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4.1 スクリーングリッド結合のみ
- 4.1.1 数値例
- 4.1.2 シミュレーション
  - 4.1.2.1 6267_sgc.cir
  - 4.1.2.2 結果

4 QUAD IIの位相反転回路

4.1 スクリーングリッド結合のみ

スクリーングリッド結合のみの等価回路は，図11のようになります．

**図 11:** スクリーングリッド結合位相反転回路の等価回路
$\begin{figure}\input{figs/equiv_sgc} \end{figure}$

等価回路より，以下の関係が成り立ちます．

e_g2	=	- (g_mg2e_g - fi_{r_p1} - fi_{r_p2}) $\displaystyle {\frac{{r_{g2}//R_{g2}}}{{2}}}$	(34)
e_p1	=	- $\displaystyle \Bigl($ g_me_g + g_m $\displaystyle {\frac{{e_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ $\displaystyle \Bigr)$ (r_p//R_L)	(35)
e_p2	=	- g_m $\displaystyle {\frac{{e_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ (r_p//R_L)	(36)
i_{r_p1}	=	$\displaystyle {\frac{{e_{p1}}}{{r_p}}}$	(37)
i_{r_p2}	=	$\displaystyle {\frac{{e_{p2}}}{{r_p}}}$	(38)

式(37), (38)に式(35), (36)を代入して，

i_{r_p1}	=	- $\displaystyle \Bigl($ g_me_g + g_m $\displaystyle {\frac{{e_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ $\displaystyle \Bigr)$ $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$
	=	- $\displaystyle {\frac{{I_p}}{{I_{g2}}}}$ g_mg2 $\displaystyle \Bigl($ e_g + $\displaystyle {\frac{{e_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ $\displaystyle \Bigr)$ $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$	(39)
i_{r_p2}	=	- g_m $\displaystyle {\frac{{e_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ ^. $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$
	=	- $\displaystyle {\frac{{I_p}}{{I_{g2}}}}$ g_mg2 $\displaystyle {\frac{{e_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ ^. $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$	(40)

式(34)に式(39), (40)を代入して，

e_g2	=	- $\displaystyle \Bigl\{$ g_mg2e_g + f $\displaystyle {\frac{{I_p}}{{I_{g2}}}}$ g_mg2 $\displaystyle \Bigl($ e_g + $\displaystyle {\frac{{2e_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ $\displaystyle \Bigr)$ $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$ $\displaystyle \Bigr\}$ $\displaystyle {\frac{{r_{g2}//R_{g2}}}{{2}}}$
	=	- $\displaystyle \Bigl\{$ g_mg2e_g $\displaystyle \Bigl($ 1 + f $\displaystyle {\frac{{I_p}}{{I_{g2}}}}$ ^. $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$ $\displaystyle \Bigr)$ +2e_g2f $\displaystyle {\frac{{I_p}}{{I_{g2}}}}$ ^. $\displaystyle {\frac{{g_{mg2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ ^. $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$ $\displaystyle \Bigr\}$ $\displaystyle {\frac{{r_{g2}//R_{g2}}}{{2}}}$
e_g2+f $\displaystyle {\frac{{I_p}}{{I_{g2}}}}$ ^. $\displaystyle {\frac{{g_{mg2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ (r_g2//R_g2) $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$ e_g2
	=	- g_mg2e_g $\displaystyle \Bigl($ 1 + f $\displaystyle {\frac{{I_p}}{{I_{g2}}}}$ ^. $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$ $\displaystyle \Bigr)$ $\displaystyle {\frac{{r_{g2}//R_{g2}}}{{2}}}$
e_g2	=	- g_mg2e_g $\displaystyle {\frac{{1+f\frac{I_p}{I_{g2}}\cdot\frac{R_L}{r_p+R_L}}}{{1 + f\fr... ...p+R_L}\cdot\frac{g_{mg2}}{\micro_{g1\mbox{\scriptsize -}g2}}(r_{g2}//R_{g2})}}}$ ^. $\displaystyle {\frac{{r_{g2}//R_{g2}}}{{2}}}$
	=	- $\displaystyle {\frac{{g_{mg2}(r_{g2}//R_{g2})e_g}}{{2}}}$ ^. $\displaystyle {\frac{{1+f\frac{I_p}{I_{g2}}\cdot\frac{R_L}{r_p+R_L}}}{{1 + f\fr... ...p+R_L}\cdot\frac{g_{mg2}}{\micro_{g1\mbox{\scriptsize -}g2}}(r_{g2}//R_{g2})}}}$	(41)

g_mg2r_g2 $\approx$ μ_g1-g2 なので， g_mg2(r_g2//R_g2) $\approx$ μ_g1-g2R_g2/(r_g2 + R_g2) より，

e_g2	$\displaystyle \approx$	- $\displaystyle {\frac{{\micro_{g1\mbox{\scriptsize -}g2}e_g}}{{2}}}$ ^. $\displaystyle {\frac{{R_{g2}}}{{r_{g2}+R_{g2}}}}$ ^. $\displaystyle {\frac{{1+f\frac{I_p}{I_{g2}}\cdot\frac{R_L}{r_p+R_L}}}{{1 + f\frac{I_p}{I_{g2}}\cdot\frac{R_L}{r_p+R_L}\cdot \frac{R_{g2}}{r_{g2}+R_{g2}}}}}$
	=	- $\displaystyle {\frac{{\micro_{g1\mbox{\scriptsize -}g2}e_g}}{{2}}}$ ^. $\displaystyle {\frac{{1+f\frac{I_p}{I_{g2}}\cdot\frac{R_L}{r_p+R_L}}}{{\frac{r_{g2}}{R_{g2}}+ 1 + f\frac{I_p}{I_{g2}}\cdot\frac{R_L}{r_p+R_L}}}}$	(42)

両相の出力の和は，

- e_p1 + e_p2 = g_me_g(r_p//R_L)

(43)

となり，通常のカソード接地のゲインと一致します．

両相のバランスは，

$\displaystyle {\frac{{e_{p2}}}{{-e_{p1}}}}$ = $\displaystyle {\frac{{-\frac{e_{g2}}{\micro_{g1\mbox{\scriptsize -}g2}}}}{{e_g+\frac{e_{g2}}{\micro_{g1\mbox{\scriptsize -}g2}}}}}$ = $\displaystyle {\frac{{-e_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}e_g+e_{g2}}}}$

(44)

より， e_g2 = - μ_g1-g2e_g/2 のときに完全になります．これは，式(42)において， R_g2 $\to$ $\infty$ の場合に成り立ちます．すなわち，スクリーングリッドに定電流回路を入れてやればバランスが取れます．

4.1.1 数値例

x	=	f $\displaystyle {\frac{{I_p0}}{{I_{g20}}}}$ ^. $\displaystyle {\frac{{R_L}}{{r_p+R_L}}}$ = 0.6477 `x` $\displaystyle {\frac{{1.086}}{{0.231}}}$ `x` $\displaystyle {\frac{{142.3}}{{3009+142.3}}}$ = 0.137565
A_x	=	g_mg2(r_g2//R_g2) = 0.2735 `x` (133.3//1000) = 32.162
A_g2	=	- $\displaystyle {\frac{{A_x}}{{2}}}$ ^. $\displaystyle {\frac{{1 + x}}{{1 + \frac{x A_x}{\micro_{g1\mbox{\scriptsize -}g2}}}}}$ = - $\displaystyle {\frac{{36.162}}{{2}}}$ `x` $\displaystyle {\frac{{1 + 0.137565}}{{1 + \frac{0.137565 \times 36.162}{36.57}}}}$ = - 16.319
A_p1	=	- g_m $\displaystyle \Bigl($ 1 + $\displaystyle {\frac{{A_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ $\displaystyle \Bigr)$ (r_p//R_L) = - 1.286 $\displaystyle \Bigl($ 1 - $\displaystyle {\frac{{16.319}}{{36.57}}}$ $\displaystyle \Bigr)$ (3009//142.3) = - 96.78
A_p2	=	- g_m $\displaystyle {\frac{{A_{g2}}}{{\micro_{g1\mbox{\scriptsize -}g2}}}}$ (r_p//R_L) = - 1.286 $\displaystyle {\frac{{-16.319}}{{36.57}}}$ (3009//142.3) = 77.99
$\displaystyle {\frac{{\vert A_{p2}\vert}}{{\vert A_{p1}\vert}}}$	=	$\displaystyle {\frac{{77.99}}{{96.78}}}$ = 0.806 = - 1.88 dB

4.1.2 シミュレーション

4.1.2.1 `6267_sgc.cir`

    1 6267 screen grid coupled phase invertor
    2 .INCLUDE 6267.lib
    3 
    4 VI 13 0 DC 0 AC 1
    5 VI2 14 0 DC 0 AC 0
    6 XV1 111 112 13 1 6267
    7 XV2 121 122 14 1 6267
    8 R1 13 0 1.5Meg
    9 R2 91 12 1Meg
   10 R3 91 22 1Meg
   11 C1 12 22 0.1u
   12 RK 1 0 762.5
   13 CK 1 0 1000u
   14 R5 11 91 180k
   15 R6 21 91 180k
   16 VIP1 11 111 DC 0
   17 VIG21 12 112 DC 0
   18 VIP2 21 121 DC 0
   19 VIG22 22 122 DC 0
   20 C2 11 31 0.1u
   21 C3 21 41 0.1u
   22 R7 31 0 680k
   23 R9 41 0 680k
   24 
   25 VB2 91 0 330
   26 
   27 .control
   28 ac dec 10 1k 1k
   29 print abs(v(31)) abs(v(41)) abs(v(112))
   30 .endc
   31 .END

4.1.2.2 結果

    1 
    2 Circuit: 6267 screen grid coupled phase invertor
    3 
    4 abs(v(31)) = 9.677640e+01
    5 abs(v(41)) = 7.798516e+01
    6 abs(v(112)) = 1.631999e+01

Next: この文書について... Up: 五極管の等価回路 Previous: 3 スクリーングリッドパスコンによる周波数特性

平成17年2月16日