Theory

AN2288

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Theory

The main objective of this application note is to demonstrate that a low cost energy meter

can be implemented without use of an external dedicated device (ASSP). Only a single 10-

bit SAR ADC is used to perform voltage and current measurements. The ADC on

ST7FLITE20 device accepts input from 0 to Vdd. Since the Vdd is +5V, the operable range

of the ADC is 0 to +5V. Since the ADC is operable in the positive range only, the AC input

signals of voltage and current have to shifted up and centered around +2.5V. This is

achieved by biasing one end of the secondaries of current transformers and one end of

potential divider with +2.5V. The input waveform of current and voltage to ADC is shown in

figure1.

Figure 1.

AC input signals to ADC

Four ADC input channels are used, three for current and one for voltage measurement. One

of the current channels is multiplexed with two current gain factors x128 and x512. The

sampling rate for each channel is 5kHz. This higher rate of sampling is there to reduce the

quantization noise. The active energy consumed is calculated based on instantaneous value

of power. The sampling rate is 5kHz, so, with a 50Hz sinewave the number of samples per

cycle is 100. The sampling time is 200µs. After every 200µs, one current and voltage sample

is taken and multiplied to get an instantaneous power sample. The discrete summation of

these power samples over time gives the active energy.

When the phase difference between voltage and current is 0, the active power will be at the

maximum which is equal to total power, and the reactive power will be 0.

If v(t) = Vsin(

ωt) and i(t) = Isin(ωt - φ), the instantaneous power:

After averaging, we get cos(2

ωt - φ)=0 for a cycle of sinewave. So, the theoretical average

power (P) is VIcos(

φ)/2. If ik and vk are respectively the instantaneous values of the current

and the voltage, the discrete formula of average power with regular, simultaneous, voltage

and current samples is:

Vdd

Vdd/2

V

t

Voltage Waveform

Current Waveform

Pt

()

vt

() it

()

×

=

VI 2

⁄

()

φ

()

cos

2

ωt φ

–

()

cos

–

()

=

P

d

1N

⁄

()

vk

ik

×

K=0

N-1

∑

=