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**OBJECTIVE:**

At the heart of an electricity meter is a transducer that at uniform time intervals senses the instantaneous voltage **V** and the instantaneous current **I** on each phase, multiplies these parameters together and cumulates the products. For calculating absorbed energy in kWh this cumulation is signed addition. For calculating kVA this cumulation occurs only if the **(V I)** product is positive.

A two element three phase transducer for delta connected circuits has two product determining elements whose outputs are summed together in the kWh and kVAh cumulation registers. A three element three phase transducer for wye circuits has three product determining elements whose outputs are summed together in the kWh and kVAh cumulation registers. In high performance power transducers each element has two dedicated analog to digital converters, one for voltage and one for current.

The voltage inputs to a power transducer are usually nominally 120 VAC RMS. The current inputs to a power transducer are usually nominally 5 A AC RMS full scale or 200 A AC RMS full scale. This web page addresses the case of 200 A AC RMS full scale which is typical for a single family residence. The calculations for other load sizes are analogous.

Input instantaneous voltage range = +/- 256 V, resolution = (1 / 128) V

Input instantaneous current range = +/- 256 A, resolution = (1 / 128) A

Desired transducer output = 4096 counts / kWh

For adequate response to the 30th harmonic of 60 Hz the minimum required sampling rate = 3600 samples / second

= (3600)^2 samples / hour

The actual sampling rate = Cx X 3600 samples / s

= 1.32560719 X 3600 samples / s

= **4772.185884 samples / second**

which gives response out to the 40th harmonic of 60 Hz and which yields an output of 4096 counts / kWh after dividing the cumulation register count by:

[2^36] = 16 (256)^4.

The sampling rate must be precisely maintained by a crystal controlled clock.

At the maximum rated instantaneous voltage and the maximum rated instantaneous current the value of (V I) per sample is:

full scale count product / sample = [(256)(128)(256)(128)]

= (256 volts) X 256 amps X [1 kW / 1000 V-A] X [1 / (Cx X 3600)]s X [1 h / 3600 s]

where:

1 < Cx < 2

or

(128)(128) = [1 / (Cx X (3.6)^2 X 10^9)] kWh

or

Cx X (3.6)^2 X 10^9 X (128)^2 = 1 kWh

or

(Cx X (1.8)^2 X 10^9)(256)^2 = 1 kWh

or

[(Cx X 3.24 X 10^9 X (256)^2) / 2^N] = 4096

or

Cx = (2^N X 4096) / [(256)^2 (3.24 X 10^9)]

= (2^N X 256 X 16) / [(256)^2 (3.24 X 10^9)]

= (2^N ) / [(16 X (3.24 X 10^9)]

Try N = 36:

Then:

2^N = [16 (256)^4]

and

**Cx** = [16 X (256)^4] / [16 X 3.24 X 10^9)]

= (256)^4 / (3.24 X 10^9)

= **1.32560719**

Thus the sampling period is:

1 s / [1.32560719 (3600)] = **209.5475793 microseconds**

Thus dividing the cumulation count by:

2^36 = 16 (256)^4

gives an output pulse rate of:

** 4096 pulses / kWh**

At a load of 1 kW the output pulse rate is:

(4096 pulses / kWh) X (1 kW) = 4096 pulses / h

This pulse rate is easy to time with a stop watch to check transducer performance.

In principal this pulse rate could easily be increased to give more resolution and accuracy. However, the higher the count per kWh the more data that must be handled by the downstream smart metering and billing systems. From that perspective there is little practical merit in a transducer output pulse rate of over 4096 pulses / kWh. If power is calculated over a 0.25 hour interval the power resolution is less than than 1 Watt.

The maximum time available for doing the multiplications, additions and logical steps related to each sample is:

209.5 microseconds

which should be sufficient if the transducer processor has a 2 byte X 2 byte hardware multiply.

The accuracy of this power transducer is limited by the absolute accuracy of the analog to digital converters that are used. These digital to analog converters should all have large signal step response times of less than 200 microseconds.

This web page last updated January 20, 2016.

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