This means that I need to have a capacitor between the sensor and my pin 1?.
The transformer is just wrapped around one side of the magent, I do not have any specs for it but as in my first post i was able to get some readings regarding the ohms, I do not know if this helps.
Its common practice to connect power supply lines to 0V via a low value capacitor. They reduce any electrical noise on the supply lines so that the ic dosn't malfunction. They also help reduce any noise the AD may make, that could cause problems elsewhere in the circuit.
Look at the LIGHT BLUE rectangles, these are decoupling.
NOTE: the caps 10uF in the RED rect are also power decoupling.
When you are testing it may be necessary to add some more conditioning to the signal, read up on the function of CF2 shown in the image.
hi,
I cannot access the datasheet for the AD736 pcb DigiKey part.
The price is high £71GBP and they are showing zero stock.?
I would think for what you have in mind it would be an overkill, unless you have a number of AD736 systems to develop.
Just buy an AD736 and a plug in project board. Remember you need a low current dual power supply for the AD736, adjustable from about +/-3V they +/-12Vdc
hi,
I cannot access the datasheet for the AD736 pcb DigiKey part.
The price is high £71GBP and they are showing zero stock.?
I would think for what you have in mind it would be an overkill, unless you have a number of AD736 systems to develop.
Just buy an AD736 and a plug in project board. Remember you need a low current dual power supply for the AD736, adjustable from about +/-3V they +/-12Vdc
I've read with interest the thread above - I'm trying to do exactly the same thing without any success.
I have a CT with 3200 Turns, and have presently loaded it with a 100R resistor. This signal I'm now trying to get into a PIC adc, by way of an ADE as previous, but readings are way off.
Can I ask did the previous questioner get sorted with the ADE and CT, and if so is there any information/ complete schematic available.
I notice a contributor posted that the ADE must get a -3.5V supply, however the app notes suggest you can use a single rail?
In addition, does one connect the CT wires to pin1 and pin2?
I've read with interest the thread above - I'm trying to do exactly the same thing without any success.
I have a CT with 3200 Turns, and have presently loaded it with a 100R resistor. This signal I'm now trying to get into a PIC adc, by way of an ADE as previous, but readings are way off.
Can I ask did the previous questioner get sorted with the ADE and CT, and if so is there any information/ complete schematic available.
I notice a contributor posted that the ADE must get a -3.5V supply, however the app notes suggest you can use a single rail?
In addition, does one connect the CT wires to pin1 and pin2?
Another question: How do I measure the output? at the moment I am connecting my multimeter to the output and to the -v supply to the chip.
Is this correct?
With regards to the image posted:
Pin 1 & 2 are connecting to my Ac current transducer.
Pin 4 to my battery negative & Pin 7 to positive battery (2 AA @ 1.5V each).
Pin 6 is my expected output which i am measuring as stated above.
Now for the blue and red line do I need those connected too?
I am trying to get 0-3V output that represents my transducer current readings.
If I need the blue and red line what should the capacitor size be at CF2 and with regards to the 33µF on the red line what does the 16V+ mean.
Any helpful feedback on this are much appreciated.
Thats a good question:
My multimeter shows nothing on DC mv and if I switch over to AC I do not see anything either but I Assumed it was becuase it is not sensitive enough to pick AC at such low voltage.
I tried the Ohms and I can see a change when I put load on the live wire, so I assumed a signal must be comming through.
If you have any suggestions I will be heading off to Maplins in the next couple of hours..
Thats a good question:
My multimeter shows nothing on DC mv and if I switch over to AC I do not see anything either but I Assumed it was becuase it is not sensitive enough to pick AC at such low voltage.
I tried the Ohms and I can see a change when I put load on the live wire, so I assumed a signal must be comming through.
If you have any suggestions I will be heading off to Maplins in the next couple of hours..