射频电路设计－理论与应用(英文版，课后习题答案.pdf

Chapter 1 Introduction

Problem 1.1

Compute the phase velocity and wavelength in an FR4 printed circuit board whose relative dielectric constant is 4.6 and where the operational frequency is 1.92 GHz.

Solution

The phase velocity is

$$ v _ { p } ^ { \cdot } = \frac { 1 } { \sqrt { \mu \varepsilon } } = \frac { 1 } { \sqrt { \mu _ { 0 } \varepsilon _ { 0 } } \sqrt { \varepsilon _ { r } } } = \frac { c } { \sqrt { \varepsilon _ { r } } } = \frac { 3 \times 1 0 ^ { 8 } } { \sqrt { 4 . 6 } } m / s $$

$$ \nu _ { p } = 1 . 3 9 9 \times 1 0 ^ { 8 } m / s $$

and the corresponding wave length

$$ \lambda = \frac { \nu _ { p } } { f } = \frac { 1 . 3 9 9 \times { 1 0 } ^ { 8 } m / s } { 1 . 9 2 \times { 1 0 } ^ { 9 } { 1 / s } } = 7 2 . 8 5 \times { 1 0 } ^ { - 3 } m $$

We notice that $\lambda / 1 0$ is equal to 7.29 mm and is comparable with typical circuit component $7 . 2 9$ dimensions. Therefore, distributed theory is applicable.

Problem 1.2

The current flowing in a microstrip line (assumed to be infinite and lossless) is specified to be $i ( t ) = 0 . 6 \cos ( \bar { 9 } \times 1 0 ^ { 9 } t - 5 0 0 z )$ A. Find the (a) phase velocity, (b) frequency, (c) wavelength, and (d) phasor expression of the current.

Solution

A traveling wave has the form cos $( \omega t - \beta z )$ as discussed on page 6. Therefore, we obtain

(a) $\nu _ { p } = \frac { \mathfrak { o } } { \mathfrak { s } } = \frac { 9 \times 1 0 ^ { 9 } ( 1 / s ) } { 5 0 0 ( 1 / m ) } = . 1 8 \times 1 0 ^ { 6 } m / s$

(b) $f = { \frac { \omega } { 2 \pi } } = 1 . 4 3 3 G H z$

(d) $i ( t ) = R e { I ( z ) e ^ { j \omega t } }$ where $I ( z ) = 0 . 6 e ^ { - j { \ B } z }$

Problem 1.3

A coaxial cable that is assumed lossless has a wavelength of the electric and magnetic fields $\cot \lambda = 2 0$ cm at 960 MHz. Find the relative dielectric constant of the insulation.

Solution

Here the wavelength, which is given as the ratio between phase velocity and frequency, can be expressed in terns of the speed of light and the relative dielectric constant (see Problem 1.1 and textbook page 8)

$$ \lambda = \frac { \nu _ { p } } { f } = \frac { c } { ( \sqrt { e _ { r } } ) f } $$

from which we find

$$ \varepsilon _ { r } = \left( \frac { c } { \lambda f } \right) ^ { 2 } = 2 . 4 4 $$

Problem 1.4

The electric wave field of a positive z-traveling wave in a medium with relative dielectric constant of $\theta _ { r }$ = 4 and with frequency of 5 GHz is given by

$$ E _ { x } = E _ { 0 x } \cos ( \omega t - k z ) \ \mathrm { V / m } $$

(a) Find the magnetic field if $E _ { 0 x } = 1 0 ^ { 6 } \mathrm { v / m } .$

(b) Determine phase velocity and wavelength.

(c) Compute the spatial advance of the traveling wave between time intervals 1, = 3μs and $t _ { 1 } =$ $t _ { 2 } = 7 \mu s .$

Solution

(a) From textbook page 6, equation (1.2), we use

$$ \frac { E _ { 0 x } } { H _ { 0 y } } = Z _ { 0 } = 3 7 7 \Omega \Big ( \frac { 1 } { \sqrt { \varepsilon _ { r } } } \Big ) $$

$$ H _ { 0 y } = \frac { E _ { 0 x } \sqrt { \varepsilon _ { r } } } { 3 7 7 \Omega } = 5 . 3 \times 1 0 ^ { 3 } A / m $$

(b) The phase velocity is $\nu _ { p } = c / ( \sqrt { \varepsilon _ { r } } ) = 1 . 5 \times 1 0 ^ { 8 } m / s$ , from which the wavelength follows as $\lambda = \nu _ { p } / f =$ 30mm.

$$ \omega t _ { 1 } = . k z _ { 1 } \Rightarrow z _ { 1 } = \frac { \omega t _ { 1 } } { k } = \frac { 2 \pi 5 G H z \times 3 \mu s } { \omega / \nu _ { p } } = 4 5 0 m $$

and for cos $( \omega t _ { 2 } - k z _ { 2 } ) = 1$ , it is concluded that

$$ \cot _ { 2 } = k z _ { 2 } \Rightarrow z _ { 2 } \Rightarrow \frac { \omega t _ { 2 } } { k } = \frac { 2 \pi 5 G H z \times 7 \mu s } { \omega / \nu _ { p } } = 1 0 5 0 m $$

Thus, the wave is advanced by 600 rn. The MATLAB routine shown below is used for the computation of the numerical results.

Problem 1.5

Find the frequency response of the impedance magnitude of the following series and parallel LC circuits:

Compare your results to the situation when the ideal inductance is replaced by the same inductance and a 5 Ω resistance connected in series. Assume that these circuits operate in the VHF/UHF frequency band of 30–3000 MHz.

RF Circuit Design: Theory and Applications

Chapter 1 • Introduction

Solution

The impedance of the parallel LC circuit is

$$ Z _ { 1 } = \frac { 1 } { 1 / ( j \omega L ) - j \omega C } = - j \left( \frac { 1 } { \omega C - 1 / ( \omega L ) } \right) . $$

The impedance of the series LC circuit is

$$ Z _ { 2 } = j ( \omega L - \frac { 1 } { \omega C } ) $$

For the modified inductance we obtain for the parallel circuit

$$ Z _ { 3 } = \frac { 1 } { j \omega C + 1 / ( R + j \omega L ) } $$

For the modified inductnce in series we get

$$ Z _ { 4 } = R + j ( \omega L - \frac { 1 } { \omega C } ) $$

Plotting the magnitude response of all four circuits is carried out in the MATLAB routine below.

Problem 1.5

clear all;

f = 30e6:1000:300@6;

$$ \ 1 = \ 1 0 e - 9 ; $$

C = 10e-12;

R = 5:

Z1=1./(j*(2*pi*f*C-1./(2*pi*f*L)));

Z2= j*(2*pi*f*L-1./(2*pi*f*C));

Z3= 1./(j*2*pi*f*C-1./(j*2*pi*f*L + R));

Z4= R + j*(2*pi*f*L-1./(2*pi*f*C));

subplot(2,2,1),plot(f/1e6,abs(Z1));grid;

title('Parallel LC circuit'),xlabel('frequency, MHz'),ylabel('|Z1|,Ohm');

subplot(2,2,2),plot(f/ie6,abs(Z2));grid;

title('Series LC circuit'),xlabel('frequency, MHz'),ylabel('|z2|,Ohm');

subplot(2,2,3),plot(f/1e6,abs(23));grid;

title('Parallel (L+R)C circuit'),xlabel('frequency, MHz'),ylabel('{Z3|,Ohm');

subplot(2,2,4),plot(f/1e6,abs(Z4));grid;

title('Series (L+R)C circuit'),xlabel('(frequency, MHz'),ylabel('{Z4|,Ohm'); print -deps 'pl_5.eps'

Executing the code results in the four impedance responses shown below.

Problem 1.6

For the circuit shown, derive the resonance frequency and plot the resonance frequency behavior as a function of the resistance R.

Solution

First we need to establish the impedance

$$ Z = { \frac { 1 } { j \Theta C + 1 / ( R + j \omega L ) } } = { \frac { R + j \omega L } { ( 1 - { \omega ^ { 2 } L C } ) + j \omega R C } } $$

and expanding by the conjugate complex denominator gives

$$ Z = { \frac { ( R + j \omega L ) ! ( 1 - \omega ^ { 2 } L C ) - j \omega R C ! } { ( 1 - \omega ^ { 2 } L C ) ^ { 2 } - \omega ^ { 2 } R ^ { 2 } C ^ { 2 } } } $$

Resonance implies zero reactance $\pmb { X } \big ( \infty , \big ) = \pmb { 0 } ,$ , i.e., we have to set the imaginary numerator

RF Circuit Design: Theory and Applications

Chapter I • Introduction

to zero

$$ \omega _ { r } L ( 1 - \omega _ { r } ^ { 2 } L C ) _ { - } \omega _ { r } R ^ { 2 } C = 0 $$

From which we find

$$ f _ { r } = \frac { 1 } { 2 \pi } \sqrt { \frac { 1 } { L C } - \frac { R ^ { 2 } } { L ^ { 2 } } } $$

The following MATLAB routine plots the above resonance frequency versus the resistance R. It is noted that at $R = 1 0 0 \Omega$ the resonance frequency is zero and any further increase results in imaginary frequencies; a physically impossible result.

t Problem 1.6

$$ \mathrm { ~ \tt ~ { ~ C ~ } ~ } = \mathrm { ~ \tt ~ { ~ 1 } ~ e ~ } - 1 2 : \mathrm { ~ \tt ~ { ~ 1 ~ } ~ e ~ } $$

R = 0:1:200;

£r= (1/(2*pi))*sqrt(-(R.^2)/(L^2)+1/(L*C));

plot(R,fr/1e6);grid;

title('Resonance frequency versus resistance $\cdot \cdot _ { i } .$

xlabel('Resistance, Ohm'); ylabel('fr, MHz');

print -deps 'p1_6.eps'

Problem 1.7

Repeat Problem 1.6 for the following circuit.

Solution

The impedance takes on the form

$$ Z \doteq j \omega L + \frac { R } { 1 + j \omega R C } = \frac { j \omega L + j \omega ^ { 3 } R ^ { 2 } L C ^ { 2 } - j \omega R ^ { 2 } C + R } { 1 + { \omega } ^ { 2 } R ^ { 2 } C ^ { 2 } } $$

Setting the imaginary part of the numerator to zero results in

$$ \omega _ { r } L + \omega _ { r } ^ { 3 } { R ^ { 2 } } L { C ^ { 2 } } - { \omega _ { r } R ^ { 2 } } C = 0 $$

and the resonance frequency is

$$ f _ { r } = \frac { 1 } { 2 \pi } \sqrt { \frac { 1 } { L C } - \frac { 1 } { R ^ { 2 } L ^ { 2 } } } $$

From this equation we also see that the resistance has to exceed R = $C / L = 0 . 1 m \Omega$ in order to produce physically meaningful results.

£r= (1/(2*pi))*Bqrt(-1./((R.^2)*(L^2))+1/(L*C));

semilogx(R,fr/1e6);grid;

title('Resonance frequency versus resistance');

xlabel('Resistance, Ohm'); ylabel('fr, MHz');

print -deps 'pl_7.eps'

RF Circuit Design: Theory and Applications

Chapter I · Introduction

Problem 1.8

For the following circuit we chose $R \propto ( \sqrt { L / C } ) / 2$

Find $\vert V _ { 0 } / V _ { i } \vert$ as a function of frequency and identify the dominant circuit portions for the low-, mid-, and high-frequency domains.

Solution

Basis of the voltage-frequency response is the transfer function

$$ \frac { V _ { 0 } } { V _ { i } } \equiv \frac { \displaystyle \frac { 1 } { j \omega C + 1 / ( j \omega L ) } } { R + \displaystyle \frac { 1 } { j \omega C + 1 / ( j \omega L ) } } = \frac { j \omega L } { ( R - \omega ^ { 2 } R L C ) + j \omega L } $$

Taking the absolute value and plotting it in MATLAB is shown below.

Problem 1.8

f = le6:le6:5*1e9;

$$ t = 1 0 \cdot 1 e \cdot 9 ; $$

C = 1e-12;

R = 50;

V = (j*2*pi*f*L)./(j*2*pi*f*L+(R-4*(pi^2)*(f.^2)*R*L*C));

plot (f/1e9,abs(V));grid;

title('Transfer response [Vo/Vi|');

xlabel('frequency, GHz'); ylabel('{vo/Vi|');

print -deps 'pl_8.eps'

Problem 1.9

One of the objectives of Chapter 1 is to sensitize the reader to high-frequency phenomena that are usually neglected in a low-frequency circuit analysis. One such phenomenon is the skin effect. To show its importance in RF calculations, (a) compute the frequency behavior of an inductor formed by 10 turns of AWG 26 copper wire on a 5 mm air core. The length of the coil is 5 mm. (b) repeat the computations by first neglecting the skin effect and then including it.

Solution

(a) The coil inductance is discussed on page 22 and yields

RF Circuit Design: Theory and Applications

Chapier 1 • Introduction

$$ L = { \frac { \pi r ^ { 2 } } { l } } \mu _ { 0 } N ^ { 2 } = { \frac { \pi { ( 2 . 5 \times 1 0 ^ { - 3 } ) } ^ { 2 } } { 5 \times 1 0 ^ { - 3 } } } 4 \pi \times 1 0 ^ { - 7 } \times 1 0 ^ { 2 } = 0 . 4 9 3 5 \mu H $$

To find the frequency dependent resistance, we first compute the DC resistance

$$ R _ { D C } = \frac { l } { \sigma _ { c u } \pi ( { d ^ { 2 } } / { 4 } ) } = \frac { 2 \pi r N } { \sigma _ { c u } \pi ( { d ^ { 2 } } / { 4 } ) } = 7 . 6 \times 1 0 ^ { - 2 } \Omega $$

and then

$$ R ~ = ~ R _ { D C } { \left( \frac { a } { 2 \delta } \right) } = ~ R _ { D C } { \left( \frac { a } { 2 } \right) } { \sqrt { \pi f | \iota _ { 0 } \mathbb { 0 } _ { c u } } } $$

A MATLAB routine that computes the frequency response is given below. Here the coil impedance without the skin effect is specified as

$$ { \cal Z } _ { 1 } = R _ { D C } + j \omega L $$

Including the skin phenomenon we need to modify this equation

$$ Z _ { 2 } = R _ { D C } + j \bigg ( \omega L + R _ { D C } \frac { a } { 2 \delta } \bigg ) $$

Problem 1.10

The leads of a resistor in an RF circuit are treated as straight aluminum wires $( { \sigma } _ { \mathrm { A l } } = 4 . 0 \times 1 0 ^ { \prime } \mathrm { S } / \mathrm { m } )$ of AWG size 14 and of total length of 5 cm. (a) Compute the DC resistance. (b) Find the AC resistance and inductance at 100 MHz, 1 GHz, and 10 GHz operating frequencies.

Solution

According to Table A4 (page 599) AWG 14 corresponds to a diameter d = 1.628mm. The DC resistance is computed as

$$ R _ { D C } = \frac { l } { \sigma _ { c u } \pi ( d ^ { 2 } / 4 ) } $$

from which we next find

$$ R ( f ) = R _ { D C } \frac { d / 2 } { 2 \delta } \nonumber = \frac { R _ { D C } } { 4 } d \sqrt { \pi f \sigma _ { A l } \mu _ { 0 } } $$

$$ L = R ( f ) { \frac { 1 } { 2 \pi f } } $$

RF Circuit Design: Theory and Applications

Chapter 1 • Introduction

The MATLAB code below implements these equations

Executing this code yields:

Problem 1.11

Compute the skin depths for copper $( \sigma _ { { \cal E } \mathrm { u } } = 6 4 . 5 1 6 \times 1 0 ^ { 6 } \mathrm { S } / \mathrm { m } ) ,$ aluminum $\mathrm { ( ) _ { A l } = 4 0 \times 1 0 ^ { 5 } S / m ) _ { \mathrm { \ell } } }$ and gold $( \sigma _ { \mathrm { A u } } = 4 \bar { 8 } . 5 4 4 \times$ 10°Š/m) at 1 GHz and 10 GHz, and find the resistance of a 10 cm wire with diameter of 1 mm.

Solution

Starting point is the skin equation as a function of frequency $( f _ { i } , i = 1 , 2 )$ and material conductivity $( \sigma _ { \alpha } , \alpha = C u , A l , A u )$ Consequently,

$$ \delta _ { \alpha } ( f _ { i } ) = \frac { 1 } { \sqrt { \pi \sigma _ { \alpha } f _ { i } \vert \boldsymbol { \mathrm { L } } _ { 0 } } } \mathrm { ~ a n d ~ } R _ { \alpha } ( f _ { i } ) = R _ { D C } ^ { \alpha } \frac { a } { 2 \delta _ { \alpha } ( f _ { i } ) } $$

where

$$ R _ { D C } ^ { \alpha } = \frac { l } { \sigma _ { \alpha } \pi ( d ^ { 2 } / 4 ) } $$

The MATLAB code below computes the skin depths and the resistances for the two frequencies.

The resulting numerical values are:

skin depth of CU at 1GHz: $\mathrm { d e l t a } = 1 . 9 8 1 4 7 \mathrm { e } { \cdot } 0 0 6$ 1/m and resistance: R = 0.248998 Ohm skin depth of CU at 10GHz: delta = 6.26594e-007 1/m and resistance: R = 0.787402 Ohm

$$ \mathbf { R } = 0 . 3 1 6 2 2 8 \mathbf { O } \mathbf { h } \mathbf { m } $$

$$ \pmb { \mathrm { R } } = 0 . 9 0 7 7 4 2 \mathbf { 0 } \mathbf { h } \mathbf { m } $$

RF Circuit Design: Theory and Applications

Chapter 1 - Introduction

Problem 1.12

A typical PCB substrate consists of $\mathbf { A l } _ { 2 } \mathbf { O } _ { 3 }$ with a relative dielectric constant of 10 and a loss tangent of 0.0004 at 10 GHz. Find the conductivity of the substrate.

Solution

From the textbook, page 11, we use

$$ \tan \Delta _ { s } = \frac { \omega \varepsilon } { \sigma _ { s u b } } $$

which gives

$$ \sigma _ { s u b } = \frac { 2 \pi f \varepsilon _ { 0 } \varepsilon _ { r } } { \tan \Delta _ { s } } = 1 . 3 9 0 8 \times 1 0 ^ { 4 } { S } / { m } $$

Problem 1.13

For the series RLC circuit with $\smash { \mathcal { R } ^ { : } = 1 }$ Ω, L = 1 nH, and $c = 1$ pF, compute the resonance frequency and quality factor at ±10% of the resonance frequency. Does the presence of the resistor affect the resonance frequency?

Solution

The resonance frequency is found by setting the imaginary part of the total impedance

$$ { \pmb Z } = { \pmb R } + j \bigg ( \omega { \pmb L } - \frac { 1 } { \omega { \pmb C } } \bigg ) $$

to zero. Thus, from $\pmb { X } ( 2 \pi f _ { r } ) = \pmb { 0 }$ , it is seen that the resonace frequency

$$ f _ { r } = { \frac { 1 } { 2 \pi } } { \frac { 1 } { \sqrt { L C } } } = 5 . 0 3 2 9 G H z $$

is unaffected by resistance R. The quality factor is defined as

$$ Q ( f ) = { \frac { | X ( f ) | } { R } } $$

Setting $X ( f _ { r } + 1 0 % f _ { r } ) = X _ { p }$ and $X ( f _ { r } { - } 1 0 % f _ { r } ) = X _ { m }$ we obtain

$$ Q ( f _ { r } + 1 0 % f _ { r } ) \ = \ \frac { \left| X _ { p } \right| } { R } \ = \ 1 1 . 4 2 5 \ \mathrm { a n d } \ Q ( f _ { r } { - } 1 0 % f _ { r } ) \ = \ \frac { \left| X _ { m } \right| } { R } \ = \ 1 4 . 1 2 4 $$

Problem 1.14

A 4.7 pF capacitor with relative dielectric constant of 4.6 and series lostmngent of 0.003 is used in a circuit operated at 10 GHz. For a combined copper lead length of 6 cm and diameter of 0.5 mm, determine (a) the lead resistance and lead reactance, and () the conductance and the total impedance. The conductivity of coppis given as $\sigma _ { \mathrm { { C u } } } = 6 4 . 5 1 6 { \times } 1 0 ^ { 6 } \Omega ^ { - 1 } { \cdot } \mathrm { m } ^ { - 1 }$

Solution

(a) From Problem 1.10, we immediately see that resistance and reactance are given via

$$ R = R _ { D C } \frac { \alpha } { 2 \delta } = \frac { R _ { D C } } { 2 } a \sqrt { \pi f \sigma _ { c u } \vert \boldsymbol { \lambda } _ { 0 } } = \boldsymbol { 0 . 9 4 4 8 0 } . $$

(b) From the definition of the conductance, see page 18, we find

$$ G _ { e } = \frac { \omega C } { \tan \Delta _ { s } } = 9 8 . 4 3 \Omega $$

and the impedance is

$$ \zeta = R + j \Theta L = R + j R + \frac { 1 } { G _ { \epsilon } + j \Theta C } = R e { Z } + j I m { Z } = 0 . 9 5 5 \Omega + j 0 . 9 4 4 8 $$

These numercial results are obtained based on the following MATLAB routine:

RF Circuit Design: Theory and Applications

Chapter 1· Introduction

Cands= 3.0*1e-3;

f = le10;

1 $= 6 \times 1 6 - 2 ;$

a * (0.5*1e-3)/2;

cond = 64.516*le6;

mu0 = 4*pi*le-7;

del = 1/sqrt(pi*f*mu0*cond);

P = (1/(cond*pi*a^2))*a/(2*del);

fprintf('Lead resistance equals lead reactance: R = wL = tg Ohm \n', R); 香

Lead resistance equals lead reactance: R = wL = 0.944882 Ohm

Conductance of capacitor: $\tt G e = 9 8 . 4 3 6 6$ Siemens

Real(Z) = 0.955041 Ohm, $\operatorname* { I m a g } ( Z ) = 0 . 9 4 4 8 5 1$ Ohm

Problem 1.15

A manufacturer data sheet records the series loss tangent of a capacitor to be $1 0 ^ { - 4 }$ at 5 GHz. For a total plate dimension of $1 0 ^ { - 2 } \mathrm { c m } ^ { 2 }$ and plate separation of 0.01 mm and a relative dielectric constant of 10, find the conductance.

Solution

The conductance, page 18, is given by

$$ G _ { e } = \frac { \omega C } { \tan \Delta _ { s } } = \frac { \omega E A } { d \tan \Delta _ { s } } = 9 2 . 7 2 1 S $$

Problem 1.16

A two-element impedance of the generic form

$$ { Z } = R + j { X } $$

has to be converted into an equivalent admittance form $\gamma = 1 / Z$ such that

$$ Y = G + j B $$

Find the conductance G and susceptance B in terms of resistance R and reactance X.

Solution

The admittance has to be equal to the inverse of the impedance, i.e.,

$$ Z = R + j X \equiv \frac { 1 } { Y } = \frac { 1 } { G + j B } $$

$$ G + j B = \frac { 1 } { R + j X } = \frac { R - j X } { R ^ { 2 } + X ^ { 2 } } $$

From this equality we can individually identify real and imaginary parts

$$ G = { \frac { R } { R ^ { 2 } + X ^ { 2 } } } { \mathrm { ~ a n d ~ } } B = - { \frac { X } { R ^ { 2 } + X ^ { 2 } } } $$

Problem 1.17

A more elaborate model of a capacitor is sometimes represented by the following circuit:

Here the loss tangent is specified as consisting of two parts involving the admittance $Y _ { p } = 1 / R _ { p } +$ + jωC with a parallel-circuit loss tangent tan $\Delta _ { p . } = \smash { \big | \mathrm { R e } { Y _ { p } } / \mathrm { I m } { Y _ { p } } \big | }$ and séries impedance $Z _ { S } = \dot { R } _ { S } + 1 / ( j \omega C )$ with a series-circuit loss tangent ta $\Im \pmb { \Delta } _ { S } = \left| \mathbb { R e } { Z _ { S } } / \mathrm { I m } { \bar { Z } _ { S } } \right|$ (it is noted that Rs is different from Example 1-4). Show that $R _ { S }$ for low-loss capacitances we approximately obtain $\tan \Delta \approx \tan \Delta _ { S } + \tan \Delta _ { p }$ ,where tan∆ $= \vert \mathrm { R e } { \bar { z } } / \mathrm { I m } { z } \vert$ and Z is the total impedance.

Solution

For the series circuit representation we obtain

$$ Z _ { s } = R _ { s } + \frac { 1 } { j \omega C } \mathrm { w i t h \ t a n } \Delta _ { s } = \frac { R _ { s } } { 1 / ( \omega C ) } = R _ { s } \omega C $$

and correspondingly for the parallel circuit

$$ Y _ { p } = G _ { p } + j \omega C { \mathrm { ~ a n d ~ t a n } } \Delta _ { p } = { \frac { G _ { p } } { \omega C } } $$

Next the impedance of the actual three-element capacitor model is written as

$$ Z = R _ { s } + \frac { 1 } { G _ { p } + j \omega C } = \frac { R _ { s } ( G _ { p } - j \omega C ) + G _ { p } - j \omega C } { G _ { p } ^ { 2 } + \omega ^ { 2 } C ^ { 2 } } $$

This yields a loss tangent of

RF Circuit Design: Theory and Applications

Chapter 1 • Introduction

$$ \tan \Delta \ = \ \frac { R e { Z } } { I m { Z } } \ = \ \frac { R _ { s } ( G _ { p } ^ { 2 } + \omega ^ { 2 } C ^ { 2 } ) + G _ { p } } { \omega C } \ = \ \frac { R _ { p } + \omega ^ { 2 } R _ { p } ^ { 2 } R _ { s } C ^ { 2 } + R _ { s } } { \omega C R _ { p } ^ { 2 } } $$

Rewriting tan∆, and tan $\Delta _ { p }$ , we obtain

$$ \tan \Delta _ { s } \ = \ \frac { \dot { R _ { p } ^ { 2 } } \dot { R _ { s } } \displaystyle C ^ { 2 } \omega ^ { 2 } } { \omega R _ { p } ^ { 2 } C } \ \mathrm { a n d } \ \tan \Delta _ { p } \ = \ \frac { R _ { p } } { \omega R _ { p } ^ { 2 } C } $$

These two terms can be combined such that

$$ \tan \Delta _ { s } \tan \Delta _ { p } ^ { 2 } = \frac { R _ { s } } { \omega R _ { p } ^ { 2 } C } $$

Thus, it becomes apparent that

$$ \tan \Delta = \tan \Delta _ { s } + \tan \Delta _ { p } + \tan \Delta _ { s } \tan \Delta _ { p } ^ { 2 } $$

which, for small loss tangents, finally yields

$$ \tan \Delta = \tan \Delta _ { s } + \tan \Delta _ { p } $$

Problem 1.18

When recording the capacitance with a measurement equipment, the user has often the choice to select a suitable circuit representation. For the series representation, the instrument attempts to predict ${ \pmb R } _ { S }$ and $c _ { s } ,$ while for the parallel representation the prediction involves $R _ { P }$ and $c _ { p }$ . Which mode should be chosen if large capacitors of more than 100 μF are to be measured? Is this mode also suitable for smal values of less than 10 μF? Explain your answers.

Solution

For large capacitors, the impedance is lower than for small capacitors $( \mathbf { C } = 1 0 0 ^ { \circ } \mu F ;$ $Z \approx 1 0 ^ { 4 } / \omega$ and ${ \bf C } = 1 0 ~ \mu F ; ~ \dot { \bf Z } = 1 0 ^ { 3 } / \omega ) .$ Consequently, the loss associated with the series resistance $R _ { s }$ is more significant than the shunt resistance $R _ { p }$ . As a result, series models are prefered for large $\cos ,$ whereas the shunt models are more suitable for small $\mathbf { C s } ,$ including most Cs constructed as surface mount devices (SMD).

Problem 1.19

The ability to store electric charge, expressed through the capacitance, depends on the operating temperature. This behavior can be quantified through the relation $C = C _ { 0 } [ 1 + \alpha ( T - 2 0 ^ { \circ } \mathrm { C } ) ]$ , where $c _ { 0 }$ is the nominal capacitance and α is a temperature

coefficient that can be positive or negative. If the capacitance $C \ { \mathrm { a t } } \ T = 2 0 ^ { \circ } { \mathrm { C } }$ is recorded to be 4.6 pF, which increases to 4.8 pF at $T = 4 0 ^ { \circ } \mathrm { C }$ , what is the temperature coefficient α? Determine the capacitance at $\scriptscriptstyle 0 ^ { \circ } C$ and 80°C.

Solution

Starting from the temperature dependent definition of capacitance, we directly observe that a $\mathrm { ~ t ~ } T = 2 0 ^ { \circ } C , C = \dot { C _ { 0 } } = 4 . 6 p \dot { F }$ . The factor α is next recovered from the $4 0 ^ { \circ } C$ recording:

$$ \alpha = \frac { 1 } { 4 0 ^ { \circ } C - 2 0 ^ { \circ } C } \bigg ( \frac { C } { C _ { 0 } } - 1 \bigg ) = 0 . 0 0 2 1 7 ^ { \circ } C ^ { - 1 } $$

With the knowledge of C0 and α, we can employ the formula $c _ { 0 }$

$$ C = C _ { 0 } [ 1 + \alpha ( T - 2 0 ^ { \circ } C ) ] $$

and compute the capacitances of the remaining two temperatures, see the MATLAB routine below.

Problem 1.20

When measuring impedance at low frequency we connect the measurement equipment to a device using a pair of wires and assurme that the reading reflects the impedance of the device under test (DUT). As we have seen in this chapter, at high frequencies we have to take into account the influence of the parasitic elements. The typical circuit representation of the measurement arrangement is as follows.

RF Circuit Design: Theory and Applications

Chapter 1 • Introduction

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