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Next: 4 両相のバランスを取るための定数 Up: オートバランス位相反転回路 Previous: 2 等価回路

3 ゲインの計算

式(1)より，

vi = i1(R1 + R3) - i2R3  
i1 = $\displaystyle {\frac{{v_i + i_2 R_3}}{{R_1 + R_3}}}$ (6)

式(2)を式(3), (4)に代入して，
$\displaystyle \mu$(i1 - i2)R3 = - i1R3 + i2(rp + R2 + R3) - i3rp  
(1 + $\displaystyle \mu$)i1R3 = i2{rp + R2 + (1 + $\displaystyle \mu$)R3} - i3rp (7)
$\displaystyle \mu$(i1 - i2)R3 = i2rp - i3(rp + RL)  
$\displaystyle \mu$i1R3 = i2(rp + $\displaystyle \mu$R3) - i3(rp + RL) (8)

これらの式に式(6)を代入して，
(1 + $\displaystyle \mu$)$\displaystyle {\frac{{v_i + i_2 R_3}}{{R_1 + R_3}}}$R3 = i2{rp + R2 + (1 + $\displaystyle \mu$)R3} - i3rp  
(1 + $\displaystyle \mu$)$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = i2$\displaystyle \Bigl\{$rp + R2 + (1 + $\displaystyle \mu$)$\displaystyle \Bigl($1 - $\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$$\displaystyle \Bigr)$R3$\displaystyle \Bigr\}$ - i3rp  
(1 + $\displaystyle \mu$)$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = i2$\displaystyle \Bigl\{$rp + R2 + (1 + $\displaystyle \mu$)$\displaystyle {\frac{{R_1 R_3}}{{R_1 + R_3}}}$$\displaystyle \Bigr\}$ - i3rp  
(1 + $\displaystyle \mu$)$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = i2{rp + R2 + (1 + $\displaystyle \mu$)(R1//R3)} - i3rp  
i2 = $\displaystyle {\frac{{(1 + \mu) \frac{R_3}{R_1 + R_3} v_i + i_3 r_p}}{{r_p + R_2 + (1 + \mu) (R_1//R_3)}}}$ (9)
$\displaystyle \mu$$\displaystyle {\frac{{v_i + i_2 R_3}}{{R_1 + R_3}}}$R3 = i2(rp + $\displaystyle \mu$R3) - i3(rp + RL)  
$\displaystyle \mu$$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = i2$\displaystyle \Bigl\{$rp + $\displaystyle \mu$$\displaystyle \Bigl($1 - $\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$$\displaystyle \Bigr)$R3$\displaystyle \Bigr\}$ - i3(rp + RL)  
$\displaystyle \mu$$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = i2$\displaystyle \Bigl($rp + $\displaystyle \mu$$\displaystyle {\frac{{R_1}}{{R_1 + R_3}}}$R3$\displaystyle \Bigr)$ - i3(rp + RL)  
$\displaystyle \mu$$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = i2{rp + $\displaystyle \mu$(R1//R3)} - i3(rp + RL) (10)

式(9)を式(10)に代入して，
$\displaystyle \mu$$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = $\displaystyle {\frac{{(1 + \mu) \frac{R_3}{R_1 + R_3} v_i + i_3 r_p}}{{r_p + R_2 + (1 + \mu) (R_1//R_3)}}}${rp + $\displaystyle \mu$(R1//R3)} - i3(rp + RL)  
$\displaystyle \mu$$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = $\displaystyle {\frac{{r_p + \mu (R_1//R_3)}}{{r_p + R_2 + (1 + \mu) (R_1//R_3)}}}$$\displaystyle \Bigl\{$(1 + $\displaystyle \mu$)$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi + i3rp$\displaystyle \Bigr\}$ - i3(rp + RL)  

x = $ {\frac{{r_p + \mu (R_1//R_3)}}{{r_p + R_2 + (1 + \mu) (R_1//R_3)}}}$ とおくと，
$\displaystyle \mu$$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi = x$\displaystyle \Bigl\{$(1 + $\displaystyle \mu$)$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi + i3rp$\displaystyle \Bigr\}$ - i3(rp + RL)  
i3{(1 - x)rp + RL} = - {$\displaystyle \mu$ - x(1 + $\displaystyle \mu$)}$\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$vi  
i3 = - $\displaystyle {\frac{{\{ \mu - x (1 + \mu) \} \frac{R_3}{R_1 + R_3}}}{{(1 - x) r_p + R_L}}}$vi  
i3 = - $\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$ . $\displaystyle {\frac{{\mu - x (1 + \mu)}}{{(1 - x) r_p + R_L}}}$vi (11)

したがって，
vo = i3RL = - $\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$ . $\displaystyle {\frac{{\{\mu - x (1 + \mu)\} R_L}}{{(1 - x) r_p + R_L}}}$vi (12)
A = - $\displaystyle {\frac{{R_3}}{{R_1 + R_3}}}$ . $\displaystyle {\frac{{\{\mu - x (1 + \mu)\} R_L}}{{(1 - x) r_p + R_L}}}$ (13)

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Next: 4 両相のバランスを取るための定数 Up: オートバランス位相反転回路 Previous: 2 等価回路
Ayumi Nakabayashi
平成19年11月17日