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Author Topic: To, or not to load up a transformer?  (Read 1401 times)
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I believe it has to do with filtering out harmonics associated with sampling the wave-form at up to 29ksps. The RC setup was recommended by the manufacturer, hence I followed their advice.
No.
The break frequency of 1K and 33pF is 9.6MHz, are you sure you have that capacitor value right?

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the Analog spec sheet shows a max input of +/-0.5V on the inputs
Can't see that can you point me at the page where it says that?
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I prefer using an isolation transformer (which I'm modeling as an AC source), you can see the simplified bipolar circuit above. What I noticed though is that the setup I'm proposing will result in a common-mode voltage whereas the Analog spec sheet shows a max input of +/-0.5V on the inputs but is mum re: common-mode voltage. Does this mean I should simply ground one leg of the AC transformer and then follow in the footsteps of the Analog AC design (i.e. a 47kOhm drop-down resistor instead of 600kOhm given that the transformer output is about 9VACrms)? Or do you think my bipolar design will work?

The circuit you provided does not define the DC common-mode voltage on the inputs of the ADE7753. Since it has an input range of +/- 0.5v for both inputs, I suggest the arrangement show in the attached schematic. This defines the common mode voltage as 0V and gives reasonable rejection of common-mode transients. Choose the ratio R1:R2 to give a little under +/- 0.5V peak at each input (remember that the transformer will deliver more than the nominal 6V rms when lightly loaded). Choose R1 + R2 to load the transformer lightly (R1 between about 1K and 10K). Then choose C to cancel out the phase shift in the transformer (see my earlier post) - it will be much larger than 33pF and will also help suppress mains-borne transients. Also I suggest grounding the transformer core.


* Scan 92.JPG (65.29 KB, 1654x1166 - viewed 12 times.)
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My apologies, gentlemen, when I replied I managed to enter the wrong multiplier for capacitors, it should have been 33nF, not 33pF. It's pretty likely that being off by three orders of magnitude had to do with Grumpy-Mike's question. smiley-eek-blue (Sorry! I'll blame our recent third kid for that malfunction.)

See p. 22-25 of the datasheet re the analog inputs discussion. That's also where Analog shows off it's +/- 0.5V input range for the ADC inside the ADE7753.

The voltage source is a nominal 230VACRMS input and 6VACRMS output transformer but the unloaded output is 9.2VACRMS per the spec sheet. Since this transformer is rated for 230VACRMS, I presume I can also run it at 115VACRMS, right? The secondary voltage would then be about 4.5 VACRMS, no?

Here is a drawing of what I think dc42 thought up for me. Could you confirm that I got the architecture right?

The drop-down resistors were chosen BTW to accomodate a 1.41 conversion from RMS to P-P as well as a 25% safety factor. (i.e. the ADE7753 will be able to read input voltages up to about 290VACRMS being applied to the primary side of the transformer.

The chip itself is OK with inputs as high as 6V without sustaining damage, well in excess of what I expect the transformer to ever survive. The transformer is on a circuit with a fuse and a MOV, hopefully that combination will kill any excesses before they can damage either the switchmode power supply or the transformer.

Thanks again for all the help.
« Last Edit: July 26, 2012, 07:54:52 am by Constantin » Logged

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Yes, that schematic describes what I suggested. You may need to increase the capacitors if you intend to compensate for the phase shift of the transformer.
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