Why resistive loads are necessary.

To fully test a high voltage power supply, it is necessary to draw current from the supply. This current may be the maximum rated current in which case the supply will output its maximum power, or it may be a portion of the rated current. In either case, to draw current it is necessary to connect the supply to a resistive load. However, a load is not an off-the-shelf item. This article will discuss how to design and build a high voltage resistive load.

A resistive load may be a useful device to have for other reasons.  In many applications, the actual load that a supply will see is not a purely resistive load. It may also contain inductive and/or capacitive elements. Occasionally, the connection to this type of a load can adversely affect the performance of the power supply. To see if this is the case, it is desirable to be able to connect the supply to a purely resistive load. If the performance is then normal with a resistive load, this test may indicate that the problem is being caused by the interaction of the actual load with the power supply.

How to Design A Resistive Load

The design of a high voltage resistive load is straightforward, but not necessarily a simple task. Although resistors are manufactured with voltage ratings in the kilovolts, they are expensive and not readily available. In addition, they may not be capable of dissipating the available power without damage. A simpler and less expensive alternative is to use a series string of resistors. Each series string consists of lower resistance, wattage, and voltage capability. Each series string is physically arranged and mounted in such a way as to avoid any paths for voltage breakdown while also allowing for the proper dissipation of heat.

Voltage Breakdown

The following curve can be used to estimate voltage breakdown levels as a function of the spacing between pointed electrodes under standard atmospheric conditions. Air contaminants, such as dust and fibers, altitude, and high humidity can significantly lower these breakdown values. These voltage breakdown levels can be increased significantly if broad and smoothly rounded electrodes are used instead of pointed electrodes. Such rounded electrodes avoid the ionization of the surrounding air and subsequent corona discharge or arcing.

Normal design precautions are adequate for resistive loads that will be used with power supplies up to approximately 10 kW. However, considerations of voltage breakdown, including the possibility of corona leakage and discharge, become increasingly important above 10 kW.

Figure 1

Our first example of the design of a resistive load is for a relatively low voltage application where voltage breakdown is not a problem but power dissipation is definitely a problem. Later we will provide examples where voltage breakdown also must be considered.

Consider the following design equations, where:

V = maximum voltage applied to load.
I = maximum current through load.
R = full load resistance = V/I.
P = power dissipated in R = V x I.

If:

Pn = power dissipated in each individual load resistor.

Then:

n = number of series load resistors required = P/Pn.
Rn = resistance of each series load resistor = R/n.
Vn = voltage drop across each series load resistor = V/n.

For example, to design a resistive load for the following conditions:

V = 10 kV
I = 300 mA
Pn = 25 W

Then:

R = V/I = 10 kV/300 mA = 33.3 k ohm.
P = V x I = 10kV x 300 mA = 3 kW.
n = P/Pn = 3 kW/25 = 120.
Rn = R/n = 33.3 k ohm/120 = 277.5.
Vn = V/n = 10 kV/120 = 83.3 V.

To construct a resistive load using standard value resistors and to provide approximately a two-to-one margin of wattage safety for each resistor, we can round off the results to a series string of 134 resistors, each 250 ohm, 50 W in value. The resulting load will draw 298 mA at 10 kV, with each resistor dissipating 22 W. Applying Kirchhoff's Law, Rp = R1 x R2 / (R1 + R2), this resistive load could also be constructed from a series string of 134 assemblies, each consisting of two paralleled 500 ohm, 25 W resistors.

Figure 2

A situation may exist where a resistive load is required to develop two or more current levels from the same supply. In the example above, consider a load design that will produce currents of either 300 mA or 150 mA at 10 kV.  Obviously, increasing the series string to 268 individual 250 ohm, 50 W loads will meet this requirement. Connecting the 10 kV supply to the top of this string will reduce the current to 10 kV/67 k ohm or 149 mA and the power drawn from the supply to 1.49 kW. Connecting the 10 kV supply to the mid-point of the string will produce the same results as in the original example. However, because of the lowered power demands on the top of the string, we can reduce both the number and wattage requirements on these upper resistors.

Considering only the top portion of the string:

V = 5 kV (only 1/2 of the 10 kV appears across this section)
I = 149 mA

Then:

R = 5 kV/149 mA = 33.6 k ohm
P = 5 kV x 149 mA= 745 W

If we dissipate 10 W in each of the resistors in this upper section, which is possible because the total power has now been reduced to 1.5 kW, then:

n = 745 W/10 W = 75
Rn = 33.6 k ohm/75 = 448 ohm
Vn = 5 kV/75 = 66.6 V

Again, to provide a power safety margin, we can round this upper section off to 67 series resistors, each 500 ohm, 25 W, to produce the following new resistive load.

Figure 3

As an example of a design where voltage breakdown also must be considered, assume the following conditions:

V = 100 kV
I = 10 mA
Pn = 5 W

Then:

R = 100 kV/10 mA = 10 M ohm
P = 100 kV x l0 mA = l kW
n = 1 kW/5 W = 200
Kn = 10 M ohm/200 = 50 k ohm
Vn = 100 kV/200 =  500 V

If we again provide a two-to-one wattage safety margin, the resulting load will consist of a series of 200 resistors, where each resistor is 50 k ohm, 10 W in value. The resulting load will draw 10 mA at 100 kV with each resistor dissipating 5 W. The voltage drop across each resistor is a safe 500 V. However, because the applied voltage is 100 kV, the problem of possible voltage breakdown must be considered for this load. We will discuss this problem in the following section that addresses the question of how to construct a resistive load.

Figure 4

Voltage breakdown considerations can arise in other ways. Consider the previous example and assume that we wish to construct a load for the same 100 kV supply but with the maximum current reduced to 1 mA. The new conditions are:

V = 100 kV
I = 1 mA
Pn = 5 W

Then:

R = 100 kV/1 mA = 100 M ohm
P = 100 kV x 1 mA = 100W
n = 100 W/5 W = 20
Rn = 100 M ohm/20 = 5 M ohm
Vn = 100 kV/20 = 5 kV

Depending on the type of resistors selected, the 5 kV drop across each series resistor could be excessive. A typical resistor may have, for example, a maximum voltage rating of 1 kV. If this is the case, we must increase the number of resistors from 20 to 100 to limit the drop across each resistor to 1 kV. Because of the increased number of resistors, both the resistance and power dissipated across each resistor are reduced by a factor of 1/5. The resulting load can then be constructed from a series string of 100 resistors, each 1 M ohm, 2 W in value.

Figure 5

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