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Figure 1029 AC circuit for the common-source MOSFET ampli er
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have: VGS (t) = and VGS (t) v(t) (1040) (R1 R2 ) v(t) R + (R1 R2 ) (1039)
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because we have purposely selected R1 and R2 to be quite large in the design of the DC bias circuit and therefore R1 R2 R Thus, we can express the load (output) voltage in terms of the input voltage v(t) as vL (t) = gm (RD RL ) v(t) (1041)
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This expression corresponds to a voltage gain of AV = vL (t) = gm (RD v(t) RL ) (1042)
Note that we have an upper bound on the voltage gain for this ampli er, AVmax , that is independent of the value of the load resistance: |AVmax | gm RD (1043)
This upper bound is achieved, of course, only when the load is an open circuit This open-circuit voltage gain plays the same role in the analysis of the MOSFET ampli er as the parameter we de ned in equation 1026 for BJT ampli ers: = gm RD (1044)
If we compute the value of gm for the MOSFET ampli er of Example 108 (gm = 01033 A/V), we nd that the open-loop voltage gain of this ampli er is = gm RD = 1033 If we computed a typical value of transconductance, a comparable BJT design could yield a voltage gain around 200 A further disadvantage of the MOSFET ampli er is, of course, the much more nonlinear drain characteristic Why then use MOSFETs as ampli ers if a BJT can supply both higher gain and improved linearity The answer lies in the large input impedance of FET ampli ers (theoretically in nite in MOSFETs) In the MOSFET common-source ampli er, the current drawn by the input is given by the expression iin = v(t) v(t) R + RG RG vL (t) AV v(t) = RL RL (1045)
while the output current is given by iL (t) = (1046)
Thus, the current gain is given by the following expression: AI = RD RL RG iL AV R G = = gm iin RL RD + R L RL
RD RG = gm RD + R L
(1047)
10
Transistor Ampli ers and Switches
Note that we have purposely made RG large (100 k ), to limit the current required of the signal source, v(t) Thus, the current gain for a common-source ampli er can be signi cant For a 100- load, we nd that the effective voltage gain for the common-source ampli er is AV = gm RD RL 5
and the corresponding current gain is AI = gm RD R G = 5,000 RD + R L
for the design values quoted in Example 108 Thus, the net power gain of the transistor, AP , is quite signi cant AP = AV AI = 25,000 This brief discussion of the common-source ampli er has pointed to some important features of MOSFETs We rst noted that the transconductance of a MOSFET is highly variable However, MOSFETs make up for this drawback by their inherently high input impedance, which requires very little current of a signal source and thus affords substantial current gains Finally, the output resistance of this ampli er can be made quite small by design Example 109 provides an illustration of the ampli cation characteristics of a power MOSFET
EXAMPLE 109 Analysis of a Power MOSFET Common-Source Ampli er
Problem
Design a common-source power MOSFET ampli er (Figure 1024(b)) using a BS170 transistor to operate at a speci ed Q point Compute the voltage and current gain of the ampli er
Solution
Known Quantities: Drain supply voltage; load resistance; MOSFET transconductance and threshold voltage; desired drain currrent and drain-source voltage Find: R1 , R2 , RD , RS , AV , and AI Schematics, Diagrams, Circuits, and Given Data: VDD = 25 V; RL = 80 ; gm = 200 mA/V for VDS = 10 V and ID = 250 mA; VT = 08 V The device data sheets for the BS170 transistor may be found in the accompanying CD-ROM Assumptions: All currents are expressed in mA and all resistors in k Analysis: Knowing the transconductance and threshold voltage of the MOSFET allows
us to compute the values of k and IDSS (see equations in Table 91 and in the following paragraph): mA 2 1 gm V 1 2002 mA k= = = 40 2 4 IDQ mA 4 250 V
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