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January 1956 Popular Electronics

January 1956 Popular Electronics Table of ContentsPeople old and young enjoy waxing nostalgic about and learning some of the history
of early electronics. Popular Electronics was published from October 1954 through April 1985. All copyrights
are hereby acknowledged. See all articles from |

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 used.

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
(R_{L}). The tube is a 6J5 and the plate supply voltage
is known. The plate current depends upon the supply voltage,
the magnitude of R_{L}., 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.

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; R_{L}.=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
R_{L}. (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 R_{L}.
With R_{L}. the only resistance left in the circuit,
the plate current must now be:

IP =

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.

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.

Quiz

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.

1. Find points A and B for the construction of the load line (Fig. 2 indicates which points these are).

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 volts?

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