How to Apply Pilot Duty Ratings

Issue:

How to Apply Pilot Duty Ratings

Solution/Resolution:

​B300, C300 Pilot Duty Rating Table – AC 50 and 60 hertz

Note 1 – Code designations are defined by industry standards such as UL 508, Industrial Control Equipment. The number following the letter indicates the maximum voltage rating of 600, 300 or 150 volts. The letter indicates the maximum continuous thermal current as follows.

Note 2 – Product qualification tests are performed to obtain pilot duty ratings. The tests are performed at the voltage and current values provided in the table (i.e. 120 vac, make 30 A, break 3 A). Special rules involving thermal current must be considered if an application is operating at levels lower than 120 vac.

Rule A – For voltages below 120 volts, the maximum make current is to be the same as for 120 volts, and the maximum break current is to be obtained by dividing the break volt-Amps by the application voltage, but these currents are not to exceed the thermal continuous test current. (1)

Example: An application is using a general purpose relay to control a solenoid operating at 24 vac. The relay is rated B300 pilot duty, what is the maximum switching current of the relay at 24 vac?

Answer: Applying Rule A the maximum make current is 30 A (the same as for 120 volts), the maximum break current is (360 volt-Amps / 24 vac) = 15 A which is higher than the thermal current, the 5 A thermal current becomes the maximum break current. Therefore with an operating voltage of 24 vac; make current is 30 A, break current is 5 A. As long as the solenoid’s make and break currents fall at or below these levels, the relay contact is suitable for controlling it.

Note 3 – The make current is applied for a very short duration during the product qualification test and is used to simulate inrush current of an inductive load. For Codes A, B, and C, the pilot-duty inrush current (make) is ten times the steady state current value (break). Special rules must be considered if an application is operating at different voltage levels than specified in the table.

Rule B – For maximum ratings at voltages between the maximum design value and 120 volts, the maximum make and break ratings are to be obtained by dividing the volt-amperes rating by the application voltage. (2)

Example: An application is using a general purpose relay with a B300 pilot duty rating to control a solenoid operating at 277 vac, what is the maximum switching current of the relay at 277 vac?

Answer: Applying Rule B the maximum make current is (3600 volt-Amps / 277 vac) = 13 A, the maximum break current is (360 volt-Amps / 277 vac) = 1.3 A. Therefore with an operating voltage of 277 vac; make current is 13 A, break current is 1.3 A. As long as the solenoid’s make and break currents fall at or below these levels, the relay contact is suitable for controlling it.

R300 Pilot Duty Rating Table – DC

Note 4 – Product qualification tests are performed to obtain pilot duty ratings. The tests are performed at the voltage and current values provided in the table (i.e. 125 vdc, make 0.22 A, break 0.22 A). Special rules must be considered if an application is operating at different voltage levels than specified in the table.

Rule C – For voltages at 300 volts or less, the maximum make and break current is to be obtained by dividing the volt-Amp rating by the application voltage, but these currents are not to exceed the thermal continuous test current. (3)

Example: An application is using a general purpose relay to control a solenoid operating at 24 vdc. The relay is rated R300 pilot duty, what is the maximum switching current of the relay at 24 vdc?

Answer: Applying Rule C the maximum make and break current is (28 volt-Amps / 24 vdc) = 1.2 A which is higher than the thermal current, the 1.0 A thermal current becomes the maximum make and break current. Therefore with an operating voltage of 24 vdc; make and break current is 1.0 A. As long as the solenoid’s make and break currents fall at or below these levels, the relay contact is suitable for controlling it.

Example: An application is using a general purpose relay to control a solenoid operating at 48 vdc. The relay is rated R300 pilot duty, what is the maximum switching current of the relay at 48 vdc?

Answer: Applying Rule C the maximum make and break current is (28 volt-Amps / 48 vdc) = 0.6 therefore with an operating voltage of 48 vdc; make and break current is 0.6 A.

Note 5 – NEMA standards provide details on the design of the test load. The DC inductive loads found in control circuits are usually electromagnetically driven relays, contactors and solenoids rated 50 Watts or less. The severity of these loads on the contacts of the control device is determined by the stored energy of the inductor, which is related to the average rate of rise of the current in the inductor or to the charging time of the inductor. It has been empirically determined that inductive loads up to 50 Watts in size usually have a charging time to 95 percent of their full current value of 6 milli-seconds per Watt or less.(4)

(1) This rule is found in the footnote in of UL 508 table rating codes for a-c control-circuit contacts.

(2) This rule is found in the footnote in of UL 508 table rating codes for a-c control-circuit contacts.

(3) This rule is found in the footnote in of UL 508 table rating codes for d-c control-circuit contacts.

(4) NEMA ICS 5-2000, General Standards for Control-Circuit and Pilot Devices