TUBE OPERATION 2
SMALL SIGNAL AMPLIFIERS
Contents of this page :-
Fig 5. Graph of 6SN7 Ra curves with load lines for 47k and 32 k.
How to find Ra for a given working point and plot loadlines in steps 1 to19.
Comment on THD and other topology outcomes.
Fig 6. Scanned Ra curves from Samuel Seely, 1958.
explanations about the Ra curves.
About gain with CCS load and µ.
6SN7 THD with CCS load calculations from data curves.
                      -------------------------------------------------------------------------------------

After you have carefully read all of 'Tube Operation 1', you might have a chance to
understand loadline analysis for the tube set up in Tube Op 1- Fig1, a 1/2 of 6SN7, with dc load = 47k and ac load of 100k.
Here is the load line analysis for the Fig 1 6SN7 triode :-
Fig 5.  


Here we have a set of anode curves for a 6J5 which I very carefully scanned from 'Electron Tube Circuits',
second edition, by Samuel Seely in printed in 1958, with a nice hard cover. This was a text book that honours students
at universities at that time would have to know off by heart and be able to correctly quote from even if asleep.
The 6J5 was a single small signal indirectly heated triode with excellent linearity and two such triodes crammed into
one glass tube gave us the 6SN7, and later gave us the 6CG7 when demand for electronics after WW2 mushroomed.
I an unaware of the degree of accuracy of the curves, but experience tells me they are accurate enough to base design
topologies upon, and to ascertain the behaviour of a signal triode. What method we use here for 1/2 a 6SIN7 can be used with any other set of curves for any other triode.

After so carefully scanning the original triode curves I was able to use the digital file copy in MS paint to produce the above loadline graph. One could try to print out a copy of the curves from the GIF image of curves without any loadlines included below on this page. And one could then use a ruler and pencil to draw the load lines, but I prefer the screen of the PC,
its more accurate, and thus saves forests which reduce the greenhouse effect.

How to plot load lines for a triode and find out the Ra, µ and gm of the triode :-

The curves are lines indicating the Ra at varying Ea and Ia values for set values of grid bias voltage.

(1)   Decide what Ea and Ia conditions look suitable for the quiescent operation point Q.
try Ea = 138V and Ia = 3.4mA.
 
(2)  Establish the grid voltage bias required for the Ea/Ia conditions and draw a short
line of the bias voltage by interpolation between Ra lines each side of the Q point.
Grid bias is -4.9V.

(3)  Draw a tangent line to the curve for Eg = -4.9V drawn in (2) so that it extends to about 2 x Ia and through Q,
and down through the Ea axis. Check that this line is also about parallel to adjoining Ra curves at about the same Ia.
Draw the line GQH.

(4)  Calculate the resistance value of the the tangent drawn in (3) with Ohm's Law.
R = Ea change / Ia change between two points on the tangent line.
We have Ea =  95V at H , Ia = 0mA, and Ea = 225V at G where Ia = 10mA.
R = ( 225 - 95 ) / 0.01 = 13,000 ohms, so Ra = 13k.
 
(5)  Decide on RL total, try Triode RL = ( Ea / Ia ) - Ra.  The result should be above 2 x Ra
RL = ( 140 / 0.0034 ) - 13,000 = 28.18k. This is barely 2 x Ra, so a value above 28k would be suitable.

(6)  Decide on the ac coupled load. The ac coupled load should be a minimum of 3 times the total RL value from (5)
Suppose we want to use a 100k volume pot. This load is OK. If we wanted to have a volume pot of 50k,
the load is too low, and we would need to parallel both sections of 6SN7 to drive it.
Similarly, if we wanted to use only 50k for the grid bias resistor for an output tube we should use the paralleled
6SN7 instead of the one 1/2 section.

(7)    After deciding on the value of the ac coupled load, calculate the suitable value for the dc supply.

Rdc =                      1                  
                  1           _             1     
              RLtotal                  Rac   
In this case, RL total from (5) = 28k
Rdc =  1 /   [ ( 1 / 28 ) - ( 1 / 100 ) ] = 38.8k.
But although Rdc could be 39k, we will choose 47k, next value up; for a preamp it is not too critical.

(8)  Determine effective B+ . It is not the B+ supply value when cathode bias is used. B+ eff  = B+ supply minus Ek, cathode bias voltage.
We have Cathode bias = 5V approx.    Therefore B+ effective = 300 - 5 = 295V.

(9)  Calculate the dc current for effective B+ across the dc RL.
I = 295 / 47k = 6.3mA.

(10)  Draw the dc load line for the dc RL = 47k.
Starting at point B at Ea = 295V on the Ea axis, draw a straight line through Q and on to point A on the Ia axis
where Ia = 6.3mA. at point A.
This line is correct if it passes through point Q.
AQB is the load line for 47k.

(11)   Confirm the total RL value for chosen RLdc and RLac in parallel.
RL total = 47k // 100k = 32k.                             

(12)  Calculate the Ia change in RLtotal for Ea = 140V
Ia change = 140 / 32k = 4.4mA.

(13)  Add the Ia quiescent current to Ia from (12),
4.4 + 3.4 = 7.8mA.

(14)   Plot the Ia value found in (13) on the Ia axis at C. Draw the straight line for the
total load line from C through Q and on to F on the Ea axis.
CQF is the load line for the total load of 32k.

(15)    Find where Eg minimum occurs = twice the grid bias voltage lies on the CQF load line.
The grid will swing +/- 4.9V peak at the maximum output voltage.
Draw the interpolated Ra line for the Eg min value to intersect the CQF line and mark the intersection as point E.
Bias = -4.9V, so Eg minimum = -9.8V.

(16)    Find where the CQF line intersects the Ra curve for where Eg = 0V and mark this point D.

(17)   Drop verticals from points D and E to the Ea axis and read off the Ea minimum anode swing under D
and Ea maximum anode swing under E.
DQE is the wanted load line for the total load at the anode of the tube.
Calculate anode negative swing = EaQ -  Eamin, we have 138 - 64 = 74V peak.
Calculate anode positive swing = Eamax - EaQ, we have 204 - 138 = 66V peak.

(18) Calculate second harmonic distortion %  =  100 x 0.5 x ( difference in peak +ve and -ve load swings )
                                                                                                       sum of peak load swings
2H % = 100 x 0.5 x 8 / 140 = 2.85%

(19)   Estimate 2H at desired signal levels.
 THD = THD max from (18) x desired maximum peak to peak voltage wanted / sum of peak-peak load swing.

Suppose we want 1Vrms max. THD = 2.85% x 2.82 x 1Vrms / 140 = 0.057 at least.
Usually THD is slightly lower than a proportional reduction would indicate, so expect about 0.05%.

The single 6SN7 would be barely able to drive an output tube in an SE amp which
had a bias voltage = -50V, but it would work fine for a preamp stage where the gain control was before the
triode so that the signal output from the anode for a power amp input would be about 0.1Vrms average
so THD would be less than 0.05% at all times.
If the triode is placed before the gain control the input signal from a CD player
could be 0.3Vrms average, so output is 4.2Vrms average and THD could be 0.2%, or much greater than
when the gain pot is before the triode.
I sometimes like to run the balance pot after preamp inputs, thus cutting the input signal 6dB and then
have the gain triode and gain pot following. This will keep THD < 0.1%.

The above list of 19 steps to plot the load lines includes both load lines for ac loads and dc loads.
The dc load may seem to be useless but should anyone decide to directly couple a cathode follower
output buffer to the anode of the tube, then the dc load line becomes relevant.
Should someone use a CCS source to supply 3.4mA to the anode then the dc load line is a horizontal one at 3.4mA
so it doesn't need to be plotted. The ac load ( if there is one ) is only to be plotted.
When a CCS is used, we can move the Q point to say Ia = 5mA plus Ea = 160V and if RL ac was
100K, THD would be less than 1/2 what we get at the original Q point.
And in fact we could have the ac coupled load at 50k because the dc RL is a CCS, and the total load is the ac coupled load.

There are better ways to make a gain stage than by just using one gain triode.
See my pages on preamps which have µ-follower gain stages. This method ensures THD is less than 0.02%
no matter what the sequence is for attenuators and triodes.

Fig 6.


This is the unblemished and tidied up image I scanned from Samuel Seely's book from 1958 with my notes on
parameters for 3 different Ea/Ia idle conditions. The gif should download easily and be able to be opened in MS paint
and worked on as a BMP image which will be monochrome, where all sorts of lines can be drawn, and magically undrawn if you make a mistake!

From these curves it is possible to calculate the 2H with a constant current source active load and µ.
Look along the Ia horizontal line for Ia = 4mA, and select Q point at Ea =169V.
Eg = 0V,     Ea minimum = 47V,
Eg = -6V,   Ea at Q point = 169V
Eg = -12V, Ea maximum = 281V

Anode V swings are - 122V and +112V.
With CCS load,
Gain = µ = peak to peak Ea change / peak to peak Eg change = ( 122V + 112V ) / 12V = 19.5.

If a tangent to the point Q chosen is drawn, Ra will be 11.3k, and gm = µ / Ra
= 19.5 / 11,300 = 1.7mA/V.

Minimum THD occurs when load line is horizontal, ie, a CCS.
2H%  = 100 x 0.5 x difference in Ea V swings    =      100 x 0.5 x ( 122 - 112 )   =  1.06%
                             sum of Ea swings                                    122 + 112


The THD contains other H products besides 2H but the oscilliscope tells me that
3H and other H are well below the 2H level and impossible to view on the CRO screen
or be able to be calculated from the tube curve readings.
But it is possible to build a differential push pull voltage amplifier
which is lightly loaded and which has buffered outputs and which makes no more than 0.1% THD
at 50Vrms outputs. This is a lot less than I could ever achieve with a pair of pentodes in a differential amp without external loop FB and where the odd numbered H in the THD spectrum are quite high and cannot be cancelled out with PP action.

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