Tube Amplifier Theory |
On this page I will publish some tube amplifier theory. Output
transformer impedance calculation is a bit tricky before you get the hang of
it. Some useful formulas for calculation of output power, distortion and impedances
will also be published. The idea to make this page came up when I bought the
Namo Web Editor, which is capable of writing maths (one of the reasons why I
bought it).
Transformer Impedance
Calculation
Primary current in at the dot gives secondary current in at the dot.
Law of induction (Faraday-Henry law):
Induced emf in a solenoid:
From this formula we get (magnetic flow is the same through both windings):
For the transformer above (1:1.33 or 3:4) this means:
The power on the primary and secondary is equal, which means:
Now the tricky stuff, the impedance calculation:
If we connect a 100R resistor on the secondary and apply 1
Volt on the primary we get:
The consequence of this is:
This leaves us with the formula for impedance calculation:
Windings in parallel does not change the number of turns,
windings in series adds the number of turns. This is a bit difficult
to understand, normally paralleled equal resistances gives half the resistance,
but not in the transformer world, and to make it even worse equal
windings in series increases the resistance with a factor of 4!
In order to understand this we rewrite the formula for induced
emf in a solenoid:
From this formula we see that the magnetic flow is proportional to the voltage
per winding-turn. If we use parallel windings, no turns
are added (acts as if the wires were connected in parallel). In series connection the turns are added (we need more voltage
when the number of turns is increased to get the same magnetic flow).
Example:
Above you see the data for the Lundahl LL1620,1623,1627 output transformers.
We are going to calculate some connection alternatives as a practice (LL1623):
Primary: 4x13.4
Secondary: 8x1
All primaries in parallel and all secondaries in parallel
gives the transformer ratio 13.4:1, the primary impedance with 8 Ohm on the
secondary is:
Two primaries in series, paralleled and all secondaries in parallel gives the
transformer ratio 26.8:1, the primary impedance with 8 Ohm on the secondary
is:
Four primaries in series and two secondaries in series, paralleled gives the
transformer ratio 13.4:1, the primary impedance with 8 Ohm on the secondary
is:
With all four primary windings in series, the number of secondaries used for
a primary load of approximately 3 kOhm at 4/8/16 Ohm secondary is listed below
together with the resulting load:
Number of secondaries (connection alternative) |
Resulting load impedance |
2 ((1+1)//(1+1)//(1+1)//(1+1)) |
2873 Ohm @ 4Ohm |
3 (((1//1)+1+1)//((1//1)+1+1)) |
2554 Ohm @ 8 Ohm |
4 ((1+1+1+1)//(1+1+1+1)) |
2873 Ohm @ 16 Ohm |
Add the primary resistance of 164 Ohm to the figures above and you have the
correct load.
Now the push-pull alternative, which is a bit difficult to
understand. We use two primary windings for each tube in the push-pull output
stage, and two windings in series on the secondary as in the example above.
The resistance from anode to anode is in this case 5745 Ohm, and this is the
figure used by transformer manufacturers. We are interested in the load on ONE
tube, which uses 2 windings in series:
This looks suspicious, but the output tubes are in reality paralleled (with
inverse phase connections), and this means that the load for each tube is 2872
Ohm (half the anode to anode load). The same applies for paralleled single end,
if you use two tubes the transformer load shall be 50% of the value that is
used for a single tube, and with four tubes the load shall be 25% (in this case
Ohms law can be used).
From Sowter I 'borrowed' this picture showing connection alternatives for their
transformers using 8 secondary windings.
The same formulas are used for interstage transformers, but these are often
specified as for example 2k:4k. I think this means that it is designed for a
source impedance of 2k, and the transformer output impedance is 4k. The transformer
ratio is then 1:1.4 (remember the square?), and the primary inductance probably
around 20-30H (see formulas below).
This concludes the transformer impedance calculation section. If anything is
unclear or wrong, please contact me.
Other Important Formulas
For Transformer Calculation
The output result from the transformer depends also on the
frequency response, and the low cut frequency is calculated from the Primary
inductance. In these calculations it is the output impedance of the driver
tube that shall be used.
The -3dB low frequency limit is calculated at the frequency where the output
impedance of the tube equals the input impedance of the transformer (at low
frequencies this depends upon the inductance):
The LL1623SE 90mA has a primary inductance of 30H, and with a single anode loaded
300B (output impedance approximately 700 Ohm) this makes the low cut (-3dB)
frequency:
Lundahl also specifies the -3dB power output point (maximum
power output reduced with 50%) as the point where the primary load impedance
equals the primary inductance, and with a 3 kOhm load this makes:
This is not as important as it looks, you will not find music signals with full
amplitude at such low frequencies.
Power Output and Distortion
Calculation
Above is the loadline of a 300B into 5.1 kOhm. I will use
this as an example how to calculate distortion and power from a curve set. The
slope of the loadline is calculated from the load impedance, and with a 5.1
kOhm load the current (I=U/R) per 100V is 19.6mA. Below is the result that the
Glass Ware SE Amp CAD has calculated for this loadline. Check the 'Tube CAD'
link for free 'True curves', each month a new tube.
Below are the values at the three interesting points for power and distortion
calculation (marked with green and red lines in the curve set).
Apart from the voltage and current readings, it is also interesting to see the
changes in mu and Rp. These can also be derived by graphical methods from the
curve set. I will describe this in the 'Other Formulas' part.
The formula used for output power calculation is:
With the values from the loadline above we get:
The SE Amp CAD got 8.35W, the difference is because of not perfect readings
above.
The formula used for calculation of the 2nd harmonic is:
With the figures from our example we get:
The result in dB is:
-20 log(0.026) = -31.7dB
For the 3rd harmonic calculation we need 2 more points from
the curve set. These points are called I_x and I_y, and are the point at the
loadline where the grid voltage is maximum positive 'grid swing' divided with
the square root of 2 (I_x), and the maximum negative 'grid swing' divided with
the square root of 2 (I_y). These readings are found below.
The formula used for calculation of the 3rd harmonic is:
With the numeric values from our example this is:
The result in dB is:
-20 log(0.0042) = -47.5dB
This calculation seems a lot more uncertain. The difference to the SE Amp Cad
result is 6dB, but it will at least give some guidance.
The THD is calculated as the mean square root of the 2nd and 3rd harmonics:
Some Other Useful
Formulas
I will write this a little later.
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