With the aid of a suitable diagram to aid your explanation, show that for a linear, non- saturated magnetic system comprising of a coil carrying some current I, the energy stored in the magnetic system, W, is
W 1 LI 2 2
Now draw a typical electromagnetic simple relay actuator arrangement, showing the soft iron, moving armature, position of the coil, airgap and return spring. Moreover, give an explanation as to how the relay functions.
A particular simple relay has a coil of N = 1200 turns, a cross sectional area of the soft iron, A = 100mm2, and airgap length X = 5mm when the armature is in the open position, and X = 2mm when the armature is in the closed position. If the spring which keeps the relay contacts open has a spring constant of 1Nm, show how the current to close the coil may be calculated, and give the current level. Furthermore, show the calculation which then gives the current at which the relay opens, stating the current level.
Explain why the current levels to close and open the contacts are not equal, and explain how this may be of benefit to the correct operation of the relay.
A transistor switch is used to energise the simple relay from a 48 V supply (Vs). The relay possesses a coil resistance of R = 30 Ω and an inductance of L = 0.06 H as shown electrically in Figure 1. The relay switches on for currents of 0.5 A or greater but does not switch off until the current has fallen below 0.2 A or less. The transistor is controlled by the drive voltage vB and you should assume that the transistor and the diode exhibits the ideal characteristics of instantaneous switching and zero on-state voltage drop. The transistor T is protected from the effects of energy stored in the relay inductance by the diode resistor network D-Rs.
Figure 1 – Relay driver