vb.net barcode reader source code RT = R1 + R2 + (R3 R4 ) = 20 + 20 + (10 50) = 4833 VT = R2 10 12 = 2 V VS = R1 + R 2 60 in Software

Generation QR Code JIS X 0510 in Software RT = R1 + R2 + (R3 R4 ) = 20 + 20 + (10 50) = 4833 VT = R2 10 12 = 2 V VS = R1 + R 2 60

RT = R1 + R2 + (R3 R4 ) = 20 + 20 + (10 50) = 4833 VT = R2 10 12 = 2 V VS = R1 + R 2 60
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Semiconductors and Diodes
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54 Diode current, mA 42 30 18 6 0 02 06 10 14 Diode voltage, V 18
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RT = 4833 iD VT = 2 V D1 + vD _
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Figure 832 1N914 diode i-v curve
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The equivalent circuit is shown in Figure 833 Next we plot the load line (see Figure 830), with y intercept VT /RT = 41 mA, and with x intercept VT = 2 V; the diode curve and load line are shown in Figure 834 The intersection of the two curves is the quiescent (Q) or operating point of the diode, which is given by the values VQ = 067 V, IQ = 275 mA
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60 54 48 42 Diode current, mA 36 30 24 21 18 12 6 0 0 02 04 06 08 10 12 Diode voltage, V 14 16 18 20 Q point Load line
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Figure 834 Superposition of load line and diode i-v curve
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To determine the battery power output, we observe that the power supplied by the battery is PB = 12 IB and that IB is equal to current through R1 Upon further inspection, we see that the battery current must, by KCL, be equal to the sum of the currents through R2 and through the diode We already know the current through the diode, IQ To determine the current through R2 , we observe that the voltage across R2 is equal to the sum of the voltages across R3 , R4 and D1 : VR2 = IQ (R3 + R4 ) + VQ = 0021 40 + 1 = 184 V and therefore the current through R2 is IR2 = VR2 /R2 = 0184 A Finally, PB = 12 IB = 12 (0021 + 0184) = 12 0205 = 246 W
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Part II
Electronics
Comments: Graphical solutions are not the only means of solving the nonlinear equations that result from using a nonlinear model for a diode The same equations could be solved numerically by using a nonlinear equation solver The code in Electronics WorkbenchTM accomplishes exactly this task
Piecewise Linear Diode Model The graphical solution of diode circuits can be somewhat tedious, and its accuracy is limited by the resolution of the graph; it does, however, provide insight into the piecewise linear diode model In the piecewise linear model, the diode is treated as an open circuit in its off state, and as a linear resistor in series with V in the on state Figure 835 illustrates the graphical appearance of this model Note that the straight line that approximates the on part of the diode characteristic is tangent to the Q point Thus, in the neighborhood of the Q point, the diode does act as a linear small-signal resistance, with slope given by 1/rD , where 1 iD = rD vD (816)
(IQ ,VQ )
iD V s R
Diode off circuit model rD 1 rD Q point V Diode on circuit model
That is, it acts as a linear resistance whose i-v characteristic is the tangent to the diode curve at the operating point The tangent is extended to meet the voltage axis, thus de ning the intersection as the diode offset voltage Thus, rather than represent the diode by a short circuit in its forward-biased state, we treat it as a linear resistor, with resistance rD The piecewise linear model offers the convenience of a linear representation once the state of the diode is established, and of a more accurate model than either the ideal or the offset diode model This model is very useful in illustrating the performance of diodes in real-world applications
Vs VD
Figure 835 Piecewise linear diode model
EXAMPLE 86 Computing the Incremental (Small-Signal) Resistance of a Diode
Problem
Determine the incremental resistance of a diode using the diode equation
Solution
Known Quantities: I0 = 10 14 A; kT /q = 0025 V (at T = 300 K); IQ = 50 mA Find: The diode small signal resistance, rD Assumptions: Use the approximate diode equation (equation 87) Analysis: The approximate diode equation relates diode voltage and current according to:
iD = I0 eqD /kT From the above expression we can compute the incremental resistance using equation 816: iD 1 = rD vD =
(IQ ,VQ )
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