Abstract:
An amplifier circuit capable of reducing load of a circuit at the previous stage by providing increased input impedance producing less noises. The amplifier circuit includes a fully-differential operational amplifier composed of an inverting input terminal, a non-inverting input terminal receiving a signal different from a signal to be input to the inverting input terminal, an inverting output terminal with the same polarity of the inverting input terminal, and a non-inverting output terminal with reverse polarity; an input impedance element with one end connected to the inverting input terminal; an input impedance element with one end connected to the non-inverting input terminal; and positive feedback impedance elements, with one end of connected to the other end of the input impedance element and the other end connected to the inverting output terminal or to the non-inverting output terminal.

Description:
FIELD OF THE INVENTION 
       [0001]    The present invention relates to an amplifier circuit, in particular, to an amplifier circuit with a fully-differential operational amplifier. 
       RELATED ART 
       [0002]    An amplifier circuit is, as is well known, a circuit having a capability of amplifying an input signal, which is connected to a different circuit makes up equipment. The amplifier circuit of this kind is usually made requests to have high input impedance, in order to reduce load in connection with power consumption of a circuit at the previous stage to which a signal is input. 
         [0003]      FIG. 5  is a view for explaining a conventional general amplifier circuit. The amplifier circuit shown in  FIG. 5  includes a fully-differential operational amplifier  104 , two impedance elements  101   a  and  101   b  connected to an inverting input terminal  104   a  and an non-inverting input terminal  104   c  of the operational amplifier  104 , respectively, and two negative feedback impedance elements  102   a  and  102   b  provided between the inverting input terminal  104   a  and the non-inverting output terminal  104   b , and between the non-inverting input terminal  104   c  and the inverting output terminal  104   d . Impedance of the input impedance elements  101   a  and  101   b  is both Z 1  and impedance of the negative feedback impedance elements  102   a  and  102   b  is both Z 2 . 
         [0004]    Also, Vip and Vin depicted in  FIG. 5  indicates voltage of a signal to be input to the amplifier circuit, Von and Vop indicate voltage of a signal to be output from the amplifier circuit. Vsn and Vsp are input voltage of the operational amplifier  104 . Subscripts “p” and “n” of a symbol indicating the above physical quantity represent a phase of the voltage. The voltage indicated by the subscript “p” and the voltage indicated by the subscript “n” are voltage of which the phase is inverted based on a DC component of an alternating electrical current respectively. Put differently, they have a phase different by 180 degrees from each other. 
         [0005]    In the general amplifier circuit shown in  FIG. 5 , a large impedance value of the input impedance elements  101   a  and  101   b  cause problems that one undergoes loud noises produced in the amplifier circuit. Thus, in such prior art, it has been encountered difficulties of incapable of increasing impedance to a sufficient degree. 
         [0006]    An example of the prior art devised for the purpose of increasing impedance of the amplifier circuit shown in  FIG. 5  includes, e.g., an amplifier circuit shown in  FIG. 6 . The amplifier circuit shown in  FIG. 6  includes a fully-differential operational amplifier  404 , input impedance elements  401  to  403 , and feedback impedance elements  410  to  413 . Out of the input impedance elements  401  to  403 , the input impedance element  401  is, as with the input impedance elements  101   a  and  101   b  shown in  FIG. 5 , connected to a non-inverting input terminal and an inverting input terminal of the operational amplifier. That is, the amplifier circuit shown in  FIG. 6  is one in which input impedance elements  402  and  403  are added to input impedance elements of the amplifier circuit shown in  FIG. 5 . 
         [0007]    With the amplifier circuit of such prior art, it enables making a ratio (Vo/Vi) between voltage Vi to be input to input terminals  406 , 407  and voltage Vo to be output from output terminals  408 , 409  higher. Incidentally, such prior art can be found in Patent Document 1. 
       PRIOR ART DOCUMENT 
       [0000]    
       
         Patent Document 1: JP 2004-320712 A 
       
     
       SUMMARY OF THE INVENTION 
     Problem to be Solved 
       [0009]    As stated above, for the load of the circuit arranged at the previous stage of the amplifier circuit to reduce to a sufficient degree, it is necessary to reduce electrical currents Iip, Iin, having a phase shifted by 180 degrees from each other, up to about “0”. The prior art shown in  FIG. 6  is, however, not to intended reducing electrical currents Iip and Iin, but to reduce electrical current i 1  flowing through the feedback impedances  410 , 411 . 
         [0010]    To reduce noises produced in the amplifier circuit, it is imperative to make impedance values Z 1  and Z 2  of the input impedance element  101 , and the negative impedance element  102  smaller. But, in the prior art shown in  FIG. 6 , it takes no account of the magnitude of an impedance value of the impedance element. Consequently, the prior art has not been able to provide an amplifier circuit which reduces load of the circuit at the previous stage as well as produces lesser noises. 
         [0011]    The present invention is made in view of such shortcoming immanent in the prior art, and its objective is to provide an amplifier circuit which is able to reduce the load of the circuit at the previous stage by increasing input impedance as well as produces lesser noises. 
       Solution to the Problem 
       [0012]    To solve the above-identified problems, an amplifier circuit of one embodiment of the present invention, comprises: a fully-differential operational amplifier (e.g., an operational amplifier  104  shown in  FIG. 1 ) including a first input terminal (e.g., an inverting input terminal  104   a  shown in  FIG. 1 ), a second input terminal (e.g., a non-inverting input terminal  104   c  shown in  FIG. 1 ) to  which a signal different from a signal to be input to the first input terminal is input, a first output terminal (e.g., an inverting output terminal  104   d  shown in  FIG. 1 ) with  the same polarity as that of the first input terminal, and a second output terminal (e.g., a non-inverting output terminal  104   b ) with reverse polarity to that of the first input terminal; an input impedance element (e.g., input impedance elements  101   a  or  101   b  shown in  FIG. 1 ), one end of which is connected to the first input terminal; and a positive feedback impedance element (e.g., positive feedback impedance elements  103   a  or  103   b  shown in  FIG. 1 ), one end of which is connected to the other end of the input impedance element and the other end of which is connected to the first output terminal. 
         [0013]    The present invention, it is preferable, in the foregoing invention, that the amplifier circuit should further comprise a negative feedback impedance element (e.g., negative feedback impedance elements  102   a  or  102   b ) to be connected between the first input terminal and the second output terminal, and being relation of: Z 3 ≧Z 2 -Z 1  among an impedance value Z 1  of the input impedance element; an impedance value Z 2  of the negative feedback impedance element; and an impedance value Z 3  of the positive feedback impedance element. 
         [0014]    The present invention is preferable, in the foregoing invention, that the amplifier circuit should further comprise a negative feedback impedance element (e.g., a negative feedback impedance element  102   a  or  102   b  shown in  FIG. 1 ) to  be connected between the first input terminal and the second input terminal, and being relation of: Z 3 ≧(Z 2 +Z 1 )×(Z 2 −2×Z 1 )÷(Z 2 +2×Z 1 ) among an impedance value Z 1  of the input impedance element; an impedance value Z 2  of the negative feedback impedance element; and an impedance value Z 3  of the positive feedback impedance element. 
         [0015]    An amplifier circuit of one embodiment of the present invention comprises: a fully-differential operational amplifier (e.g., an operational amplifier  104  shown in  FIG. 1 ) including  a first input terminal (e.g., an inverting input terminal  104   a  shown in  FIG. 1 ), a second input terminal (e.g., a non-inverting input terminal  104   c  shown in  FIG. 1 ), a first output terminal (e.g., a non-inverting output terminal  104   d  shown in  FIG. 1 ) with the same polarity as that of the first input terminal, and a second output terminal (e.g., a non-inverting output terminal  104   b  shown in  FIG. 1 ) with reverse polarity to that of the first input terminal; a first input impedance element (e.g., an input impedance element  101   a  shown in  FIG. 1 ) to one end of which an input signal is input and to the other end of which the first input terminal of the operational amplifier is connected; a second impedance element (e.g., an input impedance element  101   b  shown in  FIG. 1 ) to one end of which an input signal is input different from the input signal and to the other end of which the second input terminal of the operational amplifier is connected; a first negative feedback impedance element (e.g., a negative feedback impedance element  102   a  shown in  FIG. 1 ), one end of which is connected to the first input terminal and the other end of which is connected to the second output terminal; a second negative feedback impedance element (e.g., a negative feedback impedance element  102   b  shown in  FIG. 1 ), one end of which is connected to the second input terminal and the other end of which is connected to the first output terminal; a first positive feedback impedance element (e.g., a positive feedback impedance element  103   b  shown in  FIG. 1 ), one end of which is connected to the first output terminal and the other end of which is connected to one end of the first input impedance element; and a second positive feedback impedance element (e.g., a positive feedback impedance element  103   a  shown in  FIG. 1 ), one end of which is connected to the second output terminal and the other end of which is connected to one end of the second positive feedback impedance element. 
         [0016]    The amplifier circuit of one embodiment is preferable, in the foregoing invention, that an impedance value of the first input impedance element and that of the second input impedance element should be equal, an impedance value of the first negative feedback impedance element and that of the second negative feedback impedance element should be equal, and an impedance value of the first positive feedback impedance element and that of the second positive feedback impedance element should be equal. 
         [0017]    The amplifier circuit of one embodiment is preferable, in the foregoing invention, that relation of: Z 3 ≧Z 2 −Z 1  should be between an impedance value Z 1  of the first input impedance element and the second input impedance element, an impedance value Z 2  of the first negative feedback impedance element and the second negative feedback impedance element, and an impedance value Z 3  of the first positive feedback impedance element and the second positive feedback impedance element. 
         [0018]    An amplifier circuit of one embodiment of the present invention comprises: a fully-differential operational amplifier (e.g., an operational amplifier  104  shown in  FIG. 3 ) including a first input terminal (e.g., an inverting input terminal  104   a  shown in  FIG. 3 ), a second input terminal (e.g., a non-inverting input terminal  104   c  shown in  FIG. 3 ), a first output terminal (e.g., a inverting output terminal  104   d  shown in  FIG. 3 ) with  the same polarity as that of the first input terminal, and a second output terminal (e.g., a non-inverting output terminal  104   b  shown in  FIG. 3 ) with  reverse polarity to that of the first input terminal; a first input impedance element (e.g., an input impedance element  101   a  shown in  FIG. 3 ) to one end of which an input signal is input and to the other end of which the first input terminal of the operational amplifier is connected; a second input impedance element (e.g., an input impedance element  101   b  shown in  FIG. 3 ) to one end of which a reference voltage is input and to the other end of which the second input terminal of the operational amplifier is connected; a first negative feedback impedance element (e.g., a negative feedback impedance element  102   a  shown in  FIG. 3 ), one end of which is connected to the first input terminal and the other end of which is connected to the second output terminal; a second negative feedback impedance element (e.g., a negative feedback impedance element  102   b  shown in  FIG. 3 ), one end of which is connected to the second input terminal and the other end of which is connected to the first output terminal; and a positive feedback impedance element (e.g., a positive feedback impedance element  103   b  shown in  FIG. 3 ), one end of which is connected to the first output terminal and the other end of which is connected to the first input impedance element. 
         [0019]    An amplifier circuit of one embodiment of the present invention comprises: a fully-differential operational amplifier (e.g., an operational amplifier  104  shown in  FIG. 4 ) including a first input terminal (e.g., an inverting input terminal  104   a  shown in  FIG. 4 ), a second input terminal (e.g., a non-inverting input terminal  104   c  shown in  FIG. 4 ), a first output terminal (e.g., an inverting output terminal  104   d  shown in  FIG. 4 ) with  the same polarity as that of the first input terminal, and a second output terminal (e.g., a non-inverting output terminal  104   b  shown in  FIG. 4 ) with  reverse polarity to that of the first input terminal; a first input impedance element (e.g., an input impedance element  101   a  shown in  FIG. 4 ) to one end of which an input signal is input and to the other end of which the first input terminal of the operational amplifier is connected; a second input impedance element (e.g., an input impedance element  101   b  shown in  FIG. 4 ) to  one end of which a reference voltage is input and to the other end of which the second input terminal of the operational amplifier is connected; a first negative feedback element (e.g., a negative feedback impedance element  102   a  shown in  FIG. 4 ), one end of which is connected to the first input terminal and the other end of which is connected to the second output terminal; a second negative feedback impedance element (e.g., a negative feedback impedance element  102   b  shown in  FIG. 4 ), one end of which is connected to the second input terminal and the other end of which is connected to the first output terminal; and a positive feedback impedance element (e.g., a positive feedback impedance element  103   a ), one end of which is connected to the second output terminal and the other end of which is connected to one end of the second input impedance element. 
         [0020]    The amplifier circuit of one embodiment is preferable, in the foregoing invention, that an impedance value of the first input impedance element and that of the second input impedance element should be equal, and an impedance value of the first negative feedback impedance element and that of the second negative feedback impedance element should be equal. 
         [0021]    The amplifier circuit of one embodiment is preferable, in the foregoing invention, that relation of: Z 3 ≧(Z 2 +Z 1 )×(Z 2 −2×Z 1 )÷(Z 2 +2×Z 1 ) should be among an impedance value Z 1  of the first input impedance element and the second input impedance element, an impedance value Z 2  of the first negative feedback impedance element and the second negative feedback impedance element, and an impedance value Z 3  of the positive feedback impedance element. 
       Advantageous Effect of the Invention 
       [0022]    According to the amplifier circuit of the embodiments as mentioned above, since the amplifier circuit includes the positive feedback impedance element provided via the input impedance element connected to the input terminal between the input terminal and the output terminal outputting an signal with a signal to be input from the input terminal of the operational amplifier, the invention may increase input impedance of the amplifier circuit, without incurring an increased impedance value of the input impedance element. On account of this, the invention permits providing an amplifier circuit with high input impedance while reducing noises to be produced in the amplifier circuit. 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         [0023]      FIG. 1  is a view for explaining an amplifier circuit of a first embodiment of the present invention; 
           [0024]      FIG. 2  is a circuit diagram for explaining the inside of an amplifier circuit shown in  FIG. 1 ; 
           [0025]      FIG. 3  is a view for explaining an amplifier circuit of a second embodiment of the present invention; 
           [0026]      FIG. 4  is a view for explaining a variant of the second embodiment of the present invention; 
           [0027]      FIG. 5  is a view for explaining a general amplifier circuit; and 
           [0028]      FIG. 6  is a view for explaining a conventional amplifier circuit based on the amplifier circuit illustrated in  FIG. 5 . 
       
    
    
     DESCRIPTION OF EMBODIMENTS 
       [0029]    Hereinafter, a description will next be made to a first embodiment and a second embodiment of the present invention with reference to the accompanying drawings. 
       First Embodiment 
       [0030]    Circuit Configuration 
         [0031]      FIG. 1  is a view for explaining an amplifier circuit of the first embodiment. The amplifier circuit shown in  FIG. 1  includes an operational amplifier  104 , two input impedance elements  101   a  and  101   b , two negative feedback impedance elements  102   a  and  102   b , two positive feedback impedance elements  103   a  and  103   b , and a common mode feedback circuit (hereunder, referred to simply as CMFB in FIG.  1 ) 105 . The common mode feedback circuit  105  is a circuit to detect common mode voltage to be output from the operational amplifier  104 , and to feedback it. The common mode feedback circuit  105  allows holding the common mode voltage to a constant value. 
         [0032]    The operational amplifier  104  is a fully-differential operational amplifier which has inverting input terminals  104   a , a non-inverting input terminal  104   c  to which a signal different from a signal to be input to the inverting input terminals  104   a , a non-inverting output terminal  104   b  having polarity identical with that of the inverting input terminal  104   a , and an inverting output terminal  104   d  with reverse polarity to that of the inverting input terminal  104   a . One end of the impedance elements  101   a  is connected to the inverting input terminal  104   a , and the other end of the impedance element is connected to the signal input terminal  106 . Further, voltage Vip is applied to a signal input terminal  106  from a circuit at the previous stage. At this time, an electrical current Iip is flown thereinto from the signal input terminal  106 . 
         [0033]    It is noted that “the signal different from the signal to be input to the inverting input terminal  104   a ” indicates a signal different in a phase and voltage value to be output simultaneously. Although a signal to be input to the inverting input terminal  104   a  and a signal to be input to the non-inverting input terminal  104   c  have an approximately anti-phase relationship between their phases with each other, there are some cases where a phase difference does not exactly amount to 180 degrees depending on a relation of the impedance, etc. 
         [0034]    As stated above, a relationship between the two terminals that are respectively allocated to two signals having an approximately anti-phase relation with each other may also be represented as “polarity is reverse”. In addition, a relationship between the two terminals that are respectively allocated to two signals having an approximately in-phase relation with each other may also be represented as “polarity is same”. 
         [0035]    Between the inverting input terminal  104   a  and the non-inverting output terminal  104   b , the negative feedback impedance element  102   a  is connected. The non-inverting output terminal  104   b  is connected to a signal output terminal  108 . In this instance, voltage of a signal to be output from the signal output terminal  108  is represented as Von. 
         [0036]    The non-inverting input terminal  104   a  is connected via the input impedance element  101   a  to the signal input terminal  106 . Moreover, voltage Vip is input through a circuit at the previous stage from the signal input terminal  106 . At this time, an electrical current Iip is flown from the signal input terminal  106 . 
         [0037]    Between the non-inverting input terminal  104   c  and the inverting output terminal  104   d , the negative feedback impedance element  102   b  is connected. The inverting output terminal  104   d  is connected to the signal output terminal  109 . At this time, voltage of a signal to be output from the signal output terminal  109  is represented as Vop. 
         [0038]    The non-inverting input terminal  104   c  is connected via the input impedance element  101   b  to the signal input terminal  107 . Further, voltage Vin is input through a circuit at the previous stage through the signal input terminal  107 . On this occasion, an electrical current Iin is flown from the signal input terminal  107 . 
         [0039]    Furthermore, in the first embodiment, the positive feedback impedance element  103   b  is provided, one end of which is connected to one end of the input impedance element  101   a  and the other end of which is connected to the inverting output terminal  104   d . Also, the positive feedback impedance element  103   a  is provided, one end of which is connected to the other end of the input impedance element  101   b  and the other end of which is connected to the non-inverting output terminal  104   b.    
         [0040]    Out of the impedance elements mentioned above, in the first embodiment, letting an input impedance value of the input impedance elements  101   a  and  101   b  to be Z 1 , an impedance value of the negative feedback impedance elements  102   a  and  102   b  to be Z 2 , and an impedance value of the positive feedback impedance elements  103   a  and  103   b  to be Z 3 . 
         [0041]    The signal input terminals  106 , 107  are terminals for inputting an input signal from the circuit at the previous stage to the amplifier circuit. A differential signal having a phase different by 180 degrees from each other is input, as an input signal, to the signal input terminals  106 , 107 . Moreover, a differential signal is output, as an output signal, from the signal output terminals  108 , 109 . 
         [0042]    Subscripts “n” and “p” shown in  FIG. 1  are ones for indicating polarity of a phase of signal. It has relation of anti-polarity between a signal indicated by the subscript “n” and a signal indicated by the subscript of “p”. Furthermore, it has relation of homopolarity between signals to which a subscript “n” is appended and signals to which a subscript “n” is appended. 
         [0043]      FIG. 2  is a circuit diagram for explaining the inside of the operational amplifier  104  shown in  FIG. 1 . In  FIG. 2 , an inverting input terminal  104   a , a non-inverting input terminal  104   c , a non-inverting output terminal  104   b , and an inverting output terminal  104   d  respectively correspond to the terminals with same name. 
         [0044]    Further, as shown in  FIG. 2 , a positive power supply voltage VDD, a negative power supply voltage VSS, Vbiasp, and Vbiasn for determining an electrical current to be flown through the operational amplifier  104  are applied to the operational amplifier  104 . 
         [0045]    Operation 
         [0046]    An explanation will next be made to an operation of the operational amplifier of the first embodiment described above by using equations. 
         [0047]    (1) Operation of Conventional Amplifier Circuit 
         [0048]    Herein, to compare with an operation of the amplifier circuit of the first embodiment, an explanation will be made first to an operation of the conventional amplifier circuit shown in  FIG. 5  by using equations. In this section, voltage Vip, Von, Vin, Vop, Vsn, Vsp, Z 1 , and Z 2  are physical quantity all depicted in  FIG. 5 , or written in the description of  FIG. 5 . 
         [0049]    In the conventional amplifier circuit shown in  FIG. 5 , a condition where the sum of electrical currents at a node to which Vsp shown in  FIG. 5  is applied is expressed following the Kirchhoff&#39;s law by the following equation (1) 
         [0000]      ( Vip−Vsp )/ Z 1+( Von−Vsp )/ Z 2=0  equation (1)
 
         [0050]    Equation (2) is obtained by changing the equation (1). 
         [0000]      (1 /Z 1+1 /Z 2) Vsp=Vip/Z 1 +Von/Z 2  equation (2)
 
         [0051]    Likewise, as for a node to which voltage Vsn in  FIG. 5  is applied the following equations (3) and (4) are obtained. 
         [0000]      ( Vin−Vsn )/ Z 1+( Vop−Vsn )/ Z 2=0  equation (3)
 
         [0000]      (1 /Z 1+1 /Z 2) Vsn=Vin/Z 1 +Vop/Z 2  equation (4)
 
         [0052]    To find out differential output voltage Vop−Von, subtracting both sides of the equation (4) from the equation (2) obtains the following equation. 
         [0000]    
       
         
           
             
               
                 ( 
                 
                   
                     
                       1 
                       / 
                       Z 
                     
                      
                     
                         
                     
                      
                     1 
                   
                   + 
                   
                     
                       1 
                       / 
                       Z 
                     
                      
                     
                         
                     
                      
                     2 
                   
                 
                 ) 
               
                
               
                 ( 
                 
                   Vsp 
                   - 
                   Vsn 
                 
                 ) 
               
             
             = 
             
               
                 
                   
                     ( 
                     
                       Vip 
                       - 
                       Vin 
                     
                     ) 
                   
                   / 
                   Z 
                 
                  
                 
                     
                 
                  
                 1 
               
               + 
               
                 
                   
                     ( 
                     
                       Von 
                       - 
                       Vop 
                     
                     ) 
                   
                   / 
                   Z 
                 
                  
                 
                     
                 
                  
                 2 
               
             
           
         
       
     
         [0053]    If a gain of the operational amplifier  104  is sufficiently high, the following equation (5) is obtained as one can be regarded voltage Vsp, Vsn as having relation if Vsp=Vsn. 
         [0000]        Vop−Von=Z 2( Vip−Vin )/ Z 1  equation (5)
 
         [0054]    From the equation (5), it can be seen that the amplifier circuit shown in  FIG. 5  is an amplifier circuit with a gain of Z 2 /Z 1 . 
         [0055]    Herein, since common voltage (Von+Vop)/2 of an output signal is controlled so as to be the analog ground (assumed to be 0) by the common mode feedback circuit  105 , the following equation is obtained from relation between the equation (5) and (Vop+Vop)/2=0. 
         [0000]        Vop−Von= 2 Vop=Z 2( ViP−Vin )/ Z 1 
         [0056]    Equation (6) is obtained by changing the above equation. Where, in equation (6), (let Vin,Vip to be) Vin=−Vip. 
         [0000]        Vop=Z 2( Vip−Vin )/2 Z 1 =Vip−Z 2 /Z 1  equation (6)
 
         [0057]    On the other hand, in order to find out the input voltage Vsp and Vsn of the operational amplifier  104 , the both sides of the equations (2), (4) are respectively added to derive the following equation. 
         [0000]    
       
         
           
             
               
                 ( 
                 
                   
                     
                       1 
                       / 
                       Z 
                     
                      
                     
                         
                     
                      
                     1 
                   
                   + 
                   
                     
                       1 
                       / 
                       Z 
                     
                      
                     
                         
                     
                      
                     2 
                   
                 
                 ) 
               
                
               
                 ( 
                 
                   Vsp 
                   - 
                   Vsn 
                 
                 ) 
               
             
             = 
             
               
                 
                   
                     ( 
                     
                       Vip 
                       + 
                       Vin 
                     
                     ) 
                   
                   / 
                   Z 
                 
                  
                 
                     
                 
                  
                 1 
               
               + 
               
                 
                   
                     ( 
                     
                       Von 
                       + 
                       Vop 
                     
                     ) 
                   
                   / 
                   Z 
                 
                  
                 
                     
                 
                  
                 2 
               
             
           
         
       
     
         [0058]    Equation (7) is obtained by changing the above equation. 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       ( 
                       
                         vsp 
                         + 
                         Vsn 
                       
                       ) 
                     
                     / 
                     2 
                   
                   = 
                   
                     
                       
                         
                           
                             ( 
                             
                               Vip 
                               + 
                               Vin 
                             
                             ) 
                           
                           / 
                           Z 
                         
                          
                         
                             
                         
                          
                         
                           
                             1 
                             / 
                             2 
                           
                           / 
                           
                             ( 
                             
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 
                                   1 
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                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                       + 
                       
                         
                           
                             ( 
                             
                               Vop 
                               + 
                               Von 
                             
                              
                             
                                 
                             
                             ) 
                           
                           / 
                           Z 
                         
                          
                         
                             
                         
                          
                         
                           
                             2 
                             / 
                             2 
                           
                           / 
                           
                             ( 
                             
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
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                                 1 
                               
                               + 
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                     
                     = 
                     
                       
                         Z 
                          
                         
                             
                         
                          
                         2 
                          
                         
                           
                             
                               ( 
                               
                                 Vip 
                                 + 
                                 Vin 
                               
                               ) 
                             
                             / 
                             2 
                           
                           / 
                           
                             ( 
                             
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                       + 
                       
                         Z 
                          
                         
                             
                         
                          
                         1 
                          
                         
                           
                             
                               ( 
                               
                                 Vop 
                                 + 
                                 Von 
                               
                               ) 
                             
                             / 
                             2 
                           
                           / 
                           
                             ( 
                             
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   equation 
                    
                   
                       
                   
                    
                   
                     ( 
                     7 
                     ) 
                   
                 
               
             
           
         
       
     
         [0059]    If a gain of the operational amplifier  104  is sufficiently high, one can be regarded voltage Vsp and Vsn as having relation of Vsp=Vsn. Further, the equation (7) becomes equation (8) as relation of Vin=−Vip and Von=−Vop are between these physical quantity. 
         [0000]        Vsp=Vsn= 0  equation (8)
 
         [0060]    Then, input impedance values Zip and Zin of the amplifier circuit shown in  FIG. 5  are found. 
         [0061]    Following the Kirchhoff&#39;s law, a condition where the total of an electrical current Iip to be flown to a node to which the voltage Vip is applied is expressed as the following equation (9). 
         [0000]        Iip +( Vsp−Vip )/ Z 1=0  equation (9)
 
         [0062]    Where, Iip is an input electrical current to be flown from the outside (circuit at the previous stage) to the amplifier circuit. 
         [0063]    Equation (10) is obtained from equations (8) and (9). 
         [0000]        Iip =( Vip−Vsp )/ Z 1 =Vip/Z 1  equation (10)
 
         [0064]    Thus, the input impedance value Zip of the amplifier circuit show in  FIG. 5  is expressed by the following equation (11). 
         [0000]        Zip=Vip/Iip=Z 1  equation (11)
 
         [0065]    Likewise, the input impedance value Zin is expressed by the equation (12). 
         [0000]        Zin=Vin/Iin=Z 1  equation (12)
 
         [0066]    From the above, it can be seen that the conventional amplifier circuit has finite input impedance of Zip and Zin. 
         [0067]    (2) Operation of Amplifier Circuit of First Embodiment 
         [0068]    Next, an explanation will be made to an operation of the amplifier circuit of the first embodiment shown in  FIG. 1 . 
         [0069]    In the amplifier circuit shown in  FIG. 1 , a condition where the total of electrical current to be flown to a node to which voltage Vsp is applied is expressed by the following equation (13) following the Kirchhoff&#39;s law. 
         [0000]      ( Vip−Vsp )/ Z 1+( Von−Vsp )/ Z 2=0  equation (13)
 
         [0070]    Likewise, as for a node to which Vsn is applied, the following equation (14) is obtained. 
         [0000]      ( Vin−Vsn )/ Z 1+( Vop−Vsn )/ Z 2=0  equation (14)
 
         [0071]    Because the equations (13) and (14) are same to the equation of the conventional amplifier circuit, the aforesaid equations (1) to (8) are established even in the amplifier circuit of the first embodiment. 
         [0072]    After that, electrical currents Iip and Iin to be input from the outside to the amplifier circuit of the first embodiment are obtained. 
         [0073]    In  FIG. 1 , a condition where the total of electrical current to be flown to a node to which voltage Vip is applied expressed by the following equation (15) following the Kirchhoff&#39;s law is. 
         [0000]        Iip +( Vsp−Vip )/ Z 1+( Vop−Vip )/ Z 3=0  equation (15)
 
         [0074]    Equation (16) is obtained by changing equation (15). 
         [0000]        Iip+Vsp/Z 1 +Vop/Z 3−(1 /Z 1+1 /Z 3) Vip= 0  equation (16)
 
         [0075]    Equation (6) is substituted for equation (16). If a gain of the operational amplifier  104  is sufficiently high, and equation (16) is as follows as one can be regarded voltage Vsp as having relation of Vsp=0. 
         [0000]        Iip+Z 2 ·Vip /( Z 1 −Z 3)−(1 /Z 1+1 /Z 3) Vip= 0
 
         [0076]    Equation (17) is obtained by changing the above equation. 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         Iip 
                         = 
                           
                          
                         
                           
                             { 
                             
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 3 
                               
                               - 
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 
                                   2 
                                   / 
                                   
                                     ( 
                                     
                                       Z 
                                        
                                       
                                           
                                       
                                        
                                       
                                         1 
                                         · 
                                         Z 
                                       
                                        
                                       
                                           
                                       
                                        
                                       3 
                                     
                                     ) 
                                   
                                 
                               
                             
                             } 
                           
                            
                           Vip 
                         
                       
                     
                   
                   
                     
                       
                         = 
                           
                          
                         
                           
                             { 
                             
                               
                                 ( 
                                 
                                   
                                     Z 
                                      
                                     
                                         
                                     
                                      
                                     1 
                                   
                                   - 
                                   
                                     Z 
                                      
                                     
                                         
                                     
                                      
                                     2 
                                   
                                   + 
                                   
                                     Z 
                                      
                                     
                                         
                                     
                                      
                                     3 
                                   
                                 
                                 ) 
                               
                               / 
                               
                                 ( 
                                 
                                   Z 
                                    
                                   
                                       
                                   
                                    
                                   
                                     1 
                                     · 
                                     Z 
                                   
                                    
                                   
                                       
                                   
                                    
                                   3 
                                 
                                 ) 
                               
                             
                             } 
                           
                            
                           Vip 
                         
                       
                     
                   
                 
               
               
                 
                   equation 
                    
                   
                       
                   
                    
                   
                     ( 
                     17 
                     ) 
                   
                 
               
             
           
         
       
     
         [0077]    Accordingly, an input impedance value Zip of the amplifier circuit of the first embodiment is expressed as the following equation (18). 
         [0000]        Zip=Vip/Iip=Z 1 ·Z 3/( Z 1 −Z 2 +Z 3)  equation (18)
 
         [0078]    Likewise, an impedance value Zin is obtained by the following equation (19). 
         [0000]        Zin=Vin/Iin=Z 1 ·Z 3/( Z 1 −Z 2 +Z 3)  equation (19)
 
         [0079]    With the above equations (18) and (19), if the impedance values Z 1 ,Z 2 ,and Z 3  are set so as to have relation of Z 3 ≧Z 2 -Z 1 , it will be possible to realize an amplifier circuit with high input impedance values Zip and Zin without oscillating the amplifier circuit. 
         [0080]    Here, if we assume Z 1 ,Z 2 ,and Z 3  to have relation of Z 3 =Z 2 −Z 1 , an input impedance will be turned into infinity. Nonetheless, in order to avoid oscillation of the amplifier circuit, it is realistic that an impedance value Z 3  is set slightly larger than impedance value Z 2 -Z 1 . For this reason, Z 3  of the first embodiment has relation of Z 3 ≧Z 2 -Z 1  in which its minimum value is Z 2 −Z 1 . 
         [0081]    Further, in the first embodiment, there is an occurrence in some cases that a design value of the impedance elements  103   a  and  103   b  is set, to realize such a condition, so as to amount to 80% or so of the impedance value Z 3  to be implemented by the impedance elements  103   a  and  103   b.    
         [0082]    Alternatively, in the above-mentioned first embodiment, the impedance elements  101   a ,  101   b ,  102   a ,  102   b ,  103   a  and  103   b  may utilize any elements, as far as they function as an impedance element in the amplifier circuit, such as a capacitance element or a resistance element. It should be noted that since variations in characteristics among these impedance elements impair circuit characteristics of the amplifier circuit of the first embodiment, it is desirable to employ an impedance element of which electrical characteristic and temperature characteristic are matched, as can as possible, as each impedance element. By way of example of elements having matched characteristics, it is desirable to employ not only elements which are fabricated according to the same design and process but also elements which are mounted on the same wafer. 
       Second Embodiment 
       [0083]    Circuit Configuration 
         [0084]      FIG. 3  is a view for explaining the amplifier circuit of the second embodiment of the present invention. Out of the circuits shown in  FIG. 3 , the same circuit having the same configuration as those shown in  FIG. 1  is denoted by the same reference numeral, and a part of descriptions thereof shall be omitted for simplification. The amplifier circuit of the second embodiment differs from that of the first embodiment in which the amplifier circuit has a fully-differential configuration, in that an input section takes a single ended configuration. In other words, in the second embodiment, the other end of the input impedance element  101   b  is connected to the analog ground that is a reference voltage. 
         [0085]    The amplifier circuit of the second embodiment includes a signal input terminal  106  to input from the outside an input signal to the other end of the input impedance element  101   a , and a positive feedback impedance element  103   b  of which one end is connected to the other end of the input impedance element  101   a  and the other end of which is connected to the inverting output terminal  104   d.    
         [0086]    Operation 
         [0087]    An explanation will next be made to an operation of the amplifier circuit of the second embodiment shown in  FIG. 3  by using equations. 
         [0088]    In the amplifier circuit shown in  FIG. 3 , a condition where the total of the electrical current to be flown becomes 0 to a node to which the voltage Vsp is applied according to the Kirchhoff&#39;s law is expressed as the following equation (20). 
         [0000]      ( Vip−Vsp )/ Z 1+( Von−Vsp )/ Z 2=0  equation (20)
 
         [0089]    Equation (21) is obtained by changing equation (20). 
         [0000]      (1 /Z 1+1 /Z 2) Vsp=Vip/Z 1 +Von/Z 2  equation (21)
 
         [0090]    Likewise, as for a node to which the voltage Vsn is applied, the following equations (22) and (23) are obtained. 
         [0000]      (0 −Vsn )/ Z 1+( Von−vsn )/ Z 2=0  equation (22)
 
         [0000]      (1 /Z 1+1 /Z 2) Vsn=Vop/Z 2  equation (23)
 
         [0091]    To find out differential output voltage Vop-Von, subtracting the both sides of equation (23) from equation (21) respectively, the following equation is obtained. 
         [0000]      (1 /Z 1+1 /Z 2)( Vsp−Vsn )= Vip/Z 1+( Von−Vop )/ Z 2 
         [0092]    If a gain of the operational amplifier  104  is sufficiently high, the following equation (24) is obtained as one can be regarded voltage Vsp, Vsn as having relation of Vsp=Vsn. 
         [0000]        Vop−Von =( Z 2 /Z 1) Vip   equation (24)
 
         [0093]    From the equation (24), it can be seen that the amplifier circuit shown in  FIG. 3  is an amplifier circuit with a gain of Z 2 /Z 1 . 
         [0094]    Herein, as common voltage (Von+Vop)/2 of an output signal is controlled so as to be the analog ground (assumed to be 0) by the common mode feedback circuit  105 , the following equation is obtained from relation between the equation (24) and (Von+Vop)/2=0. 
         [0000]        Vop−Von= 2 Vop =( Z 2 /Z 1) Vip    
         [0095]    Equation (25) is obtained by changing the above equation. 
         [0000]        Vop={Z 2/(2 Z 1)} Vip   equation (25)
 
         [0096]    In the meanwhile, to find out input voltages Vsp and Vsn of the operational amplifier  104 , the both sides of equations (21) and (23) are respectively added to derive the following equation. 
         [0000]      (1 +/Z 1+1 /Z 2)( Vsp+Vsn )= Vip/Z 1+( Von+Vop )/ Z 2 
         [0097]    Equation (26) is obtained by changing the above equation. 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       ( 
                       
                         Vsp 
                         + 
                         Vsn 
                       
                       ) 
                     
                     / 
                     2 
                   
                   = 
                   
                     
                       
                         
                           Vip 
                           / 
                           Z 
                         
                          
                         
                             
                         
                          
                         
                           
                             1 
                             / 
                             2 
                           
                           / 
                           
                             ( 
                             
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                       + 
                       
                         
                           
                             ( 
                             
                               Vop 
                               + 
                               Von 
                             
                             ) 
                           
                           / 
                           Z 
                         
                          
                         
                             
                         
                          
                         
                           
                             2 
                             / 
                             2 
                           
                           / 
                           
                             ( 
                             
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 
                                   1 
                                   / 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                     
                     = 
                     
                       
                         Z 
                          
                         
                             
                         
                          
                         
                           2 
                           · 
                           
                             
                               Vip 
                               / 
                               2 
                             
                             / 
                             
                               ( 
                               
                                 
                                   Z 
                                    
                                   
                                       
                                   
                                    
                                   1 
                                 
                                 + 
                                 
                                   Z 
                                    
                                   
                                       
                                   
                                    
                                   2 
                                 
                               
                               ) 
                             
                           
                         
                       
                       + 
                       
                         Z 
                          
                         
                             
                         
                          
                         1 
                          
                         
                           
                             
                               ( 
                               
                                 Vop 
                                 + 
                                 Von 
                               
                               ) 
                             
                             / 
                             2 
                           
                           / 
                           
                             ( 
                             
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   equation 
                    
                   
                       
                   
                    
                   
                     ( 
                     26 
                     ) 
                   
                 
               
             
           
         
       
     
         [0098]    If a gain of the operational amplifier  104  is sufficiently high, as one can be regarded voltage Vsp and Vsn as having relation of Vsp=Vsn, equation (26) is deformed to equation (27). Further, as Von−Vop, equations (26) is simplified to equation (27). 
         [0000]        Vsp=Vsn=Z 2 ·Vip/ 2/( Z 1 +Z 2)  equation (27)
 
         [0099]    An electrical current Iip to be input from the outside to the amplifier circuit of the second embodiment is then found. 
         [0100]    In  FIG. 3 , a condition where the total of the electrical current Iip to be flown to a node to which the voltage Vip is applied is expressed according to the Kirchhoff&#39;s law as the following equation (28). 
         [0000]        Iip +( Vsp−Vip )/ Z 1+( Vop−Vip )/ Z 3=0  equation (28)
 
         [0101]    Equation (29) is obtained by changing the equation (28). 
         [0000]        Iip+Vsp/Z 1 +Vop/Z 3−(1 /Z 1+1 /Z 3) Vip= 0  equation (29)
 
         [0102]    Substituting the equations (25) and (27) for the equation (29), the equation (29) is as follows. 
         [0000]        Iip+Z 2 ·Vip/{ 2( Z 1 +Z 2) Z 1 }+Z 2 ·Vip /(2 Z 1 ·Z 3)−(1 /Z 1+1 /Z 3) Vip= 0
 
         [0103]    Equation (30) is obtained by changing the above equation. 
         [0000]    
       
         
           
             
               
                 
                   Iip 
                   = 
                   
                     
                       
                         [ 
                         
                           
                             
                               1 
                               / 
                               Z 
                             
                              
                             
                                 
                             
                              
                             1 
                           
                           + 
                           
                             
                               1 
                               / 
                               Z 
                             
                              
                             
                                 
                             
                              
                             3 
                           
                           - 
                           
                             Z 
                              
                             
                                 
                             
                              
                             
                               2 
                               / 
                               
                                 { 
                                 
                                   2 
                                    
                                   
                                     ( 
                                     
                                       
                                         Z 
                                          
                                         
                                             
                                         
                                          
                                         1 
                                       
                                       + 
                                       
                                         Z 
                                          
                                         
                                             
                                         
                                          
                                         2 
                                       
                                     
                                     ) 
                                   
                                    
                                   Z 
                                    
                                   
                                       
                                   
                                    
                                   1 
                                 
                                 } 
                               
                             
                           
                           - 
                           
                             Z 
                              
                             
                                 
                             
                              
                             
                               2 
                               / 
                               
                                 ( 
                                 
                                   2 
                                    
                                   
                                       
                                   
                                    
                                   Z 
                                    
                                   
                                       
                                   
                                    
                                   
                                     1 
                                     · 
                                     Z 
                                   
                                    
                                   
                                       
                                   
                                    
                                   3 
                                 
                                 ) 
                               
                             
                           
                         
                         ] 
                       
                        
                       Vip 
                     
                     = 
                     
                       
                         [ 
                         
                           
                             ( 
                             
                               
                                 2 
                                  
                                 
                                     
                                 
                                  
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 
                                   1 
                                   2 
                                 
                               
                               + 
                               
                                 2 
                                  
                                 
                                     
                                 
                                  
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 
                                   1 
                                   · 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 3 
                               
                               + 
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 
                                   2 
                                   · 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 3 
                               
                               + 
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 
                                   1 
                                   · 
                                   Z 
                                 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                               - 
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 
                                   2 
                                   2 
                                 
                               
                             
                             ) 
                           
                           / 
                           
                             { 
                             
                               2 
                                
                               
                                   
                               
                                
                               Z 
                                
                               
                                   
                               
                                
                               
                                 1 
                                 · 
                                 Z 
                               
                                
                               
                                   
                               
                                
                               3 
                                
                               
                                 ( 
                                 
                                   
                                     Z 
                                      
                                     
                                         
                                     
                                      
                                     1 
                                   
                                   + 
                                   
                                     Z 
                                      
                                     
                                         
                                     
                                      
                                     2 
                                   
                                 
                                 ) 
                               
                             
                             } 
                           
                         
                         ] 
                       
                        
                       Vip 
                     
                   
                 
               
               
                 
                   equation 
                    
                   
                       
                   
                    
                   
                     ( 
                     30 
                     ) 
                   
                 
               
             
           
         
       
     
         [0104]    Accordingly, the input impedance value Zip of the amplifier circuit of the second embodiment is expressed as the following equation (31). 
         [0000]    
       
         
           
             
               
                 
                   Zip 
                   = 
                   
                     
                       Vip 
                       / 
                       Iip 
                     
                     = 
                     
                       
                         { 
                         
                           2 
                            
                           
                               
                           
                            
                           Z 
                            
                           
                               
                           
                            
                           
                             1 
                             · 
                             Z 
                           
                            
                           
                               
                           
                            
                           3 
                            
                           
                             ( 
                             
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               + 
                               
                                 Z 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                             ) 
                           
                         
                         } 
                       
                       / 
                       
                         ( 
                         
                           
                             2 
                              
                             
                                 
                             
                              
                             Z 
                              
                             
                                 
                             
                              
                             
                               1 
                               2 
                             
                           
                           + 
                           
                             2 
                              
                             
                                 
                             
                              
                             Z 
                              
                             
                                 
                             
                              
                             
                               1 
                               · 
                               Z 
                             
                              
                             
                                 
                             
                              
                             3 
                           
                           + 
                           
                             Z 
                              
                             
                                 
                             
                              
                             
                               2 
                               · 
                               Z 
                             
                              
                             
                                 
                             
                              
                             3 
                           
                           + 
                           
                             Z 
                              
                             
                                 
                             
                              
                             
                               1 
                               · 
                               Z 
                             
                              
                             
                                 
                             
                              
                             2 
                           
                           - 
                           
                             Z 
                              
                             
                                 
                             
                              
                             
                               2 
                               2 
                             
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   equation 
                    
                   
                       
                   
                    
                   
                     ( 
                     31 
                     ) 
                   
                 
               
             
           
         
       
     
         [0105]    Selecting the impedance value Z 3  so as to close to 0 within the limits where a denominator of the above equation (31) becomes positive enables actualizing high input impedance. 
         [0106]    That is, setting the impedance values Z 1 ,Z 2 ,and Z 3  to have relation of Z 3 ≧(Z 2 +Z 1 )·(Z 2 −2Z 1 )/(Z 2 +2Z 1 ) enables implementing an amplifier circuit with high input impedance values Zip Zin without oscillating the amplifier circuit. 
         [0107]    Herein, if it is impedance values Z 1 ,Z 2 ,and Z 3  to have relation of Z 3 =(Z 2 +Z 1 )·(Z 2 −2Z 1 )/(Z 2 +2Z 1 ), this will allow input impedance to be infinity. However, in order to avoid oscillation of the amplifier circuit, it is realistic that an impedance value Z 3  is set slightly larger than the impedance values (Z 2 +Z 1 )·(Z 2 −2Z 1 )/(Z 2 +2Z 1 ). On this account, Z 3  of the second embodiment has relation of Z 3 ≧(Z 2 +Z 1 )·(Z 2 −2Z 1 )/(Z 2 +2Z 1 ) in which its minimum value is (Z 2 +Z 1 )·(Z 2 −2Z 1 )/(Z 2 +2Z 1 ). 
         [0108]    In the second embodiment as described above, one end of the positive feedback impedance element  101   a  is connected to the other end of the input impedance element  103   b  and the other end of which is connected to the inverting output terminal  104   d . However, it is to be aware of that the second embodiment is not necessarily limited to such circuit configuration. For example, instead thereof, it doesn&#39;t matter that one end of the positive feedback impedance element  103   a  may be connected to the other end of the input impedance element  101   b  and the other end of which may be connected to the positive feedback impedance element  103   a , as shown in  FIG. 4 . 
       INDUSTRIAL APPLICABILITY 
       [0109]    The amplifier circuit of the embodiment as described above may be applied to any amplifier circuit, as long as a differential output amplifier which is urged to have high input impedance while mitigating noises to be produced inside of the amplifier circuit. 
       REFERENCE SIGNS LIST 
       [0000]    
       
           101   a , 101   b : input impedance element 
           102   a , 102   b : negative feedback impedance element 
           103   a , 103   b : positive feedback impedance element 
           104 : operational amplifier 
           104   a : inverting input terminal 
           104   b : non-inverting output terminal 
           104   c : non-inverting input terminal 
           104   d : inverting output terminal 
           105 : common mode feedback circuit 
           106 , 107 : signal input terminal 
           108 , 109 : signal output terminal