Electronic – MCP 3909 energy metering chip problem

From the Reference Design User's Guide, page 42:

"The meter constant is typically given in units of impulses per kilo-watt hour. As an example, the calibration output frequency of CF, METER_CONSTANT = 3200 imp/kWh or 6400 imp/kWh"

Section 5.1.1 on page 41 describes the calibration procedure.

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Also had a look at the datasheet, and on the same page 20 Mike refers to there's this equation:

Channel 1 is the voltage sense channel, channel 0 is the current sense. Multiplied, including the current sense amplification you get instantaneous power. Input voltages are converted relative to a reference voltage, so you have to divide both channels by that (the denominator). HFc is a constant you can program by setting F0, F1 and F2, as per the table in Mike's answer (page 20 of the datasheet).

Note that this gives you a frequency proportional to power. Integrating this over time means counting periods of that frequency gives you a measure of consumed energy, and that's where the 3200 imp/kWh from the Reference Design User's Guide comes in.

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A concrete example. Let's suppose you want to measure energy of a 230 V/ 10 A device. Maximum input for the voltage channel is 660 mV, so we choose a 1:1000 divider for that. At a gain of 1 for the current channel maximum input voltage is 470 mV, so we can choose a 10 mΩ shunt for that, which gives us 10 mV/A. We use the internal Vref of 2.4 V, and set F0, F1 and F2 to 0. Then for a 230 V/ 10 A input we get

\$ HF_{OUT} = \dfrac{8.06 \times 230 mV \times 100 mV \times 1 \times 109.25 Hz}{(2.4 V)^2} = 3.516 Hz \$

That's for 2300 W, or 1.529 Hz per kW. For 1 kWh we'll have

\$ \text{pulse count} = 3600 s \times 1.529 Hz = 5503 \$

That's 5503 pulses/kWh.

Your circuit link was bizarre but I think I got the idea: -

This is the nearest I could find and this works but yours won't (very accurately) because you've put all the gain into your one and only amplifier (equiv to A1 with R2 and R4 at 67kohm and R1 and R3 at 1kohm). Think about what the output will be when no current flows i.e. the input voltages are the same.

Theoretically it will be 0V but a 1% change in the value of one of these resistors will generate a large error voltage on the op-amp output that gives you a false reading because the common-mode rejection ratio of A1 totally relies on perfect resistor matching then, imperfections are multiplied by gain (=67)

This is why the circuit I've shown works a lot better - errors in the matching of the 1st stage will be multiplied by a gain of 0.1111 (approximately) i.e. there will be minor errors for a 1% mismatch in resistors.

A2 does the main amplification and is set for a gain of 250. Overall, gain is 27.78 and smaller than yours but A2's gain can be increased.

Should you still have common-mode problems there is this: -

Other related TI products. I'm also aware the Linear technology produce a bundle of high-side current monitors.