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and by assuming that no current ows into the gate The voltage VGSQ , which controls the gate-to-source bias, is then given by the expression VGSQ = VGG IDQ RS (1049)

programming in C#, you can also try using the cast operator. However, if you are using VB .NET you should know that it does not have the cast option.

so that the desired Q point may be established by selecting appropriate values of R1 , R2 , and RS (Example 1010 will illustrate a typical design) The AC equivalent circuit is shown in Figure 1031 Since it is possible to select R1 and R2 arbitrarily, we see that the input resistance, RG , of the source follower can be made quite large Another observation is that, for this type of ampli er, the voltage gain is always less than 1 This fact may be veri ed by considering that VGS = vin (t) gm VGS RS which implies (1050)

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VGS 1 = vin (t) 1 + g m RS Since the AC output voltage is related to vout = gm VGS RS it follows that vout = gm RS VGS Thus, the open-circuit voltage gain, , may be obtained as follows: vout vout 1 = = (gm RS ) vin VGS 1 + g m RS = gm RS <1 gm R S + 1 VGS by

(1051)

Table 5-2

(1052)

+ vin ~ _

The C# cast operator as another way to do type conversions. You can think of operators as methods that use a symbol instead of a name. In this case, the symbol you use is (), with the data type you want to convert to indicated in the middle. You will see the cast operator used frequently in the C# examples of many books. Like the ToString() method, the cast operator is also part of the data types themselves. Using cast will not always work the same, depending on the type you are trying to convert to. Once again, if the type you are using does not convert the data in the manner you want, you must use another choice. To create a cast, you place the parentheses before the type you want to convert and the type you want to convert to within the parentheses. C# double x = 123.4; int y; y = (int)x; //cast double to int. Note that the .4 will be lost Console.WriteLine(y.ToString()); //Displays 123

(1053)

(1054)

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

Find the small-signal voltage gain and the input resistance of the enhancement MOSFET source follower ampli er of Figure 1032

count that functionality type) The number of occurrences is then multiplied with the multipliers, and next the total sum of function points is calculated This estimation method is rated as a rough size measure

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

The VB .NET CType()

+ v(t) ~ _

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

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