# Understanding CS5463's acquisition circuit

We can consider this question to be the sequence of this previous one.

Now I'm working on the current and voltage acquisition part (red marker) of the CS5463 (datasheet).

The CS5463 is an integrated power measurement device which combines two analog-to-digital converters, power calculation engine, energy-to-frequency converter, and a serial interface on a single chip. It is designed to accurately measure instantaneous current and voltage, and calculate VRMS, IRMS, instantaneous power, apparent power, active power, and reactive power for single-phase, 2- or 3-wire power metering applications.

First of all, one question about the spec:

• The datasheet page 16 contains: "the full-scale differential input voltage for the current and voltage channel is +/- 250 mV". From this, I assumed this channels can receive voltages from -250mV up to 250mV, and the reference for this values is the AGND pin. Am I right? (This information is kid of confirmed on pages 7 and 8.)

About the circuit itself, this time, I'm having more trouble to figure it out. My intuition says it's a simple voltage divider to measure the voltage and a high power resistor (shunt) to measure the current. The circuit is almost that, but I don't know the purpose of the extra components. And without knowing it, I can't define their values correctly.

• First question: why all the circuit considers the "line" as the ground and not the "neutral"? This way, all the waves are inverted to the originals. What's the advantage of doing this?

• What happens if I plug it inverted, in a way that the IC's "neutral" connects to the "line" of the outlet and vice versa?

• The R1 and R2 is a voltage divider block. The peak voltage in my region is 311V (220V RMS). Aiming for a 250mV peak, I chose the values 10 ohms and 12.4 kohms (if this last one doesn't exist, I'll put an approximate one or some resistors in series).

• The shunt resistor may be of 25 ohms. If a maximum current of 10A passes through it, it'll dissipate 2.5W and generate a 250mV voltage.

• How many mV should I leave as a safe range in this case? These values can be rezised to leave a safe range of 50mV above and bellow the waveform.

The remaining components:

• I can't identify their purposes. So, I can't size them correctly. The Cv+ capacitor has the grounds between its terminals, so what's it goal? This capacitor and the Rv- resistor appear to be a RC filter. But I've never seen one plugged on the ground. The Cv- and Cvdiff… I have no idea what they do. I can say the same for the equivalent components of the current sampling circuit, which also has the Ri+ resistor (without an analogue component on the voltage circuit).

The questions are scattered all over the question, but any doubts you have I try to summarize here below.

• Your schematic is confusing. What are the bold lines on the left? What do some of the signals end at a vertical line at the right, although they don't seem to be connected to it? You mention a datasheet, but datasheet of what? There is nothing in your circuit except a few resistors and capacitors. It feels like it will take too much back and forth to get the basic information, which is why I skipped over this question originally. Commented Sep 10, 2012 at 14:56
• The bold lines are the power grid lines. The image is a portion of the circuit we're interested (the link for the complete circuit image is just below this picture). I putted in this way, so it doesn't confuse things, because there are other parts that are not the focus of this question (but apparently this backfired). On the first paragraph I say that this is the circuit of the IC CS5463 (with the link to the datasheet). Commented Sep 10, 2012 at 15:10
• @OlinLathrop I've edited the post. Commented Sep 10, 2012 at 18:47

Differential voltage means exactly that - the difference of the two input lines (Vin+/Vin- and Iin+/Iin-). The CS5463 can measure up to 250 difference between them (or 50mV with a higher gain). Both signals are referenced to AGND, and can range from 0.25V below AGND up to Vref (max. 2.5V). This is stated on page 7 in the data sheet (the maximum rating are a little bit higher, see page 13).

Your first question is answered partly in the explanation of your example circuit (Figure 18 on page 41). The data sheet says In this type of shunt-resistor configuration, the common-mode level of the CS5463 must be referenced to the line side of the power line. (Though it doesn't state why). One reason might be that you then don't need to break the neutral wire (which should not be done, neutral should be as close to earth as possible).

If you revert the plug, nothing bad will happen to the circuit - as long as you isolate any outputs. When you interface with other circuits which might be ground-referenced, you can create short circuits. Depending on the cabling of you house and the appliance you want to measure, you might trigger a RCD. (I'm not an expert here, hopefully someone more knowledgeable can explain this better)

Your resistor calculation seems OK. The input voltages can actually be higher that 250mV, it is just that this is the maximum voltage you can measure (anything higher will read as 250mV). So if you want to be able to read higher voltages (where I live we have 230Vrms) or detect over-load conditions, you should leave a safety margin (50mV seems fine).

The filter are connected to ground because you have different kinds of noise. You might have noise only on one line, or on both lines. For the voltage input, I would size Rv- like R1, and for the current input significantly larger than Rshunt. Then the capacitors can be sized to have a cut-off frequency of several kilo-hertz, so you won't influence the measured signal. You might also want to look at the evaluation board schematics. There are examples for the values (though it's schematic looks a little bit different).

• Brilliant answer, Thanks! The link to the evaluation board will certainly help me a lot. Now just one thing to enlighten me... Since I obey the maximum limits, there would not be a problem if the value of Vin+ - Vin-, for example, is equal to 300mV, right? The only problem would happen is that the value would be read as 250mV, and this could be remedied leaving a security band. Right? Commented Sep 12, 2012 at 0:32
• Right. You get the measured values as percent of the full scale (see part 5.4 in the data sheet), so 300mV would be read as 100%, as would 250mV. To detect that, a security band would be needed.
– hli
Commented Sep 12, 2012 at 7:27