Resistance in an AC relay

Discussion in 'General Electronics Chat' started by smellySamsquanch, May 9, 2018.

1. smellySamsquanch Thread Starter New Member

May 9, 2018
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0
Hello,
It has been a while since linear electric circuits and i think that my brain is missing some theory! I am looking up a mechanical relay's data sheet. Here are its specs:
120VAC operating coil voltage
3900 ohm coil impedance
Rated coil power of 1.35VA

https://www.mouser.com/datasheet/2/418/NG_DS_1308242_K10_0413-918463.pdf

I am unsure how the impedance calculates into the rated coil power. V^2/R does not compute... or my brain does not compute. I hook this relay right up to the mains supply and it operates properly, but wouldn't that be about 3.1A?

Any quick help would be appreciated!

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• NG_DS_1308242_K10_0413-918463 Relay.pdf
File size:
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Jul 18, 2013
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Measure the actual current when operating on 120vac
Max.

3. crutschow Expert

Mar 14, 2008
20,838
5,922
You are confusing AC impedance with DC resistance (the coil DC resistance, not impedance, is 3900Ω).
Thus the coil AC current is 1.35VA/120V = 11.25mA.
The coil impedance (due to the vector sum of the coil inductive reactance and resistance) is then 120V/11.25mA = 10.7kΩ (nearly 3 times the DC resistance).

-live wire- likes this.
4. smellySamsquanch Thread Starter New Member

May 9, 2018
11
0
Hello Sir,
Thanks for the prompt response! It is not registering at all on my multimeter. It is a FLUKE that can read down to 1mA. The fuse is good still as well.

5. smellySamsquanch Thread Starter New Member

May 9, 2018
11
0
Ok, thanks for the help! I need to focus more on its inductance then?

Jul 18, 2013
16,853
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The specs for the AC versions can be misleading, you are better off to go with the rated coil power value.
Compare it to the DC version for 110vdc!
Max.

7. smellySamsquanch Thread Starter New Member

May 9, 2018
11
0
Thank you very much for your help Max!

8. MrSoftware Senior Member

Oct 29, 2013
1,230
366
Make sure you have it set for AC current, and make sure the leads are plugged into the correct holes on the meter.

9. smellySamsquanch Thread Starter New Member

May 9, 2018
11
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Yep, sure did that. Didn't want to sound like an idiot around you gentlemen!

May 9, 2018
11
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11. crutschow Expert

Mar 14, 2008
20,838
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Then if you have the current terminals of the meter connected in series with the relay then you should measure some AC current.
What AC current range are you using?

12. crutschow Expert

Mar 14, 2008
20,838
5,922
You need to include the inductance, as the coil's inductive reactance is the majority of the coil AC impedance.
Are you familiar with inductive reactance and AC impedance calculations?

13. smellySamsquanch Thread Starter New Member

May 9, 2018
11
0
It has been a little while since i've gone thru the paces (and i never used it afterwards). I am very interested if you would like to throw some knowledge at me. I will try to follow along. Ive been digging around in my notes but am still confused. Is there a formula you could point me to? I noticed the reactance is the derivative of current W/R/T Time? I also suppose it transfers into the phasor realm somehow. but that's all i can remember at the moment.

Thanks.

P.S. This is just one of those things I stumble across while working that doesn't really impact my ability to finish the project, but not knowing the correct answer is killing me!

14. thespackler New Member

Apr 20, 2018
1
2
You're dealing with reactive power here because it's AC. And because it's a coil the total impedance will be primarily inductive reactance and a bit of ohmic resistance from the coil itself. The formula for inductive reactance is XL = 2πfL where f is the frequency (60 Hz in your case) and L is the inductance of the coil, which you may find on the data sheet. Here's an article that gives some more detail on AC power and reactance: https://circuitcrush.com/electric-power-primer-2/

Feb 20, 2016
2,274
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And just to make things a bit interesting, the inductance will be less when the relay is open then operated
AC can be fun!

strantor likes this.
16. strantor AAC Fanatic!

Oct 3, 2010
4,988
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Yeah, pull the plunger out and apply power. Bet you get an amp reading then!
(Just kidding, don't do that)

17. smellySamsquanch Thread Starter New Member

May 9, 2018
11
0
Thanks everybody for your assistance! Very knowledgeable group!
I am just going to sum up things below so if anybody reads this they can get straight to it.

Problem: Making my tiny brain understand how the given values on a relay's datasheet relate to one another.
Given:
120VAC @ 60Hz (possibly need to change this to Vrms = 84.9Vrms Ill let someone else chime on this)
Coil Resistance = 3900 ohms
Rated Coil Power = 1.35 VA (I am thinking this is derived with Irms and Vrms since it is VA, but again someone else should chime)

Current = V/I = 1.35VA/120VAC = .011A
Total Impedance = Z = V/I = 120VAC/.01125A = 1.07E4 Ohms
Reactance = Xl = Sqrt(Z^2 - R^2) = Sqrt(10666.7^2 - 3900^2) = 9.93E3 Ohms
If : Xl = 2pifL
Inductance = L = Xl/(2pif) = 9928.13063/(2*pi*60Hz) = 26.3H

Notes:
I was mistakenly treating the relay's coil as pure resistance. It needs to be calculated as a resistive and inductive series circuit. I am still confused as to if i should be using the rms voltage and current. That would seem more propper. I would also like to be able to wrap it all back together in full circle of equations. I am thinking power was supposed to be P=Vrms*Irms*cos(60). This does not wrap back around with their supplied values however. May just be coincidence, but when you calculate 120/coil resistance, it is dangerously close to the above power calc.

Jul 18, 2013
16,853
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The simplest away around it is to use DC coil relays!
Apart from the instant at switch on, the DC are the more efficient and less likely to fail if the armature does not seal completely.
A simple bridge is all that is needed for a supply.
Max.

19. smellySamsquanch Thread Starter New Member

May 9, 2018
11
0
Agreed! This is just what i had lying around.

jeff

20. crutschow Expert

Mar 14, 2008
20,838
5,922
No.
120Vac is RMS already, not peak voltage.
Where did you get 60?
That should be the phase angle between voltage and current, not frequency.

A simple way to calculate the coil power here is I²R, where I is the current and R is the coil resistance.
Thus, the (real) relay power used would be 0.011² * 3900 = 0.472W.

Last edited: May 10, 2018