This time, I'll be talking about the power factor of AC power sources.
In the case of a DC power supply, if the current I (A) is the current when the power supply voltage E (V) is applied across the resistor R, then the power consumption is...
The power used is EI (W), which can be easily calculated, but in the case of AC power, it is more difficult to calculate.
It's not that simple.
The AC power supply load is just resistance.
Inductive loads (coils) and capacity
quantitative loads (capacitors), etc.
When a reactance component is included
(Impedance Z is the load)
The instantaneous voltage and instantaneous current of AC
A phase difference will occur.
Here, instantaneous voltage v (t), instantaneous current
i (t), phase difference θ Therefore, instantaneously
electric power p (t)teeth;
Figure 1: Current flowing through the impedance
················· （1）
Here,
V: Amplitude of instantaneous voltage
I: Amplitude of instantaneous current
ω: Angular frequency of AC voltage (= 2πf)
f: frequency of AC voltage
Generally, AC power is the time average of the product of instantaneous voltage and instantaneous current, so p(t) is the period.
T (=2 π / ω When averaged over time, the second term of equation (1) becomes 0 due to averaging, so
The current P is;
·················（2）
This is the result. Alternatively, v (t)and i (t The effective values of ) Vr, Ir Then equation (2) becomes:
···················(3)
;
In other words, AC power is the product of the magnitude (RMS value) of the AC voltage and AC current, plus the phase difference between them.
It is equal to the difference multiplied by its cosine cos(θ). Power factor It is called and the phase difference intersects
This is important for determining the current. Furthermore, this power P must be the power actually consumed.
from, Effective power (active power) The unit is watts (W). This is what we pay to the power company.
Electricity bills are also determined based on the cumulative effect of this effective power consumption.
In contrast, the product of the magnitude (effective value) of the AC voltage and AC current Vr ・ Ir of apparent power He said,
The unit is volt-ampere (VA). Also, the apparent power multiplied by sin(θ) Reactive power
It is called that.
Generally, AC power can be represented as a complex number, as shown in Figure 2.
Figure 2 Complex number representation of power
To summarize these relationships:
Effective power P: Real part of apparent power = S cos(θ)
Reactive power Q: Imaginary part of apparent power = S sin(θ)
Power factor cos (θ): (effective power)/(apparent power)
Now, how can quantities such as AC power and power factor be determined using an FFT analyzer?
Is that so?
As preparation, input the current waveform to Ch1 and the voltage waveform to Ch2 of the FFT analyzer.
(Method 1) When calculated using time axis calculations
As defined, the product of the current waveform and the voltage waveform is averaged over time to obtain the value (Figure 3).
Figure 3: Calculation of effective power on the time axis
The upper graph shows the instantaneous power waveform, and the "MEAN" value is the effective power calculated using the time-axis statistical calculation function.
In this example, we can see that it is 1.281 (W).
(Method 2) When calculated on the frequency axis
If Ch1 is the current waveform and Ch2 is the voltage waveform, then their cross spectrum is the spectrum of AC power.
This corresponds to; that is;
Cross-spectrum Mag: Apparent power
Cross-spectrum Real: Effective Power
Cross-spectrum image: reactive power
Cross-spectral phase: θ (Phase of power factor)
This is the result (Figure 4).
Figure 4 Cross-spectrum and complex power
Reading the real part, we get 1.281 (W), which is the same as Method 1.
Unlike the time-axis method, the method using the power factor allows for easy calculation of reactive power, power factor, etc.
It is possible.
Incidentally, the same process of inputting current waveform to Ch1 and voltage waveform to Ch2 into an FFT analyzer is used for transmission.
By measuring the function, we can determine Z (complex impedance) in Figure 1.
(Excerpt from the email newsletter issued on August 25, 2008)
