January 1956 Popular Electronics
Wax nostalgic about and learn from the history of early electronics. See articles
published October 1954 - April 1985. All copyrights are hereby acknowledged.
Use of load line charts is a fast way of
selecting bias (operating) point and operational ranges for nonlinear devices. Notice that I didn't
specifically say for transistors because this particular article deals with load lines for vacuum tubes.
Almost nobody has any need for tube load line charts anymore, but the skill needed to interpret load
lines for transistors is fundamentally the same as for tubes. Substitute Vce (collector-to-emitter voltage)
for Plate Volts and Ic (collector current) for Plate milliamperes and you have equivalence.
After Class - Using Load Lines
The practical usefulness of the average plate characteristic curves of vacuum tubes as they appear
in tube manuals and reference handbooks is restricted to static conditions in which there is no plate
load or input signal. These curves enable the user to investigate the interrelationships between plate
current (Ip), plate voltage (Ep), and grid voltage (Eg), and to find anyone of the three if the other
two are known. Different tubes have, of course, different "families" of curves.
As soon as a load is inserted in the plate circuit, however, the entire picture changes. Since a
load is necessary in practical designs - which may be a relay coil, a resistor, a choke, or the primary
of a transformer in amplifier applications - it becomes very difficult to predict the behavior of a
given tube unless a load line is constructed on the average plate characteristic graphs and properly
Need for a load line is evident from a consideration of the effects that take place in a circuit
such as that of Fig. 1. Imagine that it is desired to find the bias voltage (Eg) needed to produce a
plate current (lp) of given value in the load resistor (RL). The tube is a 6J5 and the plate
supply voltage is known. The plate current depends upon the supply voltage, the magnitude of RL.,
and the effective d.c. plate resistance of the tube which, in turn, is determined by the amount of bias
applied to its control grid. Since the bias - the factor we want to find - is unknown, it follows that
the plate resistance of the tube is also an unknown quantity. With this missing from the data, the plate
current cannot be stated. Thus, we encounter an impasse.
Fig. 1. Basic circuit of a typical amplifier stage using a 6J5 triode.
The construction of a load line, in one graphical step, sweeps away all the unknowns. It immediately
makes possible the precise determination of the tube's plate resistance, voltage drop from cathode to
plate, and the plate current, by what amounts to simple inspection. It solves problems which would otherwise
demand tedious calculations.
To illustrate the principle which underlies the load line, the following typical circuit constants
may be assumed: Ebb = 240 volts; RL.=20,000 ohms; and the tube is a 6J5. The construction
of the load line is easier to grasp if it is divided into three distinct steps:
(1) Let us first consider the tube as an open circuit; the circuit may be said to be open if the
control grid bias is made so negative that the tube is cut off and no plate current flows. Under such
circumstances, there can be no voltage drop across RL. (voltage drops take place only when
current flows), and the full plate supply voltage must then appear across the tube as its plate voltage,
Ep; i.e., Ep = Ebb. In our example, when Ip =0, Ep = 240 volts. This locates point A on the horizontal
axis. (See Fig. 2.)
(2) Next, imagine that the bias is made so positive that the tube acts as a perfect conductor (short
circuit) having zero internal resistance. In this case, the voltage drop across the tube (its plate
voltage, Ep) would have to be zero while the full plate supply voltage appears as a drop across RL.
With RL. the only resistance left in the circuit, the plate current must now be:
Thus, a second important point, B, on the vertical axis is determined.
(3) Let us stop for an instant and examine the results thus far. Having assumed the two most extreme
conditions - the tube first as an open circuit and then as a short circuit - we have found, in the first
case, a plate voltage of 240 volts across the tube at zero plate current, and a plate current of 12
ma. with zero plate voltage for the second assumption. Since the plate load is resistive, it follows
Ohm's law at all times, so that for any other assumptions which involve a partially conducting tube
between the two extremes, points must be located along a straight line connecting A and B. This is an
outcome of the linearity of a resistor in which current and voltage are always directly proportional
to each other. Thus, points A and B are the terminals of the straight load line shown in Fig. 2, which
is unique for a 6J5 having a 20,000-ohm plate load and a supply voltage of 240 volts.
Fig. 2. A set of average plate characteristics for a type 6J5 tube with a load-line
representing a 20,000·ohm load superimposed.
The examples that follow show how this load line is used.
Problem: Find the grid bias which permits 5 ma. of plate current to flow in a 6J5 having circuit
constants as given. Solution: The load line intersects the 5-ma. line on the -4 volt bias curve. Hence,
a bias of -4 volts is just right if the plate current is to be 5 ma. Moreover, if the intersection is
brought down to the voltage axis (C), it is seen that the plate voltage of the tube is now 140 volts.
This means that the voltage drop across the plate load resistor must be 100 volts. Note how, by mere
inspection, the load line provides information on the magnitude of the plate current, the actual plate
voltage, and the fall of potential along the load resistor.
Problem: If this tube is run at zero bias, what is its actual plate voltage? Solution: Locate the
intersection of the load line with the zero bias curve. This occurs at a plate current of 8 ma. Moving
down along the vertical line at the intersection, we cross the plate voltage axis at 80 volts - the
true plate voltage; the remaining 160 volts is the drop across the load resistor.
Problem: A relay having a 10,000-ohm coil is used as a plate load in series with
a 10,000-ohm resistor. If its pull-in current is 2 ma., what grid bias is required to activate its armature?
Solution: The relay, together with the series resistor, constitutes a plate load of 20,000 ohms for
which this load line has been drawn. The load line crosses the 2-ma. marker halfway between the -8 and
-10 volt bias curves. Thus, a bias voltage of -9 volts will be just right for pull-in. With these circuit
constants, the voltage across the tube is 200 volts and the potential applied to the series combination
is 40 volts.
Load lines are very useful in predicting stage gain on the basis of static curves since they permit
the determination of output voltage swing as compared with input voltage variations. (See After Class,
December, 1955). Using the methods outlined above, simply draw the appropriate load line for the particular
tube and load resistor, and find the voltage drop across the plate load resistor for the two extremes
of signal input. The difference between the output potentials is the output swing which may then be
compared with the input swing to determine the gain.
Answer the following questions for a circuit using the triode section of a 6AQ6 double-diode high-mu
triode having a 300-volt plate supply and a load resistor of 50,000 ohms. Curves for the 6AQ6 are given
to the right in Fig. 3.
Fig. 3. Average plate characteristics of triode section of a 6AQ6 for use in quiz.
1. Find points A and B for the construction of the load line (Fig. 2 indicates which points these
2. What grid bias is required to obtain a 50-volt drop across the load resistor?
3. What plate current in ma. flows during the quiescent period when the drop across the tube is 150
4. What grid bias in volts is needed to bring about the conditions described in question No.3?
4. What Is the voltage drop across the plate load resistor when the tube operates at zero bias?
Load Line Quiz Answers
1. Point A falls on 300 volts on the plate voltage axis; point B is the 6-ma. marker on the plate
current axis. 2. -3 volts. 3.. 3 ma. 4. -0.3 volts, approx. 5. 170 volts, approx.
Posted November 25, 2014