Figure 12.87 shows a model used by one manufacturer for air-core rf coils. The resistance rL accounts for skin effect and is given by rL=k√f, where k is an experimentally determined parameter. In Fig. 12.87, let RC=6 Ω, RS =1 mΩ, ,C=68 fF=68×10^−15 F, L=1 nH, and k=3.4×10−6 ΩHz^−1/2.

Question

Figure 12.87 shows a model used by one manufacturer for air-core rf coils. The resistance [latex]r_L[/latex] accounts for skin effect and is given by [latex]r_L=k \sqrt{f}[/latex], where k is an experimentally determined parameter.

In Fig. 12.87, let [latex]R_C=6 \Omega, R_S=1 m \Omega[/latex], [latex]C=68 fF =68 imes 10^{-15} F , L=1 nH , [/latex] and [latex]k=3.4 imes 10^{-6} \Omega Hz ^{-1 / 2}[/latex]. Construct a graph of the magnitude in dB of the equivalent impedance [latex]Z_{ab}[/latex] at the terminals a-b versus frequency [latex]f[/latex] , for [latex]100 MHz \leq f \leq 100 GHz[/latex]. Use a logarithmic scale for frequency and normalize the impedance to the impedance at the self-resonant frequency. Show also a graph of the magnitude in dB of the impedance of an ideal inductor having inductance L , also normalized to the impedance at the self-resonant frequency.

Accepted Answer

From (12.88), the self-resonant frequency of the circuit is given by

[latex] \omega_r=\frac{1}{\sqrt{L C}} \sqrt{\frac{R_L^2 C-L}{R_C^2 C-L}} [/latex] (12.88)

[latex] \begin{aligned} f_r &=\frac{1}{2 \pi \sqrt{L C}} \sqrt{\frac{r_L^2\left(f_r ight) C-L}{R_C^2 C-L}} \\ &=\frac{1}{2 \pi \sqrt{L C}} \sqrt{\frac{k^2 f_r C-L}{R_C^2 C-L}} \end{aligned} [/latex]

which yields

[latex] \begin{aligned} f_r &=\frac{-k^2 C+\sqrt{k^4 C^2-16 \pi^2 L^2 C\left(R_C^2 C-L ight)}}{8 \pi^2 L C\left(L-R_C^2 C ight)} \\ & \cong 19.3 GHz . \end{aligned}[/latex]

From Fig. 12.87, the equivalent impedance at the terminals a-b is given by

[latex] \begin{aligned} Z &=R_S+\frac{1}{\left[R_C+(j 2 \pi f C)^{-1} ight]^{-1}+[k \sqrt{f}+j 2 \pi f L]^{-1}} \ &=R_S+\frac{1}{\left[R_C+(j \omega C)^{-1} ight]^{-1}+[k \sqrt{\omega /(2 \pi)}+j \omega L]^{-1}} . \end{aligned} [/latex] (12.121)

The equivalent impedance at the selfresonant frequency [latex] f_r [/latex] is

[latex] Z_r=Z\left(j \omega_r ight)=2.27 k \Omega [/latex]

where [latex] \omega_r=2 \pi f_r [/latex] . Figure 12.88 shows graphs of the magnitude of the normalized impedance

of the model in Fig. 12.87 (solid line) and of the normalized impedance of an ideal inductor having inductance L (dashed line), where Z is given by (12.121). The normalized impedance (in dB ) of the model and that of the associated ideal inductor are given by

[latex] \begin{aligned} |Z|( dB ) &=20 \log \left|\frac{Z(j \omega)}{Z\left(j \omega_r ight)} ight| \\ \left|Z_L ight|( dB ) &=20 \log \left|\frac{j \omega L}{Z\left(j \omega_r ight)} ight| \end{aligned}[/latex]

respectively. Figure 12.88 indicates that the physical inductor approximates an ideal inductor for [latex]10 MHz < f <10 GHz[/latex] . Below 10 MHz , the series resistance [latex] R_S [/latex] becomes significant and above 10 GHz , the shunt capacitance is significant.

Question 12.51: Figure 12.87 shows a model used by one manufacturer for ai...

Related Answered Questions

In Fig. 12.38, i(t)=I1cos(ω1t)+I2cos(ω2t+θ) , where ω1≠ω2. Use superposition to obtain an expression for the voltage ν . ...

Verified Answer:

Verified Answer:

Design a gyrator-based tuned circuit subject to the following specifications: resonant frequency =f0=400Hz , quality factor =Q=10, impedance at resonance =Z0=100Ω. ...

Verified Answer:

The temperature coefficient for a certain 10kΩ resistor is 500 ppm at 20°C . Construct a graph showing resistance versus temperature T for 20°C≤T≤200°C , expressed as a percent of the nominal resistance. ...

Verified Answer:

A manufacturer’s specifications for one of his offerings are L=22μHQ=50,fT=2.52MHz , and SRF=13MHz . (a) What is the ac resistance of the coil at the operating frequency? (b) What (approximately) is the shunt (stray) capacitance of the coil? ...

Verified Answer:

In Fig. 12.69, RL=2.2kΩ and νS=VS[cos(ω1t)+cos(ω2t)] with f1=100kHz and f2=200 kHz. The circuit is intended to block the 200 kHz component and pass the 100 kHz component of the source voltage. You have available a selection of 200 and 100 kHz coils with inductances ranging from 10 to 200 ...

Verified Answer:

btain an expression for the resonant frequency of the circuit shown in Fig. 12.65 ...

Verified Answer:

Refer to Fig. 12.63. Draw a qualitative phasor diagram for the currents IR,IL,IC when the frequency of the sinusoidal source equals the resonant frequency of the parallel L C portion of the circuit. Use the source current I for the phase reference. ...

Verified Answer:

In Fig. 12.91(a), C=100 μF, ν(t)=Vcos(ωt) with f=60 Hz and V=15 V . The dissipation factor for the capacitor is 0.03 . The ESL and parallel resistance Rp can be neglected. Find the power dissipated in the capacitor (in the ESR ). ...

Verified Answer:

Refer to Fig. 12.56(a), where i=Icos(ωt). Obtain an expression for the phasor voltage across the inductor. Check the result by obtaining expressions for the voltage for f=0 and f→∞. ...

Verified Answer: