August 1947 Radio News
Wax nostalgic about and learn from the history of early
electronics. See articles from
Radio & Television News, published 1919-1959. All copyrights hereby
This is the third and final installment of a 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. After the tease implying that you will learn
something about reactors here, I have to admit that they must be
covered in the previous parts of the article series. Hopefully,
both part 1 and part 2 will soon be posted to fill the gap. In the
mean time, part 3 reviews the basics of transformer turns ratio
and current transformation.
Practical Transformer Design and Construction
By C. Roeschke
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
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 =
For example, consider a transformer which is to work out of a
5000 ohm circuit into a load of 50 ohms. Then
Turns ratio =
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
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 =
To calculate turns ratio:
Turns ratio =
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.
inductance = 10 henrys
Primary wire size = No. 33
impedance = 6000 ohms
Secondary d.c. = 70 ma.
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
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.
And from Fig. 9** if
= 24.3 then
= 0.5 x 10-2 = 0.005
By transposition if
(V = 6 cu. in.),or
= 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"
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)
- 60 ma. for each tube
Output Watts - 7.5
- 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 =
= 4.47 for 500 ohm winding, and
= 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
0.122 amp, or 122 ma. and the current in the 8 ohm winding is:
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
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 interleaved
as in a power transformer core. Thus our design is a follows:
Core 1" iron 1" stack - laminations interleaved
turns No. 33 wire, center tapped
500 ohm winding-604 turns No.
8 ohm secondary-77 turns No. 22 wire
Only one secondary
is to be used at one time.
When the approximate specifications of a transformer are known,
it is a simple matter to determine its construction in more accurate
*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 October 13, 2014