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.

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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 hard cover. This was a text book that honors 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 am unaware of the exact degree of accuracy of the

curves, but experience tells me they are accurate enough to base design

topologies upon, and to ascertain the behavior of a signal triode.

What method we use here for 1/2 a 6SN7 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, it is 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 Total RLa. 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. 100k is more than 3 x than load in (5), and 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 properly.

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

resistance value from anode to B+ rail to provide the dc supply to the triode.

Rdc = 1

1 _ 1

RLtotal Rac

In this case, Total RLa from (5) = 28k. And cap coupled RL = 100k.

Rdc = 1 / [ ( 1 / 28 ) - ( 1 / 100 ) ] = 38.8k, so a standard value 39k

could be used. Where convenient, a higher value may be used.

Let us try RLdc 47k, next value up. For a small signal preamp it is not too critical.

(8) Determine effective B+. This is not the Vdc rail voltage above 0V if

cathode bias is used.

B+ eff = B+ supply minus Ek, cathode bias voltage. We have Cathode bias = 5V approx.

Therefore B+ effective = +300 - 5 = +295Vdc. Where cathode bias voltage

is less than ( actual rail voltage to 0V) / 50 the load lines may be drawn assuming

effective B+ = actual B+, all without changing conclusions from the load line analysis.

But for this example, it is best we act with purity, and consider effective B+.

(9) Calculate the maximum dc current for effective B+ across the dc RL.

Idc = 295Vdc / 47k = 6.3mAdc.

(10) See above graph. Draw the dc load line for the dc RL = 47k.

Starting at point point A on the Ia axis where Ia = 6.3mA, and draw a straight

line to B at Ea = 295V on the Ea axis.

The triode will be connected in series with the RLdc of 47k. The triode Ea, ie,

Vdc between anode to cathode and the Vdc across the RL may be read from

the graph line AB, after choosing the triode anode Ea idle Vdc.

In this case, Ea has been chosen for +138Vdc, and Point Q is plotted on line

AB where Point Q is vertically above +138Vdc on the Ea axis. The idle dc

current flow may be read off by drawing a horizontal line from Q to the Ia

axis. Iadc = 3.4mAdc.

This current also flows in the 47k RLdc.

The Vdc across the 47k = Effective B+ minus Ea at point Q = 295V - 138V

= 157Vdc. As a general guide for most small signal triodes with RLdc the Ea

should be under 1/2 x Effective B+.

(11) Confirm the total RL value for chosen RLdc and RLac in parallel.

Total RL = 47k // 100k = 32k.

(12) Calculate the possible maximum Ia change in total RL for Ea = 138V,

Ia change = 138V / 32k = 4.3mA.

(13) Add the idle Iadc quiescent current to Ia max from (12), 4.3 + 3.4 = 7.7mA.

(14) Plot the Ia value found in (13) on the Ia axis at C.

Draw the straight line for the total RL load 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 what the Eg grid bias voltage is at the idle condition of Ea = 138Vdc

and Ia = 3.4mA.

The two Ra curves for Eg = -4V and Eg = -6V lie each side of the Point Q.

From careful visual estimation, the grid voltage bias for point Q will be -4.9Vdc.

Calculate the maximum peak to peak grid voltage swing without the grid becoming

positive. Total peak-peak Vg may be always assumed to be twice the bias voltage

at point Q.

In this case, Vg pk-pk = 2 x 4.9V = 9.8V.

This means the maximum allowable grid voltage changes from -4.9V at idle to

between 0V, the maximum most positive Eg, and -9.8V, the maximum most

negative Eg.

(16) The value of Ea for where Eg = 0V may be found where the Ra curve

for Eg = 0 passes through load line the line CQF, and and may be plotted

at Point D.

(17) The value of Ea where Eg = -9.8V may be found by interpolation.

The Ra curves for Eg = -8V and -10V are inspected where they intersect

the line CQF. The Ra value for where Eg = -9.8V has been plotted at Point E,

which is slightly to the left of where Eg = -10V crosses CQF.

(18) Drop vertical lines from points D and E to the Ea axis and read off the

two Ea values. The lesser Ea is Ea minimum peak V anode swing and the

higher Ea is the maximum peak V anode swing during each voltage wave cycle.

The line DQE is the wanted load line for the total load at the anode of the triode.

Calculate anode negative swing = EaQ - Eamin, we have 138 - 64 = 74V peak.

Calculate anode positive swing = Eamax - EaQ, we have 204 - 138 = 66V peak.

(19) Calculate THD at voltage swings in step (18)...

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 / ( 74 + 66 ) = 2.85%

Output voltage in Vrms = sum of peak load swings / 2.82

= ( 74 + 66 ) / 2.82 = 49.64Vrms.

THD = 2.82% at 49.64Vrms.

(20) Estimate 2H at desired signal levels.

At low wanted voltages, THD = THD max from (19) x lower wanted Vrms

Vrms from step (18)

THD for 1Vrms = 2.85% x 1Vrms / 49.6Vrms = 0.017%.

Usually THD is slightly lower than a proportional reduction would indicate,

so expect about 0.014%.

(21) Calculate voltage gain from load lines.

From the graph, Vg pk-pk = 9.8V,

Va pk-pk = 140V

Voltage gain = Va swing / Vg swing = 140 / 9.8 = 14.28 x.

The single 1/2 6SN7 would be barely able to drive an output tube in an SE

amp which had a bias voltage = -50V. 35Vrms or 100Vpk-pk is needed for

such an output tube.

However, the 6SN7 2H would tend to cancel the 2H of the output triode.

But some IMD products produced in the cancelling process so it is better

to try to use the 6SN7 with both halves paralleled and with higher Ea and Ia

and with a higher number of ohms for RLa to achieve a bigger possible Va

swing and much better linearity. The driver tube for an output stage should

able to make twice the signal voltage required for onset of grid current in the

output stage, and if output tube has Eg1 = -50V, then driver should make

100V pk, ie, 200V pk-pk, ie, 70Vrms.

For a line level preamp, consider the gain control pot was placed before the

1/2 6SN7 used for gain. Also consider the anode output is direct coupled to

a second 1/2 6SN7 arranged as a cathode follower, so that the only load on the

gain triode is 47k.

Usually power amps need about 1Vrms for clipping, and for normal listening

levels an average of only 0.1Vrms is needed.

If the 6SN7 preamp has gain = approx 14, then input needed at its grid

= 0.1 / 14 = 0.007Vrms and if the signal is from a CD player giving average

levels of 0.35Vrms, the pot setting will be very low at the 34dB position.

The outcome is that the SNR between output signal and noise is high,

but THD is extremely low.

If the gain control pot is placed after the gain triode the average 0.35Vrms

from CD player is amplified about 14 x to become 4.9Vrms at the anode.

This is applied to gain control pot and adjusted down to be about 0.11Vrms

and at the CF output the signal applied to the power amp is 0.1Vrms.

The volume setting is the same as before at about -34dB, and SNR is far better.

But THD might be 0.3%. THD for given tube samples vary, and 1% THD

for each 10Vrms produced by many triodes would be average.

To overcome problems of either arrangement, it is better to use a loop of

shunt FB around a gain triode to reduce its gain from 16 to 4, so that the

maximum of 1.4Vrms from CD player becomes 4.2Vrms with THD <0.1%.

The gain control setting will be at a higher rotation position, but perhaps still

too low. If there is a switched set of say 5 input sockets, one may be devoted

to CD player with a fixed resistance divider using 6k8 plus 3k3 2k2 to reduce

CD input levels by 1/3, or by -10dB, so normal listening levels are at the

12 o'clock pot position, SNR is fine and THD should always be under 0.05%.

Load lines for cathode follower tubes may be done exactly the same way as for

a gain tube with the high anode RL and grounded cathode. The CF tube with fixed

Ea and load placed between cathode and 0V operates exactly the same way as

the plate loaded gain triode. CF Gain is below unity, and THD is very low.

Another option is the use of a switch to bring CD input levels directly to a volume

control pot after bypassing the gain triode input stage which would only be used

where source signals are very low.

But even for most low signals nominally 200mV which was the standard before

CD players produced 1.4vrms after 1985, the input stage would still need a shunt

FB loop to limit its gain to about 5x, or +14dB.

In all options there should be a cathode follower buffer at the preamp output.

If there is balance pot after select switch for preamp inputs, the the input signals

may be cut by about 6dB before the gain triode and/or gain pot following.

The above list of 21 steps may also be used to plot dc loads with or without ac

coupled loads for cathode follower stages. If a cathode follower grid is directly coupled

to an anode of a gain triode, the gain triode has only the RLdc load.

Should someone use a constant current source, CCS, to supply 3.4mA or much more

Iadc if wanted to the anode then the dc load line is a horizontal one at the Idc value,

so it doesn't need to be plotted.

Then ac load ( if there is one ) is the only load to be plotted through the Ia x Ea

working idle point. When a CCS is used, we can move the Q point to say

Ia = 5mA with Ea = 160V. If there was a cap coupled ac load of say a 100k

volume pot, THD would be less than 1/2 what was achieved at the original Q point.

1/2 a 6SN7 can easily drive a 50k gain pot at low THD if the anode has a

CCS dc supply.

There are better ways to make a gain stage than by just using one gain triode.

See my preamplifier pages 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 monochrome image, ie, just black and white. All sorts of load lines

can be drawn, and magically un-drawn if you make a mistake!

From these curves it is possible to calculate the 2H with a constant

current dc source. 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.

Gm = µ / Ra = 19.5 / 11,300 = 1.7mA/V.

Minimum THD occurs when load line is horizontal, ie, a CCS.

THD % = 100 x 0.5 x difference in Ea V swings

sum of Ea swings

= 100 x 0.5 x ( 122 - 112 ) = 2.1%

122 + 112

Max Vrms output = 83Vrms, and at 49Vrms THD = 1.24% which is

less than half the THD for 49Vrms with 3.4mAdc feed via a resistance

as shown in Fig 5.

The THD contains other H products besides 2H but test gear tells me

that 3H and other H are well below the 2H level and impossible to view

on the CRO screen or able to be calculated from the tube Ea vs Ia

"characteristic curves" on data sheets. The 3H plus other H become a

smaller % of the THD as Vo is reduced. It is possible to build a differential

push pull voltage amplifier which is loaded with anode load values above

5Ra and which has buffered outputs and which makes no more than 0.1%

THD at the two phases of 50Vrms outputs.

This is a lot less THD 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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