Datasheet

STPM10 Theory of operation
Doc ID 17728 Rev 4 41/53
Equation 12
[see Figure 26 - 10]
After these two operations, another stage performs the subtraction between the results p
2
and p
1
and a division by 2, obtaining the active power:
Equation 13
[see Figure 26 - 11]
In this way, the AC part V•I•cos(2
ωt + ϕ)/2 has been removed from the instantaneous power.
The absence of any AC component allows for a very fast calibration procedure. It requires
only the setting of (using the internal device programming registers) the voltage and current
sensor conversion constants, using the effective voltage and current (V
rms
, I
rms
) readings
provided by the device’s built-in communication port, avoiding the time-averaged readings of
the active power or the need for line synchronization.
7.23.2 Reactive power
The reactive power is produced using the previously-computed signals. In case of shunt
sensor the voltage signal is derived while the current signal is not. A first computation is to
multiply the DS value of the integrated voltage channel with the value of the integrated
current channel, which yields:
Equation 14
The second is to multiply the filtered DS value of the voltage channel with the value of the
filtered current channel:
Equation 15
From the above results, Q1(t) is proportional to 1/
ω, while Q2(t) is proportional to ω. The
correct reactive power would result from the following formula:
p
2
t() vt() it()⋅
VI ϕcos⋅⋅
2
---------------------------------
VI 2wtϕ+()cos⋅⋅
2
-------------------------------------------------------––==
pt()
p
2
t() p
1
t()–()
2
------------------------------------
VI ϕcos⋅⋅
2
---------------------------------==
Q
1
t() vt()dt I⋅ t() vt() It()⋅= V ωtsin()
I
ω
--
ωt ϕ+()cos–
⎝⎠
⎛⎞
VI
2
ω
------
ϕsin 2ωt ϕ+()sin–()⋅=⋅=
∫
=
Q
2
t() vt() I⋅ t() vωωtcos I⋅= ωt ϕ+()sin
VI
2
------
ωϕsin((⋅ 2ωt ϕ+())sin+⋅==