Programmable logic controllers (PLCs) and computer-based controllers are great for handling logic and data-collection tasks for industrial controls. But when it comes to driving motors, heaters, valves, lamps, or power supplies, PLCs don’t output enough current. A better idea is to let PLCs drive plug-in relays.
In addition to switching current, relays have other benefits. A relay does an effective job of electrically isolating a PLC from power mains, transients generated by the load, and effects of a malfunctioning device. A typical PLC output drives the coil in a relay and the relay contacts switch the power loads. There is no electrical connection between the coil and contacts of a relay. Extra relay contacts can be used to report the relay status to the PLC.
Control systems de-signed with plug-in relays are easy to repair when a system is damaged by an external malfunction such as a power surge, lightning strike, or a brown-out. What’s more, the skill level required to replace a plug-in relay is considerably less than the skills required to rewire a damaged PLC.
Some relays also have the advantage of smart operation. These units have status lamps showing power to the coils and mechanical flags to indicate closed contacts. Also included for diagnostic testing are mechanical buttons to operate the output of the relays to test the driven devices. And a lock-out actuator on the mechanical button can hold the relay in the operating position.
PLCs with relays can greatly simplify design, operation, and reliability of a control system when relays are chosen properly. Consider these factors when choosing a relay for use in industrial controls:
- Voltages driving loads are the first concern. The voltage rating of a relay must be greater than or equal to the voltage driving the load. The frequency of the switched voltage is also critical. Because ac current fluctuates from positive to negative crossing through zero, the switched voltage will vary between the maximum voltage and zero. Dc voltage, on the other hand, is always at the maximum value, causing maximum wear on the contacts with every switch. Typically, a relay rated at 240 Vac will be rated for only 24 Vdc.
- The current required depends on the type of load. Most loads don’t draw a constant current. In fact, the current demand of most loads varies somewhat predictably.
It is also important to avoid switching currents that are too small for the relay to operate reliably. Proper operation of a switch relies, to some extent, on the switching of some minimum current. This current is often referred to as a wiping current because it will burn off traces of contaminants that may build up on the relay contacts. The lower limit of current that can be reliably switched is a function of several factors such as contact material, contact geometry, and mechanical sliding of the contact surfaces.
Relays with gold-plated contacts can reliably switch currents as low as 10 mA. Relays with bifurcated (split) contacts also switch lower-level currents in the 10-mA range. Sealed relays, reed relays, and mercury-wetted reed relays are intended for low-level applications. The advantages of gold-plated contacts are that the relay can be used for mixed loads where one set of contacts can switch rated currents and another set of contacts can switch low-level currents. Gold plating protects the contacts while the relay is on the shelf and while low-level currents are switched. The gold disappears from the surface when the full-rated current is switched.
- Resistive loads exhibit no surge at turn-on. Like ideal resistors, they have the same value of resistance all the time. The most common example of a resistive load is a simple heater. If it is specified at 10 A, it can be switched safely with a relay rated at 10 A. Unfortunately, there are very few purely resistive loads. Most are a combination of two or more types.
- Lamp loads have high input surges. The filament of an ordinary incandescent bulb has a high temperature coefficient. When hot, the resistance of a lamp filament is often 20 times the resistance of a cold lamp. Perhaps the most severe load fluctuation is an ordinary incandescent lamp. A common 75-W light bulb draws 0.625 A during normal operation. But when the filament is cold and the lamp is first turned on, the inrush current hits 13 A. This surge only lasts about a tenth of a second but the current must be accounted for if the lamp is turned on and off many times. A relay rated at 10 A resistive can safely switch a lamp that draws 0.625 A when hot.
- Motor loads also exhibit high input surges. An ordinary singlephase, 110-V, 1⁄3-hp synchronous motor normally draws 4.1 A. But at start up or locked rotor, the same motor can draw over 24 A. If the mechanical load is released from the motor and it runs unloaded, the motor can draw 6 A. A relay with a resistive rating of 10 A has been tested by UL and rated for 1⁄3 hp at 120 Vac.
- Capacitive loads exhibit high-current surges at turn-on. A capacitor tries to maintain a constant voltage. Switching a voltage to a capacitor which is at zero volts, the capacitor tries to short out the voltage to maintain the initial zero volts. This high current at turnon can weld contacts shut. Typical capacitive loads include dc power supply outputs and other filtered power sources.
- Inductive loads have a soft turnon, which means the current rises slowly on turn on, but a voltage surges across the contacts when the load is switched off. An inductor tries to maintain a constant current. But when the load is switched off, the inductor tries to maintain the current by increasing the voltage across the contacts. The voltage can increase across the contacts so much that it arcs. Arcing can melt the contacts and degrade them with every occurrence. Typical loads with high inductance include solenoids and electrically operated valves.
Of course, components can be derated. This improves the life of a system by choosing components with more strength than that theoretically required to do the job. The derating equation for relays is:
where Rd = derating factor, Io = actual current, and IR = rated current.
Therefore, a 12-A relay being used at 10 A has a failure rate of 100/144 or about 70% of that of a 10-A relay operating at 10 A.