
Practical Applications of Simple Math
July 1944 QST Article


July 1944 QST
[Table
of Contents]These 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 (if any) are 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).
See all available
vintage QST articles.
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



