June 1944 QST Article

## June 1944 QSTThese articles are scanned and OCRed from old editions of the ARRL's QST magazine. Here is a list of the QST articles I have already posted. All copyrights are hereby acknowledged. |

When you read a lot of tutorials about introductory electronics on the Internet, most are the same format where stoic, scholarly presentations of the facts are given. Those of you who don't have enough fingers and toes to count all of the college textbooks like that which you have read know of what I speak. When hobby articles are written in a similar fashion, it can quickly discourage the neophyte tinkerer or maybe even a future Bob Pease. QST has printed a plethora of articles over the years that are more of a story than just a presentation of the facts. My guess is the reason is because often the authors are not university professors who have forgotten how to speak to beginners. This article on basic calculations for AC series and parallel circuits is a prime example.

**Solving for Current at Voltage by Means of Admittance
**

**By S.E. Spittle, * W4HSG **

The problem of finding the resultant impedance of a group of impedances in parallel is not usually discussed in the more elementary texts on alternating currents, probably because the solution of such problems is rather lengthy and difficult without the use of complex algebra. The term "complex algebra" may have a mysterious and rather terrifying sound to those who have never heard of it, but actually the process is comparatively simple for anyone having a knowledge of the elements of plain algebra.

The recent QST article entitled "Meet Mr. j"^{1} provides
a good explanation of the application of complex algebra to alternating-current
circuits and should be read as an introduction to this discussion.
In that article one method of computing the impedance of parallel
circuits by means of their phase angles was described. An alternative
method, described here, makes use of the resistive and reactive
components without requiring a knowledge of the phase angles as
such, and therefore can be applied without the use of trigonometric
tables.

Since the laws of alternating-current circuits are merely extensions
of the laws governing direct-current circuits, taking into account
the effects produced by the storage of energy in the electric and
magnetic fields, it is logical to explain the solution of a.c. problems
in terms of the familiar operations used with d.c. problems. For
example, take a circuit consisting of two resistances, R_{1}
and R_{2} in parallel, as shown in Fig. 1.

**Fig.
1 - Parallel resistances.**

I_{1} = EG_{1}
I_{3} = EG_{3}

I_{2} = EG_{2}
I_{p} = EG_{p}

The usual formula for finding the parallel resistance of this combination is

However, this formula is only a special case of the more general formula

1/R_{p} =
1/R_{1} + 1/R_{2} + 1/R_{3} ... etc.,

applied to the case of only two resistances. In the special case of two resistances, the second formula is converted into the first by means of a few simple transpositions; When there are more than two resistances, the general formula is more practical, as will readily be seen if we apply the same process to the case of three resistances. Thus

1/R_{p} =
1/R_{1} + 1/R_{2} + 1/R_{3}

when transposed becomes

It may be interesting to note that the process of finding the resulting resistance is the same as finding the total current in the parallel circuit, using any assumed value of applied e.m.f. Thus

I_{p} = E/R_{p} =
E/R_{1} + E/R_{2} + E/R_{3}.

Changing the value of E will change I_{p} but not R_{p.}
Therefore we can assume E to be one volt, giving the formula for
the parallel resistance, which is usually expressed as

The same formula can be applied to a number of parallel impedances in an a.c. circuit, giving

The catch in this is that the currents in the various impedances
are usually not in phase with each other, so that the phase difference
must be taken into account in obtaining the correct value of parallel
impedance. In the d.c. case the quantity 1/R is known as the conductance,
represented by G whose unit of measurement is the mho. Thus, a resistance
of 5 ohms corresponds to a conductance of 1/5 or 0.2 mho. The total
conductance of a parallel circuit is the sum of the conductances
of the individual branches. Thus, making 1/R_{1} = G_{1},
etc., we have for three resistances in parallel

G_{p} = G_{1} + G_{2}
+ G_{3},

and the equivalent or parallel resistance, R_{p}, is
equal to 1/G_{p}. The conductance, G, is numerically equal
to the current which would flow with one volt applied to the circuit.
Hence, in finding the total conductance of a parallel circuit by
adding the individual conductances we are merely following the same
process as in finding the total current by adding the individual
currents. If this can be done for d.c. circuits we should likewise
be able to do the same thing for a.c. circuits. In the latter case,
1/Z is called the admittance and is represented by Y. Also, the
total admittance of a parallel circuit is

Y_{p} = Y_{1}
+ Y_{2} ... etc.

The value of Y is also measured in mhos, just as resistance, reactance and impedance all are measured in ohms.

Here we remember our forgotten friend, the phase-angle. We know that if we apply an alternating voltage to a circuit having several parallel branches, the absolute values of the branch currents will add up to more than the absolute value of total current unless all the currents happen to be in phase. Since the admittance is a measure of the value of current that would flow in an impedance when one volt is applied, it is necessary to add admittances in the same manner as we add currents in a parallel circuit. The only practical ways of adding a number of currents, voltages or impedances having various phase angles are by drawing scale diagrams or by splitting each one into two components at right angles to each other, adding the two sets of components separately and then combining the two sums by the familiar right-triangle rule, or Pythagorean theorem. One component represents the condition of current in phase with voltage, or 100 per cent energy consumption, and is often called the real component . The other component represents the condition of current and voltage 90 degrees out of phase, or 100 per cent energy storage, and is often called the imaginary component. This component is usually prefixed by the letter j to show that it has been rotated 90 degrees with respect to the real component. The letter j has been given the value of √(-1) and when it occurs in a computation it is treated as a multiplier having this value, as has already been explained in the article mentioned previously. Algebraic numbers having √(-1) as a factor are known as imaginary numbers, which explains the name given the components of voltage, etc., to which the letter j is applied. The method of adding the real and imaginary components of voltage or current also is covered very well by Mr. Noll and therefore only its application to parallel circuits will be discussed here.

In many problems of parallel impedance the mathematical solutions can be simplified by the use of admittances. In this way frequent reference to trigonometric tables is unnecessary, since phase angles no longer are factors in the computations. Practical application of the method is discussed in this article and illustrated with typical examples.

The preceding paragraph implies that we must split up each of our admittances into two components before we can add them. To obtain the components of the admittance we make use of the complex expression for impedance, which means the impedance when split into its components of resistance and reactance. Expressed thusly,

Z = R + jX,

the j means that, if a diagram of the components of impedance is drawn, the reactance, X, will be drawn at right angles to the resistance, R. Then,

since Y = 1/Z, the corresponding admittance would be

In order to get the complex expression out of the denominator we multiply the fraction by

giving

since j^{2} = -1. Now we have

or

However, we know that

R^{2} + X^{2}
= Z^{2}.

Therefore the two parts of the admittance, Y, are R/Z^{2}
and - jX/Z^{2}. If Z happened to be a pure resistance, X
would equal zero, and R/Z^{2} would become equal to R/R^{2}
or 1/R, which is called G, or conductance in d.c. circuits. The
term R/Z^{2} is therefore called the a.c. conductance and
is also represented by the letter G. The term X/Z^{2}, which
is the reactive or imaginary part of the admittance, is known as
the susceptance and is represented by B. Thus,

Y = G -jB,

where G = R/Z^{2} and B = X/Z^{2}. Note that
the phase angle of the admittance is of sign opposite to that of
the corresponding impedance. ^{2}

Fig. 2 shows an impedance diagram at A and the corresponding admittance diagram at B for a coil having 4 ohms resistance and 3 ohms reactance (C). The diagrams are shown in terms of the components of voltage and current, and an applied e.m.f. of 25 volts is assumed, so that both diagrams will be to the same scale.

To find the resultant impedance of a number of impedances in parallel, we add the admittances of the individual branches to obtain the total admittance. The impedance of the combination is then the reciprocal of the total admittance, or

Z = 1/Y_{t}.

The application of this method will be shown by a couple of examples.

For the first example we shall take the tuned circuit of Fig.
3, consisting of a condenser of 400-μμfd. capacity with negligible
resistance, and a 100-microhenry inductance coil having a resistance
of 20 ohms. A circuit with these values will be series resonant
at 795.58 kilocycles, as can be verified by the formula for resonance,

We shall now calculate the impedance of the circuit at this frequency, assuming a voltage to be applied between the points A and B, making it a parallel-resonant circuit. Since the factor ω = 2πf is used in calculating both inductive and capacitive reactances, we start by computing its value as follows:

ω = (2) (3.1416) (795,580) = 5,000,000 (electrical radians).

The impedance of the capacity branch is R_{c} - jX_{c}

R_{c} = 0 and

(There are 1,000,000,000,000 micromicrofarads in one farad). Therefore Z = 0 - j500 ohms. The admittance of this branch is

Y_{c} = G_{c} - jB_{c}

G_{c} = R_{c}/Z_{c}^{2} = 0

Therefore.

Y_{c} = 0
- (- j0.002) = 0 + j0.002 mhos.

Since the impedance in this particular case is a pure reactance
we could have found B directly, since 1/X_{c} = ωC.
This short cut cannot be used, however, if the impedance also has
a resistive component.

The impedance of the inductive branch is

Z_{L} = R_{L}
+jX_{L}

R_{L} = 20 ohms and

X_{L} = ωL
= (5,000,000) (0.0001) henries = 500 ohms.

so

Z_{L} = 20
+ j500 ohms.

The admittance

Y_{L} = G_{L} -jB_{L}

and

The admittance of the parallel circuit is

Y_{ab }= Y_{c} + Y_{L}
= (0 + j0.002) + (0.00008 - j0.001997),

which is added thusly,

0 + 0.00008 + j0.0002 - j0.001997 = 0.00008 + j0.000003 mho.

The impedance then is

The result shows that the inductive and capacitive reactances
in a parallel circuit do not cancel out in the same manner as they
do in a series circuit. This results from the fact that the resistance
in the inductive branch shifts the phase of the current slightly,
so that it is not exactly 180 degrees out of phase with the current
in the capacity branch. The current for a given applied voltage
is equal to E/Z or EY, since Y = 1/Z. Therefore a diagram of the
relative values and phases of currents in various parts of the circuit
can be drawn by using the values of Y to represent the currents,
since current is proportional to Y. Fig. 4 is such a diagram, with
the conductance and total current exaggerated to show the effect
of resistance in the circuit. Actually the frequency at which the
resultant reactance is zero is so close to the frequency at which
X_{L} = X_{c} that for all practical purposes they
can be considered identical except in circuits having lower values
of Q than are ordinarily used in radio work. Such low-Q circuits
are, however, found in television amplifiers and are also likely
to be found in audio-frequency work.

**Fig. 4** - Vector diagram of the currents in the
various branches of the circuit of Fig. 3.

A second example, illustrating a series-parallel circuit, is the resistance-coupled amplifier shown in Fig. 5. An exact calculation for such an amplifier, taking into account all possible current paths, is quite complicated, so that it is customary to reduce the amplifier to a simplified circuit which approximately represents the conditions existing for the frequency of interest. For low frequencies the circuit of Fig. 5 can be represented by Fig. 6, where the applied voltage is the a.c. voltage developed between plate and cathode by a signal, and is equal to μ times the a.c. voltage applied between grid and cathode.

The voltage e_{g2} between grid and cathode of the following
tube is practically equal to the voltage drop across the grid leak,
since the voltage across the cathode bypass condenser is negligible
if the bypass condenser has a fairly large capacity (considering
the drop caused by the applied grid voltage only). This voltage
will, therefore, be a portion of the voltage between A and B and
will then depend upon the frequency as well as the constants of
the circuit. Assuming the frequency of the applied signal to be
53 cycles per second, the reactance of the coupling condenser will
be 200,000 ohms. With an applied signal of 1 volt and an amplification
factor of 20 the amplified a.c. voltage applied to the network of
Fig. 6 will be 20 volts. This voltage will be divided between the
internal resistance of the tube, R_{p}, and the impedance,
Z_{ab}, between points A and B.

The admittance Y_{ab}, will be the sum of the admittances
of the two branches, one being the plate-coupling resistance, R_{1},
and the other consisting of the grid leak, R_{2}, in series
with the coupling condenser. For the first branch

Z_{1} = R_{1}
= 100,000 + j0 ohms and

Y_{1} = 0.00001 - j0 mho.

For the second branch

Z_{2} = R_{2}
- jX_{2} = 250,000 - j200,000 ohms and

The calculations for a high impedance such as this can be performed more conveniently by expressing the impedance in megohms. The corresponding admittance will then be in micromhos. Thus,

and Y_{1} = 10 - j0 micromhos.

Then

Y_{ab} =
Y_{1} + Y_{2} = 10 + j0 + 2.44 + j1.95 + 12.44 +
j1.95 micromhos.

The impedance between points A and B will be

In order to find the voltage e_{g2} we must first find
the voltage between the points A and B. The impedance Z_{ab}
in series with the plate resistance R_{p} forms a voltage
divider. Therefore, the voltage across Z_{ab} will be equal
to

R_{p} is 10,000 ohms resistance or 0.01 + j0 megohms.
Therefore,

Z_{ab} + R_{p} = 0.0884
- j0.0123 megohms.

e_{ab}=(20) (0.876- j0.0165) =17.52- j0.33 volts. The
voltage e_{ab} is the sum of e_{g2} and the voltage
drop across the coupling condenser since they are in series. Therefore,

e_{g2} = (e_{ab})
(R_{g}/Z_{2})

The absolute value of this voltage is

The overall amplification or ratio of voltage at grid No. 2 to that at grid No. 1 will then be 13.6, since one volt was assumed to be applied to grid No. 1. The voltage at grid No.2 will lead the voltage at plate No.1 by an angle whose tangent is

It is not necessary to find any phase angles to determine the
voltage or current anywhere in the circuit. In fact, the relative
phase of any voltage in the circuit can be found by graphical methods
with sufficient accuracy for ordinary purposes. The fact that the
voltages are expressed as complex quantities makes it easy to draw
a scale diagram of the various voltages in the circuit. Such a diagram
is useful for checking the accuracy of the calculations. A diagram
of the voltages in this circuit is shown in Fig. 7, and a diagram
of the relative currents is shown in Fig. 8. The values of alternating
current in microamperes can be found by multiplying the admittance
in micromhos by the applied e.m.f. of 20 volts (I = EY). The voltages
and currents considered in this problem are, of course, only the
alternating parts of the pulsating voltages and currents in the
actual amplifier. The d.c. values have no other effect than to establish
the values of R_{p} and μ and to determine the amplitude
of voltage that can be applied before distortion begins.

In practical work the performance of amplifiers is ordinarily computed by means of graphs or approximate formulas rather than such detailed calculations. The examples given above were used instead of the usual problems concerning miscellaneous collections of inductance and capacity because they are typical of the circuits actually found in radio equipment, and thus should indicate that the methods of calculations illustrated have practical applications.

One desirable feature of using complex notation is that it eliminates the necessity for extracting square roots except for one such operation as the last step in the calculations, when the absolute value of current or voltage is desired, as is usually the case. A vector diagram of all the values in the problem should be drawn as a check on the calculations and to make sure that plus signs have not been slipped in where there should be minus signs or vice versa. Such errors are usually readily apparent as soon as construction of the diagram is attempted. The vector diagram need only be drawn roughly to scale, and may be in terms of voltages, currents, impedances or admittances, whichever is most convenient.

* ·12 Griggs St.. Allston 34, Mass.

1 Noll, "Meet Mister
j," QST, October, 1943, p. 21.

22 B is sometimes defined
as - X / Z^{2}, making Y = G + jB, This of course, is equivalent
to the relation given above.

Posted October 5, 2012