Calculation of Singe
Ended Triode Amplifier
One of the easiest way to put
up a schematic for simple triode power amplifier is to open a
tube manual, or some other source, containing the data of the
tube and read what is written in it for the triode you have or
intend to use. There might be an example for the chosen tube with
all precalculated /suggested by the manufacturer/ optimum elements
such as Rp, Rk, Rg, Ck, Cg - given for class A - single tube.
All you have to do is to choose a proper transformer with proper
tube load and ratio to match impedance of the speakers and the
final stage is ready.
for common cathode SE power amplifier
is shown here. First "autobias" schematic requires some
more passive elements and Ug is fixed only for standing tube current,
for signal conditions Eg (Ug) is "floating" with the
input signal. Fixed bias schematic requires separate voltage for
feeding the -Eg on the grid of the tube and with the increase
of the signal tube may go to A2 class of operation. Some authors
(including me) say that autobias is better /musically/ others
like fixed bias better. No matter what you choose you have to
stay within tube limiting value parameter for Eg - optimum for
If you are not happy with what
the books say, or perhaps you may like to achieve higher power,
or reduced distortions you may try to calculate your SE triode
amp from the scratch. You need a family of curves related to the
tube, or if you don't have it - try to find it. Otherwise
you need a testing equipment to measure and draw this curves yourself
- it is a time consuming job and it is not for beginner to start
with. But if impossible to find diagram with family of plate curves
all you have to do is to measure plate current for any given voltage
on the plate and the grid and make notes. With all this data you
plot the curves on paper and make diagram (if you have the nerves
to do that). See below how it looks the diagram I made for my
"made up tube". My goal is to make it as simple as possible.
Determining slope of
the load line
Suppose we have already power
tube chosen and we have also family of curves for this particular
tube, taken from the manual or the dealer we obtain this particular
tube. Next step is to determine Rp for this tube and my
goal here is to simplify calculations. The following diagram -
family of curves is made up from me and it is not for specific
type power triode - it's an example for tubes from the type 2A3,
6B4G, 300B, AD1 and others used in DHT SE amps.
For example my "made
up tube" have plate voltage of Up 0 = 250 V, amplification
factor u=4, maximum plate dissipation Pp=15 Watt. Rg1 max= 750
K, Rp=800 Ohm
To read the diagram: on vertical
axis we put plate current - Ip, horizontal axis - plate voltage
Up, black curves shows Ip and Up for some values of grid voltage
- Ug. Blue curve
shows plate dissipation of the tube - above this curve tube may
become overloaded and burned. Yellow
line/on some monitors orange/ between
is for helping us to determine the slope of the load
line. The red
line between points Q-S paralleled
/with same slope/ to M-N help line is the needed loadline. But any line paralleled
with M-N help line may
be chosen for loadline for this chosen load. How to draw this?
All colors here are
just to make things easy to spot and understand. Usually diagrams
in the books are given in black and white.
Let's say the tube need reflected impedance Rl=2.4
K plate load as stated. To make it more clear: for example we
decide to run the tube with load of Rl=2.3 K plate impedance to
achieve a bit more power /few percentage points/ we have to see
how this will be reflected on the family of curves.
Fastest way to determine the slope of the load line
is to draw M-N help line. I choose
plate current of 100 mA (0.1A) and look for corresponding Up to calculate Rl - which
here is the impedance of the transformer primary. More for the output transformers and reflected
load - click here.
Rl must be equal to our
chosen load of 2300 Ohm (2.3K)
that is true and only one line(or line paralleled to
it) you may draw for
Ua=230 V at Ia=0.1 A (100 mA)
/Reason I choose Ia=0.1 A is for faster
calculation - otherwise you may choose different current, slope
Rl = 230 V /
0.1 A = 2.3K /Kilo Ohm/
Next is to see for Ua 0=250 V we make a vertical line
(Rl = 0) to the crosspoint with the plate dissipation
limit (the blue curve) and determine the necessary Ug /Eg/.
After that we draw line Q-S (in red)
paralleled with M-N help line.
For maximum power Q-S must have same crosspoint at blue
curve with Ua=250V. Keep Q-S line slightly in touch with the blue
curve to achieve maximum power from the given tube for
this particular load - and here we go - the line
Q-S - is the loadline for our triode at
If Q-S goes over the blue curve - that means we have exceeded
power dissipation of the tube and have to choose other parameters
- Rl, Up, Ug. If you keep this line
below the blue curve you are safe
to continue with your design and the calculations. If too far
below you will get lower power. So keep it in touch without crossing
it - that's the optimum without blowing the tube.
You may use this
fast method to calculate loadlines for preamp tubes as well!
Above crosspoint is very important for the regime
we put tube in. For class A the standing plate current Ip0 should
be around half the maximum plate current for the tube. On the
vertical axis from the crosspoint draw horizontal line (Rl = ~) we note that the standing plate current
is Ip0=60 mA (or 0.06 A). The maximum plate current reading is
Ip max=120 mA (0.12 A) taken from Q point
connected with right angle horizontal
line to the vertical axis of the diagram. From same reading
for Q make right angle vertical
line with the horizontal axis and for Ip max (maximum signal plate
current) we measure on the diagram that Up min = 110 V.
The achieved output power will be:
Pout max=(Ip max - Ip0). (Ua0 - Up min)/2
And for the data taken from this specific diagram (for
my made up tube) that makes for Pout max:
Pout max = (0.12-0.6)0.06.140/2 = 4.2 Watt
Ug from the crosspoint at the diagram at Uao=
250 V we have reading for Ug= - 45 V. That means that we must
have grid - cathode potential of - 45 V. For fixed bias - make
additional power supply to the grid of - 45 V and you are ready.
For selfbias we cannot make the grid negative due to
lack of additional power supply, we have to make the cathode more
positive. So we have to make voltage drop on Rk exactly 45 Volts.
Next is to calculate Rk for selfbias schematic. There
is no current drawn from the grid so cathode current Ik0=Ip0=0.06
Rk = 45V /0.06 A = 750 Ohm
(Or choose closest standard value if you don't have
standard result. You can combine several resistors to make the
exact value as well. Another way is to wound the resistor yourself
(bifilar or non inductive) from resistive wire if you know how
to do it.
Dissipation for this resistor is P= Ip0 x Ip0 x Rk=
2,7 Watt /choose 5-10 W resistor minimum/. If not - excessive
heat may be build up around this resistor and that's really bad
for the construction and neighboring elements.
Rg1 is given as minimum and maximum value by the tube
manufacturer. Rg1 is usually between 100 K to 1 M and it is also
important for forming the load for the preamp-driver tube. Rg1
should be made as higher as permissible by the grid requirement
of the power tube without introducing extra load on the driver
Rg2 - that's resistor you can omit in some cases. Rg2
is suppose to protect the grid of the power tube from high frequency
parasitic and oscillations. You may choose it anywhere between
1K to 10 K max.
I've seen some amp designers choosing very big capacitors
for Ck and Cg. They probably have designed transistor equipment
before getting into tubes, where big capacitors are heavily used.
Transistors handles large currents so the capacitors have to be
large. For tubes this is wrong - the real output transformer
cannot pass very low frequencies without some distortion. Tubes
and large capacitors can. That easily adds up extra distortions
at all frequencies once you overload the transformer. So
be careful choosing capacitors - not too big, but sufficient.
Choose the middle way. Click
here to read more on tubes/transistors.
Ck - choose it between 20 uF and 200 uf the higher
Rk the higher Ck you choose. Keep in mind that if you choose very
high value, that may overload the output transformer. For our
design Ck=50 - 100 uF / 100V should suffice.
Cg choose between 0.02 uF to 0.33 uF keeping in mind
same thing we said about Ck - do not overload the transformer
with too much lower frequency. That may give additional harmonic
distortions if Cg is too big. If Cg value is too small the lower
frequency of interest may suffer from that.
For our design Cg= 0.05 uF/ 400 V should be enough.
But if you can find not very expensive (let say Japanese made)
large output transformers (most probably you won't be reading
this if you have it) with lower frequency starting at 10 Hz /-0.5
dB/ you may at least triple chosen capacitors value and that should
Performance and distortion
After all elements are calculated or chosen we have
to see how this amp will perform on paper and what kind of distortion
we should expect from it. As you most probably noticed we have
two blue triangles (with Pout in the middle)
on the red loadline of diagram. They represent the output power
for each half period of the input signal. If they have visually
equal surface - zero distortion is introduced. But that's not
happening in real life. So he have to calculate what distortion
we should have at maximum rated power. Let's compare each triangle
by plate currents. To have D tot = 0 %, we must have Ip max
- Ip0 = Ip0 - Ip min. In our case:
Ip max - Ip0 = 120 mA - 60 mA
= 60 mA
that is the current for positive
half period of the input and
Ip0 - Ip min = 60 mA - 5 mA =
that the current for negative
half period of the input. Here we spot that those currents are
not equal. So some distortion might be present.
To see the % distortion use the
following basic formula:
After calculation for our example
is performed /do it for exercise/ we find that D tot = 4.54 %
- that all distortions at maximum chosen and rated power for our
readings from the diagram.
How will perform? Build it and listen to it. That's
best way to see that. Too many factors and parts will be involved
in this process and each is equally important. Evaluate it with
your hearing playing the music you know and love.
All rights reserved. © 2003
BY SYSTEM 11 Ltd.
2002, 2003 All right reserved by © P.G.Doynov