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The OnLine Valve Amp Magazine
DATE: AUGUST 1995

Class A Tube Amp Design Overview
Author: Kurt Strain
Date: 28 AUGUST 95
This is a brief (sort of) description on how to design a class A output
stage for triodes in either pushpull or singleended from notes I put
together. I hope this proves informative.
Single Ended Class A Triode Output Stage:
This is as simple as it gets in transformer coupled tube output design, in
concept. The transformer must be designed to handle the DC current at the
bias point, so a special output transformer for single ended use must be
used. The output transformer features an air gap that optimizes the
coupling at low frequencies and the DC current that serves to lower the
permeability of the core. Without the air gap, the iron will saturate under
too little DC bias to accommodate the needs for a single ended triode. Too
much gap will reduce the primary inductance so that the lowest frequencies
will not pass without severe attenuation.
A schematic for a single ended triode amp output stage is as follows:
B+ o
 ____ 8
)(
)(____ 4 OUTPUT
)(
)(____ 0

anode
>grid
cathode

++
 
Rk Ck
 
= =
There are two recognized approaches for biasing this tube: for max power
output and for best sound. They are not necessarily the same.
To maximize power, with a bypassed self biased stage, you need to make all
the voltage swing available and all the current swing available meet at the
same time. The voltage at the anode is B+ VDC, neglecting the primary
resistance drop. The max primary swing will be 2 times that voltage, peak
to peak, neglecting losses in the plate resistance and bias voltage. With
the tube self biased and bypassed, most the AC voltage swing will end up
appearing across the primary output impedance plus the plate resistance.
Usually triodes have plate resistances about onefourth that of the
transformer primary impedance, so most ends up across the primary. If B+ is
350V, then the peak to peak voltage is 700V.
The current that you can bias this tube to is the plate dissipation rating
in watts divided by the anodecathode voltage. The voltage at the cathode
that self biases the tube will be stated for the tube, and is typically
something like 60V. Thus, if the plate dissipation is 30W max and the anode
to cathode voltage is 350V  60V = 290V, then the max bias current is
30/290 = 0.103A.
The bias current represents the peak current going through the transformer,
and twice that will be the peak to peak current, or 0.206A. The current
will swing from 0 to 0.206A while the voltage swings from 700V to 0V,
roughly. This means that near optimum for power Rprimary = 700V / 0.206A =
3400 ohms. For a speaker load of 8 ohms, the turns ratio would thus be the
square root of the impedance ratio, or the square root of 3400/8 = 20.6:1,
20.6 turns on the primary for each secondary turn.
Now, assume the plate resistance, rp, to be 900 ohms, for example. The
damping factor would thus be equal to 3400 ohms of load divided by 900 ohms
of source impedance, approximately since rp is a rather dynamically
shifting thing. That would make the damping factor 3400/900 = 3.8. If
pentodes are used, the higher rp's of pentodes reduces damping factors
considerably and global feedback is usually employed to help.
A more accurate power output could be said to be Vout^2/Rprimary where Vout
= (700/2*SQRT(2))*(3400/3400+900), the voltage division of the RMS voltage
into the load, Vout = 196Vrms. The power is thus 196^2/3400 = 11.3W.
From a current standpoint, the max RMS current is 0.103/SQRT(2) =
0.073Arms. The power output from max current is Iout^2*Rprimary =
0.073^2*3400 = 18.0W. There is more current than voltage available, so if
we dropped Rprimary a little we could raise the max power available a
little, or we could use the higher Rprimary to our advantage in terms of
greater damping and lower distortion due to a lighter load. Besides,
without more trial and error, we don't really know for sure how much
voltage swing we really have available. After all, we could operate with a
little positive grid voltage if we needed to. So we could stop right here
and empirically find out how much we'll get. Or, when we go to look for an
available transformer the closest we may find is a 3000 ohm 80mA unit,
which may be optimum as found by prior experimenters despite our
calculations. A little consulting from a transformer designer can help.
In the case of parallel singleended, pay particular attention since it
very closely models a pushpull class A amp, with polarities switched
around.
A schematic for a parallel single ended triode amp output stage is as
follows:
B+ o
 ____ 8
)(
)(____ 4 OUTPUT
)(
)(____ 0

++
 
anode anode
>gridgrid
cathode cathode
 
+++
 
Rk Ck
 
= =
The current available is doubled, but the voltages remain the same. The
effective load Rl = 2*Rprimary, or a half as heavy a load per tube. For a
given Rprimary, the effective plate resistance is rp/2 (= 450 ohms).
Using the same 30W tubes before at B+ = 350V, Ibias_total = 60W/(35060V) =
0.207A, or Ipp= 0.414A. Rprimary = 700V/0.414A = 1690 ohms. So you can see
that parallel SE amp output transformers are lower in primary impedance,
and hence have a smaller turns ratio. They also must accommodate larger
primary DC bias current, 0.2 A in this case.
But the power output is now about (350V/SQRT(2)*(1690/450+1690))^2 / 1690 =
= 28.6 W, about twice as before. The damping factor is now 1690/400 = 4.2,
almost the same.
Class A PushPull (PP):
What's the difference in the circuit model of a two tube SE amp and a two
tube class A PP? Not very much, really. Here's a side by side comparison:
B+ o o B+ o o
.. ____ 8 . ____ 8
) ) ( ) ) (
) ) (____ 4 OUTPUT ) ) (____ 4
) ) ( ) ) (
) ) (____ 0 ) ).(____ 0
   
+++  ++
   
anode anode anode anode
Vin >gridgrid Vin >grid Vin >grid
cathode cathode cathode cathode
   
++ ++ ++ ++
       
Rk Ck Rk Ck Rk Ck Rk Ck
       
= = = = = = = =
These are nearly identical circuit models, just shifts in polarities. Note
the polarity differences in the inputs to the tubes and in the polarity
dots of the primary windings. That's about all the differences there are.
If Rprimary = 1500 ohms in the case of one primary winding shown and two
tubes as in the left case shown above, then the effective load in every
case for every tube would be Rl = 3000 ohms. The difference is in how we
total the primary impedances.
For pushpull, the windings are reversed in polarity. Now, the number of
turns from platetoplate just doubled, and there's an effect due to mutual
coupling, like an autotransformer. N more turns produces N^2 more
impedance, platetoplate. So doubling the number of turns makes a four
times more platetoplate impedance. The platetoplate impedance will thus
be 4*1500 = 6000 ohms. But since in a pushpull each tube is assisted in
driving the load by the other tube, the effective load is 2*1500 = 3000
ohms, or Rpp/2. The power delivered to the speakers is the same.
To calculate Rpp, or sometimes Raa, find Rprimary for the single ended case
and multiply by 4. Rprimary is the same as using half the full
centertapped primary winding of a pushpull. Rpp uses both Rprimary
halves, which quadruples the impedance.
The Class B Case:
Class B is more difficult to compute power for than class A. In class A,
bias is going through the output tubes all the time. Power is a virtual
constant, so the power supply load moves very little. This is a great
advantage, and probably one of the sonic advantages to class A. You can
design a power supply filter with less concern for massive changes in
current demands. Not true for class B. Class B allows only one side of the
transformer to power the load at any time. The voltage across the full
primary winding will look the same because the autotransformer effect
induces the full voltage swing despite only one device conducting. But the
lower power dissipation of idling class B or class AB allows more current
swing to take place. And these large swings will amplitude modulate the
power supply. It is adviseable to be have a more overdesigned power supply
for class AB as a result, with more capacitance and more current capacity.
Unfortunately, this limits your use of the type of power supply capacitors
as a result, probably requiring large electrolytics. Also, in class B,
since you do not have the assistance of the other tube during conduction,
the effective load then will be Rprimary, not 2*Rprimary as in class A, and
virtually no load when not conducting.
Class A push pull still has the disadvantage of requiring a split phase
driver, as well as more turns in the transformer, and SE has the
disadvantage of requiring DC bias through the transformer.
OK, you cover the case of biasing to maximize power, but I couldn't find
where you explain how to bias, for best sound. Could you elaborate?
First, I did mention that raising the primary impedance above max power
level can help improve damping and lower distortion. But beyond that,
experimentation and experience will tell better. Mike LaFevre of Magnequest
reported that "Stereo Sound" magazine, a Japanese publication, listed the
"optimum" operating points for various direct heated triodes in a single
ended amp, in terms of sound performance. They are listed below:
* Vaic VV30B:
Plate dissipation rating = 65W max
Vp = 430V
Ip = 70mA
Vg = 70V
Rprimary = 3K
Pout = 10W
Pd = 30W, 46% of max
Note: this is best sounding when the tube is run at medium to high
power.
* WE 300B:
Plate dissipation rating = 27W max
Vp = 400V
Ip = 60mA
Vg = 87V
Rprimary = 3.5K
Pout = 10.5W
Pd = 24W, 89% of max
Note: this is best sounding when the tube is run at high power.
* 2A3:
Plate dissipation rating = 15W max
Vp = 250V
Ip = 60mA
Vg = 45V
Rprimary = 2.5K
Pout = 3.5W
Pd = 15W, 100% of max
Note: this is best sounding when the tube is run at high power.
* 845:
Plate dissipation rating = 100W max
Vp = 430V
Ip = 62mA
Vg = 51V
Rprimary = 5K
Pout = 4W
Pd = 27W, 27% of max
Note: this is best sounding when the tube is run at low power.
* 211:
Plate dissipation rating = 75W max
Vp = 480V
Ip = 40mA
Vg = 20V
Rprimary = 10K
Pout = 2.6W
Pd = 19.2W, 26% of max
Note: this is best sounding when the tube is run at low power.
So, the best sounding operating point can be very much different from the
best power output operating point. But it is a matter of opinion. Maybe you
can see reasons why the 300B is popular. I happen to think the 2A3 is an
exceptional sounding tube, from my experiments comparing it to the 300B,
and I prefer it.
Here's how I run each of three parallel 2A3's:
Vp = 305V
Ip = 47mA
Vg = 65V
Rprimary (effective) = 4.8K
Pout = 4.3W (at clipping)
Pd = 14.3W, 96% of max
It just turns out I need to do this to fit a 1600 ohm 150mA transformer
with 3 2A3's, and the results are better than if I dropped Vp, soundwise.
Run 2A3's and 300B's hot, seems to be the best rule.

tubes@hillier.demon.co.uk
© M.J.Hillier 1995