{"id":2159,"date":"2018-05-08T07:08:21","date_gmt":"2018-05-08T07:08:21","guid":{"rendered":"https:\/\/www.tina.com\/resources\/?page_id=2159"},"modified":"2018-06-22T08:13:39","modified_gmt":"2018-06-22T08:13:39","slug":"10-fet-amplifier-design","status":"publish","type":"page","link":"https:\/\/www.tina.com\/resources\/home\/field-effect-transistor-amplifiers-2\/10-fet-amplifier-design\/","title":{"rendered":"10. FET Amplifier design"},"content":{"rendered":"<div class=\"top-navigation\">\n<div class=\"current-page\"><img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/people-location-30.png\" \/>CURRENT &#8211; 10. FET Amplifier design<\/div>\n<div class=\"previous-page\"><a href=\"https:\/\/www.tina.com\/resources\/home\/field-effect-transistor-amplifiers-2\/9-fet-amplifier-analysis\/\"><img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/leftarrow5.png\" \/>PREVIOUS- 9. FET Amplifier analysis<\/a><\/div>\n<div class=\"next-page\"><a style=\"font-size: 20px;\" href=\"https:\/\/www.tina.com\/resources\/home\/field-effect-transistor-amplifiers-2\/11-other-devices\/\">NEXT-11. Other devices<img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/rightarrow5.png\" \/><\/a><\/div>\n<div class=\"clear\"><\/div>\n<\/div>\n<p><script>\nfunction setStyle(){ \nheight = document.getElementsByClassName('entry-header')[0].clientHeight;\ndocument.getElementsByClassName('entry-header')[0].style.marginTop = height +\"px\";\ndocument.getElementsByClassName('top-navigation')[0].style.top = Math.round(-(height-20)*2) +\"px\";\n}\nsetStyle();\nwindow.onresize = function(event) {\nsetStyle();\n};\n<\/script><\/p>\n<p align=\"justify\"><strong><em><span style=\"font-size: 22px;\">FET Amplifier design<\/span><\/em><\/strong><\/p>\n<p>We now explore the extension of the FET amplifier analysis presented earlier in this chapter to the design of FET amplifiers. We will attempt to define the unknowns in the design problem, and then develop equations for solving for these unknowns. As in most electronics design, the number of equations will be less than the number of unknowns. The additional constraints are established to meet certain overall objectives (e.g., minimum cost, less variation in performance due to parameter changes).<br \/>\n<b><\/b><\/p>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">10.1 The CS Amplifier<\/span><\/b><\/span><\/em><\/h6>\n<p>The design procedure of a CS amplifier is presented in this section. We shall reduce JFET and the depletion MOSFET amplifier design to an organized procedure. While this may appear to<\/p>\n<p>reduce design to a very routine process, you must convince yourself that you understand the origin of each step since several variations may be subsequently required. If all you do to design a CS amplifier is to thoughtlessly &#8220;plug in&#8221; to the steps we present, you are missing the whole point of this discussion. As an engineer, you are seeking to do things that are <em>not<\/em> routine. Reducing theory to an organized approach is what you will be doing. You will not simply apply the approaches others have already done for you.<\/p>\n<p>Amplifiers are designed to meet gain requirements assuming the desired specifications are within the range of the transistor. The supply voltage, load resistance, voltage gain and input resistance (or current gain) are usually specified. The designer&#8217;s job is to select the resistance values <em>R<sub>1<\/sub><\/em>, <em>R<sub>2<\/sub><\/em>, <em>R<sub>D<\/sub><\/em>, and <em>R<sub>S<\/sub><\/em>. Refer to Figure 40 as you follow the steps in the procedure. This procedure assumes that a device has been selected and that its characteristics are known.<\/p>\n<div id=\"attachment_2171\" style=\"width: 497px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2171\" class=\"wp-image-2171 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-40.png\" alt=\"\" width=\"487\" height=\"205\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-40.png 487w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-40-300x126.png 300w\" sizes=\"auto, (max-width: 487px) 100vw, 487px\" \/><p id=\"caption-attachment-2171\" class=\"wp-caption-text\">Figure 40 JFET CS amplifier<\/p><\/div>\n<p>First, select a Q-point in the saturation region of the FET characteristic curves. Refer to the curves of Figure 40(b) for an example. This identifies <em>V<sub>DSQ<\/sub><\/em>, <em>V<sub>GSQ<\/sub><\/em>, and <em>I<sub>DQ<\/sub><\/em>.<\/p>\n<p>We now solve for the two resistors in the output loop, <em>R<sub>S<\/sub><\/em> and <em>R<sub>D<\/sub><\/em>. Since there are two unknowns, we require two independent equations. We begin by writing the <em>dc<\/em> KVL equation around the drain-source loop,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2174\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-58.png\" alt=\"\" width=\"173\" height=\"29\" \/>\u00a0(58)<\/p>\n<p>Solving for the sum of the two resistors yields<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2175\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-59.png\" alt=\"\" width=\"207\" height=\"48\" \/>\u00a0(59)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2176\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-60.png\" alt=\"\" width=\"273\" height=\"60\" \/>\u00a0(60)<\/p>\n<p>The resistance, <em>R<sub>D<\/sub><\/em>, <em>is the only unknown in this equation.<\/em> Solving for <em>R<sub>D<\/sub><\/em> results in a quadratic equation having two solutions, one negative and one positive. If the positive solution results in <em>R<sub>D<\/sub><\/em> &gt; <em>K<sub>1<\/sub><\/em>, thus implying a negative <em>R<sub>S<\/sub><\/em>, a new Q-point must be selected (i.e., restart the design). If the positive solution yields <em>R<sub>D<\/sub><\/em> &lt; <em>K<sub>1<\/sub><\/em>, we can proceed.<\/p>\n<p>Now that <em>R<sub>D<\/sub><\/em> is known, we solve for <em>R<sub>S<\/sub><\/em> using Equation (59) , the drain-to-source loop equation.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2177\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-61.png\" alt=\"\" width=\"146\" height=\"58\" \/>\u00a0(61)<\/p>\n<p>With <em>R<sub>D<\/sub><\/em> and <em>R<sub>S<\/sub><\/em> known, we need only find <em>R<sub>1<\/sub><\/em> and <em>R<sub>2<\/sub><\/em>.<\/p>\n<p>We begin by rewriting the KVL equation for the gate-source loop.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2180\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-62-1.png\" alt=\"\" width=\"125\" height=\"30\" \/>\u00a0(62)<\/p>\n<p>The voltage, <em>V<sub>GS<\/sub><\/em>, is of opposite polarity from <em>V<sub>DD<\/sub><\/em>. Thus the term <em>I<sub>DQ<\/sub>R<sub>S<\/sub><\/em> must be greater than <em>V<sub>GSQ<\/sub><\/em> in magnitude. Otherwise, <em>V<sub>GG<\/sub><\/em> will have the opposite polarity from <em>V<sub>DD<\/sub><\/em>, which is not possible according to Equation (62).<\/p>\n<p>We now solve for <em>R<sub>1<\/sub><\/em> and <em>R<sub>2<\/sub><\/em> assuming that the <em>V<sub>GG<\/sub><\/em> found has the <em>same polarity <\/em>as <em>V<sub>DD<\/sub><\/em>. These resistor values are selected by finding the value of <em>R<sub>G<\/sub><\/em> from the current-gain equation or from the input resistance. We solve for <em>R<sub>1<\/sub><\/em> and <em>R<sub>2<\/sub><\/em>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2182\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-63.png\" alt=\"\" width=\"112\" height=\"109\" \/>\u00a0(63)<\/p>\n<p>Suppose now that Equation (62) results in a <em>V<sub>GG<\/sub><\/em> that has the <em>opposite polarity <\/em>of <em>V<sub>DD<\/sub><\/em>. It is not possible to solve for <em>R<sub>1<\/sub><\/em> and <em>R<sub>2<\/sub><\/em>. The practical way to proceed is to let <em>V<sub>GG<\/sub><\/em> = 0 V. Thus,\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2183\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-63V.png\" alt=\"\" width=\"53\" height=\"23\" \/> \u00a0. Since <em>V<sub>GG<\/sub><\/em> is specified by Equation (62) , the previously calculated value of <em>R<sub>S<\/sub><\/em> now needs to be modified.<\/p>\n<div id=\"attachment_2188\" style=\"width: 245px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2188\" class=\"wp-image-2188 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-41.png\" alt=\"\" width=\"235\" height=\"231\" \/><p id=\"caption-attachment-2188\" class=\"wp-caption-text\">Figure 41 \u2013 CS amplifier<\/p><\/div>\n<p>In Figure 41, where a capacitor is used to bypass a part of <em>R<sub>S<\/sub><\/em>, we develop the new value of <em>R<sub>S<\/sub><\/em> as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2185\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-64.png\" alt=\"\" width=\"160\" height=\"71\" \/>\u00a0(64)<\/p>\n<p>The value of <em>R<sub>Sdc<\/sub><\/em> is <em>R<sub>S1<\/sub><\/em> + <em>R<sub>S2<\/sub><\/em> and the value of <em>R<sub>Sac<\/sub><\/em> is <em>R<sub>S1<\/sub><\/em>.<\/p>\n<p>Now that we have a new <em>R<sub>Sdc<\/sub><\/em>, we must repeat several earlier steps in the design. We once again determine <em>R<sub>D<\/sub><\/em> using KVL for the drain-to-source loop.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2187\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-65.png\" alt=\"\" width=\"93\" height=\"23\" \/>\u00a0(65)<\/p>\n<p>The design problem now becomes one of calculating both <em>R<sub>S1<\/sub><\/em> and <em>R<sub>S2<\/sub><\/em> instead of finding only one source resistor.<\/p>\n<p>With a new value for <em>R<sub>D<\/sub><\/em> of <em>K<\/em><sub>1<\/sub><em> &#8211; R<sub>Sdc<\/sub><\/em>, we go to the voltage gain expression of Equation (60) with <em>R<sub>Sac<\/sub><\/em> used for this <em>ac<\/em> equation rather than <em>R<sub>S<\/sub><\/em>. The following additional steps must be added to the design procedure:<\/p>\n<p>We find <em>R<sub>Sac<\/sub><\/em> (which is simply <em>R<sub>S1<\/sub><\/em>) from the voltage gain equation<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2190\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-66.png\" alt=\"\" width=\"114\" height=\"48\" \/>\u00a0(66)<\/p>\n<p><em>R<sub>Sac<\/sub><\/em> is the only unknown in this equation. Solving for this, we find<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2192\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-67.png\" alt=\"\" width=\"137\" height=\"51\" \/>\u00a0(67)<\/p>\n<p>Suppose now that <em>R<sub>Sac<\/sub><\/em> is found to be positive, but less than <em>R<sub>Sdc<\/sub><\/em>. This is the desirable condition since<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2193\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-68.png\" alt=\"\" width=\"105\" height=\"25\" \/>\u00a0(68)<\/p>\n<p>Then our design is complete and<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2195\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-69.png\" alt=\"\" width=\"88\" height=\"25\" \/>\u00a0 (69)<\/p>\n<p>Suppose that <em>R<sub>Sac<\/sub><\/em> is found to be positive but <em>greater<\/em> than <em>R<sub>Sdc<\/sub><\/em>. The amplifier cannot be designed with the voltage gain and Q-point as selected. A new Q-point must be selected. If the voltage gain is too high, it may not be possible to effect the design with any Q-point. A different transistor may be needed or the use of two separate stages may be required.<\/p>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">10.2 The CD Amplifier<\/span><\/b><\/span><\/em><\/h6>\n<p>We now present the design procedure for the CD JFET amplifier. The following quantities are specified: current gain, load resistance, and <em>V<sub>DD<\/sub><\/em>. Input resistance may be specified instead of current gain. Refer to the circuit of Figure 39 as you study the following procedure. Once again, we remind you that the process of reducing the theory to a set of steps is the important part of this discussion &#8211; not the actual steps.<\/p>\n<p>First select a Q-point in the center of the FET characteristic curves with the aid of Figure 20 (&#8220;Chapter 3: Junction field-effect transistor (JFET)&#8221;). This step determines <em>V<sub>DSQ<\/sub><\/em>, <em>V<sub>GSQ<\/sub><\/em>, <em>I<sub>DQ<\/sub><\/em> and <em>g<sub>m<\/sub><\/em>.<\/p>\n<p>We can solve for the resistor connected to the source by writing the <em>dc<\/em> KVL equation around the drain-to-source loop.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2200\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-70.png\" alt=\"\" width=\"125\" height=\"26\" \/>\u00a0(70)<\/p>\n<p>from which we find the <em>dc<\/em> value of <em>R<sub>S<\/sub><\/em>,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2201\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-71.png\" alt=\"\" width=\"112\" height=\"44\" \/>\u00a0(71)<\/p>\n<p>We next find the <em>ac<\/em> value of resistance, <em>R<sub>Sac<\/sub><\/em>, from the rearranged current gain equation, Equation (55).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2202\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-72.png\" alt=\"\" width=\"166\" height=\"54\" \/>\u00a0(72)<\/p>\n<p>where <em>R<sub>G<\/sub><\/em> = <em>R<sub>in<\/sub><\/em><sub>. <\/sub>If the input resistance is not specified, let <em>R<sub>Sac<\/sub><\/em> = <em>R<sub>Sdc<\/sub><\/em> and calculate the input resistance from Equation (72) . If the input resistance is not high enough, it may be necessary to change the Q-point location.<\/p>\n<p>If <em>R<sub>in<\/sub><\/em> is specified, it is necessary to calculate <em>R<sub>Sac<\/sub><\/em> from Equation (72). In such cases, <em>R<sub>Sac<\/sub><\/em> is different from <em>R<sub>Sdc<\/sub><\/em>, so we bypass part of <em>R<sub>S<\/sub><\/em> with a capacitor.<\/p>\n<p>We now turn our attention to the input bias circuitry. We determine <em>V<sub>GG<\/sub><\/em> using the equation,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2204\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-73.png\" alt=\"\" width=\"120\" height=\"25\" \/>\u00a0(73)<\/p>\n<p>No phase inversion is produced in a source follower FET amplifier and <em>V<sub>GG<\/sub><\/em> is normally of the same polarity as the supply voltage.<\/p>\n<p>Now that <em>V<sub>GG<\/sub><\/em> is known, we determine the values of <em>R<sub>1<\/sub><\/em> and <em>R<sub>2<\/sub><\/em> from the Thevenin equivalent of the bias circuitry<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2205\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-74.png\" alt=\"\" width=\"109\" height=\"106\" \/>\u00a0(74)<\/p>\n<p>There is usually enough drain current in an SF to develop the opposite polarity voltage needed to offset the negative voltages required by the JFET gate. Therefore, normal voltage division biasing can be used.<\/p>\n<div id=\"attachment_2207\" style=\"width: 425px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2207\" class=\"wp-image-2207 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-44-1.png\" alt=\"\" width=\"415\" height=\"220\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-44-1.png 415w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-44-1-300x159.png 300w\" sizes=\"auto, (max-width: 415px) 100vw, 415px\" \/><p id=\"caption-attachment-2207\" class=\"wp-caption-text\">Figure 44 \u2013 CD amplifier with part of RS bypassed<\/p><\/div>\n<p>We now return to the problem of specifying the input resistance. We can assume that part of <em>R<sub>S<\/sub><\/em> is bypassed, as in Figure 44, which leads to different values of <em>R<sub>Sac<\/sub><\/em> and <em>R<sub>Sdc<\/sub><\/em>. We use Equation (71) to solve for <em>R<sub>Sdc<\/sub><\/em>. Next, we let <em>R<sub>G<\/sub><\/em> equal the specified value of <em>R<sub>in<\/sub><\/em>, and use Equation (72) to solve for <em>R<sub>Sac<\/sub><\/em>.<\/p>\n<p>If the <em>R<sub>Sac<\/sub><\/em> calculated above is smaller than <em>R<sub>Sdc<\/sub><\/em>, the design is accomplished by bypassing <em>R<sub>S2<\/sub><\/em> with a capacitor. Remember that <em>R<sub>Sac<\/sub><\/em> = <em>R<sub>S1<\/sub><\/em> and <em>R<sub>Sdc<\/sub><\/em> = <em>R<sub>S1<\/sub><\/em> + <em>R<sub>S2<\/sub><\/em>. If on the other hand, <em>R<sub>Sac<\/sub><\/em> is larger than <em>R<sub>Sdc<\/sub><\/em>, the Q-point must be moved to a different location. We select a smaller <em>V<sub>DS<\/sub><\/em> thus causing increased voltage to be dropped across <em>R<sub>S1<\/sub><\/em> + <em>R<sub>S2<\/sub><\/em>, which makes <em>R<sub>Sdc<\/sub><\/em> larger. If <em>V<sub>DS<\/sub><\/em> cannot be reduced sufficiently to make <em>R<sub>Sdc<\/sub><\/em> larger than <em>R<sub>Sac<\/sub><\/em>, then the amplifier cannot be designed with\u00a0\u00a0the given current gain, <em>R<sub>in<\/sub><\/em>, and FET type. One of these three specifications must be changed, or a second amplifier stage must be used to provide the required gain.<\/p>\n<h6><em><span style=\"color: #000000; font-family: Bodoni MT,serif;\"><b><span lang=\"en-US\">10.3 The SF Bootstrap Amplifier<\/span><\/b><\/span><\/em><\/h6>\n<p>We now examine a variation of the CD amplifier known as the <em>SF (or CD) bootstrap FET amplifier<\/em>. This circuit is a special case of the SF called the <em>bootstrap circuit <\/em>and is illustrated in Figure 45.<\/p>\n<p>Here the bias is developed across only a part of the source resistor. This reduces the need for a capacitor bypass across part of the source resistor and thus attains a much larger input resistance than normally can be attained. This design allows us to take advantage of the high impedance characteristics of the FET without using a high value of gate resistor, <em>R<sub>G<\/sub><\/em>.<\/p>\n<p>The equivalent circuit of Figure 46 is used to evaluate the circuit operation<\/p>\n<div id=\"attachment_2213\" style=\"width: 425px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2213\" class=\"wp-image-2213 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-45.png\" alt=\"Bootstrap source follower\" width=\"415\" height=\"233\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-45.png 415w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersFigure6-45-300x168.png 300w\" sizes=\"auto, (max-width: 415px) 100vw, 415px\" \/><p id=\"caption-attachment-2213\" class=\"wp-caption-text\">Figure 45 \u2013 Bootstrap source follower<\/p><\/div>\n<p>We assume that <em>i<sub>in<\/sub><\/em> is sufficiently small to approximate the current in <em>R<sub>S2<\/sub><\/em> as <em>i<sub>1<\/sub><\/em>. The output voltage is then found to be<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2214\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-75.png\" alt=\"\" width=\"140\" height=\"31\" \/>\u00a0(75)<\/p>\n<p>where<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2215\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-76.png\" alt=\"\" width=\"101\" height=\"25\" \/>\u00a0(76)<\/p>\n<p>If the assumption about <em>i<sub>in<\/sub><\/em> is not valid, \u00a0is replaced by the expression<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2216\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-77.png\" alt=\"\" width=\"115\" height=\"46\" \/>\u00a0(77)<\/p>\n<p>A KVL equation at the input yields <em>v<sub>in<\/sub><\/em> as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2218\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-78.png\" alt=\"\" width=\"190\" height=\"25\" \/>\u00a0(78)<\/p>\n<p>The current, <em>i<sub>1<\/sub><\/em>, is found from a current-divider relationship,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2219\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-79.png\" alt=\"\" width=\"101\" height=\"46\" \/>\u00a0(79)<\/p>\n<p>Combining Equations (79) and (78) yields,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2220\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-80.png\" alt=\"\" width=\"198\" height=\"40\" \/>\u00a0(80)<\/p>\n<p>A second equation for <em>v<sub>in<\/sub><\/em> is developed around the loop through <em>R<sub>G<\/sub><\/em> and <em>R<sub>S2<\/sub><\/em> as follows.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2222 size-full\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-81V.png\" alt=\"\" width=\"420\" height=\"50\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-81V.png 420w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-81V-300x36.png 300w\" sizes=\"auto, (max-width: 420px) 100vw, 420px\" \/>\u00a0(81)<\/p>\n<p>We eliminate <em>v<sub>in<\/sub><\/em> by setting Equation (80) equal to Equation (81) and solve for <em>i<sub>in<\/sub><\/em> to obtain<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2224\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-82.png\" alt=\"\" width=\"260\" height=\"53\" \/>\u00a0(82)<\/p>\n<p>The input resistance, <em>R<sub>in<\/sub><\/em> = <em>v<sub>in<\/sub><\/em>\/<em>i<sub>in<\/sub><\/em>, is found by dividing Equation (81) by Equation (82) with the result,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2225\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-83-300x50.png\" alt=\"\" width=\"300\" height=\"50\" srcset=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-83-300x50.png 300w, https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-83.png 323w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/>\u00a0(83)<\/p>\n<p><em>R<sub>G<\/sub><\/em> is the only unknown in this equation, so we can solve to obtain,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2226\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-84.png\" alt=\"\" width=\"269\" height=\"53\" \/>\u00a0(84)<\/p>\n<p>The current gain is<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2227\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-85.png\" alt=\"\" width=\"115\" height=\"47\" \/>\u00a0(85)<\/p>\n<p>We can now use the equations derived earlier along with the observation that <em>R<sub>S <\/sub><\/em>&#8211; <em>R<sub>S2 <\/sub><\/em>= <em>R<sub>S1<\/sub><\/em> in order to solve for the current gain.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2228\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-86.png\" alt=\"\" width=\"194\" height=\"42\" \/>\u00a0(86)<\/p>\n<p>The voltage gain is<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2229\" src=\"https:\/\/www.tina.com\/resources\/wp-content\/uploads\/2018\/05\/Field-effectTransistorAmplifiersEq6-87.png\" alt=\"\" width=\"280\" height=\"46\" \/>\u00a0(87)<\/p>\n<p>Note that the denominator in Equation (84) is larger than the numerator, thus showing that <em>R<sub>G <\/sub><\/em>&lt; (<em>R<sub>in<\/sub><\/em>&#8211;<em>R<sub>S2<\/sub><\/em>). This proves that a large input resistance can be attained without having the same order of size as <em>R<sub>G<\/sub><\/em>.<\/p>\n<div class=\"previous-page\"><a href=\"https:\/\/www.tina.com\/resources\/home\/field-effect-transistor-amplifiers-2\/9-fet-amplifier-analysis\/\"><img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/leftarrow5.png\" \/>PREVIOUS- 9. FET Amplifier analysis<\/a><\/div>\n<div class=\"next-page\"><a style=\"font-size: 20px;\" href=\"https:\/\/www.tina.com\/resources\/home\/field-effect-transistor-amplifiers-2\/11-other-devices\/\">NEXT-11. Other devices<img decoding=\"async\" src=\"https:\/\/www.tina.com\/images\/resource\/rightarrow5.png\" \/><\/a><\/div>\n<div class=\"clear\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>CURRENT &#8211; 10. FET Amplifier design PREVIOUS- 9. FET Amplifier analysis NEXT-11. Other devices FET Amplifier design We now explore the extension of the FET amplifier analysis presented earlier in this chapter to the design of FET amplifiers. We will attempt to define the unknowns in the design problem, and then develop equations for solving [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1764,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"yst_prominent_words":[],"class_list":["post-2159","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/2159","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/comments?post=2159"}],"version-history":[{"count":43,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/2159\/revisions"}],"predecessor-version":[{"id":3456,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/2159\/revisions\/3456"}],"up":[{"embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/pages\/1764"}],"wp:attachment":[{"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/media?parent=2159"}],"wp:term":[{"taxonomy":"yst_prominent_words","embeddable":true,"href":"https:\/\/www.tina.com\/resources\/wp-json\/wp\/v2\/yst_prominent_words?post=2159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}