http://www.winson.com.tw/Data%20Sheet/WCS1700.pdf (this link is only the sensor i have the module)
Hi. I have this hall sensor module . How to connect with arduino and which code should I use!
Vcc - to 5volts
Dout- ?
Gnd- to Gnd
Aout- ?
I Want for DC.I will put it on the positive cable? thank you.
You must provide a schematic of your module, there is at least a chip on the module which probably is some ADC but we must know which one it is and how it's wired to the sensor and the outgoing interface pins.
Where will I find them?
They should be supplied by the shop where you bought the part. Never buy parts if you don't get the schematics or you pay much more (in time) for it because you have to reverse engineer the needed schematics yourself.
Are shown in the photos?
No, at least not in these photos (much to blurry). Don't expect us to do your job. If you don't get the schematics from the supplier of the device, follow each trace on the board, identify all the chips and other components and draw the schematics yourself. And yes, this is time-consuming, that's why cheap boards are often more expensive in the end than the counterparts from the quality sites as Sparkfun, Adafruit and the like.
Is Chinese is excluded to have it..!
i found only this !!
product information:
Product Name: WCS1700 overcurrent detection sensor overcurrent / short circuit protection detection module
Features
First, the size: length 37mmX width 31mmX high 16.5mm
Second, the main chip: LM393 op amp, WCS1700
Third, the working voltage: DC DC 5V
Fourth, the characteristics:
1, with signal output indication;
2, the current detection range wide DC ± 70A AC: 50A;
3, the current detection resolution of 32mV / A;
4, the output over-current signal instructions;
5, over-current signal set the critical point adjustable, set the resolution of 3A;
6, with mounting holes, easy to install the firmware;
7, the sampling current conversion analog signal output, can be ADC, TTL level signal output, can be connected to the microcontroller IO port control.
8, the output signal is: analog signal and level signal.
Application: intelligent car motor overcurrent detection, stall protection current detection, short circuit protection testing, demonstration teaching experiments and other occasions.
Second, the main chip: LM393 op amp
LM393 is not an op amp but a comparator. I guess that one together with the potentiometer (blue) produces the Dout signal as soon as the measured current is above the adjusted limit.
This means you might get the output of the WCS1700 directly to the Aout pin. Using the standard analog input you get a resolution of about 0.2A and an accuracy of about 1A. Using the internal ref voltage you can push the resolution below 0.1A and the accuracy to about 0.2A.
As the description already says you can use it for an overcurrent detection but don't expect to measure current exactly. What do you want to use it for?
You can't use the internal reference as the signal is centred mid-rail, ie 2.5V for 0A.
Alas the device is not ratiometric so you'll have to recalibrate it if you change supply voltage.
Quite a nice looking little sensor - is there somewhere that sells just the sensor?
You can't use the internal reference as the signal is centred mid-rail, ie 2.5V for 0A.
That's for AC. Are you sure that this is also the case for DC? The datasheet might be interpreted like this but it doesn't explicitly mention the DC case.
pylon What do you want to use it for?
Hi. I want to stop the motor jet drive If the pump sticks to algae, bags, net, etc.The motor consumes continuously 1300W 24v 55A , if take over 60A to stop the motor.
i found this photo I do not know if it helps.
MarkT:
You can't use the internal reference as the signal is centred mid-rail, ie 2.5V for 0A.Alas the device is not ratiometric so you'll have to recalibrate it if you change supply voltage.
Quite a nice looking little sensor - is there somewhere that sells just the sensor?
Connect Aout to A0 on Arduino, then run this program and tell us what numbers you get with ZERO current. Also, which Arduino?
void setup()
{
Serial.begin(9600);
}
void loop()
{
Serial.println(analogRead(A0));
delay(666);
}
I found this.can somebody help me?
ADC0809.pdf (317 KB)
I found this.can somebody help me?
What help do you need?
The wiring is already done. The output of the analog input gives you the current with the following formula: I = (readA0 - 508) * 78 / 512. The 508 is empirically identified (see above), the 78 are simply 2.5V / 32mV and the 512 are the steps the Arduino returns for 2.5V on the analog input.
pylon:
What help do you need?The wiring is already done. The output of the analog input gives you the current with the following formula: I = (readA0 - 508) * 78 / 512. The 508 is empirically identified (see above), the 78 are simply 2.5V / 32mV and the 512 are the steps the Arduino returns for 2.5V on the analog input.
Hi! must put this I = (readA0 - 508) * 78 / 512 to a code?The potentiometer How will I use it?I turned the potentiometer right to the left and gave 7 amps but does not the led ligth on or of.I have a simple one hall sensor module if i go near a magnet the led turn on ,Should not it work like that?
must put this I = (readA0 - 508) * 78 / 512 to a code?
Yes, sure!
The potentiometer How will I use it?
If you use above formula you don't need it at all. The potentiometer is used to set a fixed level where the digital output is activated. I understood your intention to use the processor of the Arduino to decide at which level you start an action.
pylon:
Yes, sure!If you use above formula you don't need it at all. The potentiometer is used to set a fixed level where the digital output is activated. I understood your intention to use the processor of the Arduino to decide at which level you start an action.
Hi.Can you help me with the code? Which will I put them?And how I will tell arduino if AMPS it is more 60A? do that
Can you help me with the code?
I can help you but I won't write it for you. Try to write the code, post the results.
And how I will tell arduino if AMPS it is more 60A?
The formula returns the current in A, so the check for the limit should be simple.
