July 1944 QST
Wax nostalgic about and learn from the history of early electronics. See articles
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 four-part
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 - Resistance-Coupled Amplifier Calculations
By Edward M. Noll* EX-W3FQJ
Fig. 1 - Triode resistance-coupled
The design of a resistance-coupled 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 vacuum-tube characteristic
curves, will permit the ready calculation of all required design values for the resistance-coupled
amplifier shown in Fig. 1.
The characteristic curves for the type 6J5 tube are shown in Fig. 2. The Ep/Ip
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
When a load is applied, such as the plate resistor of a resistance-coupled 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 .plate-current 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
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 peak-to-peak 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 plate-current 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 8-volt 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 -8-volt 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 plate-voltage 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 resistance-coupled voltage amplifier has a high value
of plate resistance in comparison to a power amplifier, where we are interested in a
large plate-current variation to develop power. Thus we find our 6J5 load line for an
8-volt signal well down on the curve; in fact, point B on the-8-volt curve was chosen
as far down as possible without moving into the region of distortion as indicated by
Using the point on the -8-volt 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 -4-volt 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 plate-voltage
and plate-current 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, Rp, 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 power-supply voltage, Eb, 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 power-supply
voltage. The position at which the load line crosses the zero plate-current axis is point
C, representing 270 volts. Therefore, Eb=270 volts.
3) Calculate the required value of the cathode resistor, Rk.
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, RL.
Since the total plate resistance includes the cathode-biasing resistance, the actual
value required for the plate resistor is the total plate resistance minus the value of
the cathode resistor, or
RL = Rp - Rk = 77,000 - 1900 = 75,100 ohms.
5) Determine the value of the grid resistor, Rg.
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, Ck.
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 Rk is equal to 1900 ohms, the
reactance of Ck for a minimum frequency of 60 cycles should be 1905 ohms.
The minimum capacity for Ck may then be determined as follows:
7) Determine the value of the coupling condenser, Cc.
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 plate-voltage
vs. plate-current characteristic curves for the Type 6J5 triode tube.
8) Determine the. peak plate-voltage and plate-current 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 grid-signal swing, the
peak-to-peak plate voltage and current can be determined by simple subtraction.
Peak-to-peak plate voltage = 175 - 40 = 135 Peak-to-peak plate current = 3 - 1.25
= 1.75 ma.
9) Determine the peak-to-peak voltage output of the tube.
Since the plate-voltage 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, Eo, of the stage is
Eo = 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 plate-supply voltage. As an example, in the circuit of Fig. 3 a
supply voltage of 200 (Eb) 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 plate-voltage 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 5-volt grid
signal without distortion, while load line AC represents a load of 110,000 ohms which
provides for a 1-volt signal. Load-line AC would be the most common, since the 65J7 is
a high-gain pentode which is designed to amplify small input signals to a much higher
Fig. 3 - Pentode resistance-coupled
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 higher-bias
curves. Thus we can obtain a large plate-voltage 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,
2) Find the proper value for the cathode resistor Rk.
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 plate-current
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
Rk is 1.25 ma. In order to secure a 4 1/2 volt drop, the value of Rk
3) What should be the value of the plate resistor, RL?
RL = Rp - Rk = 110,000 - 3600 = 106,400 ohms.
4) Determine the value of the cathode condenser, Ck.
5) What should be the value of the coupling condenser, Cc,
when using a 1-meqohm grid resistor, Rg?
Fig. 4 - Plate-voltage vs. plate-current
characteristic curves for the Type 65J7 pentode tube.
6) The value of the screen-dropping resistor, Rs, 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 screen-dropping resistor adequately, the
reactance of the bypass condenser, Ck, should be not more than 1/10th the
resistance of the screen-resistor at the lowest frequency.
8) If the resistance-coupled amplifier is employed in an audio system
which has three or more stages, it may be necessary to employ a decoupling network, RfCf
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 Rf. The value of Rf often employed is 1/10 of the value
Rf = (0.1) (106,400) = 10,600 ohms.
9) The condenser Cf bypassing Rf, should have
a reactance, at the lowest frequency to be passed, of not more than 10 percent of the
resistance of Rf.
10) The new supply voltage would, of necessity, be 200 volts plus
the voltage drop across RF.
E = 200 + (Is + Ip) Rf = 200 + (0.00025 + 0.001)
11) The total plate-voltage swing as determined by the perpendiculars
of Fig. 4 is 130-30 = 150 volts. From the ratio,
, we know
that 97 percent of the output voltage or 97 volts peak-to-peak appears across the plate
resistor. Since a 1-volt peak-to-peak 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 two-stage audio amplifier, an approximate
method will be outlined to convert the curves to a lower screen potential.
Posted December 4, 2018 (original 6/11/2013)