# 8 Hardware

A smart meter performs both current and voltage measurements to compute power.

**Current Measurement**

The resistor used for current measurement is placed in series with the incoming current. Hence, this resistance is chosen as small as possible. Power loss from this resistor is I^{2}R. For instance, if the resistance was 300 micro Ohms and the maximum current is 100 Amps. The power lost from this measurement is 3 W. The voltage across the resistor is V= I*R = 100Amps * 300 mico Ohms = 30 mV. Before performing an ADC (Analog to Digital Conversion) in a system with a maximum input voltage of 5 volts, the signal would need to be amplified by 167 = 5V/30 mV.

**Voltage Measurement**

The voltage measurement can be performed by putting a resistor(s) in parallel with the line. This is sometimes called a shunt resistor. For instance, if the resistance was 1 M Ohm and the transmission line was less than 1 Ohm, then only 1/10^{6} of the current would go through the 1 M Ohm resistor to perform the voltage measurement. If the maximum current was 100 Amps, the maximum loss would be P = I^{2}R = (100A/10^{6} )^{2} * 1 MOhm = 10 Watts.

**Analog-Digital Conversion Examples**

If a smart meter is measuring a 60 Hz signal and wants to get information up to the 10th harmonic, what frequency would it need to sample at? The 10th harmonic is 10*60Hz = 600 Hz. The Nyquist theorem states that the minimum sampling frequency has to be twice as high as the highest harmonic. Hence, the ADC sampling rate should be at least 1.2 kHz.

If a smart meter has a 16 bit ADC, what is the maximum error of real power due to quantization if the maximum voltage is 124 volts and the maximum current is 100 A?

The resolution of the ADC for 124 volts is 124v/2^{16} = 1.9 mV. The resolution of the ADC for 100 amps is 100/2^{16 }= 1.53 mA. Hence the maximum error would be I*V = 1.9mV * 1.53 mA = 2.89 micoWatts. (Ekanayake, 86)

**Works Cited**

Ekanayake, J. B. *Smart Grid : Technology and Applications*. Chichester, Wiley, 2012.