Relay voltage specification
Relay voltage specification
(OP)
I have the manufacturing data on several relays (coil windings, wire gauge, core type, etc.)used in elevator controls. Is there a way to estimate the voltage they were intended to work at?
RE: Relay voltage specification
1. Number of turns
2. Wire gauge
3. Wire/coil resistance for DC relay or impedance for AC relay
4. AC/DC type of relay
5. Wire type and its insulation voltage rating
6. Elevator standards, it is intended for
RE: Relay voltage specification
RE: Relay voltage specification
One example relay is the replacement for the Otis O21A20Y1 which uses an Otis 222CY1 coil. The coil uses 15,310 turns of #39 gauge magnet wire with polyurethane or SFV insulation. The resistance is 3,384 ohms +- 10%. The wire is wound on a bobbin with a 0.773 inch outer diameter core. This sits on a 5/8 inch E magnet iron core. The core length is 1.498 inches and weighs about 0.148 lbs. The main mass that it is pulling is the 1.975x1.250x0.156 armature clapper. The clapper is made of E magnet iron. It is pulling against a 5 oz spring. The relay pulls the clapper a distance of 0.171 inches. It is non-latching.
My best guess is that this is a 135VDC coil but I would like to know for sure.
RE: Relay voltage specification
References:
1. Giacoletto L. J. "Electronics Designers' Handbook," 2nd Edition, McGraw-Hill Book Co., 1977
2. Warrington A. R. van "Protective Relays Their Theory and Practice," Volume I, Chapman & Hall, Ltd., 1971
Relay voltage range estimation:
1. Constraint: AWG B&S Wire Gage 39 has an operating current of Idc=0.01264 Amp listed in Table 3.14, Reference 1. This implies DC voltage (assuming that the relay has a solid iron core suitable for DC magnetization/operation):
Edc=Rdc x Idc=3,384 x 0.01264=42.77VDC
2. Constraint: The DC coil imput parameters Idc and Edc must develop relay operating force greater than 5oz relay spring force.
Elementary formula for the force F, magnetic induction (flux desity), length of wire L, and current I in the coil:
F=B x I x L which has to be properly applied to the relay coil magnetic geometry.
Reference 2 indicates "pull force F" Equation (2.10) on page 2.4 as
F=(2 * (pi) * (N * Idc)**2 * A)/(x**2)
where
F=force
Pi=3.14
N=coil turns
Idc=coil current
A=pole face gap area
x=air gap at the pole center
**2=exponent 2
Take-off appropriate relay parameters and substitute them into the above formula to make sure that the Idc is adequate to develop F > 5 oz