analog curent sensor

I bought this sensor :
 
 
 
http://www.ebay.com/itm/230991057211
 
 
How do I convert the number(in my case 525) to the real amp comsuption? I triied a lot of thing I never got an accuracy and a  valid answer.
 
 
This is the A/Q from the seller :
 
Range: AC-75A RMS, DC-100A
Epoxy bonded 
Class F insulation coating
 
Q & A:
 
1. Q: What types of current can be measured?
     A: DC, AC(up to several KHz) , PWM
 
2. Q: What is the reading out of this sensor?
     A: DC voltage,the range is from 0 - Vcc 
         * Vcc : 4.5 V -- 6 V
 
3. Q: How do I convert the voltage reading to the unknown current value?
     A: Formula: Unknown Current (A) = (V1-V0) / CF
          e.g. V0 = 2.5 V , CF = 0.0224
          V1 is the volt reading @ unknown current    
             
         Example 1:
         V1=3.5 V
         Unknown Current (A) = (3.5-2.5) / 0.0224 = 44.64 A
         
         Example 2:
         V1=2.0 V
         Unknown Current (A) = (2.0-2.5) / 0.0224 = -22.32 A
 
4. Q: What are V0 and CF?
     A: V0 is the output voltage at 0 Amp
          CF is increase of volt reading when the current increase 1 Amp
       *  All sensors will be shipped along with calibrated values of V0 and CF   
        
5. Q: How can I measure V0 and CF?
     A: To measure V0:
          Power on the sensor and read the output voltage while no current is 
          applied. This voltage is V0.
          To measure CF:
          While applying some current(e.g. 20 amp)  to sensor (passes through 
          the toroid core), read the output  voltage V1. The CF can be calculated using 
          the formula: CF=(V1-V0) / 20
 
Sample of Test Report 1 Output: Voltage Meter Reading@ Input: DC Current)
S/N:  #053101
2.500 V @ 0 Amp, 
2.714 V @ 10 Amp, 
2.926 V @ 20 Amp,
3.139 V @ 30 Amp, 
3.353 V @ 40 Amp,
3.567 V @ 50 Amp,
3.780 V @ 60 Amp, 
 
V0 = 2.50 V
CF = 0.021 V / Amp
 
 
 
 

First off, you're going to want an amplifier to get any decent resolution.
The webcontrol theoretical resolution is 9.76mV/bit, but with noise and jitter, it's nowhere near that in reality.
I'm surprised you're seeing a value of 525. Assuming you don't have any sort of amplifier and you're using a standard WC8, you can convert 525 back to a voltage simply by dividing it by 1024 then multiplying by 10V, this 525 = 5.127V in.
 
It looks to me like you have a wiring error or a faulty sensor.

I tried 2 brand new sensor. Connect neg, pos and the hite wire to the aip1. I use wc32 board.
 
 
I guest I could make something like that: result var2:
 
START
                DIV AIP1 1024 VAR1
                MUL VAR1 10 VAR2
END
 
 
But the result var2 will show 5.12. But it's supposed to show 0
 
I can read value 525 while no current is applied.....

BTCAD said:

I tried 2 brand new sensor. Connect neg, pos and the hite wire to the aip1. I use wc32 board.
 
But the result var2 will show 5.12. But it's supposed to show 0
 
I can read value 525 while no current is applied.....

Click to expand...

 
ok, WC32 is a different animal, I believe it's 5V full scale.
In which case, 525 = 2.56 volts on the input, which would be about right for 0 amps.
Sounds like you don't quite understand how the device works. 0A != 0V

yes it's 5v full scale, but I just plug the current sensor to the board without any cables trough the hole, So it should be 0v but the output of the sensor is 2.56v. Sould'nt show 0v????

Sounds like zero current output is set to mid-range so sensor is able to measure positive or negative current.
 
/tom

From what datasheet of that sensor, input amp zero == output 2.5V with 5V supply.  If you got different reading with zero current input, your sensor has problem, or test room is very close to a big current transformer.

BTCAD said:

Sample of Test Report 1 Output: Voltage Meter Reading@ Input: DC Current)
S/N:  #053101
2.500 V @ 0 Amp, 
2.714 V @ 10 Amp, 
2.926 V @ 20 Amp,
3.139 V @ 30 Amp, 
3.353 V @ 40 Amp,
3.567 V @ 50 Amp,
3.780 V @ 60 Amp, 
 
V0 = 2.50 V
CF = 0.021 V / Amp

Click to expand...

Do you even read your own posts?

BTCAD said:

yes it's 5v full scale, but I just plug the current sensor to the board without any cables trough the hole, So it should be 0v but the output of the sensor is 2.56v. Sould'nt show 0v????

Click to expand...

 
The current will be 0A, the output will be (as specified in the docs you linked to) "V0" (presumably Offset Voltage).
As I said earlier, you simply don't understand what they've told you, and compounded the problem by initially not telling us the full scenario.
 
~2.5V is exactly what I would expect from the device, and on a 10-bit ADC with 5V FSD a count of around 512 is exactly what I'd expect.

rossw said:

As I said earlier, you simply don't understand what they've told you, and compounded the problem by initially not telling us the full scenario.

Click to expand...

 
 
Sorry, I don't speak in english, so explain technical situation when you don't speak in english is very hard. So i try my best to be clear in each post. 
 
 
 
First of all, I asked you :
 
How do I convert the number(in my case 525) to the real amp comsuption? I tried a lot of thing I never got an accuracy and a  valid answer. 
 
I want to use this sensor to read the real amp that is in the cable. NOT 525. I never done this before, I'm not a programmer so I don't know how to proceed.
 
 
I tried to mesure an oven (1440w) that is plug to 120vac. Only the 120v wire is in the hole of the sensor, I never been able to read 12 (amp) from a var on the web page. I tried a lot of thing that probably make no sense.
 
 
is any body have an idea to convert this sensor reading into a good AMP value?

Tschmidt said:

Sounds like zero current output is set to mid-range so sensor is able to measure positive or negative current.
 
/tom

Click to expand...

 
 
Yes that make sense beause this sensor is 75a AC and 100a DC

BTCAD said:

How do I convert the number(in my case 525) to the real amp comsuption? I tried a lot of thing I never got an accuracy and a  valid answer. 

Click to expand...

 
You measure the zero-current output voltage and store this as the offset. You then take the difference between this offset and the instantaneous readings you take later, multiply it by the scale factor and you have an answer in amps (plus or minus, depending if it's above or below the initial offset)
 

BTCAD said:

I want to use this sensor to read the real amp that is in the cable. NOT 525. I never done this before, I'm not a programmer so I don't know how to proceed.

Click to expand...

 
As above, and demonstrated with math earlier.
 
 

BTCAD said:

I tried to mesure an oven (1440w) that is plug to 120vac. Only the 120v wire is in the hole of the sensor, I never been able to read 12 (amp) from a var on the web page. I tried a lot of thing that probably make no sense.

Click to expand...

 
You are going to need additional circuitry, because you have no way to synchronise the webcontrol analog reading to the mains cycle, so you will measure current at some random point in the cycle. If you take a lot of readings and average them, you will get 0, because it's AC.
 
You can probably get away with a peak-and-hold circuit, and just measure the positive half-cycle maximums, and assume your load is symmetrical.
If you are going to the effort to do that, I would also use an opamp level-shifter to offset the 2.5V close to 0V.

I also noticed that the aip1 value varied from 477 to 580 when the oven use full 1440w, is it normal? That's why I should use a additional circuitry?

rossw said:

You can probably get away with a peak-and-hold circuit, and just measure the positive half-cycle maximums, and assume your load is symmetrical.
If you are going to the effort to do that, I would also use an opamp level-shifter to offset the 2.5V close to 0V.

Click to expand...

 
 
okay, but is there an other sensor I should use to read amp from a cable?

BTCAD said:

I also noticed that the aip1 value varied from 477 to 580 when the oven use full 1440w, is it normal? That's why I should use a additional circuitry?

Click to expand...

 
477-525 = -48
580-525 = +55
 
That's close enough to equal "above and below" the zero for my money.