DC relay problem

[JatTiw]

Hi Every body
I Had a circuit where three components, connected in series (2000-ohm resistor then Indication lamp & then relay coil) across the 125V DC source. These components are rated for 120 V DC. Under this situation the relay coil didn’t get energized because of not sufficient voltage to pick-up the relay coil.
Recently I replaced existing relay with different type of relay having same voltage rating. Now I had the problem, as this relay is getting energized when the circuit is powered-up.
Other data I have is original relay coil wattage was 4.5 W & new coil wattage is 7.8 W.
I want to know why the new relay is picking-up.

 

[analogkid2digitalman]

Hi JatTiw:

Check the relay specs in terms of coil resistance and pull in current, I suspect they are different. Since the 2K ohm and lamp are in series with the coil, there may not be enough current to pull in the contacts. 125V/2000 ohms will limit current through the coil to 63mA (this assumes lamp and relay coil resistance is zero. If the coil turns cannot produce enough force with this current, the contacts will not make. Pull in current/voltage is typically higher than steady state conditions. Hope this helps

 

[jbartos]

Suggestion: A variable resistor, e.g. rheostat, might be used to find out the resistance in series that is appropriate for your circuit specification.

 

[DanDel]

If both relay coils are rated 120VDC, and the old coil wattage is 4.5 and the new is 7.8, then the new coil has less resistance than the old. (1846 ohms compared to 3200 ohms). The difference is causing higher current flows through the circuit and may be causing enough of a magnetic field to close in the relay.
To reproduce the original circuit, you may have to increase the series resistance to compensate.

 

[JatTiw]

HiEvery body
Thanks for all suggestions. I think when ever a current of 63mA passes through this circuit, whole voltage shall drop across 2000 ohm resistor & lamp. At input of coil, there should be only 3-4 volts. Is it possible that this much volts(3-4 Volts) & 63mA shall be able to pick-up this relay?
Other experiment carried out was, I replaced new relay with another relay coil (of resistance 20 ohm) & circuit was not able to pick-up this relay

 

[electricpete]

I don't think you can get 63mA dc through the circuit. You could if the 2000 ohm resistor was the only thing in the circuit

As pointed out above the 4 and 8 w coil (round numbers) have resistance
R = (120v)^2/(watts) ~ 1800 or 3600 ohms

The highest current you can get is using the higher wattage lower resistance coil which is 1800 ohms

I = 120/(1800+2000) = ~ 30 milliamps.

The voltage across coil you can compute by voltage divider:
V = 120*(1800/[1800+2000]) ~ 56v

All above has assumed the lamp has no significant resistance above the 2000ohms.

 

[JatTiw]

Hi Electripete,
You are right. Actually I was trying to say that with 56V & 3mA, is it possible to pick-up the relay rated for 120V & 7.8W

 

[afterhrs]

Jattiw,
I would also pay close attention to the polarity, alot of older dc relays operated the coil with out regard to polarity. That is not the case for the newer generation of dc relays I have replaced lately, The polarity has be correct!

Regards,
Afterhrs

 

[afterhrs]

I would also trace out the positive and negative lines that is energizing this relay and all other componets in this circiut and ensure the polarity is correct all the way back to the power supply.

Regards,
Afterhrs

 

[JatTiw]

Dear Afterhrs
I checked the polarity & any ground fault. Everything is fine.

 

[buzzp]

There is some misinformation here. If you had a 4.5W coil and it did not pull in, changed to 7.8W and it does pull in, then changed to coil with 20 ohms resistance and it did not pull in, assuming the voltage is the same, there is something wrong with your statements or your measurements. The current is what will pull in the coil not the voltage.
Please double check your statements and/or measurements. Also, verify the voltages of the relay coils you used were all the same.

 

[JatTiw]

Hi Buzzp,
The infromation I put is correct. The situation I had is very confussing too. Any-way I am also trying to contact the relay manfacturer (which is GE) to resolve this issue.

 

[OperaHouse]

Is there any special purpose to having the resistor in series. I've done it to reduce current coil but not to this extent. Was the old relay modified so that mechanically the core was never allowed to close, like in the old voltage regulators on generators. Is the indicator neon or does this also put a load on the circuit? Given that a resistor in series with the relay would give less reliable operation, there must be a purpose. A mechanical shock could cause this relay to drop out. Could you indicate what this relay controls. I've seen pull in currents for even the same relay vary more than 40% in the same lot. You may have just gotten lucky.

 

[JatTiw]

Hi OperaHouse
This arrangment is to check "breaker trip circuit healthy" indication. The Indication lamp (Incandescent Type) along with resistance is connected in series with trip coil (20 ohm resiatance). In parallel to trip coil auxiliary relay is connected. These relays shall pick-up only when there is trip signal (from relay etc). Recently auxiliary relay was replaced. The orignal relay (4.5W) was replaced by new 7.8W relay

 

[electricpete]

My 2 cents in response to Buzz' question: yes current determines pickup for a given relay. But when comparing two different relays, all bets are off.

It sounds to me like you need to add more series resistance to keep your coil from picking up when you don't want it to. Maybe relay manufacturer can tell you at what current/voltage pickup is expected. Or you can measure it on the bench. Or if you tell us the part number maybe someone will look it up on the internet.

 

[BJC]

Check the circuit. Soudnd like your running a coil through a GE ET-16 indicating lamp. I would guess there is a wiring error and the lamp is supposed to energize through a contact on the relay. The coil should energize via some other path.

 

[electricpete]

I guess I do have a question about the circuit as described by the original poster. Under what circumstances can the aux relay ever deenergize?

Here's the way I picture it

Vin
|
-------
| |
RL TripContact
| |
R |
| |
-------
| |
TC AuxRelay
| |
------- Ground.

Where:
Input voltage applied between Vin and Ground
RL = Red Light
R = resistor
TripContact - the contact that closes to initiate trippign
TC = Trip Coil
Aux = Aux Relay to sense tripping

Maybe there are some tc b contacts in series with TC.

Is this right or missing something?

 

[electricpete]

I withdraw the question at beginning of previous post. It was a moment of stupidness.

Assuming the drawing is correct:
- Normal operation with TripContact open the aux relay is deenergized since current is limited.
- When TripContact closes the aux relay picks up.

My proposed fix as mentioned is to increase resistance.
I don't understand why BJC suggests there is a miswiring.

 

[JatTiw]

I will re-write the whole question:
+ VE
---------------
| |
RL Trip Contact
| |
LAMP |
| |
-------
| |
Bkr NO |
Contact |
| |
TC Aux.Relay
| |
--------------- -VE



RL-Resistance 2000 ohm
Lamp- Indication Lamp
TC - Trip Coil
Aux. Relay- Orignal relay has 4.5W & new relay has 7.8W coil
(Trip & Aux. relay rated for 120 V DC)

The problem is as Under:

Under Normal conditions the indication lamp is ON (Shows bkr trip circuit is healthy) & aux. Relay & trip relay will not pick-up. In case of fault the trip contact (from protective relay) will provide 125 V DC & the trip relay & aux. Relay will pick-up. The problem is after replacing Aux. Relay with new relay, the new aux. relay is picking-up under normal condition (without trip signal).
I feel to pick-up a relay, a potential difference is required (Which should be rated voltage) & this will cause flow of current. With the circuit I had the 2000-ohm resistor is giving me sufficent voltage drop that the relay should not pick-up. But actually it is picking-up.
I could not understand why it is picking up?

 

[electricpete]

Looks like the same diagram. It still sounds like you need to increase the series resistance to get this relay to work as desired (or go back to the original coil style). What other choice do you have?

With some very simple measurements you can determine the coil pickup current (or voltage), confirm the coil resistance and other component resistances, and select resistance accordingly. Or take a shot in the dark and add another 2000ohm resistor in series and see if it works.