barcode reader vb.net codeproject Figure 1030 MOSFET source follower and DC circuit in Software

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Figure 1030 MOSFET source follower and DC circuit
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D gm VGS + vout
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Figure 1031 AC equivalent circuit for the MOSFET source follower
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We can see that the open-circuit voltage gain will always be less than 1 The output resistance can be found by inspection from the circuit of Figure 1031 to be equal to RS In summary, the source-follower ampli er has an input resistance, ri = RG = R1 R2 , which can be made arbitrarily large; a voltage gain less than unity; and an output resistance, ro = RS , which can be made small by appropriate design
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EXAMPLE 1010 Analysis of MOSFET Source Follower
Problem
Find the small-signal voltage gain and the input resistance of the enhancement MOSFET source follower ampli er of Figure 1032
Part II
Electronics
Solution
Known Quantities: Drain, source and gate resistors; drain supply voltage; MOSFET k and threshold voltage parameters Find: VGSQ ; VDSQ ; IDQ AV and ri Schematics, Diagrams, Circuits, and Given Data: RG = 10 M ; RD = 10 k ; RL = 10 k ; VDD = 15 V; VT = 15 V; k = 0125 mA/V2 Assumptions: All currents are expressed in mA and all resistors in k Analysis: To determine the Q point of the ampli er we use the DC equivalent circuit
R1 C
+ v(t) ~ _
VDD RD C D G S RL vout +
shown in Figure 1033 We immediately observe that, since the input resistance of the transistor is in nite, the current IGQ must be zero Thus, there is no voltage drop across resistor RG If there is no voltage drop across RG , then the gate and drain must be at the same potential, and VGSQ = VDQ = VDSQ , since the source is grounded The drain current equation for the MOSFET in the active region is: IDQ = k VGSQ VT
= 0125 VGSQ 15
= 0125 VDSQ 15
RD + VSDQ VDD
The drain circuit imposes the condition: VDD = VDSQ + RD IDQ or 15 = VDSQ + 10IDQ Solving the two above equations we obtain the quiescent operating point of the ampli er: IDQ = 106 mA; VDSQ = 44 V Note that for these quiescent values, the MOSFET is, indeed, operating in the saturation region, since vDS > vGS VT and vGS > VT Next, we compute the transconductance parameter from equation 1034: mA V With the above parameter we can nally construct the AC equivalent circuit, shown in Figure 1034, and derive an expression for the output voltage: gm = 2k VGSQ VT = 2 0125 (44 15) = 0725 vout = and since VDS = gm VGS (RD ||RL )
RG + VGSQ
Figure 1033 DC equivalent circuit for Example 1010
D + RL vout
VGS = vin , the AC small signal voltage gain of the ampli er is: vout = vin V VDS = gm (RD ||RL ) = 0725 5 = 3625 VGS V vout vin vin (1 AV ) RG
G + vin ~ _ gm VGS S ri
AV =
To determine the input resistance of this ampli er, we determine the input current: iin = vin vout vin = RG RG 1 =
Since the input resistance is de ned as the ratio of input voltage to input current, we have: Ri = RG 106 vin vin = = = 216 M = v in iin (1 AV ) 4625 (1 AV ) RG
Comments: By selecting a very large value for RG (with no effect on the Q point) it is possible to design MOSFET ampli ers that have remarkably high input impedance, though their voltage gain is modest This is a very useful feature in designing input stages for multistage ampli ers, as we shall see in the next section
10
Transistor Ampli ers and Switches
Check Your Understanding
108 Compute the quiescent drain current and the small-signal transconductance parameter for a MOSFET with VT = 14 V, k = 2 mA/V2 , for the following values of VGS : 18 V, 20 V, 22 V, 24 V, 26 V, 28 V, 30 V 109 Find the open-circuit AC voltage gain, , for the ampli er design of Example 108 What is the effective voltage gain of the ampli er if the load resistance is 1 k 1010 Repeat Exercise 109 for the values of gm found in Exercise 108
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