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Making QR Code ISO/IEC18004 in Software Comments: Note how the resistance of the coil wire is relatively insigni cant This is

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true because the inductor is rather large; wire resistance can become signi cant for very small inductance values At high frequencies, a capacitance should be added to the model because of the effect of the insulator separating the coil wires
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EXAMPLE 413 Impedance of a More Complex Circuit
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Find the equivalent impedance of the circuit shown in Figure 438
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R1 L 100 10 mH ZEQ 50 R2 C 10 F
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Known Quantities: = 104 rad/s; R1 = 100
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; L = 10 mH; R2 = 50
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, C = 10 F
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Find: The equivalent impedance of the series-parallel circuit Analysis: We determine rst the parallel impedance of the R2 -C circuit, Z||
Z|| = R2 =
1 R2 j C 1 R2 = = 1 j C 1 + j CR2 R2 + j C
50 50 = = 192 j 962 1 + j 104 10 10 6 50 1 + j5
= 981 13734 Next, we determine the equivalent impedance, Zeq : Zeq = R1 + j L + Z|| = 100 + j 104 10 2 + 192 j 962 = 10192 + j 9038 = 1362 0723 Is this impedance inductive or capacitive in nature
Comments: At the frequency used in this example, the circuit has an inductive
impedance, since the reactance is positive (or, alternatively, the phase angle is positive)
Part I
Circuits
Capacitive Displacement Transducer
Earlier, we introduced the idea of a capacitive displacement transducer when we considered a parallel-plate capacitor composed of a xed plate and a movable plate The capacitance of this variable capacitor was shown to be a nonlinear function of the position of the movable plate, x (see Figure 46) In this example, we show that under certain conditions the impedance of the capacitor varies as a linear function of displacement that is, the movable-plate capacitor can serve as a linear transducer Recall the expression derived earlier: 8854 10 3 A C= pF x where C is the capacitance in pF, A is the area of the plates in mm2 , and x is the (variable) distance in mm If the capacitor is placed in an AC circuit, its impedance will be determined by the expression 1 ZC = j C so that x ZC = j 8854A Thus, at a xed frequency , the impedance of the capacitor will vary linearly with displacement This property may be exploited in the bridge circuit of Figure 47, where a differential pressure transducer was shown as being made of two movable-plate capacitors, such that if the capacitance of one increased as a consequence of a pressure differential across the transducer, the capacitance of the other had to decrease by a corresponding amount (at least for small displacements) The circuit is shown again in Figure 439, where two resistors have been connected in the bridge along with the variable capacitors (denoted by C(x)) The bridge is excited by a sinusoidal source
d R1
+ ~
FOCUS ON MEASUREMENTS
Cdb(x) b
vS(t)
Vout +
R2 c
Cbc(x)
Figure 439 Bridge circuit for capacitive displacement transducer
Using phasor notation, we can express the output voltage as follows: ZCbc (x) R2 Vout (j ) = VS (j ) ZCdb (x) + ZCbc (x) R1 + R 2
4
AC Network Analysis
If the nominal capacitance of each movable-plate capacitor with the diaphragm in the center position is given by A C= d where d is the nominal (undisplaced) separation between the diaphragm and the xed surfaces of the capacitors (in mm), the capacitors will see a change in capacitance given by A A and Cbc = Cdb = d x d +x when a pressure differential exists across the transducer, so that the impedances of the variable capacitors change according to the displacement: d x d +x and ZCbc = ZCdb = j 8854A j 8854A and we obtain the following expression for the phasor output voltage: d +x R2 j 8854A Vout (j ) = VS (j ) d x d +x R1 + R 2 + j 8854A j 8854A = VS (j ) = VS (j ) x R2 1 + 2 2d R1 + R 2
x 2d if we choose R1 = R2 Thus, the output voltage will vary as a scaled version of the input voltage in proportion to the displacement A typical vout (t) is displayed in Figure 440 for a 005-mm triangular diaphragm displacement, with d = 05 mm and VS a 25-Hz sinusoid with 1-V amplitude
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