AC DC Ampere Rating

Follow the equipment manufacturer's ratings information.

This approximation uses the arching point -E.field strength- between the contacts.

Take the ac current as i = I_peak.Cos(ω.t) then the charges those form the electrical field between contacts, will be

Q_ac = 2.∫i (from t=0 to T/4 seconds )

Q_ac = 2.(Ipeak / ω).|Sin(2∏/4) - 0| since ω = 2∏ / T then Q_ac = I_peak coulomb.

Now assume that the dc supply is an unregulated voltage source as a rectifier output, then this charge will be

Q_dcu = 2.(Ipeak / ω).2.|Sin(2∏/4) - 0| = 2.I_peak clmb.

If the dc supply is a regulated - like an accumulator - then the above calculation yields:

Q_dcr = ∏.I_peak clmb.

Result: with a true dc current, the contact rating of the contactor decrease down to one third ( 1/∏) of that ac rating.

I think the relay contact's rating depends also on the contact's voltage .

If you loock at the data sheet of Omron relay part # G5V-1-DC24 you find the contacts rating :

1A at 24 VDC

0.5A at 115 VAC

Where the contact DC current rating is twice the AC rating.

Hello zacky ,

I think you are confused a bit.

1-It is essential to presume that the two systems are at the same power rating with thislike comparisons. I also do.

2- Please note that i used I_peak for calculations, not I_eff.

Now let's play with your example, Omron's specs:

0,5A is the effective current. I_peak = √2 .I_eff = √2 . 0,5 = 0,707 A ≈ 0,7A. (Our sample current is i= 0,7.Cos(ω.t) ).

Since voltage ratings are different we will compare "powers" not currents only.

1A at 24V yields 24W.

0,7A at 115V yields 80W.

24 / 80 = 1 / 3,33 ! This is even more what i said (1/∏). Omron also allocates a bit room for safety...

now okay?

regards

Seems overly complex. With the contacts closed equivalent RMS and DC current values will stress the contacts and conductors in the relay similarly ( P = I^2 * R). As the contacts open the current either peak or DC is only effecting the performance based on the inductance of the conductors and traces in the system. The stray inductance will try to keep the current flowing and can generate very high voltage arcs across the opening contacts. The system voltage also contributes to the arcing. But neglecting the stray inductance (because we do not know what it is) RMS and DC current will effect the contacts similarly. Assuming the inductance is figured in and the driving factor for current rating the DC level could be higher because the peak current for an RMS waveform would have to have been considered in the RMS rating. The only other thing I see is the RMS waveform will create an sinusoidal (sine wave current assumed) heat profile on the contacts and conductors with an average heating effect equal to the DC heating effect, but it will have peak temperatures higher than the constant DC heating effect.

I am not a relay guy but that is pretty basic analysis. Please correct me if I am missing something

Hello Guest, thanks for your reply.

I will try to clear my way of solution (#2).

1- Cause of using I_peak instead of I_eff is just getting rid of (√2) crowd in equations. This is no effect on results. If u want to use I_eff then multiply the currents of ac with a √2, that's all.

2- The kind of load - resistive or inductive- also not important for this comparison. Of course with an inductive load it would be important the self emk effect of inductances and I^2*R thermal effects on contacts. But whatever the conditions are, those will be valid both AC and DC.

Think of a loaded contactor is just opening his contacts. Voltages are voltages and currents are currents. How to differentiate the AC and DC?

This is why i have used an approximation concerning waveform of the energy and have got into account the available charges those will ionize the media. The effect of stray inductances - but load inductance! - could be neglected relative to this ionizing field.

This is not a complete analyzing of arching with a specific circuit, this is only a comparison between ac and dc, under the same power and environmental conditions.

Isn't this what the S. Murali is asking for?

Best regards

For the example of the Omron relay mentioned above , how come the relay contacts depen on the voltage ( 24V or 110V) while the contacts voltage is zero when the contacts are closed.

Dear Guest, i'm surprized with this question..

As long as the contacts are closed the i²*R dissipation is in action. This is no matter provided that the actual current is not bigger than what the spec say.

Arching phenomenon will be question while you just open these contacts!

okay?