * (7.837V &#8211; 0.7V)\u00b2<br>(W\/L)6 = 25 \/ [25 * (7.137)\u00b2] = 1 \/ 50.94 \u2248&nbsp;<strong>0.0196<\/strong><\/p>\n\n\n\n<p><strong>b. If M6 is implemented as a diode-connected nFET instead of pFET, what W\/L ratio would be required?<\/strong><\/p>\n\n\n\n<p>If M6 is an nFET, for it to conduct current from M4 to M1, its drain (higher potential) must connect to M4&#8217;s drain and its source (lower potential) to M1&#8217;s drain. For it to be diode-connected, its gate must be tied to its drain. Therefore, V_G6 = V_D6. The voltage at M6&#8217;s drain is V_D4, and the voltage at its source is V_D1.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>V_D4 = V_DD &#8211; V_SG4 = 5V &#8211; 1.147V = 3.853V<\/li>\n\n\n\n<li>V_D1 = V_SS + V_GS1 = -5V + 1.016V = -3.984V<\/li>\n<\/ul>\n\n\n\n<p>The gate-source voltage for the nFET M6 is V_GS6 = V_G6 &#8211; V_S6 = V_D4 &#8211; V_D1 = 3.853V &#8211; (-3.984V) = 7.837V.<\/p>\n\n\n\n<p>Now we calculate the required W\/L ratio for the nFET M6:<br>25\u00b5A = \u00bd * (100 \u00b5A\/V\u00b2) * (W\/L)6 * (7.837V &#8211; 0.7V)\u00b2<br>(W\/L)6 = 25 \/ [50 * (7.137)\u00b2] = 0.5 \/ 50.94 \u2248&nbsp;<strong>0.0098<\/strong><\/p>\n\n\n\n<p><strong>c. Which implementation, pFET or nFET, would take up the most area, assuming W is fixed at 1\u00b5m?<\/strong><\/p>\n\n\n\n<p>The area of a transistor is approximately A = W \u00d7 L. Since W is fixed at 1\u00b5m, the area is directly proportional to the channel length L. The length can be expressed as L = W \/ (W\/L). This means the implementation with the smaller W\/L ratio will require a larger channel length L, and therefore will occupy more area.<\/p>\n\n\n\n<p>Comparing the two results:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(W\/L)pFET \u2248 0.0196<\/li>\n\n\n\n<li>(W\/L)nFET \u2248 0.0098<\/li>\n<\/ul>\n\n\n\n<p>Since (W\/L)nFET is smaller than (W\/L)pFET, the&nbsp;<strong>nFET implementation<\/strong>&nbsp;will have a larger channel length and would take up the most area.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-243.jpeg\" alt=\"\" class=\"wp-image-47471\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-243.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-243-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-243-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-243-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Now we replace the resistor in the circuit of problem 1 with a diode-connected PMOS device as shown The Correct Answer and Explanation is: Here is the correct answer. To solve this problem, we must assume typical process parameters for the MOSFETs, as they are not provided. Let&#8217;s assume the following values for a standard [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-47462","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47462","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=47462"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47462\/revisions"}],"predecessor-version":[{"id":47472,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47462\/revisions\/47472"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47462"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47462"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47462"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}