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The MOSFET Common-Source Ampli er It is useful at this stage to compare the performance of the common-source ampli er of Figure 1024(b) with that of the BJT common-emitter ampli er The DC equivalent circuit is shown in Figure 1028; note the remarkable similarity with the BJT common-emitter ampli er DC circuit The equations for the DC equivalent circuit are obtained most easily by reducing the gate circuit to a Th venin equivalent with VGG = R2 /R1 + R2 VDD e and RG = R1 R2 Then the gate circuit is described by the equation VGG = VGSQ + IDQ RS and the drain circuit by VDD = VDSQ + IDQ (RD + RS ) (1036)
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Figure 1028 DC circuit for the common-source ampli er
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D + R1 G VDSQ RD
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(1035)
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+ VGSQ S R2 VDD RS
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Note that, given VDD , any value for VGG may be achieved by appropriate selection of R1 and R2 Example 108 illustrates the computation of the Q point for a MOSFET ampli er
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EXAMPLE 108 Analysis of MOSFET Common-Source Ampli er
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Design a common-source MOSFET ampli er (Figure 1024(b)) to operate at a speci ed Q point
Solution
Known Quantities: Drain supply voltage; MOSFET threshold voltage and k; desired gate-source and drain-source voltages Find: R1 , R2 , RD , RS
VDD = 10 V VT = 14 V; k = 95 mA/V2
Schematics, Diagrams, Circuits, and Given Data: VGSQ = 24 V; VDSQ = 45 V; Assumptions: All currents are expressed in mA and all resistors in k Analysis: First, we compute the quiescent drain current for operation in the saturation
region; we know that the MOSFET is operating in the saturation region from the equations of Table 91, since vDS > vGS VT and vGS > VT IDQ = k VGSQ VT
= 95 (24 14)2 = 95 mA
Applying KVL around the gate loop requires that: VGG = VGSQ + IDQ RS = 24 95RS while the drain circuit imposes the condition: VDD = VDSQ + IDQ (RD + RS ) = 45 + 95 (RD + RS ) or 10 = 45 + 95 (RD + RS ) (RD + RS ) = 0058 k
10
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Since the choice of drain and source resistors is arbitrary at this point, we select RD = 11 and RS = 47 Then we can solve for VGG in the rst equation: VGG = 6865 V To achieve the desired value of VGG we need to select resistors R1 and R2 such that VGG = R2 VDD = 6865 V R1 + R 2
To formulate a problem with a unique solution, we can also (arbitrarily) impose the condition that R1 ||R2 = 100 k Then we can solve the two equations to obtain R1 = 319 k , R2 = 146 k The nearest standard 5 percent resistor values will be: R1 = 333 k , R2 = 150 k , resulting in R1 ||R2 = 1034 k
Comments: Since the role of the resistors R1 and R2 is simply to serve as a voltage divider, you should not be overly concerned with the arbitrariness of the choice made in the example In general, one selects rather large values to reduce current ow and therefore power consumption On the other hand, the choice of the drain and source resistances may be more delicate, as it affects the output resistance and gain of the ampli er Focus on Computer-Aided Solutions: The calculations carried out in the present
example may also be found in electronic form in a MathcadTM le in the accompanying CD-ROM
Substituting the small-signal model in the common-source ampli er circuit of Figure 1024(b), we obtain the small-signal AC equivalent circuit of Figure 1029, where we have assumed the coupling and bypass capacitors to be short circuits at the frequency of the input signal v(t) The circuit is analyzed as follows The load voltage, vL (t), is given by the expression vL (t) = ID (RD where ID = gm VGS (1038) RL ) (1037)
Thus, we need to determine VGS to write expressions for the voltage and current gains of the ampli er Since the gate circuit is equivalent to an open circuit, we
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