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Long tail phase splitter (also known as Cathode Coupled Phase Splitter) is widely used in push pull tube amp driver stage, it is more convenient and simple but the
output voltge has about 10% difference between V1 and V2, to use a larger value resistor as V2 plate resistor is a norm
in long tail driver for output voltage balance. The practice is adjusting a variable pot while watching a oscilloscope
for a best sine wave form.
Please see picture: 
Please see the equivalent circuit 
From the equivalent circuit :
(R2 + Ra )Rk
Let Rx = R1 + Ra + ----------------------------------- <--------- F1
Rk+ Ra + R2
(R1 + Ra )Rk
Let Ry = R2 + Ra + ----------------------------------- <----------F2
Rk+ Ra + R1
Then :
µVgk1
I1 = ----------------------------------------------- <-------------------- F3
Rx
µVgk2
I2 = ----------------------------------------------- <-------------------- F4
Ry
(R2 + Ra )
Let RAo = ----------------------------------- <--------- F5
Rk+ Ra + R2
(R1 + Ra )
Let RBo = ----------------------------------- <----------F6
Rk+ Ra + R1
Then:
Io1 = I1RAo <---------------- F7
Io2 = I2RBo <---------------- F8
From F3 and F7
µVgk1RAo
Io1 = ----------------------------------------------- <-------------------- F9
Rx
From F4 And F8
µVgk2RBo
Io2 = ----------------------------------------------- <-------------------- F10
Ry
From the equivalent circuit Then :
Vgk2 = (Io1-Io2)Rk <--------------------- F11
From F9, F10 , F11
µVgk1RAo µVgk2RBo
Vgk2 = (------------------------ - ------------------------)Rk <--------------- F12
Rx Ry
µVgk1RAoRk µVgk2RBoRk
Vgk2 = ------------------------------- - ----------------------------- <--------------- F13
Rx Ry
µVgk2RBoRk µVgk1RAoRk
Vgk2 + ------------------------------- = ----------------------------- <--------------- F14
Ry Rx
µRBoRk µVgk1RAoRk
Vgk2 (1 + ------------------------------- ) = ----------------------------- <--------------- F15
Ry Rx
µVgk1RAoRkRy
Vgk2 = -------------------------------------------------- <--------------- F16
(Ry+µRBoRk)Rx
From the equivalent circuit Then :
Vgk1 = Vi - Io1Rk + Io2Rk <-------------------- F17
From F9, F10, F17
µVgk1RAoRk µVgk2RBoRk
Vgk1 = Vi - --------------------------------- + -------------------------------- <-------------------- F18
Rx Ry
From F16, F18
µVgk1RAoRk µRBoRk µVgk1RAoRkRy
Vgk1 = Vi - --------------------------------- + -------------------------------------------------- <-------------------- F19
Rx Ry(Ry+µRBoRk)Rx
Vi
Vgk1 = -------------------------------------------------------------------------------------- <------------------- F20
µRAoRk µ2RAoRBoRk2
1 + ----------------------- - ------------------------------------
Rx (Ry+µRBoRk)Rx
1
Let (F) = -------------------------------------------------------------------------------------- <------------------- F21
µRAoRk µ2RAoRBoRk2
1 + ----------------------- - ------------------------------------
Rx (Ry+µRBoRk)Rx
From F20, F21
Vgk1 = Vi(F) <--------------------------------- F22
From F16, F22
µVi(F)RAoRkRy
Vgk2 = -------------------------------------------------- <--------------- F23
(Ry+µRBoRk)Rx
By current ratio in Rk path and R2+Ra path
I1Rk
IL1 = -------------------------------------------------- <--------------- F24
Rk+R2+Ra
By current ratio in Rk path and R1+Ra path
I2Rk
IL2 = -------------------------------------------------- <--------------- F25
Rk+R1+Ra
Then the out put currents are:
Iout1 = I1+ IL2 <----------------------- F26
Iout2 = I2+ IL1 <----------------------- F27
From F3, F4, F25 anf F26
µVgk1 µVgk2Rk
Iout1 = ------------------------ + ------------------------------------------- <----------------- F28
Rx Ry (Rk+R1+Ra)
From F22, F23, F28
µVi(F) µRkµVi(F)RAoRkRy
Iout1 = ------------------------ + ------------------------------------------------------------------ <----------------- F29
Rx Ry(Rk+R1+Ra)(Ry+µRBoRk)Rx
1 µRkRAoRk
Iout1 = µVi(F) (------------------------ + ---------------------------------------------------------- ) <----------------- F30
Rx (Rk+R1+Ra)(Ry+µRBoRk)Rx
Since triode 1 output voltage = Iout1R1 so that From F30
1 µRkRAoRk
Iout1R1 = µVi(F)R1 (--------------------- + ---------------------------------------------------------- ) <----------------- F31
Rx (Rk+R1+Ra)(Ry+µRBoRk)Rx
Then :
Iout1R1 1 µRAoRk2
-------------- = µ(F)R1 (---------------------- + ---------------------------------------------------------- ) <----------------- F32
Vi Rx (Rk+R1+Ra)(Ry+µRBoRk)Rx
And the voltage gain of triode 1 is equal to output voltage devided by input voltage Vi so that :
1 µRAoRk2
V1 GAIN = µ(F)R1 (---------------------- + ---------------------------------------------------------) <------------- F32A
Rx (Rk+R1+Ra)(Ry+µRBoRk)Rx
From F3, F4, F24, F27
µVgk2 µVgk1Rk
Iout2 = ------------------------ + ------------------------------------------- <----------------- F33
Ry Rx (Rk+R2+Ra)
From F22, F23, F33
µµVi(F)RAoRkRy µVi(F)Rk
Iout2 = ------------------------------------------- + ------------------------------------------- <----------------- F34
Ry (Ry+µRBoRk)Rx Rx (Rk+R2+Ra)
µRAoRkRy Rk
Iout2 = µVi(F) (-------------------------------------- + -------------------------------------------) <----------------- F35
Ry (Ry+µRBoRk)Rx Rx (Rk+R2+Ra)
Iout2R2 µRAoRk Rk
---------------- = µ(F)R2 (-------------------------------------- + -------------------------------------------) <----------- F36
Vi (Ry+µRBoRk)Rx Rx (Rk+R2+Ra)
And the voltage gain of triode 2 is equal to output voltage devided by input voltage Vi so that :
µRAoRk Rk
V2 GAIN = µ(F)R2 (-------------------------------------- + ----------------------------------------) <----------- F36A
(Ry+µRBoRk)Rx Rx (Rk+R2+Ra)
F32A and F36A are the fomula for gains of tube V1 and tube V2. Please note that Rx Ry RAo RBo and (F) are group of resistance and constants please refer to F1, F2, F5, F6 and F21 respectively. Please see Marantz 9 phase splitter circuit diagram 
If the adjusting pot is set at 4K and 6K each side so that R1 will be 46.4+4=50.4K while R2 will be 46.4+6=52.4k , we have: R1 = 50.4K R2 = 52.4K Rk = 38.3K Ra = 2.64K µ = 33 By F32A V1 gain = 15.7 By F36A V2 gain = 15.7This is in balance and the adjusting pot can be replaced with one 4K and one 6K fixed resistor if you desire but I don't suppose anybody would. This calculation could be a guide line for adjustment, if the balance point is too far away from that point (4K, 6K), the 6DJ8 should be replaced.
Written by Terry Tam I.Eng MIEEIE MIET (U.K.) |
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