A few months ago, a colleague asked whether he needed a fuse in series with the 5-volt supply on an OEM-supplied printed wiring board. The 5-volt supply to the board was rated about 40 amps. My colleague wanted assurance that the board would not compromise the safety of the product.
I recognize that, in the case of a user-accessible board, some standards require an 8-amp inherent limit, or a 5-amp fuse, or a 240- V A limit, or a 150-watt limit. However, readers of this column will recognize that I address the hazards involved rather than the standards requirements. Standards presume certain hazards already exist in the equipment regardless of whether such hazards actually exist. The design of a product can obviate the hazards from high current or high power by means other than by making such circuits inaccessible. To do so requires that the hazard be identified, e.g., fire, and that the way the hazard arises, i.e., the fire starts, be identified.
Safeguards can then be designed into the product so as to prevent the hazard from arising. Such a process yields a safe product, but not necessarily a certifiable product. Such is the sorry state of affairs of the product safety discipline.
My colleague was dealing with a 5-volt, 40-amp, 200-watt (minimum) source. Two hundred watts is more than enough power to start a fire if the power dissipation is not properly managed and controlled. In terms of whether the safety of the product might be compromised, we would first look at whether this power could be dissipated in such a fashion as to cause a fire.
Let’s review the necessary conditions for an electrically-caused fire.
Electrically-caused fire only occurs under fault or mis-use conditions. (I believe it is obvious that fires in electronic equipment do not occur under normal or normal-use conditions.)
Electrically-caused fire occurs when electrical heating raises the temperature of a fuel material to ignition temperature.
Three elements are crucial to this statement. First, virtually all materials will burn if the temperature is high enough and if heat is continuously applied to the material. (For many materials, the combustion process produces sufficient thermal energy to sustain flaming and burning.) Second, the temperature of the heating element must be greater than the ignition temperature of the fuel material. Third, there must be sufficient electrical energy converted to thermal energy and transmitted to the fuel material to raise its temperature to ignition temperature.
The reason electric heaters don’t ignite and burn is that the ignition temperature of the materials is greater than the temperature produced by the heating element. In addition, some electric heaters are built in such a fashion as to limit transmission of thermal energy to nearby external materials.
Electrical heating only occurs when electrical energy is converted to thermal energy (i.e., power dissipated in some device which produces significant heat).
Since electrically-caused fires occur only under fault conditions, the first job is to identify those parts of the circuit which, under fault conditions, would operate as heating elements and could convert a significant amount of power to thermal energy such that:
- the heating element temperature will be greater than the ignition temperature of nearby fuel materials, and
- there will be sufficient electrical energy converted to thermal energy so as to raise the temperature of candidate fuel material to ignition temperature, and
- there will be sufficient time to transfer the necessary thermal energy to the candidate fuel material.
To get some clues as to what parts could dissipate significant power under fault conditions, we can examine the variables which would cause power in a circuit to increase by a significant amount. Power dissipation is expressed in three ways:
- P = E x I
- P = E x E / R
- P = I x I x R
where
P is power in watts,
E is potential in volts,
I is current in amperes,
R is resistance in ohms.
Now, which parts on the circuit board can change to affect E, I, and R in such a way as to increase power dissipation?
In equations (1) and (2), if we increase E, we will increase P. Since the voltage source is external to the circuit board, no circuit board faults can increase the value of E supplied to the board. So, we need to look further.
With a constant-voltage power supply, if we decrease R, we will increase P. Let’s get some idea of the values of P related to the values of R. (Table 1)
|
P = E2 ÷ R |
|||
|
Power |
Potential | ° | |
| Watts | Volts | E2 | Ohms |
| 12.5 | 5 | 25 | 2 |
| 25 | 5 | 25 | 1 |
| 50 | 5 | 25 | 0.5 |
| 125 | 5 | 25 | 0.2 |
| 160 | 4 | 16 | 0.1 |
| 80 | 2 | 4 | 0.05 |
| 32 | 0.8 | 0.64 | 0.02 |
| 16 | 0.4 | 0.16 |
0.01 |
Table 1
(Note that, if we were dealing with a constant-current source, increasing R would decrease P. However, in practice, the vast majority of circuits are constant-voltage.)
In equation (3), if we increase R, we will increase P. But, since R is in series with the source, if we increase R we will reduce I. Again, let’s look at some values of R and I. (Table 2)
| P = I2 × R | |||
| Power | Current | ° | Resistance |
| Watts | Amperes | I2 | Ohms |
| 12.5 | 2.5 | 6.25 | 2 |
| 25 | 5 | 25 | 1 |
| 50 | 10 | 100 | 0.5 |
| 125 | 25 | 625 | 0.2 |
| 160 | 40 | 1600 | 0.1 |
| 80 | 40 | 1600 | 0.05 |
| 32 | 40 | 1600 | 0.02 |
| 16 | 40 | 1600 | 0.01 |
Table 2
In a constant-voltage circuit, as R increases, I decreases, and, because I is squared in the equation, I dominates the total effect on P. Thus, increasing R does not increase P. Instead, the total effect is just the opposite: as R decreases, P increases!
(Note that if we were dealing with a constant-current source, increasing R would indeed increase P just as predicted by the equation. However, in practice, the vast majority of circuits are constant-voltage.)
These data suggest that if we can identify a part failure in which the part resistance goes down to less than one ohm, then we have a candidate part or circuit for converting electric power to thermal energy. The “ideal” part failure is where the fault resistance is sufficiently low to draw maximum current from the supply.
These data dictate a second parameter for candidate parts: The part must be capable of carrying the fault current for an extended time interval without fusing.
Under normal conditions, we usually ignore the resistances inherent to wires, circuit board conductors, and connector contacts. However, when we are dealing with fault conditions where total circuit resistances are fractions of an ohm, and where currents are ten, a hundred, or even a thousand or more times normal current values, we can no longer ignore conductor resistances and connector contact resistances. So, when we are looking for candidate parts for converting electrical energy to thermal energy, we must now include printed-wiring conductor resistances and connector contact resistances.
Now we’ve got some bounds on what we’re looking for. We’re looking for that single component which is connected between a high-current voltage source and the return for that source (usually ground). The component doesn’t have to dissipate the power, but it does need to carry the high current for an extended time without fusing open. As an example, this could be a bypass capacitor located near the power pin of an IC. Such capacitors can short, resulting in a very low resistance and very high current-carrying capacity — ideal conditions for dissipating power in the fractional-ohm resistance of the conductors to and from the capacitor.
Or, we’re looking for two components in series, one of which is a low resistance under normal conditions. As an example, this could be a decoupling circuit comprised of a low-value series resistor and a capacitor to ground. If the capacitor should short, then excessive power would be dissipated in the resistor. Many metal-film and carbon-film resistors initially decrease in value when subjected to heating as from over-power conditions. Again, we have ideal conditions for dissipating power in the form of thermal energy.
Finally, we’re looking for candidate fuel materials and their proximity to the power dissipating components. The coatings on capacitors and resistors can be fuel, but there’s not a lot of it. Such coatings, when heated to ignition temperature, will burn only for a very short time — less than 30 seconds-before the fuel is consumed. Because of this short burning time, not much thermal energy will be transmitted to other near-by candidate fuels.
For a printed wiring board, the most obvious candidate fuel material is the epoxy of the printed wiring board. And, the quantity of fuel is relatively high. It is in intimate contact with board conductors, and is very near to heat dissipating components mounted on the board.
(At this point, some of you may be saying that your boards are flame-rated by UL and therefore won’t burn. Recall that virtually all materials will burn if the temperature is high enough and if heat is continuously applied to the material. The flame tests in UL 94 and its clones are measures of the time of burning after removal of the source of heat. The UL 94 flame tests do not address what happens to the material in the presence of a high-temperature source of heat! V-rated boards burn very nicely in the presence of a source of heat.)
And, for a printed wiring board, the most obvious candidate power dissipating component is the conductor itself.
Now that we’ve reviewed the necessary conditions for an electrically-caused fire, let’s look at the case for electrically-caused fire in multilayer circuit boards.
Remember, we are looking for means for dissipating power. To dissipate power from a voltage source, we are looking for low-value resistances that occur or become significant under fault conditions. The resistance value must not be so low as to cause the voltage source to go into current-source mode.
There are two broad categories of devices which have low-value resistances which maximize power dissipation from the voltage source. First are board-mounted components such as resistors, semiconductors, and connectors. (Capacitors are not included as a shorted capacitor is very-low resistance and does not itself dissipate power.)
Resistors would be limited to low-value or high-power types which, with a fault elsewhere in the circuit, could be caused to dissipate excessive power. Semiconductors would be diodes, transistors, and power ICs which, with a fault else-where in the circuit, could be caused to turn on in a fashion to continuously dissipate power in the semi-conductor forward resistance.
The second broad category are board conductors. In a multilayer board, the inner layers often are used for power distribution and return (ground). These are usually full copper sheets with holes for vias and interconnects.
Because there is so much copper, the cross-sectional area is high and the resistance is very, very low. So, there is low likelihood that the inner layer sheet conductors will overheat. The interconnects from the holes to these planes often have thermal isolation to make soldering easier. While these interconnects have a small cross-sectional area, their total resistance is not high because they are physically short in length and they are well heat-sunk to the inner planes.
With such a construction, we want to look for the longest power supply and ground-return conductors on the top and bottom of the board. These conductors will have small cross-sectional area for the available current, and sufficient length to have enough resistance to dissipate significant power.
In the case of my colleague’s board, we found such construction with a bypass capacitor at the end of a 5-volt supply conductor on the top side of the board. We shorted the capacitor, and connected the board to a 5-volt, 40¬-ampere power supply, and monitored both the voltage and the current.
In a few minutes, the board turned brown, and then black in the region of the capacitor. A few minutes later we had smoke. A few more minutes and we had glowing, and then burning. For the next hour we had alternating glowing and burning as the glowing and burning followed the board conductor to its origin at the board edge connector. The current ranged between 7 and 20 amperes, dissipating from 35 to 100 watts. Then, as it reached the connector area, the current dropped to less than 4 amps, and the glowing and flaming stopped.
We examined the board in the burned area, and we found the epoxy had been burned away, the glass fiber remained, and the circuit was open at the shorted capacitor! What, then, was sustaining the current?
There are a number of possible explanations, especially when such a fire destroys insulation and could cause additional shorts. But such shorts would need to be robust low-resistance shorts to carry the 7 to 20 amperes without fusing. In addition, the low-impedance short would need to be through the fiberglass of the board. None of these explanations is highly plausible.
The one explanation that agrees with the evidence is that the epoxy becomes conductive as it liquefies. Not only does it become conductive, it is a low-value resistance which dissipates power. This power dissipation melts more epoxy, and the process continues until the epoxy is consumed.
We tested this hypothesis by threading nichrome wire through the holes of a bare multilayer board. We connected the board power supply and ground planes to a 5-volt, 10-ampere power supply, and monitored the voltage and current. We gradually increased the power to the nichrome wire until the epoxy started to turn color and then to liquefy. Initially, the current to the board was zero.
At the moment the epoxy liquefied, the current rapidly increased. In about 10 seconds, the current was the full 10 amperes our supply could provide! A few moments later, the epoxy was glowing.
This board had numerous holes for ICs. The glowing slowly progressed among the holes for six hours before we terminated the test! We had no flame. We attribute the lack of flame to the limited power available from the power supply. The glowing appeared to be between the 5-volt plane and the ground plane in the center two layers of the 4-layer board. Sometimes the glowing was on the top of the board, sometimes on the bottom of the board. Throughout, the power input to the board remained constant.
The point is, printed wiring board epoxy is conductive when it is heated to the temperature at which it liquefies. At that temperature, and if sufficient current is available, its resistance dissipates sufficient power to heat additional epoxy which sustains the production of liquefied epoxy. Depending on the available current, the liquefied epoxy will smoke, glow, or flame.
Next, we must address the question: How can we provide protection against such fires?
To answer this question, we must review the three conditions necessary for the fire to start:
- the heating element temperature will be greater than the ignition temperature of nearby fuel materials, and
- there will be sufficient electrical energy converted to thermal energy so as to raise the temperature of candidate fuel material to ignition temperature, and
- there will be sufficient time to transfer the necessary thermal energy to the candidate fuel material.
We can provide protection by obviating or preventing any one of these three conditions. We’ll look at these in reverse order.
The conventional means for providing protection against electrically caused fire is to control the parameter of time (3) through the use of a fuse or other automatic disconnect in the event of an overcurrent situation.
Since the current involved with this particular fire is proportional to the geometry of the supply and return conductors, a fuse will likely provide protection if the fire evolves between large-area inner layers where the volume of molten epoxy will be large and the current through the epoxy correspondingly large. But the fuse will not provide protection between small-area top or bottom conductors and the inner conductors where the volume of molten epoxy will be small and the current through the epoxy correspondingly small. This is because the differential current between normal mode and fault mode is large for a large-area conductor, and small for a small-area conductor.
Another means for providing protection against electrically caused fire is to control the parameter of conductor resistance (2) through the use of large cross-sectional-area conductors such that the conductors will not overheat with the available fault current.
If we increase the cross-sectional area of supply and return conductors, then we have limited the dissipation of electrical energy in the form of thermal energy.
Lastly, we can heat-sink all power dissipating devices so as to limit their temperature rise (1). On printed wiring boards this is often done by using large area conductors for each of the terminals of resistors and power semiconductors.
What about two-sided and single-sided boards?
I haven’t seen evidence of extensive fires for two-sided and single-sided boards as I have for multilayer boards. There is some rationale for the phenomenon not being prevalent on two-sided and single-sided boards. If we review the process by which the fire arises, it starts with melting the epoxy with some electrical source of heat. If we consider conductors on opposite sides of a two-sided board, then we have at least 1/16-¬inch of material to heat to melting temperature. This takes a lot more heat than the 1/32-inch or less in a multilayer board. Even if this occurs, the supply and return conductors need to be opposite each other, which is not usually the case.
If we consider the case of adjacent conductors on the same side of the board, we have a similar situation. We must heat the epoxy between the two conductors to melting temperature, which, again, takes more heat because of the greater distance and the loss of heat to the air.
Should this occur, the geometry of the conductors likely contributes to a variable distance between conductors and variable resistance of the epoxy — which may reduce the power dissipated in the epoxy
which in turn reduces the power and the heating. The fire may not be sustained.
Well… there is opportunity for more study of power dissipated in heated and liquefied printed wiring board epoxy.
Acknowledgments
I want to acknowledge colleague Joe Thomas, Hewlett Packard, Greely, Colorado, who first demonstrated this burning phenomenon to me, and then repeated the test with identical results.
Thanks to Kevin Cyrus, HP Vancouver, who set up the test that conclusively demonstrated the conductivity of the liquefied epoxy.
Another party presented me with additional evidence of this burning process. Though the party must remain anonymous, I want to acknowledge the contribution and thank the party for providing the evidence.
Mr. Nute is a Life Senior Member of the IEEE, a charter member of the Product Safety Engineering Society (PSES), and a Director of the IEEE PSES Board of Directors. He was technical program chairman of the first 5 PSES annual Symposia and has been a technical presenter at every Symposium. Mr. Nute’s goal as an IEEE PSES Director is to change the product safety environment from being standards-driven to being engineering-driven; to enable the engineering community to design and manufacture a safe product without having to use a product safety standard; to establish safety engineering as a required course within the electrical engineering curricula.
