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Differential transformer how to measure this signal

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sardineta

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hello
I am using a differential transformer to sense unbalanced lines.
When it happens the toroid (differential transformer) induce a current and so a voltage accross the terminals of the differential transformer.

for currents as low as 26 mA ac induce a voltage of 129 mV ac and for currents of 830 mA, it induces a voltage of 7 V ac.

I would like to measure this signal to prevent electrical damage on communication equipments. But I have no idea of this kind of electrical considerations, I have never done a design with this kind of transformers.
When I put a resistor of 98Kohms between the two terminal in ordet to measure just the voltage so in this way I supposed not to demand current to the transformer, the measured values changed.

Asking about this matter, someone told me that for this toroids large resistors can saturate the core of the toroid, but a lower resistor can avoid it.

Does any one can help me how to read the induced voltage in a differential transformer.

Thanks in advanced
 
sardineta said:
hello
I am using a differential transformer to sense unbalanced lines.
When it happens the toroid (differential transformer) induce a current and so a voltage accross the terminals of the differential transformer.

for currents as low as 26 mA ac induce a voltage of 129 mV ac and for currents of 830 mA, it induces a voltage of 7 V ac.

I would like to measure this signal to prevent electrical damage on communication equipments. But I have no idea of this kind of electrical considerations, I have never done a design with this kind of transformers.
When I put a resistor of 98Kohms between the two terminal in ordet to measure just the voltage so in this way I supposed not to demand current to the transformer, the measured values changed.

Asking about this matter, someone told me that for this toroids large resistors can saturate the core of the toroid, but a lower resistor can avoid it.

Does any one can help me how to read the induced voltage in a differential transformer.

Thanks in advanced

I dont understand your problem. Maybe if you send a schematic, that can help.

As far as core saturation goes. Flux density is what determines core saturation. If the flux density gets too high (see mfg toroid datasheet), the core can saturate. What generates flux density? Volt-second product on the windings.
 
Optikon

Inside the section I pass one wire from the 127v outlet, then I conect a resistor to reduce or change the current through it, so in this way by using a multimeter i read the voltage in the two output terminals. But when I tried to design an amplifier stage, I had problems with my readings becasue the values changed, so I woulf like to design an amplifier stage becasue when a current of 30mA pass through the resistive load in the diferential transformer a voltage in induced around milivolt is induced and I want to amplify it.

here I have found on the wab some specifications, but they are in spanish

WG-35/70/105/140/210:
- Para funcionar con relés serie G.
- La sensibilidad (corriente diferencial de disparo) del conjunto transformador-relé vendrá fijada por el relé.
- Se caracteriza por su linealidad, y una mayor sensibilidad, reduciéndose la relación ruido-señal.
- Dos bobinados sobre el mismo núcleo. (Bobinado de test y de trabajo), permiten el chequeo real de todo el conjunto toro + relé.


- Toroidal.
· Sección útil: Ø 35 ÷ 210 mm.
· Dimensiones: 100x79 ÷ 299x284 mm.
· Peso: 0,15 ÷ 2,50 kg.

Características técnicas:
- Tensión de trabajo: 720 V c.a.
- Tensión de aislamiento. 3.000 V c.a.
- Temperatura de trabajo: -10/+50°C.
- Construidos en carcasa de plástico tipo ABS, autoextinguible, de color negro, con material antichoque.
- Soportes metálicos para su fijación en panel.
- Bornes niquelados protegidos con tapa de plástico transparente recintable.

- Normas: IEC 185, UNE 21088, UL 94, VDE 0414, IES 41-1, protección IP20.
 

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sardineta said:
Optikon

Inside the section I pass one wire from the 127v outlet, then I conect a resistor to reduce or change the current through it, so in this way by using a multimeter i read the voltage in the two output terminals. But when I tried to design an amplifier stage, I had problems with my readings becasue the values changed, so I woulf like to design an amplifier stage becasue when a current of 30mA pass through the resistive load in the diferential transformer a voltage in induced around milivolt is induced and I want to amplify it.

here I have found on the wab some specifications, but they are in spanish

WG-35/70/105/140/210:
- Para funcionar con relés serie G.
- La sensibilidad (corriente diferencial de disparo) del conjunto transformador-relé vendrá fijada por el relé.
- Se caracteriza por su linealidad, y una mayor sensibilidad, reduciéndose la relación ruido-señal.
- Dos bobinados sobre el mismo núcleo. (Bobinado de test y de trabajo), permiten el chequeo real de todo el conjunto toro + relé.


- Toroidal.
· Sección útil: Ø 35 ÷ 210 mm.
· Dimensiones: 100x79 ÷ 299x284 mm.
· Peso: 0,15 ÷ 2,50 kg.

Características técnicas:
- Tensión de trabajo: 720 V c.a.
- Tensión de aislamiento. 3.000 V c.a.
- Temperatura de trabajo: -10/+50°C.
- Construidos en carcasa de plástico tipo ABS, autoextinguible, de color negro, con material antichoque.
- Soportes metálicos para su fijación en panel.
- Bornes niquelados protegidos con tapa de plástico transparente recintable.

- Normas: IEC 185, UNE 21088, UL 94, VDE 0414, IES 41-1, protección IP20.

when you use multimeter to read output terminals, are you referring to another loop of wire wound around toroid? I understand that you have 120V source with series resistor making one loop around toroid. You then read voltage induced (in another loop) around same toroid? Please clarify this.

Your AC source is 50Hz? and the toroid is a low frequency type rated for that? I can't read spanish but based on the numbers given, they don't look like very useful specs anyways..
 
Yes your right,

I measure the AC voltage (120 Vac) with the multimeter at the outlet, when I put a resistor then this loop is closed, if I connect the multimeter to measure the current then I put it in serie, when I have already measure the current (the voltage is known), I move the multimeter to the terminals of the diferential transformer.

This transformer has two terminal where in there I measure the induced voltage, with the multimeter. This is not a loop but I have to close it with an OpAmp or a resistor in order to sense the voltage by an ADC.

The problem here is this; when I connect a resistor in the terminals of the Differential transformer the voltage values change, I mean the voltage at these two terminal is not the sme as when I measure it in a open loop, as you can see in the figure.

The figure shows an open loop, a measurement just with a multimeter, but instead of doing it by an instrument I want to read this values by an ADC, it's here where I have problems, because I do not know what to do. How I should measure the induced voltage?

When connected a resistor of 100K in this to terminal the readings changed, I think the toroid was saturated
I have no electrical parameters of this diferential transformer, it is a WG-70 or 35 transformer from circutor
there is just physical parameters, size, weight etc.

What can I do?
 
sardineta said:
Yes your right,

I measure the AC voltage (120 Vac) with the multimeter at the outlet, when I put a resistor then this loop is closed, if I connect the multimeter to measure the current then I put it in serie, when I have already measure the current (the voltage is known), I move the multimeter to the terminals of the diferential transformer.

This transformer has two terminal where in there I measure the induced voltage, with the multimeter. This is not a loop but I have to close it with an OpAmp or a resistor in order to sense the voltage by an ADC.

The problem here is this; when I connect a resistor in the terminals of the Differential transformer the voltage values change, I mean the voltage at these two terminal is not the sme as when I measure it in a open loop, as you can see in the figure.

The figure shows an open loop, a measurement just with a multimeter, but instead of doing it by an instrument I want to read this values by an ADC, it's here where I have problems, because I do not know what to do. How I should measure the induced voltage?

When connected a resistor of 100K in this to terminal the readings changed, I think the toroid was saturated
I have no electrical parameters of this diferential transformer, it is a WG-70 or 35 transformer from circutor
there is just physical parameters, size, weight etc.

What can I do?

Try this, keep the resistor in the 120V loop and put the same value resistor in the measure look (where you want to sense induced voltage). Now, all you have created is a 1:1 transformer. We still don't know for sure if the core is saturating but once you have this arrangement things will be easier to predict.

With your 1:1 transformer and ignoring Xm (magnitization inductance) and leakage parasitics for the moment, you should see the same current flow in your output (Vinduced) loop. You can measure this with a scope or possible RMS meter. Make sure the measuring device input impedance is much much greater than the resistor values in the loop (usually is ok)

If you do not see the same current and voltage measurements (approximately) from the input loop compared to the output loop, then something else is happening like saturation. If it is saturation, the core will get warm(is this happeining) and you will get different values.

If this is not the case, just use this method for sensing. Since you are wishing to amplify the signal, why not get some amplifying with turns ration on the transformer? If you have one turn on input side and 10 turns on output side, you will get a 10X on voltage (approx).

There are alot of other considerations to make about the design but for now, try and learn something about the toroid you have to use by performing the simple 1:1 experiment.
 
other basic info about the transofrmer is:

relation 500/1
it sense current, as we know.
In the terminal of the transformer, where I measure the induced voltage, I read almost 8 ohms with the multimeter


What you have told me, how it can be possible if the transformer is not a voltage/current source?

thanks in advanced.
 
sardineta said:
other basic info about the transofrmer is:

relation 500/1
it sense current, as we know.
In the terminal of the transformer, where I measure the induced voltage, I read almost 8 ohms with the multimeter


What you have told me, how it can be possible if the transformer is not a voltage/current source?

thanks in advanced.

Oh, well this changes things.. it is 500:1 then.

You should be able to measure volts on output accordingly.

Check this:

http://www.coilwinder.com/current_transformer_design_and_theory.htm


Use the resistor load and measure voltage. You need to first discover if core is saturating. Knowing 500:1 will let you evaluate.

Consider starting with small current on input and measure output. Slowly increase curren ton input and keep measuring output. Each tim eyou increase input current, output current (voltage) should remain proportional (and almost linear). Keep increasing input current until you begin to see output current failing to increase by the same amount. This is where you are entering core saturation. In saturation, a large change in input current will result in only a small change in output current. The core should get warm too.

If you start with sufficiently small currents, you can be sure you are not saturating the core. You know the ratio, input voltage (and current). You will only be limited in your measurement accuracy on output.
 
Ok I have already done the following

WITHOUT LOAD
LOAD I(from load) Voltage(induced)
131ohm 830 mA 7.17 V
290 mA 3.71 V
130 mA 1.11 V
100 mA 732 mV
70 mA 452 mV
63 mA 390 mV
45 mA 256 mV
35 mA 188 mV
31 mA 164 mV
>800ohm 26 mA 129 mV

WITH LOAD of 100kohm
at terminal from DT
LOAD I(from load) Voltage(induced)
131ohm 830 mA 4.43 V
290 mA 3.19 V
....
31 mA 101 mV
>800ohm 26 mA 80 mV


As you see, values changes, What we are looking for is the way to connect the two terminal from the DT to an amplifying circuit just to amplify the low voltage, and reduce the maximum voltage induced at 600 mA in order to read it by an embbeded ADC in a uC (PIC16F870)

See you
 
sardineta said:
Ok I have already done the following

WITHOUT LOAD
LOAD I(from load) Voltage(induced)
131ohm 830 mA 7.17 V
290 mA 3.71 V
130 mA 1.11 V
100 mA 732 mV
70 mA 452 mV
63 mA 390 mV
45 mA 256 mV
35 mA 188 mV
31 mA 164 mV
>800ohm 26 mA 129 mV

WITH LOAD of 100kohm
at terminal from DT
LOAD I(from load) Voltage(induced)
131ohm 830 mA 4.43 V
290 mA 3.19 V
....
31 mA 101 mV
>800ohm 26 mA 80 mV


As you see, values changes, What we are looking for is the way to connect the two terminal from the DT to an amplifying circuit just to amplify the low voltage, and reduce the maximum voltage induced at 600 mA in order to read it by an embbeded ADC in a uC (PIC16F870)

See you

830mA flowing in 131 Ohms on the output side (131 Ohm is across terminals)? And you measured current in 131 Ohm how? The numbers by themselevs are not making sense let alone the changing values. Dont worry about the changing values yet. First you have to get a predictable output so you can understand how it is working. You are sure it is 500:1?

Also, what value are you using on input side (700>R>400) what is its value?
 
Sorry, the value of 400 ohm in the figure should be less.

Here is what I am triyng to say:
Voltage source = 112 Vac @ 60 Hz
6 Resistors of =~ 131 ohms
840 ohm 5W in parallel
Current = 830 mA

Induced voltage = 7.14 V


Voltage source = 112 Vac @ 60 Hz
10 Resistors of = ~386 ohms
840 ohm 5W in parallel + 4 of 840 in parallel
Current = 290 mA

Induced voltage = 3.71 V



and so on, The AC voltage is not constante, it depends on the quality, in theory it should be 120 Vac.

I change The value of the resistor and current changes is getting lower, I mean if R is bigger then current should be less as you see in the table

This current is through the load, not in the terminals where is induced the voltage.


When I put a resistor of 100K 1/2W at the terminal where the induced voltage is induced I got the following

Voltage source = 112 Vac @ 60 Hz
10 Resistors of = ~131 ohms
840 ohm 5W in parallel
Current = 830 mA

Induced voltage = 4.43 V


Voltage source = 112 Vac @ 60 Hz
10 Resistors of = ~386 ohms
840 ohm 5W in parallel + 4 of 840 in parallel
Current = 290 mA

Induced voltage = 3.19 V

and so on
I write to circutor and they say me 500/1. They do not send electrical datasheets, just physical characteristics


Now I measure again without resistor in the terminals of the transformer and the voltage values are as if the resistor of 100K was there.
 
sardineta said:
Sorry, the value of 400 ohm in the figure should be less.

Here is what I am triyng to say:
Voltage source = 112 Vac @ 60 Hz
6 Resistors of =~ 131 ohms
840 ohm 5W in parallel
Current = 830 mA

Induced voltage = 7.14 V


Voltage source = 112 Vac @ 60 Hz
10 Resistors of = ~386 ohms
840 ohm 5W in parallel + 4 of 840 in parallel
Current = 290 mA

Induced voltage = 3.71 V



and so on, The AC voltage is not constante, it depends on the quality, in theory it should be 120 Vac.

I change The value of the resistor and current changes is getting lower, I mean if R is bigger then current should be less as you see in the table

This current is through the load, not in the terminals where is induced the voltage.


When I put a resistor of 100K 1/2W at the terminal where the induced voltage is induced I got the following

Voltage source = 112 Vac @ 60 Hz
10 Resistors of = ~131 ohms
840 ohm 5W in parallel
Current = 830 mA

Induced voltage = 4.43 V


Voltage source = 112 Vac @ 60 Hz
10 Resistors of = ~386 ohms
840 ohm 5W in parallel + 4 of 840 in parallel
Current = 290 mA

Induced voltage = 3.19 V

and so on
I write to circutor and they say me 500/1. They do not send electrical datasheets, just physical characteristics


Now I measure again without resistor in the terminals of the transformer and the voltage values are as if the resistor of 100K was there.

With your 131 ohms on input side (830mA flowing), short circuit output terminals and measure current though it. What is output short circuit current? If we assume it is 500:1 then you should see about 44uA flowing if your core is not saturated. The other thing is the magnitizing current may be a significant fraction of your primary current.

Remember, this thing is a transformer so when you add 100k on output terminals and "load" the transformer, this impedance is reflected back into primary side and that current will change. 100k is much greater of a load than open circuit (no load just volt-meter) so the changing values makes sense to me but I am still wondering if it is not saturated or even if the 500:1 is correct. You need to find an operating point that shows the turns ratio without a doubt.
 
Please give me a hit for the design

Optikron considering what I have done, can you imagine what would be the input impedance of an operational amplifier?

or which configuration you recommend me? how I should read this signal by the ADC.

You got it too, I read 44.1uA in the terminals where the induced voltage is generated when a current of 830mA passed inside the toroid.
This current was obteined with the resistor of 100K 1/2W and the multimeter in series to measure the current.

I haven't measure the current in short circuit (where the induced voltage is generated), just the voltage in open circuit.

Also voltage and current in close loop with the resistor.
how did you get 44 uA

Thanks a lot Optikron
 
In the right upper corner, are the two terminal where I measure the voltage.

inside the circle I pass one wire from the monophase AC voltage to generate the current with a load (Resistors of 840 ohm in series and in parallel to change the value)
 

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Re: Please give me a hit for the design

sardineta said:
Optikron considering what I have done, can you imagine what would be the input impedance of an operational amplifier?

or which configuration you recommend me? how I should read this signal by the ADC.

You got it too, I read 44.1uA in the terminals where the induced voltage is generated when a current of 830mA passed inside the toroid.
This current was obteined with the resistor of 100K 1/2W and the multimeter in series to measure the current.

I haven't measure the current in short circuit (where the induced voltage is generated), just the voltage in open circuit.

Also voltage and current in close loop with the resistor.
how did you get 44 uA

Thanks a lot Optikron

What you need for sensing is an I-V converter. The simplest I-V converter is a resistor. Let the transformer drive your resistor and just measure the voltage of that resistor for sensing. Use your opamp for that.

The input impedance of an op amp is very high but also dependent on the process used.

In the specs for op amp if front-end is made with bipolar transistors, then the bias currents they need will be in microamp range. Feedback will boost the input impedance by order fo magnitude.

You can expect 10's to 100's of megaohms at worst.

If you use a "FET" input type, this number goes up to Giga-Ohms even Tera-Ohms (for electrometer types).

It's safe to say that your opamp circuit will not load the transformer at all. But you will delibrately put a resistor load on the transformer making it drive 100k or whatever you end up with.

Will one end be grounded? Be careful with common mode voltage range in your design. also, consider using an instrumentation amplifier if you can live with less than MHz bandwidth.

Also for highest signal sensing level choose the resistor to provide the amp the largest signal it will allow for the worst case sensing condition. Then in subsequent stage(s) attenuate and level translate for appropriate input range of ADC. So that you do not start off with a signal so small that noise prevents you from sensing properly.

Also be sure to account for worst case line conditions. If that phase is normally 120V you can bet there will be times when it is much higher.

Also keep your opamp circuits away from the high current side of the xformer because there will be strong magnetic fields present and your wires will pickup 50/60Hz form that easily in the 10's of mV or worse levels. And also wire the circuits so that you do not create any large loop areas with supplies and grounds because those will also act like another loop of wire around the xformer also picking up the field.

try it!
 
Optikon
I read this post this week-end and I got sick. Now I feel better and I was thinkg about your ideas.

By the way, I really do not understand this part:
Let the transformer drive your resistor and just measure the voltage of that resistor for sensing. Use your opamp for that.

What resistor refers to? an external resistor or the resistive value the transformer has

This too:
It's safe to say that your opamp circuit will not load the transformer at all


I am thinking on the design, I suppose it is better to connect directly the xformer to the opamp, using the two inputs, I realise that instrumental amplifiers can handle it, the book I am reading about this, one terminal is connected to the ground vie a resistor.

I do not know if it makes to one of the wire that comes from the xformer be grounded.


I know that measuring the voltage is easier that measuring current, that is the the purpose. However, something cause me noise. the impedance you talked, it will affect the voltage value as I told you before, when I put an external resistor between the two terminal the voltage drop from 7.17@830mA to 4.43@830mA

I did it becasue I wanted to measure the current.

PS:I am still extracting ideas from what you said me in your post
thanks a lot
 
What you're asking is the same thing as ground fault circuit interrupter (GFCI, GFI), isn't it? Why not use an existing off-the-shelf device?

I wondered how these work. I've never taken one apart, but I assume there's a toroid with a pair of opposing, isolated primary windings with just a few turns for each leg. Then there would be a lot of secondary windings. If the current is exactly the same, the induced flux from both legs in the same core will cancel each other out and there is no voltage on the secondary. If they're different, then the core will have flux which induces secondary voltage. It can even generate enough current to trip a breaker's element if it's got a specially isolated element. This makes everything electrically isolated, there's no resistors to create heat and drop the voltage under load.

Sooner or later I need to take one apart and see if that's how it works. I'm pretty sure that's what they do, they're simple cheap devices so they can't have complicated electronics to compare voltage drops with totally dissimilar high voltage grounds. If not, I don't see why YOU can't do it that way.
 
Oznog said:
What you're asking is the same thing as ground fault circuit interrupter (GFCI, GFI), isn't it? Why not use an existing off-the-shelf device?

I wondered how these work. I've never taken one apart, but I assume there's a toroid with a pair of opposing, isolated primary windings with just a few turns for each leg. Then there would be a lot of secondary windings. If the current is exactly the same, the induced flux from both legs in the same core will cancel each other out and there is no voltage on the secondary. If they're different, then the core will have flux which induces secondary voltage. It can even generate enough current to trip a breaker's element if it's got a specially isolated element. This makes everything electrically isolated, there's no resistors to create heat and drop the voltage under load.

Sooner or later I need to take one apart and see if that's how it works. I'm pretty sure that's what they do, they're simple cheap devices so they can't have complicated electronics to compare voltage drops with totally dissimilar high voltage grounds. If not, I don't see why YOU can't do it that way.

well I was under the impression (which could be incorrect) that the author wanted to measure the imbalance not necessarily detect a large one.
 
Optikon yes you right, I want to measure an induced voltage due to unnbalanced lines (300 or 600mA) and fault currents (30mA)

(corrientes de fuga)
 
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