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Tons of articles talking about SRPP tube amplifier but very rare of them touching the key points. Please see picture: 
In most cases we use two identical triodes and RK=RAK so that the bias are equal for two tubes.
How about if some body told you that the voltage gain of a SRPP amp is :
-µ(Rp + µRk)
Gain = ------------------------------------------
(µ+1)Rk+2Rp
Again, another body told you that the voltage gain of a SRPP amp is :
-µ(Rp + µRk)
Gain = -----------------------------------------------
(µ+1)Rk+2Rp+(Rp+Rk)Rp/RL
Where does it come from, believe it or not ?
Actually there are two key points for SRPP amplifier for achieving excellent result :
1. Static balance :- V1 and V2 should share 1/2 B+ supply each in case of no input signal.
2. Dynamic balance :- An input signal to V1 grid caused a delta Ip1 for V1, this delta Ip1 will
cause a delta Ip2 for V2 because of RAK , it is known as dynamic
balance if delta Ip1= delta Ip2 .
Both static and dynamic balance ensure the maximum non-distorted output voltage swing
in load resistor RL.
The static balance is a easy job that V1 and V2 identical and RK =RAK for same bias, so same Ip and
shares 1/2 B+ each tube.The dynamic balance is more important because this is a push pull operation, this will ensure low distortion voltage output. How to achieve dynamic balance is more complicated, but if we can go step by step with patient we will get it. Ok let's go.
Firstly a balance resistor R should be added at the plate of V2, 
Where R= RAK= RK , and RL is load resistor. 
There are two tubes work together, we can assume only V1 is working while V2 signal is short circuited and then
only V2 is working while V1 signal is short circuited, then calculate them one by one and add the results
together, the sum will be equal to that result of two tubes work together simultaneously. (superposition theorem)
By Ohm's law :
µVi
I1 = ------------------------------------------------------- <--------------------------F 1
Ra+R + (Ra+R)RL
-------------------------
Ra+R+RL
I1(Ra+R)
I1o = ----------------------------------------- <--------------------------F 2
Ra+R+RL
I1RL
I1s = --------------------------------------- <--------------------------F 3
Ra+R+RL
V2 working V1 signal short circuited please see picture : 
µVgk
I2 = ---------------------------------------------------------- <--------------------------F 4
Ra+R + (Ra+R)RL
--------------------------------
Ra+R+RL
I2(Ra+R)
I2o = ------------------------------------------- <--------------------------F 5
Ra+R+RL
I2RL
I2s = ------------------------------------------ <--------------------------F 6
Ra+R+RL
Now V1 and V2 working together please see picture :
For V2 :
Vgk = I1R - I2sR <--------------------------F 7
From F 6 and F 7 :
I2RLR
Vgk = I1R - ----------------------------------------------- <--------------------------F 8
Ra+R+RL
If we set :
(Ra+R)RL
Ry = Ra+R + ---------------------------------- <--------------------------F 9
Ra+R+RL
And also set :
RLR
Rx = ----------------------------------------------- <--------------------------F 10
Ra+R+RL
From F 1 and F 9 :
µVi
I1 = ----------------------------------------- <--------------------------F 11
Ry
From F 2 and F 9 :
µVgk
I2 = ----------------------------------------- <--------------------------F 12
Ry
From F 8 and F 10 :
Vgk = I1R - I2Rx <--------------------------F 13
From F 12 and F 13 :
µ(I1R - I2Rx)
I2 = ---------------------------------------------- <--------------------------F 14
Ry
Then :
µI1R µI2Rx
I2 = ------------------------- - ---------------------------- <--------------------------F 15
Ry Ry
From F 11 and F 15 :
µ2ViR µI2Rx
I2 = ------------------------ - ------------------------ <--------------------------F 16
Ry2 Ry
Then :
µI2Rx µ2ViR
I2 + -------------------------- = ------------------------------- <--------------------------F 17
Ry Ry2
Then :
Ry + µRx µ2ViR
I2( ------------------------------- ) = ----------------------------- <--------------------------F 18
Ry Ry2
Then :
µ2ViR
I2 = --------------------------------------- <--------------------------F 19
Ry2+ µRxRy
We set :
Ra+R
Rz = ---------------------------------------------- <--------------------------F 20
Ra+R+RL
From F 2 and F 20 :
I1o = I1Rz <--------------------------F 21
From F 11 and F 21 :
µViRz
I1o = --------------------------------------------- <--------------------------F 22
Ry
From F 5 and F 20 :
I2o = I2Rz <--------------------------F 23
From F 19 and F 23 :
µ2ViRRz
I2o = --------------------------------------- <--------------------------F 24
Ry2+ µRxRy
From F 22 and F 24 :
µViRz µ2ViRRz
I1o+I2o= ------------------------- + --------------------------- <----------------F 25
Ry Ry2+ µRxRy
Since I1o+I2o is the output current so that - (I1o+I2o)RL is the output voltage :
µViRzRL µ2ViRRzRL
- (I1o+I2o)RL = --------------------------- + -------------------------------------- <-------------F 26
Ry Ry2+ µRxRy
Then the voltage gain is :
- (I1o+I2o)RL µRzRL µ2RRzRL
---------------------- = --------------------------- + ------------------------------------- <------------F 27
Vi Ry Ry2+ µRxRy
And so :
µRzRL µ2RRzRL
A = - ( ---------------------------- + --------------------------------------- ) <------------F 27a
Ry Ry2+ µRxRy
This is the correct formula for SRPP amplifier voltage gain in terms of µ , Ra , R and RL. Since the I1o and I2o are the output current of V1 and V2, for dynamic balance, we need : I1o = I2o <---------------------------------F 28 From F 22 , F 24 and F 28 :
µViRz µ2ViRRz
--------------------------------- = ------------------------------------------ <--------------------F 29
Ry Ry2+ µRxRy
We have :
1 µR
---------------------------------- = --------------------------------------------- <--------------------F 30
Ry Ry2+ µRxRy
And :
µR
1 = --------------------------------------------- <--------------------F 31
Ry+ µRx
And :
µR = Ry+ µRx <--------------------F 32
From F 9, F 10 and F 32 we have :
(Ra+R)RL µRRL
µR = Ra+R + ----------------------------- + ---------------------------- <--------------------F 33
Ra+R+RL Ra+R+RL
Then :
Ra(Ra+R+RL) + (Ra+R)RL + µRRL
µR - R = ------------------------------------------------------------------------- <-------------------F 34
Ra+R+RL
Then :
R(µ - 1)(Ra+R+RL) = Ra(Ra+R+RL) + (Ra+R)RL + µRRL <--------------------------F 35
Then :
R(µ - 1)Ra +R2(µ - 1)+R(µ - 1)RL = Ra(Ra+R+RL) + (Ra+R)RL + µRRL <------------------F 36
And :
(µ -1)R2 +(µRa - Ra+µRL -RL)R = Ra2 +RaR + RaRL + RaRL+RRL + µRRL <---------------------F 37
Then :
(µ - 1)R2 +(µRa - Ra+µRL -RL)R = Ra2+ 2RaRL +(Ra+ RL + µRL)R <---------------------F 38
And :
(µ - 1)R2 + { µRa - 2(Ra + RL) }R = Ra2+ 2RaRL <--------------------------F 39
Thus :
(µ - 1)R2 + { µRa - 2(Ra + RL) }R - (Ra2+ 2RaRL) = 0 <--------------------------F 40
Yes , this is exactly the second order equation :
AX2 + BX + C = 0
- B ± \/ B2 - 4AC
Where X = --------------------------------------------------------
2A
To find R :
2(Ra + RL) - µRa + \/ { µRa - 2(Ra + RL) }2 + 4(µ - 1)(Ra2+ 2RaRL)
R = -------------------------------------------------------------------------------------------------------------- <-------F 41
2(µ - 1)
Find B2 - 4AC :
{ µRa - 2(Ra + RL) }2 + 4(µ - 1)(Ra2+ 2RaRL)
= µ2Ra2 - 4µRaRL -4µRa2 +4RL2 +8RLRa + 4Ra2 + 4µRa2 +8µRaRL - 4Ra2 - 8RaRL
= µ2Ra2 + 4µRaRL + 4RL2
= (µRa + 2RL)2 <-------------------------------------F 42
From F 41 and F 42 we have :
2(Ra + RL) - µRa + \/ (µRa + 2RL)2
R = -------------------------------------------------------------------------- <---------------------------F 43
2(µ - 1)
Then :
2Ra + 2RL - µRa + µRa + 2RL
R = -------------------------------------------------------------------- <---------------------------F 44
2(µ - 1)
Then :
Ra + 2RL
R = ------------------------------------------ <---------------------------F 45
µ - 1
So simple that this is the relationship between µ , Ra , R and RL for dynamic balance.
If V1 and V2 are identical triode and R = RK = RAK we get static balance. Please see picture The sound quality is very clear in high frequency and the low is good. So that this is another choice
for driver stage.
Written by Terry Tam I.Eng MIEEIE MIET (U.K.) |
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