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Sensing current through AC motor

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premkumar9

Member
Hi,
I need to sense the AC current taken by a motor. I need only to know zero current, around 10A and around 30A (these 3 values give me the required information). I plan to convert it to 0 to 5V DC to give to a comparator. Can anybody give some hints on how to do that?
 

Sceadwian

Banned
You sense AC current the same was as DC, with a sense resistor. also known as a shunt. When the voltage across the sense resistor is zero there is no current flowing through it.
 

MikeMl

Well-Known Member
Most Helpful Member
Sensing the current is one thing. Isolating the AC circuit from your low-voltage control circuitry makes it harder. Current transformers provide both at the same time with zero voltage drop. Hall-Effect devices do too.

Methods employing sense resistors introduce a voltage drop to the motor and are not intrisically isolated. How much drop can you tolerate?
 

Willbe

New Member
How long is the power cord to the motor? 10A through a 50' #14 AWG conductor will give you 1.3vac.
 

premkumar9

Member
How long is the power cord to the motor? 10A through a 50' #14 AWG conductor will give you 1.3vac.
But I need isolation. Can a few turns of wire around the main conductor give me the required voltage OR shall I have to use a current transformer?
(of course I can experiment it, but advice from experienced persons will save time)
 

Sceadwian

Banned
Look up the wikipedia entry on Current transformer, it has an example of a winding, you can't just wrap the wire around the conductor I don't think.
 

Diver300

Well-Known Member
Most Helpful Member
You should use a current transformer. That can be AC which is just like a conventional transformer or a Hall Effect one that will work at DC

You cannot just wind a few turns around the conductor. That is going in the wrong direction. The magnetic field goes around the conductor, so to induce a current the wires should be parallel to the main conductor.

The most usual form of a current transformer is a toroidal with a single turn primary. You could use any toroidal as a current transformer as long as you can fit the primary down the middle.

You must put a suitable load on at least one of the windings or there will be big voltages.

If you use a toroidal transformer of about 30 VA, it will have about 0.25 V per turn, so on 120 V primary that is about 480 turns, and on a 6 V secondary that is about 24 turns.

If you have a single 30 A turn as your new primary, that will give you about 1.166 A on the 6 V winding or 20 mA on the 120 V winding. You shouldn't exceed the rated voltages as a load, so either 5 Ω on the 6 V winding or 1920 ohms on the 120 V winding. However, as all you want to do is sense the current, a 250 Ω load on the 120 V winding will give you a 5 V signal.

There will be a voltage drop, but it will be tiny. If the secondary voltage is 5 V, the primary will be about 10 mV for that sort of turns ratio.
 

ecerfoglio

New Member
Hi,
I need to sense the AC current taken by a motor. I need only to know zero current, around 10A and around 30A (these 3 values give me the required information). I plan to convert it to 0 to 5V DC to give to a comparator. Can anybody give some hints on how to do that?
I agree with the other replies, current transformer is the best method to sense AC current.

Is it an induction motor (AC motor) or an "universal" (AC/DC) one? The following applies only if it is an induction motor:
Have you measured tne motor's current (with an ammeter)? AC (indution) motors ussually have a very high "no load current" (about one half of the "full load current"), so current measurement is nearly useless as an indicator of the motor's output power (or torque)​
 

premkumar9

Member
I agree with the other replies, current transformer is the best method to sense AC current.

Is it an induction motor (AC motor) or an "universal" (AC/DC) one? The following applies only if it is an induction motor:
Have you measured tne motor's current (with an ammeter)? AC (indution) motors ussually have a very high "no load current" (about one half of the "full load current"), so current measurement is nearly useless as an indicator of the motor's output power (or torque)​
This is part of a project for a customer who wants to know from remote place (through mobile communication) whether the motor, which is used for water pumping is really pumping water. I am told that it is a 20HP three phase motor and when pumping the current measured through each phase is 30A and if the motor is working and there is no water pumping due to some problems, then the current is 10A. These two conditions plus the idle condition of motor (no current) are to be displayed on a control panel. For this I wanted to sense the currents.
 

premkumar9

Member
You should use a current transformer. That can be AC which is just like a conventional transformer or a Hall Effect one that will work at DC

You cannot just wind a few turns around the conductor. That is going in the wrong direction. The magnetic field goes around the conductor, so to induce a current the wires should be parallel to the main conductor.

The most usual form of a current transformer is a toroidal with a single turn primary. You could use any toroidal as a current transformer as long as you can fit the primary down the middle.

You must put a suitable load on at least one of the windings or there will be big voltages.

If you use a toroidal transformer of about 30 VA, it will have about 0.25 V per turn, so on 120 V primary that is about 480 turns, and on a 6 V secondary that is about 24 turns.

If you have a single 30 A turn as your new primary, that will give you about 1.166 A on the 6 V winding or 20 mA on the 120 V winding. You shouldn't exceed the rated voltages as a load, so either 5 Ω on the 6 V winding or 1920 ohms on the 120 V winding. However, as all you want to do is sense the current, a 250 Ω load on the 120 V winding will give you a 5 V signal.

There will be a voltage drop, but it will be tiny. If the secondary voltage is 5 V, the primary will be about 10 mV for that sort of turns ratio.
Thank you. I will learn your suggestions and come back.
 

user_88

Member
If you have one pump/motor condition where the phase current is 30 A, and the second condition is 10A, then the difference in electrical power loss that is incurred by the resistance of the motor windings ... the Joule heat or I²R loss... is a factor of 900 to 100. Consequently, the surface temperature of the motor casing will be significantly higher when it is working properly, as opposed to when it is not. Devise a method to measure the temperature of the motor casing ... thermistor, RTD, or maybe a thermocouple ... set up a comparator to turn on at an appropriate level.

Another possibility would be to test the temperature of the air flow through the motor ... probably easier to measure than the casing temperature .... more indicative of the motor winding temperatures.
 
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premkumar9

Member
If you have one pump/motor condition where the phase current is 30 A, and a second condition is 10A, then the difference in electrical power incurred by the resistance of the motor windings is a factor of 900 to 100, or nine times more power flow through the motor when it is working than when it is not. Consequently, the surface temperature of the motor casing will be significantly higher when it is working properly, as opposed to when it is not. Devise a method to measure the temperature of the motor casing ... thermistor, RTD, or maybe a thermocouple ... set up a comparator to turn on at an appropriate level.
If the motor is in the second condition (10A), it should be reported immediately. Also if the motor is taking 30A and suddenly it falls to 10A due to some problem, that also has to be reported immediately. Temperature building up and cooling down may not be that fast. Am I right?
 

Hero999

Banned
The most usual form of a current transformer is a toroidal with a single turn primary. You could use any toroidal as a current transformer as long as you can fit the primary down the middle.

You must put a suitable load on at least one of the windings or there will be big voltages.

If you use a toroidal transformer of about 30 VA, it will have about 0.25 V per turn, so on 120 V primary that is about 480 turns, and on a 6 V secondary that is about 24 turns.

If you have a single 30 A turn as your new primary, that will give you about 1.166 A on the 6 V winding or 20 mA on the 120 V winding. You shouldn't exceed the rated voltages as a load, so either 5 Ω on the 6 V winding or 1920 ohms on the 120 V winding. However, as all you want to do is sense the current, a 250 Ω load on the 120 V winding will give you a 5 V signal.

There will be a voltage drop, but it will be tiny. If the secondary voltage is 5 V, the primary will be about 10 mV for that sort of turns ratio.
That's a good idea.

Wouldn't it be better to use one with a 230V secondary or two 120V secondariness?

What about the size of the torroid?

Isn't smaller better as there'll be less of an air gap between the conductor and the windings?
 

user_88

Member
If the motor is in the second condition (10A), it should be reported immediately. Also if the motor is taking 30A and suddenly it falls to 10A due to some problem, that also has to be reported immediately. Temperature building up and cooling down may not be that fast. Am I right?
For a fast response notification .... as you describe ..... you would want to try to measure the temperature of the forced convection air current that circulates across the motor windings ...
In this case, the heat source is the phase current flowing through the stator or rotor winding .... whichever is present.
The relevant mass element is that of the copper wire which makes up the coils.
It is difficult to say with any precision what the time constant is in this case .... a guess would be from 10 to 60 seconds. It may be worth your effort to set up a test or experiment, in order to confirm the time and temperature values .... under actual operating conditions.

Measurement of the motor casing surface temperature would not be practical, if an immediate temperature response were required.
 

Diver300

Well-Known Member
Most Helpful Member
That's a good idea.

Wouldn't it be better to use one with a 230V secondary or two 120V secondariness?
I don't think it makes a lot of difference. The bigger the rating of the primary, the bigger the turns ratio, so the smaller output current of the current transformer. In this case, I think that the ratio will be big so the current will be less than 50 mA whatever.

What about the size of the torroid?

Isn't smaller better as there'll be less of an air gap between the conductor and the windings?
I think not. I know that is counter-intuitive, but I don't think that an air gap matters. If you buy a purpose-built current transformer, there is often a big gap between the primary conductor and the toroid.
 

premkumar9

Member
I understand that the right solution in my case is to use a current transformer. I have got options of getting a off-the-shelf item OR make my own. But in the second case I think I have to select the core material and size to avoid saturation and accommodate my turns. Am I right?
 

Diver300

Well-Known Member
Most Helpful Member
How did you reach 0.25V per turn?

how is this calculated?
That is what I have found for medium sized toroidal transformer. The number is larger for larger transformers.

Quote:
If you have a single 30 A turn as your new primary, that will give you about 1.166 A on the 6 V winding or 20 mA on the 120 V winding.
how is this calculated?
If the transformer is 0.25 V per turn (but it will vary depending on the design) then the 6 V winding will be 24 turns. So if the primary is 30 A and 1 turn, then the secondary will be 30 x 1 / 24 = 1.166 A

I think that I got the wrong figure for the 120 V winding.

The 120 V winding will be about 480 turns, so if that is used as the secondary, it will be about 30 x 1 / 480 = 62.5 mA
 
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