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
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.
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?
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.
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
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.
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
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.
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