# MCP3909 Energy metering IC , How to use the HFOUT in it?

I'm using the Microchip MCP3909 Energy metering IC, How to use the HFOUT in it ?

note:
HFout supplies a frequency output proportional to the average active real power .

This frequency should be sent to the meter on HFOUT pin which access a Timer/Counter in MC and according to number of Pulses I should measure power and use it

The problem is that I couldn't reach the correct relation which determine each pulse represents what amount of watt?

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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.

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Ok , so in MCU I will count a pulse then count the time till next pulse , this will be the power ??? –  xsari3x Jul 11 '12 at 16:24
I need to use this HFOUT to get a value from this chip , can you help ?? It's my graduation project ....will be delivered in 2 days & I'm stuck in this –  xsari3x Jul 11 '12 at 16:26
The energy, not the power. Power is watts, energy is kWh. If it's calibrated at 3600 pulses/kWh, then a 1000 W load will generate these 3600 pulses in 1 hour, or 1 per second. A 500 W load will give 1 pulse per 2 seconds. And so on. –  stevenvh Jul 11 '12 at 16:26
this is the device data sheet , ww1.microchip.com/downloads/en/DeviceDoc/22025a.pdf what u saw was application note , .... sorry Is there something in the datasheet that shows the setting of hfout to imp/kwh ? –  xsari3x Jul 11 '12 at 16:41
I added a concrete example with realistic values to my answer. –  stevenvh Jul 12 '12 at 5:04

The datasheet for the MCP3909 device has a table (table 4-3 on page 20) which shows how the HFout pin may be programmed with the F0, F1 & F2 pins for various sensitivities.

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I don't understand this –  xsari3x Jul 11 '12 at 16:23
Is this related to imp/kwh some how ? –  xsari3x Jul 11 '12 at 16:42