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barcode reader vb.net codeproject Part II in Software
Part II Quick Response Code Recognizer In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. QRCode Generator In None Using Barcode generation for Software Control to generate, create Quick Response Code image in Software applications. Electronics
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Create ECC200 In None Using Barcode maker for Software Control to generate, create Data Matrix ECC200 image in Software applications. Painting Universal Product Code Version A In None Using Barcode creator for Software Control to generate, create UPCA Supplement 5 image in Software applications. (1035) Print UCC  12 In None Using Barcode maker for Software Control to generate, create GTIN  128 image in Software applications. Code 39 Full ASCII Creation In None Using Barcode printer for Software Control to generate, create Code39 image in Software applications. + VGSQ S R2 VDD RS
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EAN13 Supplement 5 Generation In Java Using Barcode creator for Eclipse BIRT Control to generate, create EAN13 image in BIRT reports applications. Bar Code Recognizer In Java Using Barcode decoder for Java Control to read, scan read, scan image in Java applications. Design a commonsource 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 gatesource and drainsource 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 ComputerAided Solutions: The calculations carried out in the present example may also be found in electronic form in a MathcadTM le in the accompanying CDROM
Substituting the smallsignal model in the commonsource ampli er circuit of Figure 1024(b), we obtain the smallsignal 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 + ~ _

