OPERATION 5. June 2013.
About Input and Driver circuits
for PP tube amps.
STAGE DIFFERENTIAL AMP.
Here I explain a differential
input amplifier, aka Long Tail Pair amplifier input stage.
It has been rarely ever been
used by any brand-name manufacturers in power amps.
But it has been used in old
fashioned scientific amplifiers and for balanced preamps.
At seen in the schematic below,
the amp stage responds mainly to the signal difference
between applied input signals.
In this case the two signals may be a signal from a CD
player and the NFB signal from
the output of an amplifier.
The CCS may have effective Rc =
5M, so any common mode input signal at both grids
will appear at each anode at
-47dB lower level than at input.
A slightly more complex CCS may be made using 2 x bjts to get
the effective collector
resistance approaching an infinite ohm value. The stray
capacitance between collector
and anything else and the C4 shunting R10 will then determine
the bandwidth of the
CCS. Extreme bandwidth is not any advantage. While many other
types of tubes could
be used, I have based my example around 6CG7 or 6SN7, and all
shown are valid for these tubes.
This Fig 1 shows a very good
input stage for a PP amp. Signal voltages shown are for 1dB
for a power amp producing
9.67Vac output signal.
The measurement accuracy is
limited by my Fluke meter accuracy, and the distortion within
the output voltages
which form the error correction
signal. However, the wave forms look substantially clean on my
and the measurements are
accurate enough to indicate what I want to point out to all
The signal voltages shown
depend on the use of an extremely high resistance current sink
to commoned cathodes.
This is easily done with MJE340
as shown and the effective resistance looking into the collector
terminal is well over
1 Megohm, with little
capacitance. The signal current flow in this current source is
so minute it may be considered
negligible and there is no need
to include it in calculations.
The schematic has equal RL load
values for both tubes, in this case, 56k // 150k = 40.8k for
Where the anode resistances are
indeed equal, then the Vo at each anode will also be equal,
current change in each triode
MUST have equal amplitude but sum to zero amps because there
any change in total of cathode
currents flowing into CCS collector of MJE340.
The differential gain = Va-a /
Vg-g and equals the gain of one tube according to the simple
Gain, A = µ x RL / ( Ra +
RL ), and where RL is the total of RLdc and Cap coupled RL in
But we must be interested
primarily in real world gain, not calculated gain because we
cannot know the exact
values of µ and Ra. These
vary with Ea and Ia and the tube data sheets give only the
approximate values for
the Ea and Ia tabled on the
gain may be calculated from measured signals, = Va-a / Vg-g = (
1.96 + 1.96 ) / ( 1.54 - 1.32 ),
/ 0.22 = 17.8.
However, while considering the
differential gain we must question what gain exists for signals
fed into only one grid.
In Fig 1, I show 1.54Vac input
to V1a which may be from a CD player or test oscillator.
The signal at V1b grid is shown
as 1.32Vac and in this case it is nominated as from the output
of a following stage
in a power amp.
This signal will have a
fraction of the various distortion voltages that exist at the
amp output. If the amp produces
0.3% THD at the levels of
signal shown then the THD voltage. VTHD at V1b grid = 0.3% of
1.32Vac = 3.96mVac.
Its not a huge voltage. But it
IS THERE. This VTHD is amplified by the whole of the amp to
produce an output
distortion signal of opposite
phase to oppose the VTHD made by the amp when NFB is not
For simplicity and easy
calculation of the amount of NFB reduction, the gain for signals
applied to one grid only
is considered to be 1/2 the
In the Fig case, if there is
3.96mVac VTHD at V1b grid only, then one grid is at 0V
Differential gain remains at
17.8x. This means 3.96mVac produces Va-a = 70.5mVac, and at each
is +35.25mVac, and -35.25mVac.
The gain between the one grid and each anode is 1/2 x 17.8 =
If we examined the amp without
any NFB connected, it would mean V1b grid needs to be grounded,
Vin at V1a grid needed for the
same Vo shown will simply be the difference between the shown
Vin and VFB
= 1.54 - 1.32 = 0.22Vac.
The voltage with NFB connected
is 1.54Vac, and thus the amount of voltage NFB applied may be
20 x log Vin with NFB applied /
Vin without NFB applied = 20 x log 1.54 / 0.22 = 20 x log 7.0 =
Most DIY constructors and many
company designers get hopelessly confused by calculations of
loading, voltage gain,
NFB application, and many other
things involved which usually require several pages of notebook
Be warned now, if you think you
understand fully after reading this once, you don't, you need to
measure an amp!
If we made the approximate
assumption that µ = 20, and Ra = 10k for 1/2 6CG7 or 1/2
6SN7 then for one triode
Gain A should be 20 x 40.8 / (
10k + 40.8k ) = 16.0. This is well short of the observed gain =
But from the real world
measurements in Fig 1, we can also measure the voltage gain of
For any triode, its gain may be
calculated Gain = Va-k / Vg-k.
If we measure the gain of each
triode in Fig1, we will find surprising differences between
their gains and load conditions,
but the load current is the
*same* amplitude for each triode. This load current = VRL / RL =
1.96V / 40.8k = 0.048mAac.
The load signal ac current is a
change of Idc and the direction of change is opposite for each
Resistance loads = 56k RLdc in
parallel with cap coupled 150k = 40k8 total.
A = Va-k / Vg-k = +3.33V /
0.17V = 19.58.
The load current = Va / 40k8 =
1.96V / 40.8 = 0.048mA.
This signal current flows in
the V1a triode.
But there is -1.37Vac at
cathode, and total signal voltage across V1a triode = 1.96 +
1.37 = 3.33Vac.
Therefore the triode has an RLa
= Va-k / IRL = 3.33V / 0.048mA = 69k4.
The "effective" load the tube
actually experiences must be calculated. We see there is 56k and
in parallel, but the
measurements and necessary calculation indicate otherwise! If
the V1a cathode was fixed
at 0Vac, then indeed its load
would be 40.8k, and Va-k would be +1.96V. But there is -1.37Vac
V1 cathode, and Va-k = 3.33Vac,
and so the effective load must be calculated as Va-k / Ia RL
= 3.33 / 0.048 = 69.38k.
For V1b :-
A = Va-k / Vg-k = 0.59 / 0.05 =
The load current = 0.048.
The effective load = Va-k / IRL
= 0.59 / 0.048 = 12.29k.
So the two triodes are each
working with very different RL, and one may think this prevents
well balanced performance, but
we have measured equal Vo at each anode, and equal IRL.
Although Va-k is very different
across each triode, The Ia is balanced. Each tube produces 2H
currents which will be found to
be similar in value and amplitude and in phase and in common
phase at the cathode and
because the cathodes connect to a very high impedance of MJE340
there are virtually no 2H
currents flowing in the anode loads. The only 2H voltage seen at
be due to the difference
between 2H currents in each triode.
V1a has 3.33Vac Ea swing across
69.38k, V2 has 0.59Vac swing across 12.29k. You might go to a
of trouble to try to calculate
and predict the 2H of each tube, but you should find that the
in 2H will be small enough to
be considered negligible. What you may find is that some 3H is
at each anode and opposite
phase, and that the voltages shown in Fig 1, perhaps THD =
3H, with 2H less than 3H.
From the measured voltages and
gains we might be able to calculate the unknown exact tube
of µ and Ra.
A = 19.58 = µ x 69.38 / (
Ra + 69.38 ),
Therefore µ = 19.58 x (
Ra + 69.38 ) / 69.38.
For V1b, A = 11.8 = µ x
12.29 / ( Ra + 12.29 ),
Therefore µ = 11.8 x ( Ra
+ 12.29 ) / 12.29.
If we assume both V1a and V1b
triodes have the same Ra and µ, then
19.58 x ( Ra + 69.38 ) / 69.38
= 11.8 x ( Ra + 12.29 ) / 12.29.
The only unknown in the
equation is Ra, which may be solved easily with school boy
0.2822 ( Ra + 69.38 ) = 0.96 x
( Ra + 12.29 ),
0.2822Ra + 19.6 = 0.96Ra + 11.8
7.8 = 0.6788Ra
Ra = 7.8 / 0.6788 kOhms =
Substituting Ra = 11.49 into
equation for V1a µ,
µ = 19.58 x ( 11.49 +
69.38 ) / 69.38 = 22.8.
If we had just one triode set
up with cathode at 0V and RL = 40.8k and Ia and Ea at same as
A = 22.8 x 40.8 / ( 11.49 +
40.8 ) = 17.79.
In a paragraph above, I
said "Circuit gain = 17.8."
So the calculated values of Ra
and µ for the 6CG7 used for this experiment MUST BE
In fact, many 6CG7 do seem to
have a higher µ and higher Ra than the figures of 20 and
in data sheets. Now for all
tubes, transconductance, Gm = µ / Ra and for my 6CG7 gm =
22.8 / 11.49 = 1.98mA/V.
If you look at the Ra curves
for GE 6CG7, and carefully draw a tangent to a curve near Ea =
122V and Ia = 5mA,
you should find Ra calculates
at about 10k0. The µ is the Ea change along the horizontal
line for Ia = 5mA and
I found µ = 21.25. My
6CG7 is AWV made in Australia. Any batch of tubes made a
different factory and or country
to used for original tests will
be found to have slightly different characteristics, and in fact
most often the gm is lower,
and Ra higher.
The GE curves include those
giving "average characteristics" for µ and Ra for 3 values
of Ea and for any value
of Ia and these nearly agree
with my reading of the Ra curves.
Regardless of what the curves
for the tubes might indicate, the important thing is that the
operation is linear, gives wide
bandwidth and low noise and
that Ra is low enough, and all requirements are well satisfied
with any NOS or newly made
samples of 6CG7, 6SN7, ECC32,
12AU7, 6DJ8, 6922, or other suitable twin triodes or individual
triodes such as 6J5,
or small pentodes strapped as
triodes such as 6AU6, or power pentodes such as EL84, EL86,
6AR5, 6M5, etc, etc, etc.
The other consideration is
longevity. Although these small signal triodes have better
characteristics at higher Ea and Ia,
the betterment in sound or THD
achieved by say running 1/2 6CG7 with Ia at 10mA is negligible.
The higher the Ea and
Ia become, the and higher the
Pda and the temperature and the shorter the tube life. Where
higher Ia and /or Ea is needed
to give wider Va swing in
circuits driver output tube grids it is better to use the two
triodes in the one tube in parallel.
In the case 6CG7 I have used
above the µ will remain at 22.25, but Ra = 5.7k, and gm =
STAGE DIFFERENTIAL AMP.
My pages about 50W, 85W, 100W
and 300W amps show my commonly used SET input triode stage often
anode CCS Iadc supply and
followed by LTP with 2 triodes, or a pair of paralleled twin
The best examples of this
include MJE350 used for CCS Idc supply element for V1 anode, and
for commoned cathode current
for LTP. Some samples also include a choke with CT used for
anode supply to LTP,
as seen at
Here are two similar driver LTP
These two driver stage
schematics are based on the driver stage shown at Reformed RCA 30W amp.
On the left hand side there is
the exact stage used for the reformed RCA amp, the right hand
side shows what
may be used if the input stage
was a single ended input stage.
For these stages, I estimated
the 6CG7 used had µ = 20.9 and Ra = 12k0.
The left schematic has cathode
current taken to a -45Vdc supply rail via 5k0. Because the there
well balanced input to both
grids, each tube operates with the same RLa total of 30.5k, from
The signal voltage at cathodes
is virtually zero. 2H harmonic currents with same phase flow in
loads and triodes, and a 2H
harmonic voltage appears at cathodes. Some 2H voltage appears at
each anode but it is same
phase, common mode, and a very small % of signal and is rejected
following output stage so its
presence has negligible effect on overall THD or IMD production.
But the common mode gain of the
stage is not low with the Rk R19 5k0 shown. If there
was 0.1V of common mode signal
to each grid, then 0.319V would appear at each anode
with same phase. The stage
above is preceded by an input LTP which produces almost zero
common mode signal or
distortion so high common mode rejection in the driver stage is
needed and hence no need for an
active constant current sink, MJE340. Although the MJE340
is not needed, you are free to
include it if you wish. The balance of output without the CCS
was found to be excellent
because resistances used were matched closer than +/-1%.
Output resistance at each anode
is a high ohm value and poor when considering each anode
separately. This is a property
of all balanced amps with commoned cathodes and high value
resistances or CCS taken to a
negative or 0V rail. However, there is no need for the Rout at
anode to be low because there
is no intention or need to drive output tube grids into positive
voltage region where the grid
input resistance suddenly reduces from megohms to say 1.5kohms
due to grid current onset.
Therefore the LTP stage cannot easily produce drive voltages
the output tube grid current
region and it causes voltage clipping at one LTP positive going
while producing a peaked
negative going voltage at the other anode. The negative going
not experience any grid
current. The result of LTP overload is entirely benign and
recovery is instant
and there is symmetrical
waveform clipping at the amp output. In most amplifiers, the
onset of grid
current in output tubes
indicates onset of output tube overload and exponential rise of
THD and IMD
and the amp has reached its
useful high fidelity power ceiling. The LTP driver should be
be able to produce at least
twice the drive voltage required by output tubes for their
The above LTP may be found to
generate 70Vac at each anode, at less than 1% THD, without
output tubes present.
The Rout at each LTP anode
where the loading of each anode is equal is much lower than
anode only is considered. The
Rout measured from anode to anode is the sum of loads on each
in parallel with the sum of the
Ra for each triode. In this case, the Rout a-a =
( 2 x 30.5k ) in parallel with
( 2 x 12k0 ) = 2 x 8k7 = 17k4. There are 2k2 "grid stopper"
in series with each KT88 grid
so total driver g-g resistance = 17k4 + 4k4 = 21k8. This is a
low enough resistance to drive
the grid to grid input impedance including the Miller
capacitance of a
pair of KT88 tubes with 20% UL
taps. The KT88 Cg-g may be estimated at approximately 40pF,
and the resultant HF cut off
pole would be at 180kHz.
Where grid current drive of
output tubes is required, then there must be a "buffer" stage
the LTP and output tubes. In
most cases this consists of an additional 6CG7 with its its
grids driven by the
above gain stage and similar
C&R coupling, with grids biased with a negative voltage
supply. The extra
6CG7 cathodes are directly
connected to OP tube grids which can then be driven to perhaps
without causing any grid
current in the cathode followers. Thus a class AB2 amp using a
pair of KT88
strapped in triode can be made
to produce nearly twice the possible class AB1 power.
Better fidelity and reliability
is available by using 4 output tubes in class AB1 rather than
driver tubes and flogging a
pair of output tubes harder.
But where a large number of
parallel output tubes need to be driven and where the drive
voltage is high
because output tubes are in
triode or where local output CFB operation is used, then the use
of just one
6CG7 for a driver LTP is
Here is a "blameless" wonderful
sounding input / driver combination for where a large anode
voltage is needed to drive an
The input LTP produces very low
THD, mainly 3H, with 2H and other H at lower levels.
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