Voltage Variation Compared to Current Inductor

The holding current is easy. In your example, with 24 Vac, the holding current is 250 mA. That means the steady-state resistance is 24 Vac/250 mA = 96 Ohms, and that looks resistive. So when you apply 30 Vac, the holding current will be 30 Vac/96 Ohms = 312.5 mA.

Inrush current is more difficult. The impedance of coils I have used is so resistive that the inductive part didn't matter. In that case, there would be no inrush current. But even if inductive effects dominate, we can calculate the inrush current.

From V = L di/dt

where,

V is applied potential in Volts,

L is coil winding inductance, in Henries,

i = current in Amperes, and

t = time in seconds,

we can see that if L and dt are held constant, and V is increased, then di increases proportionately.

So inrush current should increase in proportion to the increase in driving potential, the same as for holding current, but for different reasons.

if there is o resistors or anything like that in the circuit. Just a simple solenoid powered by AC.

from above deduction of 1#, yoiu can also use this simple formula,

I = V x t / L

but for practice ac relay, situation will be more complex.

I just discussed this topic in a different perspective today---

Big SCRs or soft starts with ammeter readouts and fuses or breakers upstream and their sizing.

This may help understand in rush to anything--

SCRs reduce the voltage during start up, right?--The ammeter pegs beyond the scale--Right?

The voltage at the SCR(and beyond) is reduced--Increasing the amperage.

The fuses or breakers before it are not seeing the amperage--Right?

The voltage drop increases the amperage, but only at the the SCR to the motor.

--I would like to see a current transformer and readout before the SCR, that would be interesting.

I can say that if you increase the voltage to a coil it would see less of an inrush current and use less amps once moved--It's insulation would in question, though.

Or not?

Cheers

Here is one issue of which to be aware with an AC relay. If you apply too little voltage to close the armature, then the inductance of the magnetic circuit stays at the open armature level and you will source a much higher current then would be observed with the armature closed. In fact, it is likely that the coil was not designed to sustain the higher current indefinitely and you can overheat the coil and cause a failure as a result. Therefore, it is important to have wiring adequate to sustain the minimum pull in voltage at the inrush (open armature) current to prevent relay failure and to ensure correct operation. Too long of a run of too small wire is a formula for disaster and to the inexperience the reason may not be readily apparent.

Hi ECM

Inrush current is the instantaneous V/R coil. At worst case this is contact at the Peak of the sine (24*1.414) = 34V/?

This pulse will be very brief and will only be seen with a scope. Then, as noted by RCapper, you will have a current limited by the X_L with the magnetic circuit open. Once the armature is in the operated position X_L will increase thus reducing the holding current.

All of these currents are dependent on voltage and thus, if you increase the voltage, you will increase the current.

I think what you have mixed up is that for the same relay mechanism a higher voltage would require a higher impedance (more turns) and so less current would flow but with the same magnetic result since the field is the product of amp * turns

Kind Regards

Chas

Yes, I mixed up the ohms law in my original post. Sorry about that. I meant to say that as the voltage is increased the current would also increase as suggested by ohms law. I'm not sure if I was thinking of power (watts) or what.

Thank you everyone for your replies.

well, your spirit would be deserved respect. I admire it. a few people could do it. Another side, you are not completely wrong. Farady law is alway basic to analysis ac circuit. Just like ohm law, the higher voltage, the large the current, no problem. AC relay or contactor or mageticelectric iron is not an ideal solenoid. the L value changes by distance of gas gap, the long the gap the small L value will get.this has been already shown by rcapper. from your deduced, the large current will be inrushed. sometimes we can calculate the u value simpely as this = Lh/lg where Lh is iron core loop length, lg is gap distance. so that the L will inverse rate to distance gap. a veriable at initial. In fact the AC relay constructure is more complex than DC one. a short circuit loop exists in the iron core to reduce noise and keep reliable. this of cuae will absorb more current at all procedure, just like induct single ac motor does.

inrush current not only depends on amplitude of voltage, but input initial phase as well. if a distribution cap is taken account, so much complex the nonlinear equation will be. thats why once there would be a gap bewteen load and it, will be burned soon during work.

An AC relay is simply DC coil and some form of rectification (full or half wave). Without rectification, the failsafe (not latching assumed) relay contacts would just chatter open and closed at the frequency of the AC source. An inductor has no inrush (current actually lags until the magnetic field is established), and with the size relay coil in your example (probably less than 100 mH) the delay would be difficult to measure. The initial current will slowly decrease until the coil reaches a steady state temperature (a minute or so - depends on the structure and mounting). This is due to the fact that the increased temperature of the coil magnet wire will increase the wire's resistance (this can be calculated). As a rule of thumb, a relay should operate at about 75% of it's rated room temperature voltage so that it will drop out and pull back in at it's highest rated temperature with it's rated voltage continuously applied.

Unless the rectification is capactive (very rare because it is not efficient), the higher or lower voltage applied would generally follow Ohms Law, but not exactly because of the typical .7 - 1.5 voltage drop on the rectifiers and the heating effect on the coil resistance. Relays can usually handle a 10% overvoltage without damage. With an overvoltage applied, the steady state current will be a little lower than Ohm's Law would imply beacuse the steady state temperature will be a little higher. A 25% overvoltage (30V on a 24V rated relay) will usually overheat and burn up the coil windings, depending on the magnet wire insulation rating. Inversely, applying a lower voltage would result in a slightly higher steady state current than what Ohm's Law would imply. If you decrease the voltage applied much less than 18V, at some point the relay will not pull in, but it will continue to draw current generally following Ohm's Law.

Dear Guest ,

You have so many errors in your appreciation of the real world of relays. I don't know where you have gained your data but it is very erroneous.

1/ AC relays are AC relays, there is no rectification,

2/ The relay will not chatter at the frequency for which it was designed.

3/ At the moment of connection there is no magnetic field and the wire behaves as a resistance. It goes against what I learned at college 40years ago but it is so.

4/ The current in the coil is limited not only by the coil resistance but also by its reactance, i.e. its impedance. Before the armature finishes its travel or the slug bottoms out (in the case of a solenoid) the coil reactance will be lower than when the magnetic circuit is closed and so will draw more current. As stated by RCapper, if the armature is prevented from closing it frequently causes the relay to overheat and generally result in the relay failing due to the deformation of the plastic coil former.

5/ while I cannot argue with your comment on the decrease in current due to the positive temperature coefficient of copper I would suggest that at approx 0.004 and a likely temperature increase of less than 20ºC its effect is usually minimal.

5/ Your comments on the percentages of over and under voltages permissible are highly speculative. Look at any manufactures catalogue and you will see that there are a very wide diversity of operating ranges and duty cycles from 100% to 10%.

Please log on if you wish to discuss this further

Regards

Chas