Coming from outside link ? - click here

 

 

Calculation of Output Transformers

The output transformer is probably the most expensive part in the amplifier and something you have to live with if you like classic tube amplification. To separate the high voltage and match the speaker impendance with the tube impendance we need audio grade /output/ transformer. Single Ended /SE/ Amplifiers also run unbalanced DC through load and even if we somehow manage to connect a special very high impedance custom made speaker - that will introduce at least significant losses /speaker most probably will overheat and burn in smoke/. Output transformers are transferring certain power to the speaker so they are closer to their cousins - the power /mains/ transformers and in many cases good outputs are way larger than mains for the same rated power. There are circuits of Output Transformerless amplifiers /OTL/ which properly made nave really great sound but are harder to maintain and less reliable /some designs use up to 16 paralleled low impedance power tubes per channel to do the job/. Most tubes have high output impedance and unfortunately all of contemporary dynamic loudspeakers have low impedance. So we have to match them with tube impedance and if we need good performance - wide bandpass, low distortions and low losses we have to take special care of preliminary calculations and design. It is not easy task and takes a lot of experience but if you have a calculator on hand you could DIY and even wind it yourself. All Equations here are for Hi-FI Class A amplifiers - that is even better if you design AB or B amps. Here is how is made the simplest practical way:

First you have to choose output tube/s/ and desired power to achieve. Determine tube/s/ Load Resistance RL or Plate Resistance - RP from the tube data sheet. The tube RL optimum is usually listed there. If you don't know it look here choosing optimum load...

We are going to perform SE transformer calculations without negative feedback and my goal is to simplify the equations. We need to determine minimal size of the necessary core E - I type lamination is most economical "no waste" iron and is also classic for outputs.

Two coils C-core transformer is probably better for reduced leakage inductance but the iron is expensive and harder to obtain. There are also ironless "AIR-CORE" transformers shown on some sites - they need many pounds of copper, have an increased losses up to 60 % and not very good bass response, but are very linear /no distortions is introduced/ that's deriving from the linear magnetic properties of the air itself.

Now - we have Pout chosen and we have to calculate and choose core size - Q in square centimeters /sq. sm./.

Where :

- Pout is the output power in VA /Watt/

- Q - core size in sq. cm. /For square inches divide the result by 6.5./

Now increase Q with 10 % and choose a standard E-I size lamination. /For cheeper Hi-FI /with feedback/ or guitar amps you may reduce the size twice/. For longer equation - look here.
 
Now lets calculate number of primary Turns/Volt - n1.
 
/for feedback guitar amps you may reduce n1 twice/ Basic longer equation for Turns/Volt - look here.
 
The AC voltage on RL in the primary is
 
N1 - number of primary turns we need to wound is
 
N1 = n1.U1
 
Next we have to to match the loudspeaker impedance to tube impedance look again here /or to determine reflected impedance ratio/ and the basic formulae for transformer without losses is:
 
 
Where:

- n1; n2 - number of Turns/Volt for primary and secondary

- N1; N2 - total number of turns for primary and secondary

- U1 - AC voltage on primary

- Z is loudspeaker impedance in Ohm

- d - wire diameter in millimeters

- I - current in Amperes

 
Now calculate N2 and to reflect the losses in copper and iron we need to increase number of secondary turns - N2 with 5% to 15 % (the lower the power - the higher the percentage). Now the transformer is almost ready to be wound.
 
Wire diameters depends mainly on Is A/sq.mm. you choose. To calculate the diameter in mm we use the equation:
 
 
Where:
I is higher tube current in Amperes
t = 0.7 when Is = 2.5 A/sq.mm;
t = 0.8 when Is = 2 A/sq. mm;
t = 0.9 when Is = 1.5 A/sq.mm
 
The thicker the wire the less power is lost and less heat is produced in both windings /primary and secondary/. So if possible choose wire thicker than calculated above, but keep in mind that physically you may not be able to fit all the numbers of turns when you start winding the coil.
 
Now we have to calculate how our transformer can be made in the real world with the calculated parameters or we have to go back and choose different core or wire size. We choose square stack with following sizes:
 
 
h - that the longest side of the coil former in millimeters.
H = / h - 2mm / we have to to fit and secure the leads with strong tape - that is why we cannot use total h size and have to reduce it with 2-5 mm. /In Millimeters/
b - shortest side /usually 1/4 of the flange size/ in mm. /Only one flange shown here/.
B =0.5 - 0.9 . (b - 1 mm) that comes from the inter turn/inter winding insulations and interleaving we need to include in the coil /mixed primary and secondary - that reduces leakage inductance and is important for band pass of the transformer/. That again reduced size - but that real life. All windings have to be well electrically insulated with paper, cloth or insulation tape which adds up significant thickness and reduces inter winding capacity. That important for the higher frequency of interest and the electrical stability. /If you are making very "high voltage" transformers B could go below 0.5./
D - wire diameter with lacquer insulation in millimeters/a bit thicker 5% - 10% aprox. more than calculated/.
N11; N22 - number of turns per layer
M11 - number of layers
 
Now we have to calculate if the number of turns we have calculated will fit our design:
 
N11 = H/D
for number of turns per layer we take closest smaller number /example if calculated 29.47 = 29/
 
M11 = B/D
 
for number of layers we take closest bigger number /example if calculated 9.47 = 10/. That the calculation for primary.
 
The same way you calculate secondary N22= H/D ; M22= B/D
 
if it calculation does not fit the coil former - select thinner wire and insulation or bigger core size and start over.
 
We may well split primary in series /high voltage side/ and split secondary in parallels /high current side/ in at least three or four equal parts each and interleave them for best frequency response - result from lower leakage induction.
Start winding - secure leads tightly with glue or insulation tape and remember turns nave to be very tight and as close as possible to each other. You may start with the secondary and finish again with it - that interleaving is good for reducing overall leakage induction; parasitic capacity to ground and also increases safety - low voltage side is on top. That's four secondaries and three primary windings. This design method is proven over times and gives extremely good results for Ultra - Fi SE amps. Finally you have to complete you transformer stacking the iron on the coil /for SE amps adding "air gap" between E and I sections of the iron is a must/. Air - gap is introducing an increased resistance in the magnetic path which shows increase in iron losses. There's complicated formulas for calculating "air-gaps" but in practice it boils down to the following:
The "air gap" is not usually "air" but precut piece of non magnetic material such as thick paper or "shoe box cardboard" /0.2 - 0.5 mm/ for reducing DC magnetic saturation of the iron and linearizing the magnetic hysteresys curve. Insert 0.3 mm thick material between E part and I part of the iron for best results /up to 15W SE output/.
 
That is practical way to do a output transformer for SE amplifier. After completing the transformer test it the simplest way - connect the primary to the mains - if does not go out in smoke or burn in flames it is ready to be connected to the tubes in your schematics. If buzzing occurs tight the screws and look at the coil former - the transformer may need few more laminations.
 
For Push-pull stages you may use same methods and equations and use the reduced core size since there is no standing current in the primary. Plate resistance here is:
 
Rpp = 2Rp.
Reflected impedance per tube is 1/4 of total impedance plate to plate. No unbalanced DC flowing so no "air gap" is used here E-I laminations piece by piece is inserted from both sides of the coil former/some folks may like to use air-gap for more linear response - that may create other problems with the low level induction and somewhat increased losses/. For Class AB or B you may need to split the coil former in two "disks" and wind first one of the "disks" interleaving primary and secondary and later take the coil out, turn it over and put it back for winding the second "disk". Here is how coil former looks in that case:
 
 
 
There are equations to check induction of primary; frequency response and other parameters - that's no object of our practical transformer calculations. and it also difficult to do without exact iron parameters /such as magnetic permeability, flux density, exact B-H curve, iron per kilo/pound/ losses which varies from steel to steel/ specified by the silicon grade steel manufacturers.
 
 
Regards P.G.
All rights reserved.
 
if you are coming from outside direct link to this page and wish to continue you may click here
 
 
 
 
 
about me