Designing resistive impedance-matched signal splitting networks
is nowadays mostly done with the assistance of computer software.
In fact, odds are pretty high that the designer either has no
idea what the formulas behind the 'magic' are, or at least it
has been a very long time since working them with pencil and
paper. There's no shame in that, though, just as there is no
reason to expect someone using a cellphone must know the intricacies
of the internal circuits or the network to which it is connected.
We've moved past that. For those of us who still appreciate
a refresher on the behind-the-scenes calculations being performed
at lightning speed within the smartphone or computer, this article
will be a welcome bit of information.

Splitting Pads

By H. C. Carmichael, Consulting Engineer

Details on how various resistor combinations can be used
to provide the proper matching networks.

It is sometimes necessary in communication and television
work to apply two alternating currents such as speech, carrier,
or music currents to one channel. This may be done by means
of a transformer, but owing to the electromagnetic fields and
frequency losses from the transformer, this method is not always
desirable, and another method using combinations of non-inductive
resistances is sometimes used. This method involves the formation
of resistance networks, called pads, from non-inductive resistances.
In the case of 600 ohm circuits it is usual to make matching
pads from 1200 to 600 ohms and to connect the 1200 ohm outputs
in parallel. There is a third method, shown in Fig. 1A, using
a network of standard non-inductive resistors which involves
a small power loss but is less expensive than the other methods.

Fig. 1 - Circuit configurations

An application of this method is shown in Fig. 1A, where
it is used to connect two sources of alternating current to
one channel or vice versa. For purposes of matching, it is necessary
to make the values R_{1} R_{2} R_{3},
shown in Fig. 1A, such that when channels 1 and 2 are terminated
in their correct impedances (600 ohms), then channel 3 will
also be 600 ohms. Fig. 1A may be simplified to the arrangement
shown in Fig. 1B. Now, when two resistances, R_{1} and
R_{2} are connected in parallel their joint resistance
may be found by the aid of the formula:

If this formula is applied to Fig. 1B, then the joint resistance
of the two paths (600 ohms and x ohms) from A to C is:

Similarly, the joint resistance of the two resistances (600
ohms and ix ohms) between Band C is:

These two joint resistance values are in series with respect
to the line and thus their total resistance is:

The other resistor from A to B (x ohms) is in parallel with
this combination as shown in Fig. 1C which is a further simplification
of Fig. 1A. Thus, if formula (1) is applied to this circuit,
then the joint resistance of the combination will be:

This, of course, must equal the resistance of channel 3 and
thus:

This may be simplified to show that x = 1800 ohms.

Figure 1D shows the resistance network to satisfy the conditions
shown in Fig. 1A.

**Power Loss of Pad **

The power loss from each channel to the mixer channel resulting
from the use of this pad can be calculated as follows. Assuming
a voltage of 10 volts across a source of 600 ohms in channel
3, then the voltage across points A and B in Fig. 1D would be
10 volts and the voltage across points A and C would be 5 volts,
as the resistance of the combined resistor AC is equal to the
combined resistor CB. The power loss equals:

and if the information supplied in the data is substituted
in this formula, then: -

Power loss =
= 20 log 2 = 6.02 decibel

**Delta to Star Conversion **

Another connection called the "star" connection has the advantage
over the "delta" connection of being a simpler combination,
and the delta circuit shown in Figs. 1D and 1E may be converted
to the star or "Y" connection shown in Fig. 1F. Now if these
circuits are to be equivalent, then the resistance between points
AB, BC, and CA in Figs. 1E and 1F must be similar. At a glance
if may be clear that in Fig. 1F the resistance from A to O will
equal 600 ohms, the resistance O to B will also equal 600 ohms,
making A to B equal to 1200 ohms as required. It may be mathematically
proved, however, that: -

For example, to convert the delta formation in Fig. 1D to
an equivalent star or "Y" formation, then: -

=
600 Ohms

and likewise, R_{B} and R_{C} will be equal
to 600 ohms respectively. The equivalent star or "Y" connection
would be then as shown in Fig. 1G.

**Practical Application **

This type of network has practical application in circuits
such as the one shown in Fig. 1L where three balanced channels
are connected together via the star connection. This circuit
enables two sources of alternating current to be fed into one
channel or one source of alternating current to be split into
two channels.

Another practical application of the star pad is shown in
Fig. 1M where the output from an amplifier may be: -

(a) Disconnected from both channels,

(b) Connected to both channels, or

(c) Connected to either channel.

This arrangement may be also used to mix the output of two
amplifiers into the one channel.

There are many and varied applications of this simple circuit
arrangement but it must be remembered that the loss of 6.02
decibel is a disadvantage. The use of standard 600 ohm non-inductively
wound bobbins considerably simplifies the construction of the
pads and carbon resistors of suitable value may also be used.

**Matching with Series Resistors **

Another method of matching is by the use of series resistances
in each branch of individual circuits as shown in Fig. 1J. In
this circuit, channels 2 and 3 are assumed to be terminated
in their correct impedances, and thus channel 1 must look like
600 ohms. Now the circuit shown in Fig. 1J may be simplified
to that shown in Fig. 1I, and Fig. 1I, in turn, simplified to
that shown in Fig. 1H. The joint resistance of two equal resistances
in parallel may be found by dividing the resistance of one by
the number of resistances connected in parallel, and in the
case of the two equal resistances shown in Fig. 1H, the joint
resistance will be: -

The impedance of the network shown in Fig. 1H must match
the impedance of channel 1 (that is, 600 ohms) and thus: -

therefore x = 200 ohms.

This value of 200 ohms represents the sum of the values of
both resistors in each branch of the network and thus each resistor
will be: -

or 100 Ohms

The completed circuit will then be as shown in Fig. 1K. This
circuit has an approximate loss of 4.44 decibels in each channel
which, however, is not a very serious disadvantage. The same
circuit arrangements may be used with this method as shown in
the previous circuits, and, while not as simple as these, it
is nevertheless simpler than the standard matching pad. Another
feature of the split pad is that the pad does not affect the
frequency response, provided that non-inductive resistors are
used.

The author wishes to thank Mr. Charles Smith for his assistance
in the mathematical development of various equations.

October 15, 2015