TUBE OPERATION 5. June 2013.

About Input and Driver circuits for PP tube amps.


INPUT 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 performance figures
shown are valid for these tubes.   

Fig 1,
LTP-basic-analysis-june-2013.GIF

This Fig 1 shows a very good input stage for a PP amp. Signal voltages shown are for 1dB below clipping
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 CRO,
and the measurements are accurate enough to indicate what I want to point out to all readers.

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 each triode.
Where the anode resistances are indeed equal, then the Vo at each anode will also be equal, because the
current change in each triode MUST have equal amplitude but sum to zero amps because there cannot be
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 formula :-

Gain, A = µ x RL / ( Ra + RL ), and where RL is the total of RLdc and Cap coupled RL in parallel.
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 sheet.

The circuit gain may be calculated from measured signals, = Va-a / Vg-g = ( 1.96 + 1.96 ) / ( 1.54 - 1.32 ),

=  3.92 / 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 connected.
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 differential gain.
In the Fig case, if there is 3.96mVac VTHD at V1b grid only, then one grid is at 0V effectively.
Differential gain remains at 17.8x. This means 3.96mVac produces Va-a = 70.5mVac, and at each anode there
is +35.25mVac, and -35.25mVac. The gain between the one grid and each anode is 1/2 x 17.8 = 8.9.

If we examined the amp without any NFB connected, it would mean V1b grid needs to be grounded, and then

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 calculated as
20 x log Vin with NFB applied / Vin without NFB applied = 20 x log 1.54 / 0.22 = 20 x log 7.0 = 16.9dB.

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 calculations.
Be warned now, if you think you understand fully after reading this once, you don't, you need to build and
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 = 17.8.

But from the real world measurements in Fig 1, we can also measure the voltage gain of each triode.

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 triode.

For V1a:-

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 150k

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 at
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 = 11.80
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 them offering

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 distortion
currents which will be found to be similar in value and amplitude and in phase and in common mode
phase at the cathode and because the cathodes connect to a very high impedance of MJE340 collector,
there are virtually no 2H currents flowing in the anode loads. The only 2H voltage seen at anodes will
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 lot
of trouble to try to calculate and predict the 2H of each tube, but you should find that the difference
in 2H will be small enough to be considered negligible. What you may find is that some 3H is produced
at each anode and opposite phase, and that the voltages shown in Fig 1, perhaps THD = 0.01%, mainly
3H, with 2H less than 3H.

From the measured voltages and gains we might be able to calculate the unknown exact tube parameters

of µ and Ra.

For V1a,

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 maths,
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 = 11.49k.

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 Fig 1,

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 CORRECT.


In fact, many 6CG7 do seem to have a higher µ and higher Ra than the figures of 20 and 8k8 given

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 = 3.9mA/V.
------------------------------------------------------------------------------------------------------------------------------
DRIVER STAGE DIFFERENTIAL AMP.

My pages about 50W, 85W, 100W and 300W amps show my commonly used SET input triode stage often with

anode CCS Iadc supply and followed by LTP with 2 triodes, or a pair of paralleled twin triodes.
The best examples of this include MJE350 used for CCS Idc supply element for V1 anode, and MJ340 CCS
for commoned cathode current for LTP. Some samples also include a choke with CT used for anode supply to LTP,
as seen at http://www.turneraudio.com.au/300w-1+2-schem-input-driver-output-jan06.htm

Here are two similar driver LTP stages,

Fig 2.
LTP-driver-stages-june-2013.GIF

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 is a

well balanced input to both grids, each tube operates with the same RLa total of 30.5k, from 41k//120k.
The signal voltage at cathodes is virtually zero. 2H harmonic currents with same phase flow in both
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 by
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 not
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 cathode
resistances or CCS taken to a negative or 0V rail. However, there is no need for the Rout at each
anode to be low because there is no intention or need to drive output tube grids into positive grid
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 beyond
the output tube grid current region and it causes voltage clipping at one LTP positive going anode
while producing a peaked negative going voltage at the other anode. The negative going anode does
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 designed to
be able to produce at least twice the drive voltage required by output tubes for their clipping level.
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 where one

anode only is considered. The Rout measured from anode to anode is the sum of loads on each anode
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" resistors
in series with each KT88 grid so total driver g-g resistance = 17k4 + 4k4 = 21k8. This is a quite
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 used between

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 +25Vdc
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 increasing the
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 inadequate.

Here is a "blameless" wonderful sounding input / driver combination for where a large anode signal

voltage is needed to drive an output stage.
Fig 3.
LTP-input-LTP-driver-choke-june2013.GIF

The input LTP produces very low THD, mainly 3H, with 2H and other H at lower levels.

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