Electronic – Understanding CS5463’s acquisition circuit

adcanalogcomponentsintegrated-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.

acquisition circuit

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.

Best Answer

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).