Voltage Drop across size five starter M coil @120V

Isaiah

Senior Member
I am concerned about the voltage drop on the "M" coil of a size 5 contactor when it energizes for a 125Hp, 460VAC, 3Ph motor (156A). The CPT is 480-120V, rated 300VA - not sure this is enough to close the contactor at the 120V level and thinking I may need to move it to the primary side of the CPT @ 480V, L1- L2.
Does anyone know how to calculate this at 120VAC?
 

ptonsparky

Senior Member
I am concerned about the voltage drop on the "M" coil of a size 5 contactor when it energizes for a 125Hp, 460VAC, 3Ph motor (156A). The CPT is 480-120V, rated 300VA - not sure this is enough to close the contactor at the 120V level and thinking I may need to move it to the primary side of the CPT @ 480V, L1- L2.
Does anyone know how to calculate this at 120VAC?
Your supplier should be able to tell you what the pull in current of that coil is. I would think 2.5 amps at 120v should be adequate.
 

Jraef

Moderator
Staff member
Your supplier should be able to tell you what the pull in current of that coil is. I would think 2.5 amps at 120v should be adequate.
Correct. The pull-in VA varies greatly from one mfr to the next, it depends on the magnet design, stroke length of the armature, spring strength etc. etc. If you know the make and model of the starter, that is data that can usually be looked up on-line now. So for example, on an A-B Bulletin 509 starter the coil inrush is 1490VA, on a Square D 8536 starter it is 2970VA for the same NEMA size 5. That makes a big difference in the size of CPT you need.
 

kwired

Electron manager
Another consideration is how much of a drop will the 480 volt supply have when the motor is starting, and whether that will impact the contactor coil, or any other items in the control circuit.
 

Ragin Cajun

Senior Member
If you are worried about the voltage drop over a long control circuit, simply add an aux relay that picks up the big M coil, have done this often.
 

topgone

Senior Member
I am concerned about the voltage drop on the "M" coil of a size 5 contactor when it energizes for a 125Hp, 460VAC, 3Ph motor (156A). The CPT is 480-120V, rated 300VA - not sure this is enough to close the contactor at the 120V level and thinking I may need to move it to the primary side of the CPT @ 480V, L1- L2.
Does anyone know how to calculate this at 120VAC?
My notes tell me (re NEMA Heavy Duty Contactors):
1) Sealed VA = 6.7 VA
2) Inrush VA = 590 VA
3) Drop out = 60% of rated voltage
4) Rated V = 110 - 127V
5) Frequency = 50/60 Hz.
Depending on your CPT, if you have a 300 VA, that thing can possibly supply about 2,000 VA on inrush with secondary output voltage dropping to 95% of rated.
 

Jraef

Moderator
Staff member
My notes tell me (re NEMA Heavy Duty Contactors):
1) Sealed VA = 6.7 VA
2) Inrush VA = 590 VA
3) Drop out = 60% of rated voltage
4) Rated V = 110 - 127V
5) Frequency = 50/60 Hz.
Depending on your CPT, if you have a 300 VA, that thing can possibly supply about 2,000 VA on inrush with secondary output voltage dropping to 95% of rated.
That's only valid for whatever brand of contactor you are looking at. They are all different, i.e. what I posted above.
 

Isaiah

Senior Member
My notes tell me (re NEMA Heavy Duty Contactors):
1) Sealed VA = 6.7 VA
2) Inrush VA = 590 VA
3) Drop out = 60% of rated voltage
4) Rated V = 110 - 127V
5) Frequency = 50/60 Hz.
Depending on your CPT, if you have a 300 VA, that thing can possibly supply about 2,000 VA on inrush with secondary output voltage dropping to 95% of rated.

I've contacted the vendor and he has stated M coil draws 1158VA on pickup and the size 5 combo-starter comes with a 300VA CPT.
The control circuit runs from the MCC to a 'Start-Stop' push-button hand-station located nearly 1000 feet away at the motor. Standard wiring for this circuitry is #14AWG,...using an online calculator, (1158VA/120V = 9.65A) amounts to 49VD, which leaves only 71V. I feel this is excessive and believe I need an auxiliary relay as mentioned above and energize the M coil on the primary side of the CPT at 480V....
Your thoughts?
 

ptonsparky

Senior Member
IIRC we had several interposing relays involved when we changed the control of a 60s era size 5 with 480v coil, to PLC control. Could we have done with less, Possibly. Everything is gone now so it's moot.
 

Jraef

Moderator
Staff member
Yes, an interposing relay is the answer. If it were me though, I would change the contactor coil to 480V and use a relay with 600V rated contacts, that way the CPT is never burdened by the contactor inrush at all.
 

myspark

Senior Member
I am concerned about the voltage drop on the "M" coil of a size 5 contactor when it energizes for a 125Hp, 460VAC, 3Ph motor (156A). The CPT is 480-120V, rated 300VA - not sure this is enough to close the contactor at the 120V level and thinking I may need to move it to the primary side of the CPT @ 480V, L1- L2.
Does anyone know how to calculate this at 120VAC?
Very important concern, however, it hasn’t been addressed despite the responses.

Since you are asking how to calculate the current draw, perhaps you can go down to basics.
If doubts still persist the following will at least something to work on.

Ohm's Law is your friend.

You know the voltage. Put the VA rating aside for now, since that is what you are not sure (ie.doubtful) of its size.

Take the resistance of the magnet coil.

Ohm's Law:

E=IR

I= E/R

Value of I based on nameplate is 300 VA divided by 120 vac. = 300/120= 2.5A
Your coil resistance based on reading from your meter is X

From what are known:
R=120/I
2.5=120/R

Therefore:

R=120/2.5=48 Ohms

Any value lower than 48 ohms (from your meter) would require a larger transformer VA rating. You can increase the VA as a safety factor.

Easy peazzy :) Physics can sometimes save you from falling into a deep precipice. How’s that for a metaphor. :).

Enlarging the control transformer is easier than adding those relays. . . and it also keeps the original concept.

Purists will argue:

But. we are talking about AC.

Well, you won’t be far off if you use the example above.
 

Isaiah

Senior Member
Very important concern, however, it hasn’t been addressed despite the responses.

Since you are asking how to calculate the current draw, perhaps you can go down to basics.
If doubts still persist the following will at least something to work on.

Ohm's Law is your friend.

You know the voltage. Put the VA rating aside for now, since that is what you are not sure (ie.doubtful) of its size.

Take the resistance of the magnet coil.

Ohm's Law:

E=IR

I= E/R

Value of I based on nameplate is 300 VA divided by 120 vac. = 300/120= 2.5A
Your coil resistance based on reading from your meter is X

From what are known:
R=120/I
2.5=120/R

Therefore:

R=120/2.5=48 Ohms

Any value lower than 48 ohms (from your meter) would require a larger transformer VA rating. You can increase the VA as a safety factor.

Easy peazzy :) Physics can sometimes save you from falling into a deep precipice. How’s that for a metaphor. :).

Enlarging the control transformer is easier than adding those relays. . . and it also keeps the original concept.

Purists will argue:

But. we are talking about AC.

Well, you won’t be far off if you use the example above.
Great post - I will definitely keep this in mind. thanks.
 

kwired

Electron manager
Yes, an interposing relay is the answer. If it were me though, I would change the contactor coil to 480V and use a relay with 600V rated contacts, that way the CPT is never burdened by the contactor inrush at all.
That is typical on irrigation wells around here. Pump panel with 480 volt coil. "Hand" operation directly switched in the 480 volt control circuit, "Auto" operation driven by the 120 volt center pivot controls to an interposing relay to switch the 480 volt pump control circuit. Sizes 3 and 4 contactors are pretty common on these wells. Maybe an occasional size 5, but mostly on older installs. New installs larger than 100 HP POCO's typically want soft start, reduced voltage starting, etc. I have only run into a few that were over 100HP.
 

Isaiah

Senior Member
Very important concern, however, it hasn’t been addressed despite the responses.

Since you are asking how to calculate the current draw, perhaps you can go down to basics.
If doubts still persist the following will at least something to work on.

Ohm's Law is your friend.

You know the voltage. Put the VA rating aside for now, since that is what you are not sure (ie.doubtful) of its size.

Take the resistance of the magnet coil.

Ohm's Law:

E=IR

I= E/R

Value of I based on nameplate is 300 VA divided by 120 vac. = 300/120= 2.5A
Your coil resistance based on reading from your meter is X

From what are known:
R=120/I
2.5=120/R

Therefore:

R=120/2.5=48 Ohms

Any value lower than 48 ohms (from your meter) would require a larger transformer VA rating. You can increase the VA as a safety factor.

Easy peazzy :) Physics can sometimes save you from falling into a deep precipice. How’s that for a metaphor. :).

Enlarging the control transformer is easier than adding those relays. . . and it also keeps the original concept.

Purists will argue:

But. we are talking about AC.

Well, you won’t be far off if you use the example above.


[/QUOTE
Myspark, did you mean to say "Any value HIGHER than 48 ohms (from your meter) would require a larger transformer VA rating"?
Thanks, Isaiah
 

topgone

Senior Member
Very important concern, however, it hasn’t been addressed despite the responses.

Since you are asking how to calculate the current draw, perhaps you can go down to basics.
If doubts still persist the following will at least something to work on.

Ohm's Law is your friend.

You know the voltage. Put the VA rating aside for now, since that is what you are not sure (ie.doubtful) of its size.

Take the resistance of the magnet coil.

Ohm's Law:

E=IR

I= E/R

Value of I based on nameplate is 300 VA divided by 120 vac. = 300/120= 2.5A
Your coil resistance based on reading from your meter is X

From what are known:
R=120/I
2.5=120/R

Therefore:

R=120/2.5=48 Ohms

Any value lower than 48 ohms (from your meter) would require a larger transformer VA rating. You can increase the VA as a safety factor.

Easy peazzy :) Physics can sometimes save you from falling into a deep precipice. How’s that for a metaphor. :).

Enlarging the control transformer is easier than adding those relays. . . and it also keeps the original concept.

Purists will argue:

But. we are talking about AC.

Well, you won’t be far off if you use the example above.


[/QUOTE
Myspark, did you mean to say "Any value HIGHER than 48 ohms (from your meter) would require a larger transformer VA rating"?
Thanks, Isaiah
Let me answer for that, if I may?
myspark used the 300VA CPT you have given as the basis: Irated = 300/120V = 2.5 amps! The resulting calculations yielded a limiting resistance of 48 ohms (120V/2.5A).
 

Isaiah

Senior Member
Explain what you are looking for.

Do you want to size the control transformer adequately for the size 5 coil?

Do you want to know what the VD will be for a size 5 coil if the transformer is located xxx feet away?
I am trying to size the control circuity for a 125Hp motor with size 5 starter, (i.e. #14's, #12's OR #10's?). The pickup on the M coil is 1158VA; the circuit contains 2 LEDs and a simple start-stop control (no load basically). The space heater is fed from an external source so we're only looking at the M coil pickup and seal-in.
Normally I would simple use Table 8 NEC resistances/calc to find the VD or an on-line calculator for ball-park estimate. The std CPT for a size 5 starter from Eaton is 300VA and I am not sure how the size of the CPT figures in the V-Drop on the control circuit.
 
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