July 1944 QST
Table
of Contents
Wax nostalgic about and learn from the history of early electronics. See articles
from
QST, published December 1915  present (visit ARRL
for info). All copyrights hereby acknowledged.

Many people were (and still are) reluctant to approach the
theoretical aspect of electronics as it applied to circuit design and analysis.
QST (the ARRL's monthly publication) often 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. This "Practical
Applications of Simple Math" piece in the July 1944 edition of QST
is the third installation of at least a fourpart tutorial covering 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 September 7, 2023 (updated from original
post on 6/11/2013)
