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Practical Transformer Design and Construction
August 1947 Radio News

August 1947 Radio News

August 1947 Radio & Television News Cover - RF Cafe[Table of Contents]

Wax nostalgic about and learn from the history of early electronics. See articles from Radio & Television News, published 1919-1959. All copyrights hereby acknowledged.

This is the third and final installment of a 1947 Radio News magazine series on iron core transformer and reactor design. You of course are familiar with transformers, but what about reactors? A reactor is an inductive element used to limit overvoltage or short circuit current from a source to a load. These days they are most often seen in electrical distribution substations, associated with transformers. I did not intend to imply that you will learn something about reactors here; they were covered in Part 1, "The design and construction of iron core transformers and reactors." Part 2 is titled "Complete details for designing and constructing your own iron core reactors." I will attempt to procure the June and July issues. In the mean time, part 3 reviews the basics of transformer turns ratio and current transformation. 

Practical Transformer Design and Construction

Practical Transformer Design and Construction, August 1947 Radio News - RF CafeBy C. Roeschke

Part 3. Concluding article covering the design and construction of iron core transformers and reactors.

In this final article on transformer design we will discuss the design and construction of a plate modulation transformer, and an audio output transformer. Several valuable hints on the practical construction of transformers are also included for the assistance of the builder.

A modulation transformer, like any audio power transformer, is designed to match the output impedance of one piece of equipment to an input impedance of a second piece of equipment. This matching is done to insure maximum power transfer and satisfactory frequency response.

When a load is connected to the secondary winding of an audio transformer, there is reflected into the primary circuit an impedance which is determined by the turns ratio of the transformer. This is the ratio of the number of turns in one winding to the number of turns in the other winding.

Mathematically, the turns ratio is equal to the square root of the ratio of the two impedances, or turns ratio = Turns ratio 1 - RF Cafe

For example, consider a transformer which is to work out of a 5000 ohm circuit into a load of 50 ohms. Then

Turns ratio = Turns ratio 2 - RF Cafe = Equation 1 - RF Cafe = 10.

This means that the primary winding must have ten times as many turns as the secondary to match the 5000 ohm primary circuit to the 50 ohm secondary load.

For the purpose of design demonstration, assume that a plate modulation transformer is required for the following conditions of operation.

1. Modulator output is push-pull pentodes requiring load impedance of 10,000 ohms and that each tube draws 60 ma. plate current.

2. Audio output power to be 7.5 watts.

3. The Class C r.f, amplifier to be plate modulated draws plate current of 70 ma. at 420 volts.

4. Voice frequencies are to be used. The secondary load resistance is equal to the r.f. amplifier d.c. plate voltage divided by the r.f. amplifier plate current. In this case then: R = 420/0.07 = 6000 ohms

To calculate turns ratio:

Turns ratio = Equation 2 - RF Cafe = 1.29

Primary inductance required - RF Cafe

Fig. 16 - Primary inductance required for good response at low frequencies. (hy. = henrys)

Therefore, the primary must have 1.29 times the number of turns used in the secondary.

When a transformer is employed in the plate circuit of a vacuum tube, it is necessary that it be designed to reflect the proper load impedance for the tube. In addition, the primary inductance will determine the low frequency response of that audio stage. For this reason, Fig. 16 is included to indicate proper primary inductance for good low frequency response at 50 cycles or 150 cycles. Fig.  9 ** can be used to' calculate the primary inductance. Since our modulator tubes are pentodes and we are interested only in voice frequencies, we see in Fig. 16 that the primary inductance should be about 10 henrys.

Fig. 5* shows that the primary could be wound with No. 34 wire to carry 60 ma. and the secondary can be wound with No. 33 wire for 70 ma.

For convenience we shall use No. 33 . for both windings.

This transformer is to handle 7.5 watts so, according to Fig.  16, we can try to design it with 1" laminations and a stack of 1".

Let us tabulate the data we have so far:

Primary impedance = 10,000 ohms Primary d.c. = 60 ma.

Primary inductance = 10 henrys

Primary wire size = No. 33

Secondary impedance = 6000 ohms

Secondary d.c. = 70 ma.

Secondary wire size = No. 33

Core size = 1" stack of 1" laminations

Turns ratio = 1.29

First let us try 2700 turns, center tapped, for the primary winding. Then the secondary must have 2700/1.29 = 2090 turns. Next we calculate the coil size as explained previously and find that the build is 81 per-cent which means that the coil will fit into the core.

The primary is for push-pull tubes and therefore must be center tapped at 1350 turns. Since "B+" is connected to the center tap, the direct current flows in opposite directions in each half of the winding which means that the core saturation effect is cancelled out as far as the primary is concerned. But the secondary has 70 ma. d.c. flowing through its entire length in one direction and this must be considered in the calculation of inductance.

Then for the secondary, NI = 2090 x 0.07 = 146. But we are interested in the primary inductance so let us convert this effect into the primary. Thus:

Since NI = 146

Then 2700 x I= 146

And I = 146/2700 = 0.054 amp.

This means that 54 ma. flowing in the 2700 turns primary would give the same core saturation effect that 70 ma. flowing in the secondary would give. For this reason we shall use the figure 54 ma. in our primary inductance calculation.

Then Turns x current / length (1) - RF Cafe = 24.3

And from Fig. 9** if Turns x current / length (2) - RF Cafe = 24.3 then L * I^2 / V (1) - RF Cafe = 0.5 x 10-2 = 0.005

By transposition if L * I^2 / V (2) - RF Cafe then L= Inductance formula (1) - RF Cafe (V = 6 cu. in.),or

L = Inductance formula (2) - RF Cafe = 10.3 henrys (primary inductance).


Therefore this design is satisfactory. In Fig. 12** we see that when NI/l = 24.3 the paper gap in the core should be about 0.0015" thick.

Audio Output Transformer

Here let us assume that our audio amplifier employs the same output tubes as used in the modulator described before, which means that the following conditions exist:

Output tubes - P.P. pentodes (10,000 ohm load)

Plate Current - 60 ma. for each tube

Output Watts - 7.5

Lowest Frequency - 150 cycles (voice frequencies)

Required Primary Inductance - 10 henrys

Here we have the same primary circuit requirements which existed when we designed the previous modulation transformer. Therefore, we can use the same core size and primary winding.

Assume that at times this amplifier will feed into a 500 ohm line and that at other times it will drive a loudspeaker having an 8 ohm voice coil. Then two secondary windings will be needed; one for a 500 ohm load and one for an 8 ohm load.

   Turns ratio = Turns ratio (3) - RF Cafe = 4.47 for 500 ohm winding, and

   Turns ratio (3) - RF Cafe = 35.3 for 8 ohm winding.

With 2700 turns in the primary we find that the 500 ohm winding must have 2700/4.47 = 604 turns. Similarly, the 8 ohm winding must have 2700/35.3 = 76.5 turns (use 77 turns).

Now, we want to be able to put the full 7.5 watt output either into the 500 ohm load or into the 8 ohm load. Then, the wire used in each of these secondary windings must be heavy enough to carry such current. At 7.5 watts, the current in the 500 ohm winding is:

W = I2Z

Then: I2 = W/Z

And Current formula (1) - RF Cafe

   I = Current formula (2) - RF Cafe = 0.122 amp, or 122 ma. and the current in the 8 ohm winding is:

   I = Current formula (3) - RF Cafe = 0.968

Fig. 5* indicates that the 500 ohm winding must be wound with No. 30 wire to carry 0.122 amp and the 8 ohm winding must have at least a No. 22 wire to carry 0.968 amp.

The coil size calculation shows a build of 83 per-cent which is satisfactory.

Since the only direct current is in the primary winding where it flows in opposite direction in each half of the coil, the core does not have to have a gap. The laminations can be inter-leaved as in a power transformer core. Thus our design is a follows:

Core 1" iron 1" stack - laminations interleaved

Primary-2700 turns No. 33 wire, center tapped

500 ohm winding-604 turns No. 30 wire

8 ohm secondary-77 turns No. 22 wire

Only one secondary is to be used at one time.

Practical Hints

When the approximate specifications of a transformer are known, it is a simple matter to determine its con-struction in more accurate detail.


*Figs. so designated appear in Part 1 of this article published in the June issue of Radio News.

**Figs. so designated in Part 2 of this article published in the July issue or Radio News.



Posted January 30, 2023
(updated from original post on 10/13/2014)

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