

Practical Applications of Simple Math
July 1944 QST Article 
July 1944 QST
Table
of Contents
Wax nostalgic about and learn from the history of early electronics. See articles
from ARRL's
QST, published December 1915  present. All copyrights hereby acknowledged.

Many
people were reluctant to approach the theoretical aspect of electronics
as it applied to circuit design and analysis, QST (the ARRL's monthly
publication) included equations and explanations in many of their
project building articles. Occasionally, an article was published
that dealt specifically with how to use simple mathematics. In the
July 1944 edition is the third installation of at least a fourpart
tutorial that covers resistance and reactance, amplifier biasing
oscillators, feedback circuits, etc. I do not have Part I from
the May 1944 edition or Part IV from the August 1944 edition,
but if you want to send me those editions, I'll be glad to scan
and post them (see
Part II here).Practical Applications of Simple Math
Part III  ResistanceCoupled Amplifier Calculations
BY EDWARD M. NOLL.* EXW3FQJ
Fig. 1  Triode resistancecoupled amplifier circuit.

The design of a resistancecoupled amplifier is a relatively simple
operation involving considerably less formula juggling and mental
exertion than computing all the deductions subtracted from your
pay check these days. The information given in previous installments
of this series, plus some added information on the practical use
of vacuumtube characteristic curves, will permit the ready calculation
of all required design values for the resistancecoupled amplifier
shown in Fig. 1.
The characteristic curves for the type
6J5 tube are shown in Fig. 2. The E_{p}/I_{p} curves,
the most common in general use, show the variations in plate current
with changes in plate voltage for various fixed values of grid bias,
the complete set of curves forming a "family" of characteristics
for the particular tube under consideration. These curves represent
static variations in tube potentials and currents when the tube
circuit is not loaded.
When a load is applied, such as the
plate resistor of a resistancecoupled amplifier, an additional
line called the load line must be drawn to represent the dynamic
variations in tube potentials and currents. It is apparent that
the .platecurrent variations through the load resistance cause
a varying voltage drop across the plate resistance, which is actually
a change in plate voltage. Thus, a change in grid potential with
the applied signal does not change the plate current without changing
the plate voltage. In fact, the resultant change in plate voltage,
caused by the variations of plate current through the load resistance,
represents the useful output of the amplifier. Therefore, a load
line representing the plate load resistance (total resistance between
plate and cathode) is drawn on the characteristic curves to show
the actual dynamic changes in tube operation.
Many excellent
articles have been written on the theory of characteristic curves
and load lines. Since this article is aimed at illustrating the
practical application of the curves, theoretical considerations
will be brought into the discussion only when necessary.
Constructing the Load Line
In constructing
the load line on the characteristic curves, the actual resistance
of the load depends upon the circumstances under which the amplifier
is to be operated. Each set of conditions may require a slightly
different value of load resistance for optimum performance. Optimum
values may be chosen, for instance, for maximum possible undistorted
voltage output with a given value of input signal, or for maximum
possible undistorted voltage output with a definite amount of plate
supply voltage. Each of these requirements may necessitate the use
of different values.
As an illustration of the method used
in arriving at the first of these objectives, let us assume that
the maximum peaktopeak signal delivered to the grid of the amplifier
of Fig. 1 from the preceding stage is 8 volts. It is necessary to
place the load line on the curves in such a position as to permit
the signal to swing over the linear portion of the characteristic
curves. Therefore the signal must not swing into the curved portions
at low platecurrent values, nor must it swing into the positive
grid distortion region. In the case of the 6J5 triode curves the
least value of bias that can be employed with an 8volt signal is
4 volts, permitting the signal to swing between 0 and 8 volts.
Bias in excess of 4 volts should not be used because it would result
in an undesirable reduction in gain. As a result, the load line
must be drawn to permit the grid signal to swing over the linear
region, between the 0 and 8volt bias curves.
The actual
load line might be drawn at any one of a number of different slopes
and in each case the plate voltage swing each side of the mean value
would be equal and therefore distortionless. However, we are interested
in obtaining a maximum platevoltage swing with a low value of average
plate current and a minimum variation in plate current. From this
consideration, it is evident that the load line should appear practically
horizontal and well down on the characteristic curves. Since a load
line which approaches a horizontal position represents a high value
of resistance (large change in plate voltage with a small change
in plate current) the resistancecoupled voltage amplifier has a
high value of plate resistance in comparison to a power amplifier,
where we are interested in a large platecurrent variation to develop
power. Thus we find our 6J5 load line for an 8volt signal well
down on the curve; in fact, point B on the8volt curve was chosen
as far down as possible without moving into the region of distortion
as indicated by excessive curvature.
Using the point on
the 8volt curve as one point of the load line, a straightedge
is moved about this point as a pivot until an equal plate voltage
is set off by the swing of the signal on each side of the average
bias value set on the 4volt curve. When this position is found,
a line is drawn along the straightedge which represents the value
of plate resistance which permits maximum nondistorted voltage output.
The value of this resistance is readily calculated by extending
the load line until it crosses the platevoltage and platecurrent
coordinates, as shown in Fig. 2. The slope of the load line, or
the resistance represented by the load line, is equal to the change
in plate voltage divided by the change in plate current.
Maximum Voltage Gain
We are now ready
to consider Some typical problems.
1) What should the total
plate load resistance, R_{p}, be for maximum undistorted
voltage gain in the amplifier of Fig. 1, using the characteristics
shown in Fig. 2?
From Fig. 2 we find that the slope of the
load line is
2) Find the plate powersupply voltage, E_{b}, required.
Since
the maximum plate voltage is applied to the plate only when no plate
current flows through the load, the plate voltage indicated at zero
plate current is the powersupply voltage. The position at which
the load line crosses the zero platecurrent axis is point C, representing
270 volts. Therefore, E_{b}=270 volts.
3) Calculate
the required value of the cathode resistor, R_{k}.
Examination of the curve shows that the average plate current
at our operating bias, point O, is equal to 2.1 ma. Therefore, the
resistance required to develop this amount of bias across the cathode
resistor is
4) Calculate the required value of the plate load resistor,
R_{L}.
Since the total plate resistance includes
the cathodebiasing resistance, the actual value required for the
plate resistor is the total plate resistance minus the value of
the cathode resistor, or
R_{L} = R_{p} 
R_{k} = 77,000  1900 = 75,100 ohms.
5)
Determine the value of the grid resistor, R_{g}.
The value of the grid resistor should be at least four times
greater than the plate load resistor of the previous stage, but
should not exceed the maximum value set by the tube manufacturer
for safe operation of the tube. In the case of the 6J5, the maximum
value set by the manufacturers when using cathode bias is 1 megohm.
In most cases the value used is in the vicinity of 500,000 ohms.
6) Determine the value of the cathode bypass condenser,
C_{k}.
The capacity of the cathode bypass condenser
is set at a value which will pass the lowest frequency to be amplified
with a gain equal to 70.7 percent of the gain over the middle range
of frequencies. (The calculation of capacity values will be elaborated
upon in the next installment. However, it is a basic rule that,
if the reactance of the condenser at the lowest frequency is equal
to the resistor value, the amplifier response will be down 70.7
per cent at this frequency.) Since R_{k} is equal to 1900
ohms, the reactance of C_{k} for a minimum frequency of
60 cycles should be 1905 ohms. The minimum capacity for C_{k}
may then be determined as follows:
7) Determine the value of the coupling condenser, C_{c}.
The coupling condenser, which also causes a loss of low
frequencies because of its reactance, is calculated in like manner
with respect to the grid resistor, or
Fig. 2  Family of platevoltage vs. platecurrent characteristic
curves for the Type 6J5 triode tube. 
8) Determine the. peak platevoltage and platecurrent variations.
By dropping perpendicular lines to the coordinates from
the points A, O, and B in Fig. 2, which represent the average bias
and the extremities of gridsignal swing, the peaktopeak plate
voltage and current can be determined by simple subtraction.
Peaktopeak plate voltage = 175  40 = 135 Peaktopeak plate
current = 3  1.25 = 1.75 ma.
9) Determine the peaktopeak
voltage output of the tube.
Since the platevoltage swing
represents the variations in potential between plate and cathode,
the portion of the variation across the cathode resistor is lost.
The actual voltage output, E_{o}, of the stage is
E_{o }= 131 volts
10) Determine the voltage
gain of the amplifier stage.
Voltage gain is equal to the
output voltage divided by the input voltage.
Maximum Power Output
Let us consider
now the case where it is desired to obtain maximum possible undistorted
output for a selected platesupply voltage. As an example, in the
circuit of Fig. 3 a supply voltage of 200 (E_{b}) is assumed.
Since the maximum value of 200 volts is applied to the 6SJ7 plate
only when no plate current flows, one point on our load line is
certain to be at point A, shown in Fig. 4, where the plate current
is zero and the plate voltage 200. From point A, load lines of various
slopes may originate; the lower the plate load resistance, the steeper
the slope. Since, as in the previous example, we are only interested
in obtaining a large platevoltage variation with a minimum variation
in plate current, the slope of our load line should be as far down
on the curves as possible and still accommodate the complete grid
swing without running into the distortion region. Therefore, two
typical load lines were drawn on the curves shown in Fig. 4. The
load line AB represents a load resistance of 13,000 ohms which provides
for a 5volt grid signal without distortion, while load line AC
represents a load of 110,000 ohms which provides for a 1volt signal.
Loadline AC would be the most common, since the 65J7 is a highgain
pentode which is designed to amplify small input signals to a much
higher level.
Fig. 3  Pentode resistancecoupled amplifier circuit.

Inspection of the curve shows that we are operating the tube at
a negative bias of 4 1/2 volts and that the negative peak of the
grid signal reaches 4 volts. In the case of a triode, such an amplifier
would not be operating under optimum conditions. However, the presence
of the screen and suppressor in the pentode permits the plate voltage
and plate current to swing to very low values without distorting
even on the higherbias curves. Thus we can obtain a large platevoltage
variation at reasonable efficiency if we do not permit the signal
to approach zero on its positive peak. From the information
available we may now proceed to calculate suitable circuit values
and some of the operating conditions: 1) Find the
total plate resistance represented by the load line, AC.
2) Find the proper value for the cathode resistor R_{k}.
Since the bias point, midway between points D and
E which represent the extremities of permissible grid swing without
distortion, is at 4 1/2 volts, the average platecurrent flow is
1 ma. and our average plate voltage is 80 volts. The screen current
is approximately 25 percent of the average plate current and, therefore,
the total current passing through R_{k} is 1.25 ma. In order
to secure a 4 1/2 volt drop, the value of R_{k} is
3) What should be the value of the plate resistor, R_{L}?
R_{L} = R_{p}  R_{k} = 110,000
 3600 = 106,400 ohms.
4) Determine the value of the cathode
condenser, C_{k}.
5) What should be the value of the coupling condenser, C_{c},
when using a 1meqohm grid resistor, R_{g}?
Fig. 4  Platevoltage vs. platecurrent characteristic
curves for the Type 65J7 pentode tube. 
6) The value of the screendropping resistor, R_{s}, is
readily calculated if the screen voltage and screen current are
known. The screen potential must be 100 volts to meet the requirements
of the characteristic curves, which are drawn for a screen potential
of 100 volts. Therefore, the voltage drop required across the series
screen resistor is 200  100 = 100 volts.
7) In order to bypass the screendropping resistor adequately,
the reactance of the bypass condenser, C_{k}, should be
not more than 1/10th the resistance of the screenresistor at the
lowest frequency.
8) If the resistancecoupled amplifier is employed in an audio
system which has three or more stages, it may be necessary to employ
a decoupling network, R_{f}C_{f} to prevent feedback
through the common plate impedance. In this case, the power supply
voltage must be increased by an amount sufficient to compensate
for the voltage drop across R_{f}. The value of R_{f}
often employed is 1/10 of the value of R_{L}.
R_{f}
= (0.1) (106,400) = 10,600 ohms.
9) The condenser C_{f}
bypassing R_{f}, should have a reactance, at the lowest
frequency to be passed, of not more than 10 percent of the resistance
of R_{f}.
10) The new supply voltage would, of necessity, be 200 volts
plus the voltage drop across R_{F}.
E = 200 + (I_{s}
+ I_{p}) R_{f} = 200 + (0.00025 + 0.001) (10,600)
11) The total platevoltage swing as determined by the perpendiculars
of Fig. 4 is 13030 = 150 volts. From the ratio,
,
we know that 97 percent of the output voltage or 97 volts
peaktopeak appears across the plate resistor. Since a 1volt peaktopeak
signal is applied at the grid, the stage gain is 97/1 = 97.
If a different screen voltage were selected the curves would
change somewhat, calling for alterations in the values.
In the next installment, covering the design of a twostage audio
amplifier, an approximate method will be outlined to convert the
curves to a lower screen potential.
Posted 6/11/2013



