Patent Publication Number: US-8536930-B2

Title: Switching circuit

Description:
CROSS-REFERENCE TO RELATED APPLICATIONS 
     This application claims priority to Provisional Application Ser. No. 61/501,504, filed on Jun. 27, 2011 and claims the benefit of Japanese Patent Application No. 2011-141988, filed on Jun. 27, 2011, all of which are incorporated herein by reference in their entirety. 
    
    
     BACKGROUND 
     1. Field 
     Embodiments of the present invention relate to a switching circuit. 
     2. Description of the Related Art 
     There are known switching circuits using switching elements such as transistors (See Japanese Patent Laid-Open No. 2006-101637). In the switching circuit disclosed in Patent Literature 1, a PWM (pulse width modulation) signal is supplied to an input terminal (e.g., a gate terminal) of the switching element. The switching element is controlled to be switched ON/OFF by a clock frequency (a switching frequency) of the PWM signal. A voltage at a connection point between an output terminal (e.g., a drain terminal) and an inductor varies by the switching ON/OFF of the switching element. As a result, driving of a load circuit whose one end is connected to the connection point can be controlled. 
     SUMMARY 
     However, an extra current may flow through the load circuit or the switching element due to a signal component that is an integer multiple of the clock frequency (the switching frequency) for switching ON/OFF the switching element in addition to a current flowing through the load circuit when the switching element is in an ON state. Thus, there is unnecessary power consumption. 
     It is an object of the present invention to provide a switching circuit which can improve power efficiency. 
     A switching circuit according to one aspect of the present invention includes: a switching element that has a first terminal and a second terminal, and is driven by a pulse signal to switch a conduction state between the first terminal and the second terminal; a power source section that supplies a voltage to the first terminal of the switching element; a load circuit that is connected in parallel with the power source section; a passive circuit section that is connected between a connection point between the power source section and the load circuit, and the first terminal of the switching element, and suppresses a current flowing from the connection point to the switching element at a frequency N times (N is an integer of 1 or more) as high as a clock frequency of the pulse signal; and a resonant circuit section that is connected between the passive circuit section and the connection point, and resonates at the frequency of N times. 
     In the configuration, the switching element is connected to the connection point via the passive circuit section and the resonant circuit section. Thus, the state of a voltage supplied to the connection point from the power source section can be controlled by a switching operation of the switching element. As a result, the state of a voltage applied to the load circuit connected in parallel with the power source section can be controlled. Meanwhile, the passive circuit section suppresses the current flowing from the connection point to the switching element at the frequency N times as high as the clock frequency of the pulse signal. Thus, extra energy consumption can be reduced. Furthermore, the resonant circuit section resonates at the frequency of N times. Thus, the N-time frequency component out of the voltage component applied to the load circuit is reduced, so that extra energy consumption can be reduced. As a result, power efficiency can be improved. 
     In one embodiment, an imaginary part of an impedance of the passive circuit section anticipated from the switching element may be zero or more, and twice or less of an absolute value of a reactance of an output parasitic capacitance of the switching element. In this case, the imaginary part of the impedance of the passive circuit section anticipated from the switching element may be zero or more, and twice or less of the absolute value of the reactance of the output parasitic capacitance of the switching element at the frequency of N times. 
     In the embodiment, the passive circuit section can more reliably suppress the current flowing from the connection point to the switching element at the frequency N times as high as the clock frequency of the pulse signal. 
     In one embodiment, a real part and an imaginary part of an impedance of the resonant circuit section anticipated from the load circuit may be smaller than an impedance of the load circuit at the clock frequency of N times. 
     In the embodiment, the voltage applied to the load circuit can be further reduced at the frequency of N times. As a result, extra energy consumption can be reduced. 
     The pulse signal may be a signal obtained by temporally modulating a duty cycle of a pulse width of the pulse signal by a signal having a lower frequency component than the clock frequency. 
     While the passive circuit section and the resonant circuit section act against the frequency N times as high as the clock frequency, the passive circuit section and the resonant circuit section do not substantially act against the lower frequency component than the clock frequency. Thus, the signal having the lower frequency component can easily pass through the passive circuit section and the resonant circuit section to be transmitted to the load circuit. As a result, the state of the voltage applied to the load circuit varies by the signal component. 
     In one embodiment, the passive circuit section may be composed of at least one reactance element, and may have a first end connected to the first terminal of the switching element and a second end connected to the connection point. 
     In one embodiment, the passive circuit section may be composed of at least one transmission line, and may have a first end connected to the first terminal of the switching element and a second end connected to the connection point. 
     In one embodiment, the resonant circuit section may have M resonant elements each having at least one reactance element connected in series. In the embodiment, at least one resonant element out of the M resonant elements may resonate at the frequency of N times. The M resonant elements may be connected in parallel. 
     In one embodiment, the resonant circuit section may have M (M is an integer of 1 or more) transmission lines. In the embodiment, at least one transmission line out of the M transmission lines may have an electrical length of ¼ of a wavelength corresponding to the frequency N times as high as the clock frequency. 
     As mentioned above, the switching circuit which can improve the power efficiency can be provided. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         FIG. 1  is a diagram illustrating a schematic configuration of a switching circuit according to one embodiment; 
         FIG. 2  is a diagram for explaining a signal that drives the switching circuit shown in  FIG. 1 ; 
         FIG. 3  is a graph showing design conditions of a passive circuit section having the minimum number of elements and designed with respect to a clock frequency; 
         FIG. 4  is a diagram illustrating one example of a configuration of the switching circuit operating at the clock frequency; 
         FIG. 5  is a graph showing design conditions of a passive circuit section having the minimum number of elements out of passive circuit sections designed with respect to first-order and second-order harmonics of the clock frequency; 
         FIG. 6  is a diagram illustrating examples of a circuit configuration of the passive circuit section designed with respect to the first-order and second-order harmonics of the clock frequency; 
         FIG. 7  is a diagram illustrating examples of a circuit configuration of a passive circuit section designed with respect to the first-order harmonic, the second-order harmonic, and a third-order harmonic of the clock frequency; 
         FIG. 8  is a diagram illustrating specific examples of a connection state of terminals A and B in the circuit configuration shown in  FIG. 7(   a ); 
         FIG. 9  is a schematic diagram illustrating a schematic configuration of a switching circuit including another example of a resonant circuit section; 
         FIG. 10  is a table showing that there are a plurality of harmonic orders at which one end-open stub can resonate at the same time; 
         FIG. 11  is a diagram illustrating arrangement options of a transmission line when a passive circuit section is composed of one transmission line; 
         FIG. 12  is a diagram illustrating arrangement options of transmission lines when the passive circuit section is composed of two transmission lines; 
         FIG. 13  is a diagram illustrating a circuit for simulations when the passive circuit section is composed of two transmission lines; 
         FIG. 14  is a graph showing a simulation result; and 
         FIG. 15  is a diagram illustrating configuration examples of the passive circuit section composed of three transmission lines. 
     
    
    
     DETAILED DESCRIPTION 
     In the following, embodiments of the present invention will be described by reference to the drawings. In the description of the drawings, the same elements are assigned the same reference numerals to omit the overlapping description. The dimensional ratios in the drawings do not necessarily correspond to those of the description. 
     A switching circuit according to one embodiment will be described based on  FIGS. 1 and 2 .  FIG. 1  is a circuit diagram illustrating a schematic configuration of a switching circuit  1  according to one embodiment.  FIG. 2  is a diagram for explaining a signal that drives the switching circuit  1 . The switching circuit  1  is a switching power source circuit or a time-varying power source circuit, for example. 
     The switching circuit  1  includes a switching element  10 . In the present embodiment, the switching element  10  is an insulated field-effect transistor (MOSFET) unless otherwise noted. Examples of the MOSFET include a power MOSFET. In this case, the switching element  10  has a source terminal  11  to be grounded, a gate terminal  12  connected to a signal source  20  and to which a signal is supplied from the signal source  20 , and a drain terminal  13  connected to a power source section  30  and to which a voltage Vdd (e.g., 16 V) is supplied. An output parasitic capacitance C ds  exists in the switching element  10  due to its configuration. The output parasitic capacitance C ds  includes an output parasitic capacitance between a drain and a source. In  FIG. 1 , the output parasitic capacitance C ds  is shown as a capacitor  14 . 
     The signal source  20  supplies a signal for switching the switching element  10  to the gate terminal  12 . The signal inputted into the gate terminal  12  from the signal source  20  is a PWM signal S P . The PWM signal S P  will be described by reference to  FIG. 2 .  FIG. 2  is a diagram for explaining the PWM signal.  FIG. 2(   a ) shows one example of two signals for generating the PWM signal S P .  FIG. 2(   b ) shows one example of the PWM signal. The PWM signal S P  is a pulse signal whose duty cycle of a pulse width is modulated by a signal (S 1 ) having a first frequency using a signal (S 2 ) having a second frequency higher than the first frequency. Example of the signal (S 2 ) is triangle-wave or sawtooth-wave signal (see  FIG. 2(   a )). A clock frequency f CK  of the PWM signal S P , that is, a switching frequency that switches the switching element  10  corresponds to the second frequency. The other end of the signal source  20  is grounded. 
     Returning to  FIG. 1 , the configuration of the switching circuit  1  will be described. The power source section  30  is a direct-current power source section including a direct-current power source  31 . The power source section  30  may include an inductor  32  in order to prevent the first frequency component of the switching element  10  from flowing into the direct-current power source  31 . A positive pole of the direct-current power source  31  is connected to the drain terminal  13  via the inductor  32 . A line connecting the drain terminal  13  and the direct-current power source  31  is referred to as signal path below. A negative pole of the direct-current power source  31  is grounded. 
     A load circuit  40  is connected in parallel with the power source section  30 . Examples of the load circuit  40  include a resistive load and an inductive load. One end  41  of the load circuit  40  is connected to the positive pole of the direct-current power source  31 , and the other end  42  of the load circuit  40  is grounded. As shown in  FIG. 1 , when the inductor  32  is provided, the one end  41  of the load circuit  40  is connected to an end of the inductor  32  on the opposite side from the direct-current power source  31 . The connection point functions as an output port P. 
     In the above configuration, when the PWM signal S P  supplied from the signal source  20  is inputted into the switching element  10 , a conduction state between the drain terminal  13  and the source terminal  11  is switched by the PWM signal S P . The state of a voltage at the output port P thereby varies, so that the state of a voltage applied to the load circuit  40  changes. As a result, a current flowing through the load circuit  40  varies. Accordingly, for example, when a high-frequency power amplifier is connected as the load circuit, an output amplitude of a high-frequency signal can be modulated with a large depth while high power efficiency is being maintained for the power source. 
     The switching circuit  1  includes a passive circuit section  50  and a resonant circuit section  60  between the switching element  10  and the output port P so as to reduce power consumption in a switching operation. The passive circuit section  50  and the resonant circuit section  60  function as a filter that cuts an N-th order harmonic of the clock frequency f CK  of the PWM signal S P  and allows a frequency lower than the clock frequency f CK , i.e., the signal S 1  having the first frequency to pass therethrough. In the following, the passive circuit section  50  and the resonant circuit section  60  will be described. 
     The passive circuit section  50  is arranged between the switching element  10  and the output port P. The passive circuit section  50  has a first end  50   a  connected to the drain terminal  13 , and a second end  50   b  connected to the output port P. The passive circuit section  50  has a configuration that satisfies next “passive circuit section conditions” at a frequency N times as high as the clock frequency f CK  of the pulse signal supplied to the switching element  10 . In the following description, an impedance of the passive circuit section  50  anticipated from the switching element  10  (or as viewed from the switching element  10 ) is Z, and an imaginary part of the impedance Z is Z img . 
     (Passive Circuit Section Conditions) 
     The imaginary part Z img  of the impedance Z is zero or more, and twice or less of an absolute value of a reactance of the output parasitic capacitance C ds . 
     When the above “passive circuit section conditions” are satisfied, a combined impedance of the output parasitic capacitance C ds  and the passive circuit section  50  becomes large at the N-th order harmonic. As a result, the passive circuit section  50  prevents a current having the N-time clock frequency component from flowing through the switching element  10 . Thus, extra power consumed in the switching element  10  in the switching operation can be reduced. As long as the passive circuit section  50  is configured to satisfy the “passive circuit section conditions”, the passive circuit section  50  may be a two-port circuit network composed of at least one reactance element. The passive circuit section  50  may be also a two-port circuit network composed of at least one transmission line (including a case of a stub). Specific examples of the passive circuit section  50  will be described later. 
     The resonant circuit section  60  is a circuit that resonates at the frequency N times as high as the clock frequency f CK  of the PWM signal S P  as the pulse signal. As an example of a resonant state, a real part and an imaginary part of an impedance of the resonant circuit section  60  anticipated from the load circuit  40  (or as viewed from the load circuit  40 ) may be smaller than an impedance of the load circuit  40  at the frequency N times as high as the clock frequency f CK . 
       FIG. 1  shows a circuit section obtained by connecting in parallel M resonant elements  61   1  to  61   M  each having an inductor and a capacitor as a reactance element connected in series as one example. One end of each of the resonant elements  61   1  to  61   M  is connected between the second end  50   b  and the output port P on the signal path connecting the output port P and the drain terminal  13 . The other end of each of the resonant elements  61   1  to  61   M  is grounded. An element value of each of the inductor and the capacitor of at least one resonant element out of the M resonant elements  61   1  to  61   M  is an element value that allows resonance at the frequency N times as high as the clock frequency f CK . 
     In the configuration, when the switching element  10  performs the switching operation at the frequency N times as high as the clock frequency f CK , resonance occurs in the resonant circuit section  60 . Due to the resonance, the real part and the imaginary part of the impedance of the resonant circuit section  60  become smaller than the impedance of the load circuit  40 . Thus, a current flows through the resonant circuit section  60  more easily than through the load circuit  40 . That is, when the switching element  10  performs the switching operation, the frequency component N times as high as the clock frequency f CK , which the voltage applied to the load circuit  40  may have, approaches zero. Accordingly, extra energy consumption by the load circuit  40  can be reduced at the N-th order harmonic of the clock frequency f CK . 
     The extra power consumption is reduced in the switching circuit  1  including the passive circuit section  50  and the resonant circuit section  60  as described above. As a result, the power efficiency can be improved. 
     In the following, the configurations of the passive circuit section  50  and the resonant circuit section  60  will be specifically described by reference to various embodiments. 
     (First Embodiment) 
     In the embodiment, the switching circuit  1  is sometimes referred to as switching circuit  1 A. The passive circuit section  50  and the resonant circuit section  60  are referred to as passive circuit section  50 A and resonant circuit section  60 A, respectively. The passive circuit section  50 A is a two-port circuit network composed of at least one reactance with no resistance. The resonant circuit section  60 A is a parallel circuit composed of the M resonant elements  61   1  to  61   M  as shown in  FIG. 1 . 
     In this case, the impedance Z of the passive circuit section  50 A is composed only of the imaginary part. The passive circuit section  50 A is designed such that the imaginary part Z img  of the impedance Z satisfies the above “passive circuit section conditions.” That is, in the present embodiment, the passive circuit section  50 A is designed such that the combined impedance of the passive circuit section  50 A and the output parasitic capacitance C ds  becomes infinity at the clock frequency f CK . In this case, the imaginary part Z img  of the impedance of the passive circuit section  50  is equal to the absolute value of the reactance of the output parasitic capacitance C ds . Thus, the “passive circuit section conditions” are satisfied. In the following, the description is made by setting the reactance of the passive circuit section  50 A with respect to the frequency of N times to X N (ω) as a function of an angular frequency ω. Although the passive circuit section  50 A is described by exemplifying a specific numerical value for N, the same reference numerals may be assigned as the reference numerals of elements. Please note that element values of respective elements are set as numerical values corresponding to the exemplified value of N. 
     First, a case of N=1 will be described. Although N and M do not necessarily correspond to each other, N=1 and M=1 in the following description.  FIG. 3  is a graph showing design conditions of the passive circuit section having the minimum number of elements in the case of N=1. In  FIG. 3 , a horizontal axis represents an angular frequency ω, and a vertical axis represents a reactance (Ω). In  FIG. 3 , a solid line represents a reactance X 1 (ω) of the passive circuit section  50 A, and an alternate long and short dash line is a reactance curve of the output parasitic capacitance C ds  based on the absolute value of the reactance of the output parasitic capacitance C ds . Since the reactance X 1 (ω) shown in  FIG. 3  is a linear function, the reactance X 1 (w) is expressed by a following expression. 
     [Expression 1]
 
 jX   1 (ω)= jωL   510   (1)
 
     The passive circuit section  50 A having the reactance curve indicated in the expression (1) may be composed of an inductor  510  as one coil as shown in  FIG. 4 .  FIG. 4  is a diagram illustrating one example of the configuration of the switching circuit in the case of N=1. The load circuit  40  is shown as a resistance. In the expression (1), L 510  is an element value (an inductance) of the inductor  510 . When an angular frequency corresponding to the clock frequency f CK  is ω CK , conditions to be satisfied by the inductor  510  as the passive circuit section  50 A are as follows. 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     2 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       X 
                       1 
                     
                     ⁡ 
                     
                       ( 
                       0 
                       ) 
                     
                   
                   = 
                   0 
                 
               
               
                 
                   ( 
                   
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                     ⁢ 
                     a 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       X 
                       1 
                     
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                       ( 
                       
                         ϖ 
                         CK 
                       
                       ) 
                     
                   
                   = 
                   
                     1 
                     
                       
                         ϖ 
                         CK 
                       
                       ⁢ 
                       
                         C 
                         ds 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     2 
                     ⁢ 
                     b 
                   
                   ) 
                 
               
             
           
         
       
     
     An expression (3) is obtained from the expressions (2a) and (2b). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     3 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     L 
                     510 
                   
                   = 
                   
                     1 
                     
                       
                         ϖ 
                         CK 
                         2 
                       
                       ⁢ 
                       
                         C 
                         ds 
                       
                     
                   
                 
               
               
                 
                   ( 
                   3 
                   ) 
                 
               
             
           
         
       
     
     When the clock frequency f CK  is 200 MHz, ω CK =2π×200 MHz. Moreover, when C ds =60 pF, L 510 =10.54 nH. 
     Here, one example of a method for calculating element values of elements constituting the resonant circuit section  60 A will be described. In the case of M=1 and N=1, the resonant circuit section  60 A may be composed of one inductor  62  and one capacitor  63  as shown in  FIG. 4 . Resonant conditions of the resonant element  61   1  are expressed by a next expression. 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     4 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       j 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         ϖ 
                         CK 
                       
                       ⁢ 
                       
                         L 
                         62 
                       
                     
                     + 
                     
                       1 
                       
                         
                           j 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             ϖ 
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                           ⁢ 
                           
                             C 
                             63 
                           
                         
                         ⁢ 
                         
                             
                         
                       
                     
                   
                   = 
                   0 
                 
               
               
                 
                   ( 
                   4 
                   ) 
                 
               
             
           
         
       
     
     When a resistance value R L  of the load circuit  40  is 10Ω, C ds =60 pF, and ω CK =2π×200 MHz, L 62 C 63 =633.26 nH·pF from the expression (4). Element values L 62  and C 63  of the inductor  62  and the capacitor  63  may be determined so as to satisfy L 62 C 63 =633.26 nH·pF. 
     Next, a case of N=2 will be described. Here, the passive circuit section  50 A capable of preventing a current having a frequency component up to a second-order harmonic (specifically, a fundamental wave and the second-order harmonic) out of harmonics of consecutive orders from the fundamental wave (the case of N=1) from flowing through the switching element  10  will be described. In this case, while N and M do not necessarily correspond to each other as described above, the resonant circuit section  60 A also needs to resonate at a frequency once and twice as high as the clock frequency f CK . Thus, N=2 and M=2.  FIG. 5  is a graph showing design conditions of the passive circuit section having the minimum number of elements in the case of N=2. In  FIG. 5 , a horizontal axis, a vertical axis, and an alternate long and short dash line are the same as those in  FIG. 3 . 
     A reactance X 2 (ω) shown in  FIG. 5  is expressed by an expression (5). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     5 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     j 
                     ⁢ 
                     
                         
                     
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                         X 
                         2 
                       
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                       ⁢ 
                       
                           
                       
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                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         a 
                         2 
                       
                       ⁢ 
                       
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                               ( 
                               
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                                 ⁢ 
                                 
                                     
                                 
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                           + 
                           
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                               z 
                               ⁢ 
                               
                                   
                               
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                             ⁢ 
                             
                                 
                             
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                   ( 
                   5 
                   ) 
                 
               
             
           
         
       
     
     In the expression (5), ω z1  is an angular frequency that satisfies X 2 (ω)=0, and a first angular frequency from a direct current (ω=0), ω p1  is an angular frequency of a first pole from the direct current (that is, ω=0), and a 2  is a free parameter determined so as to satisfy expressions (6a) to (6c) as the design conditions of the passive circuit section  50 A. 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     6 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         X 
                         2 
                       
                       ⁡ 
                       
                         ( 
                         0 
                         ) 
                       
                     
                     = 
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                     ⁢ 
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                         2 
                       
                       ⁡ 
                       
                         ( 
                         
                           ϖ 
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                         ) 
                       
                     
                     = 
                     
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                           ϖ 
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                         ⁢ 
                         
                           C 
                           ds 
                         
                       
                     
                   
                   , 
                 
               
               
                 
                   ( 
                   
                     6 
                     ⁢ 
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                     ⁡ 
                     
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                           ϖ 
                           CK 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     1 
                     
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                       ⁢ 
                       
                         ϖ 
                         CK 
                       
                       ⁢ 
                       
                         C 
                         ds 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     6 
                     ⁢ 
                     c 
                   
                   ) 
                 
               
             
           
         
       
     
     The passive circuit section  50 A may be designed as described below in the case of N=2. First, the circuit configuration is determined by executing topology search of the circuit configuration based on the expression (5). 
     For example, when jω=s in the expression (5), the expression (5) is expressed by an expression (7). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     7 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
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                               ⁢ 
                               
                                   
                               
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                           ( 
                           
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                       + 
                       
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                   = 
                   
                     
                       
                         
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                         ⁢ 
                         
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                           ϖ 
                           
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                             ⁢ 
                             
                                 
                             
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                           1 
                         
                         2 
                       
                     
                   
                 
               
               
                 
                   ( 
                   7 
                   ) 
                 
               
             
           
         
       
     
     An expression (8) is obtained by transforming the expression (7). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     8 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         
                           s 
                           3 
                         
                         ⁢ 
                         
                           a 
                           2 
                         
                       
                       + 
                       
                         
                           sa 
                           2 
                         
                         ⁢ 
                         
                           ϖ 
                           
                             z 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                           2 
                         
                       
                     
                     
                       
                         s 
                         2 
                       
                       + 
                       
                         ϖ 
                         
                           p 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           1 
                         
                         2 
                       
                     
                   
                   = 
                   
                     
                       sL 
                       510 
                     
                     + 
                     
                       1 
                       
                         
                           sC 
                           520 
                         
                         + 
                         
                           1 
                           
                             sL 
                             511 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   8 
                   ) 
                 
               
             
           
         
       
     
     In the expression (8), L 510 , C 520 , and L 511  are element values of the inductor  510 , a capacitor  520 , and an inductor  511  of the passive circuit section  50 A. The element values L 510 , C 520 , and L 511  are provided by expressions (9a) to (9c) in the transformation process from the expression (7) to the expression (8). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     9 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       L 
                       510 
                     
                     = 
                     
                       a 
                       2 
                     
                   
                   , 
                 
               
               
                 
                   ( 
                   
                     9 
                     ⁢ 
                     a 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       C 
                       520 
                     
                     = 
                     
                       1 
                       
                         
                           
                             a 
                             2 
                           
                           ⁢ 
                           
                             ϖ 
                             
                               z 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                             
                             2 
                           
                         
                         - 
                         
                           
                             a 
                             2 
                           
                           ⁢ 
                           
                             ϖ 
                             
                               p 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                             
                             2 
                           
                         
                       
                     
                   
                   , 
                 
               
               
                 
                   ( 
                   
                     9 
                     ⁢ 
                     b 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     L 
                     511 
                   
                   = 
                   
                     
                       
                         
                           a 
                           2 
                         
                         ⁢ 
                         
                           ϖ 
                           
                             z 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                           2 
                         
                       
                       - 
                       
                         
                           a 
                           2 
                         
                         ⁢ 
                         
                           ϖ 
                           
                             p 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                           2 
                         
                       
                     
                     
                       ϖ 
                       
                         p 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                       
                       2 
                     
                   
                 
               
               
                 
                   ( 
                   
                     9 
                     ⁢ 
                     c 
                   
                   ) 
                 
               
             
           
         
       
     
     The expression (7) can be also transformed into an expression (10). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     10 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         
                           s 
                           3 
                         
                         ⁢ 
                         
                           a 
                           2 
                         
                       
                       + 
                       
                         
                           sa 
                           2 
                         
                         ⁢ 
                         
                           ϖ 
                           
                             z 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                           2 
                         
                       
                     
                     
                       
                         s 
                         2 
                       
                       + 
                       
                         ϖ 
                         
                           p 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           1 
                         
                         2 
                       
                     
                   
                   = 
                   
                     
                       1 
                       
                         
                           1 
                           
                             sL 
                             510 
                           
                         
                         + 
                         
                           1 
                           
                             
                               sL 
                               511 
                             
                             + 
                             
                               1 
                               
                                 sC 
                                 520 
                               
                             
                           
                         
                       
                     
                     . 
                   
                 
               
               
                 
                   ( 
                   10 
                   ) 
                 
               
             
           
         
       
     
     In the expression (10), L 510 , L 511 , and C 520  are provided by expressions (11a) to (11c) in the transformation process from the expression (7) to the expression (10). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     11 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       L 
                       510 
                     
                     = 
                     
                       
                         
                           a 
                           2 
                         
                         ⁢ 
                         
                           ϖ 
                           
                             z 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                           2 
                         
                       
                       
                         ϖ 
                         
                           p 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           1 
                         
                         2 
                       
                     
                   
                   , 
                 
               
               
                 
                   ( 
                   
                     11 
                     ⁢ 
                     a 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       L 
                       511 
                     
                     = 
                     
                       
                         a 
                         2 
                       
                       
                         1 
                         - 
                         
                           
                             ϖ 
                             
                               p 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                             
                             2 
                           
                           
                             ϖ 
                             
                               z 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                             
                             2 
                           
                         
                       
                     
                   
                   , 
                 
               
               
                 
                   ( 
                   
                     11 
                     ⁢ 
                     b 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     C 
                     520 
                   
                   = 
                   
                     
                       1 
                       - 
                       
                         
                           ϖ 
                           
                             p 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                           2 
                         
                         
                           ϖ 
                           
                             z 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                           
                           2 
                         
                       
                     
                     
                       
                         a 
                         2 
                       
                       ⁢ 
                       
                         ϖ 
                         
                           z 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           1 
                         
                         2 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     11 
                     ⁢ 
                     c 
                   
                   ) 
                 
               
             
           
         
       
     
     The expressions (8) and (10) represent circuit configurations shown in  FIGS. 6(   a ) and  6 ( b ). Since the resonant circuit section  60 A resonates at the frequency once and twice as high as the clock frequency f CK , the second end  50   b  is shunted (that is, grounded) at the frequency. Therefore, a circuit configuration in  FIG. 6(   c ) is obtained as a modification of  FIG. 6(   a ), and a circuit configuration in  FIG. 6(   d ) is obtained as a modification of  FIG. 6(   b ). 
     Element values of respective elements in circuits shown in  FIGS. 6(   a ) and  6 ( b ) can be calculated based on the expressions (9a) to (9c) and the expressions (11a) to (11c) by providing ω z1 , ω p1  and a 2  so as to satisfy the expressions (6a) to (6c). In the case of N=2 and M=2, the resonant circuit section  60  is composed of the two resonant elements  61   1  and  61   2  connected in parallel. Element values of the inductor and the capacitor constituting each of the resonant elements  61   1  and  61   2  can be calculated in a similar manner to the case of N=1. 
     In a case of N=3, the passive circuit section  50 A can be also designed in a similar manner to the case of N=2. Here, the passive circuit section  50 A capable of preventing a current having a frequency component up to a third-order harmonic (specifically, the fundamental wave, the second-order harmonic, and the third-order harmonic) out of the harmonics of consecutive orders from the fundamental wave (the case of N=1) from flowing through the switching element  10  will be described.  FIGS. 7(   a ) to  7 ( n ) are diagrams illustrating configuration examples of the passive circuit section  50 A in the case of N=3. In the case of N=3, the passive circuit section  50 A may be composed of a combination of three inductors  510 ,  511 , and  512  and two capacitors  520  and  521  (see  FIGS. 7(   a ) to  7 ( k )). In the case of N=3, the passive circuit section  50 A may be also composed of a combination of two inductors  510  and  511  and three capacitors  520 ,  521 , and  522  (see  FIGS. 7(   l ),  7 ( m ), and  7 ( n )). In  FIGS. 7(   a ) to  7 ( n ), reference numerals are assigned for the sake of convenience so as to distinguish the inductors and the capacitors. Element values of the inductors and the capacitors of each circuit are set such that the inductors and the capacitors function as the passive circuit section  50 A in the circuit. 
     In  FIGS. 7(   a ) to  7 ( n ), a terminal A and a terminal B are connected to the second end  50   b  or grounded. The case of a circuit configuration in  FIG. 7(   a ) will be specifically described as one example.  FIGS. 8(   a ) to  8 ( d ) show configurations in which the terminals A and B in  FIG. 7(   a ) are connected to the second end  50   b  or grounded. 
       FIG. 8(   a ) shows a configuration in which both terminals A and B are grounded.  FIG. 8(   b ) shows a configuration in which both terminals A and B are connected to the second end.  FIG. 8(   c ) shows a configuration in which the terminal A is grounded and the terminal B is connected to the second end.  FIG. 8(   d ) shows a configuration in which the terminal A is connected to the second end and the terminal B is grounded. Although the configuration in  FIG. 7(   a ) has been specifically described, the same applies to  FIGS. 7(   b ) to  7 ( n ). Accordingly, in the case of N=3, the passive circuit section  50 A may have 38 circuit configurations. 
     In the case of N=3, the resonant circuit section  60 A is composed of the three resonant elements  61   1  to  61   3  connected in parallel as M=3. Element values of the inductor and the capacitor constituting each of the resonant elements  61   1  to  61   3  can be calculated in a similar manner to the case of N=1. 
     Here, the configuration of the passive circuit section  50 A capable of preventing a current having a frequency component up to the N-th order harmonic (that is, the first-order to N-th order harmonics) with the exemplified N as a maximum order from flowing through the switching element  10  has been mainly described with respect to each of the cases of N=1 to 3. Even when N is 4 or more, the passive circuit section  50 A can also similarly prevent the current having the frequency component up to the N-th order harmonic with N as a maximum order from flowing through the switching element  10 . When N is 4 or more, element values of the inductor and the capacitor constituting each of the resonant elements  61   4  to  61   M  of the resonant circuit section  60 A can be also calculated in a similar manner to the cases of N=1, 2, and 3. Please note that N and M do not necessarily correspond to each other as described above. 
     In the circuit configurations shown in  FIGS. 6(   a ) to  6 ( d ), by adjusting the element values of the inductor elements and the capacitor elements, the passive circuit section  50 A capable of preventing a current having a frequency component of not only the harmonics of consecutive orders but of any two harmonics (e.g., the first-order and third-order harmonics) from flowing through the switching element  10  can be obtained. Similarly, in the circuit configurations shown in  FIGS. 7(   a ) to  7 ( n ), by adjusting the element values of the inductor elements and the capacitor elements, the passive circuit section  50 A capable of preventing a current having a frequency component of not only the harmonics of consecutive orders but of any three harmonics (e.g., the first-order harmonic, the third-order harmonic, and a fifth-order harmonic) from flowing through the switching element  10  can be obtained. 
     As described above, when the passive circuit section  50 A and the resonant circuit section  60 A are composed of the so-called lumped constant elements, higher efficiency can be achieved with smaller physical dimensions with respect to the same frequency as compared to a case in which the passive circuit section uses a distributed constant element when the clock frequency f CK  is relatively low (e.g., 100 MHz or less). 
     (Second Embodiment) 
       FIG. 9  is a schematic diagram illustrating a schematic configuration of a switching circuit including another example of a resonant circuit section. A switching circuit  1 B may have the same configuration as that of the switching circuit  1  except for the configuration of a resonant circuit section  60 B. Thus, the configuration of the resonant circuit section  60 B will be mainly described. 
     The resonant circuit section  60 B has M end-open stubs of first to M-th end-open stubs  64   1  to  64   M  whose one end is connected between the output port P and the second end  50   b  on the signal path. The first to M-th end-open stubs  64   1  to  64   M  are so-called distributed constant elements. In other words, the first to M-th end-open stubs  64   1  to  64   M  are transmission lines each having a predetermined impedance Z S  and a predetermined electrical length. The electrical length of each of the first to M-th end-open stubs  64   1  to  64   M  is determined according to a frequency at which each of the first to M-th end-open stubs  64   1  to  64   M  resonates. While the number M of lines may be determined regardless of a clock frequency or a harmonic order, at least one of the first to M-th end-open stubs  64   1  to  64   M  has an electrical length of ¼ of a signal wavelength λ with respect to the frequency N times as high as the clock frequency f CK . In this case, the electrical length of any end-open stub out of the M end-open stubs may be ¼ of a signal wavelength λ with respect to any harmonic. In the following description, the first to M-th end-open stubs  64   1  to  64   M  may be sometimes referred to as end-open stub  64  for the convenience of description. 
       FIG. 10  is a table showing that there are a plurality of harmonic orders at which one end-open stub can resonate at the same time. For example, the first end-open stub  64   1  resonating at the fundamental wave (the first-order harmonic) in the case of N=1 also resonates at third-order, fifth-order, seventh-order, . . . , and (2k−1)-th order harmonics at the same time in addition to the first-order harmonic (k is an integer of 1 or more). Similarly, the second end-open stub  64   2  resonating at the second-order harmonic also resonates at sixth-order, tenth-order, fourteenth-order, . . . , and (2k−1)×2-th order harmonics at the same time. The M-th end-open stub  64   M  resonating at a 2 M−1 -th order harmonic also resonates at 3×2 M−1 -th order, 5×2 M−1 -th order, 7×2 M−1 -th order, . . . , and (2k−1)×2 M−1 -th order harmonics at the same time. The number N of harmonic orders at which the end-open stub  64  resonates is indicated by a “◯” mark in  FIG. 10 . From  FIG. 10 , when the first to M-th end-open stubs  64   1  to  64   M  are provided in parallel, the consecutive number N of harmonic orders at which the end-open stub  64  can resonate is 2 M −1.  FIG. 10  shows a combination in which the end-open stub  64  continuously resonates through N=1 to 15 by using n=1 to 4 as one example. However, the combination of n and N is not limited thereto. “n” in  FIG. 10  is an index number of the end-open stub  64  in the total number (M) of end-open stubs  64 . The end-open stub  64  does not always need to resonate at harmonics of consecutive orders. For example, when harmonic components of even orders included in a switching voltage waveform are smaller than harmonic components of odd orders, it is effective to allow the end-open stub  64  to resonate at the harmonic components of odd orders. In this case, only one end-open stub  64   1  is required since the end-open stub  64  needs to resonate only at the harmonic components of odd orders. 
     The points above will be specifically described. To allow the resonant circuit section  60 B to continuously resonate through, for example, N=1, 2, and 3, that is, up to the third-order harmonic, the first end-open stub  64   1  and the second end-open stub  64   2  are required. To allow the resonant circuit section  60 B to continuously resonate through N=1, 2, 3, 4, 5, and 6, that is, up to the sixth-order harmonic of the clock frequency f CK , the first to third end-open stubs  64   1 ,  64   2 , and  64   3  need to be connected in parallel. The harmonics of the plurality of orders N can be covered at the same time by one end-open stub  64  as described above since a standing wave formed in the end-open stub  64  has a periodic shape, and as a result, impedances as viewed from the point P toward the resonant circuit section  60 B are the same (0Ω). 
     Since each of the first to M-th end-open stubs  64   1  to  64   M  is the so-called transmission line, the first to M-th end-open stubs  64   1  to  64   M  can be formed as a conductive pattern on a printed board. That is, the element can be formed without using separate components such as an inductor element and a capacitor element. Accordingly, when the resonant circuit section  60 B is composed of the first to M-th end-open stubs  64   1  to  64   M  as described in the second embodiment, the manufacturing efficiency, component cost, reliability and power durability of the switching circuit  1 B can be improved. Since each of the first to M-th end-open stubs  64   1  to  64   M  can be composed only of the transmission line, i.e., the conductive pattern on the printed board, its element value has high accuracy, so that the switching circuit  1 B requires no adjustment after being manufactured. 
     (Third Embodiment) 
     In the first embodiment, the passive circuit section  50 A ( 50 ) is composed of the so-called lumped constant element. However, the passive circuit section  50  may be also composed of a distributed constant element. The case in which the passive circuit section  50  is composed of the distributed constant element will be described. Here, the passive circuit section  50  is referred to as passive circuit section  50 B 
     The passive circuit section  50 B including a transmission line  530  as the distributed constant element may be designed as described below. First, the case of N=1 will be described. In one embodiment, the transmission line  530  may be a stub. 
     The passive circuit section  50 B is a two-port circuit network. An impedance matrix Z and an inverse matrix (an admittance matrix) Y are known for the two-port circuit network. The impedance matrix and the admittance matrix Y are expressed as in expressions (12a) and (12b) as a function of an angular frequency ω. 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     12 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   Z 
                   = 
                   
                     [ 
                     
                       
                         
                           
                             
                               z 
                               11 
                             
                             ⁡ 
                             
                               ( 
                               ϖ 
                               ) 
                             
                           
                         
                         
                           
                             
                               z 
                               12 
                             
                             ⁡ 
                             
                               ( 
                               ϖ 
                               ) 
                             
                           
                         
                       
                       
                         
                           
                             
                               z 
                               21 
                             
                             ⁡ 
                             
                               ( 
                               ϖ 
                               ) 
                             
                           
                         
                         
                           
                             
                               z 
                               22 
                             
                             ⁡ 
                             
                               ( 
                               ϖ 
                               ) 
                             
                           
                         
                       
                     
                     ] 
                   
                 
               
               
                 
                   ( 
                   
                     12 
                     ⁢ 
                     a 
                   
                   ) 
                 
               
             
             
               
                 
                   Y 
                   = 
                   
                     
                       Z 
                       
                         - 
                         1 
                       
                     
                     = 
                     
                       [ 
                       
                         
                           
                             
                               
                                 y 
                                 11 
                               
                               ⁡ 
                               
                                 ( 
                                 ϖ 
                                 ) 
                               
                             
                           
                           
                             
                               
                                 y 
                                 12 
                               
                               ⁡ 
                               
                                 ( 
                                 ϖ 
                                 ) 
                               
                             
                           
                         
                         
                           
                             
                               
                                 y 
                                 21 
                               
                               ⁡ 
                               
                                 ( 
                                 ϖ 
                                 ) 
                               
                             
                           
                           
                             
                               
                                 y 
                                 22 
                               
                               ⁡ 
                               
                                 ( 
                                 ϖ 
                                 ) 
                               
                             
                           
                         
                       
                       ] 
                     
                   
                 
               
               
                 
                   ( 
                   
                     12 
                     ⁢ 
                     b 
                   
                   ) 
                 
               
             
           
         
       
     
     A current having an angular frequency ω=0, that is, a direct current needs to flow from the end  50   a  to the end  50   b  without a voltage drop. To satisfy the condition, when the end  50   b  is shunted to a ground terminal, the end  50   a  may be also regarded to be shunted. When expressed in an expression, this is expressed by an expression (13a). Moreover, the direct current should not leak from the end  50   a  to the ground terminal. To satisfy the condition, when the end  50   b  is opened, the end  50   a  may be also regarded to be opened. This is expressed by an expression (13b). Furthermore, a current having an angular frequency ω=ω CK , that is, a current having the clock frequency is prevented from flowing from the drain terminal  13  toward the passive circuit section  50 B (the right side in  FIG. 1 ). That is, a parallel combined admittance of the end  50   a  and the output parasitic capacitance C ds  may become zero when the end  50   b  is shunted to the ground. This is expressed by an expression (13c). 
     [Expression 13]
 
 y   11 (0)=∞  (13a),
 
 z   11 (0)=∞  (13b),
 
 y   11 (ω CK )+ jω   CK   C   ds =0  (13c)
 
     In the expressions (13a) and (13b), a sign ∞ has a meaning that an absolute value of a complex number becomes infinity. 
     In the case of N=1, the required degree of freedom is 3 as indicated in the expressions (13a) to (13c). However, the expressions (13a) and (13b) are obtained at the same time when the transmission line  530  is one transmission line connected in series. Thus, the minimum required number of transmission lines is 1. 
     The total number of topologies of the passive circuit section  50 B composed of one transmission line  530  is 3 as shown in  FIGS. 11(   a ) to  11 ( c ). One that satisfies at least one of an incompatibility condition (I) and an incompatibility condition (II) described below is eliminated from the above three topologies. 
     Incompatibility Condition (I): The end is grounded with respect to the direct current. 
     Incompatibility Condition (II): An element that directly shunts the second end  50   b  is provided. 
     The reason why the circuit satisfying the incompatibility condition (I) is incompatible as the circuit of the passive circuit section  50 B is that the circuit satisfying the incompatibility condition (I) contradicts the expression (13b). The reason why the circuit satisfying the incompatibility condition (II) is incompatible as the circuit of the passive circuit section  50 B is that the second end  50   b  corresponds to be grounded at the clock frequency f CK , so that the voltage of the clock frequency f CK  is not applied to the transmission line  530  in the configuration of the incompatibility condition (II) and there is not enough degree of freedom. 
     When the inappropriate topology is eliminated by applying the incompatibility condition (I) and the incompatibility condition (II) to the three types of topologies, the passive circuit section  50 B composed of the distributed constant element has a configuration shown in  FIG. 11(A)  in the case of N=1. 
     By applying such a condition that a combined impedance Z c1  of the output parasitic capacitance C ds  and the passive circuit section  50 B becomes infinity at the clock frequency f CK , that is, the expression (13c) to the passive circuit section  50 B shown in  FIG. 11(   a ) in a similar manner to the first embodiment, an electrical length of the transmission line  530  is determined. The electrical length of the transmission line  530  can be represented by a phase difference θ 530  at the clock frequency f CK . Thus, the electrical length is also referred to as electrical length θ 530  below. 
     To be more specific, the expression (13c) as the condition that the combined impedance Z c1  of the output parasitic capacitance C ds  and the passive circuit section  50 B becomes infinity at the clock frequency f CK  (or that the combined admittance becomes zero at the clock frequency f CK ) is expressed by an expression (14). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     14 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       j 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         ϖ 
                         CK 
                       
                       ⁢ 
                       
                         C 
                         ds 
                       
                     
                     + 
                     
                       1 
                       
                         j 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           Z 
                           0 
                         
                         ⁢ 
                         tan 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           θ 
                           530 
                         
                       
                     
                   
                   = 
                   0 
                 
               
               
                 
                   ( 
                   14 
                   ) 
                 
               
             
           
         
       
     
     In the expression (14), Z 0  is a characteristic impedance of the transmission line  530 . An expression (15) is obtained from the expression (14). 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     15 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     θ 
                     530 
                   
                   = 
                   
                     arg 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       tan 
                       ⁡ 
                       
                         ( 
                         
                           1 
                           
                             
                               Z 
                               0 
                             
                             ⁢ 
                             
                               ϖ 
                               CK 
                             
                             ⁢ 
                             
                               C 
                               ds 
                             
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   15 
                   ) 
                 
               
             
           
         
       
     
     For example, when C ds =60 pF, ω CK =2π×200 MHz, and Z 0 =50Ω, θ 530 =0.2593 radians. Thus, when the clock frequency f CK  is 200 MHz and the output parasitic capacitance C ds  is 60 pF, the transmission line  530  in the case of N=1 may be configured such that the impedance Z 0  is 50Ω and the electrical length (the phase difference) is 0.2593 radians. 
     In the case of N=2, the passive circuit section  50 B is designed so as to satisfy the following four conditions by re-using an element of the admittance matrix Y indicated in the expression (12b). Here, the passive circuit section  50 B capable of preventing the current having the frequency component up to the second-order harmonic (that is, the fundamental wave (the case of N=1) and the second-order harmonic) out of the harmonics of consecutive orders of the clock frequency f CK  from flowing through the switching element  10  will be described as the passive circuit section  50 B in the case of N=2. 
     [Expression 16]
 
 y   11 (0)=∞  (16a),
 
 z   11 (0)=∞  (16b),
 
 y   11 (ω CK )+ j  ω     CK   C   ds =0  (16c),
 
 y   11 (2ω CK )+ j 2  ω   CK   C   ds =0  (16d)
 
     It is the same as the case of the expressions (13 a) and (13c) that y 11 (ω) in the expressions (16a), (16c), and (16d) is a first element of the admittance matrix Y. 
     In the case of the passive circuit section  50 B corresponding to up to the second-order harmonic, the required degree of freedom is 4 as indicated in the expressions (16a) to (16d). Thus, the minimum number of transmission lines  530  of the passive circuit section  5013  is 2. When the two transmission lines  530  are separately described, the two transmission lines  530  are referred to as transmission lines  531  and  532 . 
     The total number of topologies of the passive circuit section  50 B composed of the two transmission lines (including the case of the stub) is 10 as shown in  FIGS. 12(   a ) to  12 ( j ). By eliminating one that satisfies at least one of the incompatibility conditions (I) and (II) from the 10 topologies, configurations shown in  FIGS. 12(   a ) to  12 ( c ) remain. 
     By applying the conditions of the expressions (16c) and (16d), that is, the condition that the combined admittance becomes zero at the fundamental wave (N=1) and the second-order harmonic (N=2) at the same time, the characteristic impedance and the electrical length of each of the transmission lines  531  and  532  can be obtained. 
     The case in  FIG. 12(   a ) will be specifically described. The characteristic impedance of the transmission line  531  is referred to as Z 531 , and the electrical length of the transmission line  531  is referred to as θ 531 . Similarly, the characteristic impedance of the transmission line  532  is referred to as Z 532 , and the electrical length of the transmission line  532  is referred to as θ 532 . In this case, a next expression is obtained. 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     17 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ϖ 
                         CK 
                       
                       ⁢ 
                       
                         C 
                         ds 
                       
                     
                     - 
                     
                       1 
                       
                         
                           Z 
                           531 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           
                             
                               
                                 Z 
                                 532 
                               
                               ⁢ 
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 θ 
                                 532 
                               
                             
                             + 
                             
                               
                                 Z 
                                 531 
                               
                               ⁢ 
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 θ 
                                 531 
                               
                             
                           
                           
                             
                               Z 
                               531 
                             
                             - 
                             
                               
                                 Z 
                                 532 
                               
                               ⁢ 
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 θ 
                                 532 
                               
                               ⁢ 
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 θ 
                                 531 
                               
                             
                           
                         
                       
                     
                   
                   = 
                   0 
                 
               
               
                 
                   ( 
                   
                     17 
                     ⁢ 
                     a 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       2 
                       ⁢ 
                       
                         ϖ 
                         CK 
                       
                       ⁢ 
                       
                         C 
                         ds 
                       
                     
                     - 
                     
                       1 
                       
                         
                           Z 
                           531 
                         
                         ⁢ 
                         
                           
                             
                               
                                 Z 
                                 532 
                               
                               ⁢ 
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                               ⁢ 
                               
                                 θ 
                                 532 
                               
                             
                             + 
                             
                               
                                 Z 
                                 531 
                               
                               ⁢ 
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                               ⁢ 
                               
                                 θ 
                                 531 
                               
                             
                           
                           
                             
                               Z 
                               531 
                             
                             - 
                             
                               
                                 Z 
                                 532 
                               
                               ⁢ 
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                               ⁢ 
                               
                                 θ 
                                 532 
                               
                               ⁢ 
                               tan 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                               ⁢ 
                               
                                 θ 
                                 531 
                               
                             
                           
                         
                       
                     
                   
                   = 
                   0 
                 
               
               
                 
                   ( 
                   
                     17 
                     ⁢ 
                     b 
                   
                   ) 
                 
               
             
           
         
       
     
     When Z 531 /Z 532 =α, a next expression is obtained from the above expressions. 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     18 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         α 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         tan 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           θ 
                           531 
                         
                       
                       + 
                       
                         tan 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           θ 
                           532 
                         
                       
                     
                     
                       α 
                       - 
                       
                         tan 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           θ 
                           531 
                         
                         ⁢ 
                         tan 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           θ 
                           532 
                         
                       
                     
                   
                   = 
                   
                     2 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       
                         
                           α 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           2 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             531 
                           
                         
                         + 
                         
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           2 
                           ⁢ 
                           
                             θ 
                             532 
                           
                         
                       
                       
                         α 
                         - 
                         
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           2 
                           ⁢ 
                           
                             θ 
                             531 
                           
                           ⁢ 
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           2 
                           ⁢ 
                           
                             θ 
                             532 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   18 
                   ) 
                 
               
             
           
         
       
     
     From the expressions (17a), (17b), and (18), Z 531  and Z 532  are expressed as below. 
     
       
         
           
             
               
                 
                   [ 
                   
                     Expression 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     19 
                   
                   ] 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     Z 
                     531 
                   
                   = 
                   
                     
                       1 
                       
                         
                           ϖ 
                           CK 
                         
                         ⁢ 
                         
                           C 
                           ds 
                         
                       
                     
                     ⁢ 
                     
                       
                         α 
                         - 
                         
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             531 
                           
                           ⁢ 
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             532 
                           
                         
                       
                       
                         
                           α 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             531 
                           
                         
                         + 
                         
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             532 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     19 
                     ⁢ 
                     a 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     Z 
                     532 
                   
                   = 
                   
                     
                       1 
                       
                         α 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           ϖ 
                           CK 
                         
                         ⁢ 
                         
                           C 
                           ds 
                         
                       
                     
                     ⁢ 
                     
                       
                         α 
                         - 
                         
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             531 
                           
                           ⁢ 
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             532 
                           
                         
                       
                       
                         
                           α 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             531 
                           
                         
                         + 
                         
                           tan 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             θ 
                             532 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     19 
                     ⁢ 
                     b 
                   
                   ) 
                 
               
             
           
         
       
     
     Here, the clock frequency f CK  is 200 MHz. In this case, ω CK =2π×200 MHz. Also, it is assumed that C ds =60 pF. One example of a solution obtained when the expressions (19a) and (19b) are numerically calculated is as follows.
 
 Z   531 =60.0347 Ω
 
θ 531 =27°
 
 Z   532 =21.4348 Ω
 
θ 532 =144°
 
     Since the characteristic impedances Z 531  and Z 532  and the electrical lengths θ 531  and θ 532  of the respective transmission lines  531  and  532  are obtained as described above, the respective transmission lines  531  and  532  can be configured. 
     A simulation was performed using a circuit configuration shown in  FIG. 13  by providing Z 531 , θ 531 , Z 532 , and θ 532  described above. In the circuit configuration shown in  FIG. 13 , it was assumed that the resonant circuit section  60  resonated at the clock frequency f CK  and the frequency twice as high as the clock frequency f CK , that is, the second end  50   b  was shunted (grounded). A signal having a frequency f (MHz) was supplied to the first end  50   a  from the signal source  20 . 
       FIG. 14  is a graph showing a simulation result. In  FIG. 14 , a horizontal axis represents the frequency f CK  supplied to the first end, and a vertical axis represents an admittance. In  FIG. 14 , a solid line represents a real part of the combined admittance of the passive circuit section  50 B and the output parasitic capacitance C ds , and a dash line represents an imaginary part of the combined admittance. As shown in  FIG. 15 , it can be understood that the real part and the imaginary part of the combined admittance become zero when the frequency f CK  is 200 MHz, and 400 MHz as the frequency twice as high as 200 MHz, and the expressions (16c) and (16d) are satisfied. It can be also understood that the imaginary part of the combined admittance becomes infinity when the frequency is 0, and the expression (16a) is also satisfied. 
     In the case of N=3 or more, the passive circuit section  50 B may be similarly composed of the distributed constant elements. For example,  FIGS. 15(   a ) to  15 ( k ) are diagrams illustrating configuration examples of the passive circuit section  50 B capable of preventing the current having the frequency component up to the third-order harmonic (that is, the fundamental wave (the case of N=1), the second-order harmonic, and the third-order harmonic) out of the harmonics of consecutive orders of the clock frequency f CK  from flowing through the switching element  10  as the passive circuit section  50 B in the case of N=3. In  FIGS. 15(   a ) to  15 ( k ), three transmission lines  530  are referred to as transmission lines  531 ,  532 , and  533  so as to be distinguished from each other. Conditions (the electrical length or the like) to be satisfied by the respective transmission lines  530  can be determined in a similar manner to the case of N=2. 
     Since the transmission line  530  is a conductive pattern on a printed board, the element can be easily formed without using separate components such as an inductor and a capacitor. As a result, the manufacturing efficiency, component cost, reliability and power durability of a switching circuit  1 C can be improved in a similar manner to the second embodiment. Since the passive circuit section  50 B can be formed only by the transmission line, i.e., the conductive pattern on the printed board, the switching circuit  1 C requires no adjustment after being manufactured because of the same reason as that in the second embodiment. 
     Here, the passive circuit section  50 B capable of preventing the current having the frequency component up to the N-th order harmonic (that is, the first-order to N-th order harmonics) with N as a maximum order out of the harmonics of consecutive orders of the clock frequency f CK  from flowing through the switching element  10  has been described. However, as long as the passive circuit section  50 B is designed with respect to the N-th order harmonic, the passive circuit section  50 B may be designed so as to prevent a current having a frequency component of harmonics of any one or more orders (e.g., odd or even orders) up to the N-th order harmonic from the fundamental wave (the case of N=1) from flowing through the switching element  10 , for example. 
     In the various embodiments of the passive circuit section  50  described above, the element values of the passive circuit section  50  are determined on the condition that the combined impedance of the passive circuit section  50  and the output parasitic capacitance C ds  becomes infinity at the frequency N times as high as the clock frequency f CK  (or the combined admittance becomes zero at the frequency N times as high as the clock frequency f CK ). However, any element value may be employed for the elements of the passive circuit section  50  as long as the passive circuit section  50  satisfies the above “passive circuit section conditions.” The same applies to the resonant circuit section  60 . That is, in the various embodiments of the resonant circuit section  60 , the elements constituting the resonant circuit section  60  are designed such that the real part and the imaginary part of the impedance of the resonant circuit section  60  become zero at the frequency N times as high as the clock frequency f CK . However, the real part and the imaginary part of the impedance of the resonant circuit section  60  may not necessarily become zero in actual circuit manufacturing. That is, the element values of the elements constituting the resonant circuit section  60  may be any value as long as the real part and the imaginary part of the impedance of the resonant circuit section  60  are smaller than the impedance of the load circuit  40  in a resonant state. 
     Moreover, the passive circuit section  50  and the resonant circuit section  60  in the various embodiments described above may be combined together. For example, in addition to the case in which both of the passive circuit section and the resonant circuit section are composed of the lumped constant elements or the distributed constant elements, the switching circuit may include the passive circuit section composed of the lumped constant element and the resonant circuit section composed of the distributed constant element. Conversely, the switching circuit may include the passive circuit section composed of the distributed constant element and the resonant circuit section composed of the lumped constant element. 
     Although the embodiments of the present invention have been described above, the present invention is not limited to the aforementioned various embodiments, and various modifications may be made therein without departing from the scope of the present invention. For example, the switching element may be a field-effect transistor other than the insulated field-effect transistor, a bipolar transistor, and an insulated gate transistor in addition to the insulated field-effect transistor. Although the signal that drives the switching element is the PWM signal, any pulse signal may be employed as long as the ON/OFF switching of the switching element can be controlled. Moreover, in the second embodiment, the resonant circuit section  60 B is composed of the end-open stub  64 . However, the resonant circuit section  60 B may be composed of the transmission line as the distributed constant element.