SCT-013 current sensor
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Many thanks Robert for you help, I've used your script code the result is:
empty power:
0.09
0.04
0.12
0.03
0.05
0.01
0.02
0.11
0.07
0.11Now for convert in amper how I do it? it sensor is at 15 A
Thanks a lot
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@StefanoF you need a known current and use that for scaling. So attach a load and a amperemeter, and then you get the number. As an estimation: the sensor is told to give 1Veff at 15A. At 1V i have had a reading of about 230. So the scaling is 230/15 = 1/15.33. that should be about the range. Do not make the measurement time too long. At the moment it is 1 second, but it could be shorter. The noise increases then, but the sensitivity to mains frequency variations decreases with shorter test times.
Edit: I repeated the test. Since my power supply is rated for 5A, i made three turns through the sensor, increasing the sensitivity by 3. At 5A, it shows 192. so the correction factor is 1/12.8. Above 12A my sensor goes slowly goes into saturation.
Edit 2:
Mains frequency is usually stabilized to be between 49.8 and 50.2 Hz. Only in very rare conditions it's outside. <50Hz means too much load. >50ZHz means too much power. Below 49.8Hz, first large industrial consumers are dropped step by step. e.g. At 47.5Hz, the power supply is completely shut off. Or at 51.5 Hz, all solar power generators are shut off, ... -
Many thanks Robert, I'm use your script code as a library :-),
the period at 500 millisecond I've good result
It also read 6 watts, incredible!!!Best regards
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@StefanoF To detail more on the dependency of the frequency I made a few tests:
Freq 1 sec 400ms 50 1.0 1.0 50.2 0.883 0.993 49.8 0.985 0.996Two results:
a) the numbers are is not symmetric. That is caused by the internal clock not being exactly in time. You can compensate for that by slightly modifying the argument to the class constructor, using e.g 50.08 (that would by the best value here) instead of 50.
For a assumed frequency of 50.08, I get:Freq 1 sec 400ms 50 1.0 1.0 50.2 0.934 0.996 49.8 0.933 0.996b) Shorter sampling times are less sensible to frequency variations.
Since you mentioned EMonlib a while ago. I looked into that code and saw, that is is almost identical to an approach which I had tested before. That has different pro's and con's. While the emonlib approach is not sensible to frequency variations, it has a large zero current reading. just because it is not frequency selective, it picks up all noise. In my test, the no current noise was equivalent to about 70 mA current. In comparison your zero current noise figures are equivalent to about 4 mA current.
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Hi Robert,
is possible know also the voltage AC 230.
Now is a constant 230 volt
Best regards
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@StefanoF I was not happy with the last change to use a virtual frequency. While it made the test symmetric, it increased the quiet current. So I looked for other possible reasons of the non-symmetry. The assumption was wrong, that the internal clock of the FiPy board I use for testing was wrong. Testing it with an SDR gave 159.998 MHz. That's just fine. But the sampling interval was wrong. It was set to 2000µs, but turned out to be 2003µs instead. So I added a correction, which keeps both the Zero current reading low and the filter symmetric. Code attached.
# import gc import math import array from utime import sleep_ms, sleep_us, ticks_us, ticks_diff from machine import ADC, Timer, idle, enable_irq, disable_irq, Pin, SPI # # acquire ADC values. The paramters of the constructor are: # 1. the mains frequency (number, typically 50 or 60) # 2. the sampling period (ms). The default is 2 ms. # class Acquire: def __init__(self, freq=50, *, sample_period=2): self.sample_period = sample_period self.omega = 2.0 * math.pi * freq / (1000 / sample_period) self.coeff = 2.0 * math.cos(self.omega) self.freq = freq self.adc = ADC(bits=9) def start(self, pin, time=200): gc.collect() # avoids automatic gc during sampling self.pin_adc = self.adc.channel(pin=pin, attn=ADC.ATTN_11DB) self.samples = time // self.sample_period # Test code for creating the artificial signal for phase noise # self.pi_f = (2 * math.pi * self.freq) / 1000000 # angle step per us # self.start_us = ticks_us() self.count = 0 self.busy = True self.q1 = 0.0 self.q2 = 0.0 self.alarm = Timer.Alarm(self.read_adc, 0, us=(self.sample_period * 1000 - 3), periodic=True) def stop(self): self.alarm.cancel() def read_adc(self, alarm): self.q0 = self.pin_adc() + self.coeff * self.q1 - self.q2 # Test code usign an artificial Signal # self.q0 = math.sin(ticks_diff(ticks_us(), self.start_us) * self.pi_f) self.q2 = self.q1 self.q1 = self.q0 self.count += 1 if self.count >= self.samples: self.alarm.cancel() self.busy = False def result(self): while self.busy == True: sleep_ms(self.sample_period) amplitude = 2 * math.sqrt(self.q1 * self.q1 + self.q2 * self.q2 - self.q1 * self.q2 * self.coeff) / self.count # if phase is required: # real = self.q1 - self.q2 * math.cos(self.omega) # imag = self.q2 * math.sin(self.omega) # amplitude = 2 * math.sqrt(real * real + imag * imag) / self.count # phase = math.atan2(real, imag) # return amplitude def reading(self, pin, time): self.start(pin, time) return self.result() # def run(period = 1000, n=10): acq = Acquire(50.0, sample_period=2) # 50 Hz sum = 0 for _ in range(n): value = acq.reading("P13", period) sum += value print (value) print ("average: ", sum/n) -
@StefanoF For the voltage you need a second sensor path, which is both independent from the current and isolating. You could use a small power transformer (like a 1 W model, 3 to 6 V AC output) or a optocoupler. I would prefer the transformer, because it is known to have a linear transfer characteristic, and the isolation is part of the design. The transformed has an AC output, just like SCT-013, which is too an transformer. You will need a voltage divider to bring that into a ~2.5Vpp range and the similar circuit like the SCT-013 to add a DC offset.
Once you have current and voltage with each an amount and a phase relative to the same reference, you can calculate the phase between current and voltage. -
@robert-hh
Hi,thank for support,
into my project there is a transformer 220 to 5 v DC
what is the wiring diagram to measure the voltage for Lopy ?Thanks a lot
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@StefanoF The 5vDC is a regulated output. That is of no help for you. You need an unregulated 50 Hz AC output.
Unregulated - because it has to follow the fluctuations of the 230V mains,
50 Hz AC - because you need the phase information to calculate the power factor cos(phi) between voltage and current. -
@robert-hh
I'm sorry, but not have space into my pcb , Can I use stc 013 variations set off 1.65 at empty?
Thanks a lot -
@StefanoF No. You cannot use anything from the SCT-013 sensor set-up, because this 1.65 reference point is derived from the regulated 3.3V supply. With two sources you could use a different interface circuit, but still you need a second "data" path for the 230V AC. I have attached a schematics which uses a common reference point for both signals. That is not my favorite one, but is simple and should work.
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@robert-hh
Hi,
Is there another alternative, just electronic components without transformer?
Example my electronic ammeter?
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@StefanoF The problem is isolation against electric shocks. The ammeter is isolated as a complete box. There is no metal from the inside that you can touch (at least you should not). And it is electrically floating, which your device is not. So you need something for both safety and waveform transfer.
As initially said, you may also use an optocoupler. But the transfer characteristic is not linear, it changes with the temperature, and they are not made for analog mode. But still that may do. I can make tests for that and see, how that works.
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@StefanoF Using an optocoupler one can be in a partially analog mode. I tested it NOT with 230V AC, but with a lower voltage variable transformer and a signal generator. The circuit is below. For 230V R6 has to be about 390kOhm.
The signal at P13 is:
The picture shows color grading. The red area is the hot one which the trace most often passes. For the the picture the function generator was used as AC source.
I measured the result with the goertzel test script, which only looks at the 50Hz component of the signal. For a constant input voltage the output readings vary by +/- 1 %.
But the output varies a lot with temperature. Cooling down the optocoupler increased the output by ~30%.
Sample picture below. The pink area is the one the trace swept over during temp variation. The numbers displayed by the goertzel script went from ~120 to ~165.
The Input/output transfer function is not linear. It is about Vout = x * Vin ** 0.7. But for a small range line 230V +/- 10% it's ok.
The blue line is Vin, the red Vout.
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@StefanoF So I made a few tests in the 230V range. Changed circuit with R=390k and a capacitor for less noise:
The transfer function is as below:
Not too bad, almost linear in the upper range. What stays is the huge temperature dependency.
Because the green and red line are almost identical, here the table:Volt Average Sdev dAvg/dVolt 100 23,56 0,55 125 40,18 0,33 0,6648 150 60,11 0,6 0,7972 175 82,32 0,83 0,8884 200 106 0,778 0,9472 210 115,5 0,68 0,95 220 126,1 0,832 1,06 230 135,3 0,88 0,92 240 145,73 0,73 1,043 250 157 0,549 1,127 260 167,88 0,638 1,088 270 179,28 0,883 1,14 -
Hi Robert,
thanks for you support,I try this way , please, the componets how many watts and tolerances?
Thanks advance
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Note. BE CAREFUL when testing to avoid an electric shock! I used an isolating variable output transformer for my tests to be safe.
Watts: the only part of interest is R6. So you have:
P = 230 * 230 / 390000 0.136 W
So that is negligible. The critical property of R6 is voltage capability. You have to get a 350V RMS type. That's typically better for higher power resistors. So get a 0.5W or 1W type.
The value of the resistor is to be determined. Since the transfer ratio of the optocoupler varies, you have to select a proper value such that at 230 V the output at R7 is about 2Vpp. You need an oscilloscope to tell. If you just have a Multimeter, use 0.7V AC. Below is a modified script capturing data from P13 and P14 (current and voltage). It returns the raw values and the phase difference between the two signals. That phase difference may have an offset, since the voltage signal is only half of the sine wave.
# import gc import math import array from utime import sleep_ms, sleep_us, ticks_us, ticks_diff from machine import ADC, Timer, idle, enable_irq, disable_irq, Pin, SPI # # acquire ADC values. The paramters of the constructor are: # 1. the mains frequency (number, typically 50 or 60) # 2. the sampling period (ms). The default is 2 ms. # class Acquire: def __init__(self, freq=50, *, sample_period=2): self.sample_period = sample_period self.omega = 2.0 * math.pi * freq / (1000 / sample_period) self.coeff = 2.0 * math.cos(self.omega) self.freq = freq self.adc = ADC(bits=9) def start(self, pin1, pin2, time=200): gc.collect() # avoids automatic gc during sampling self.pin_adc1 = self.adc.channel(pin=pin1, attn=ADC.ATTN_11DB) self.pin_adc2 = self.adc.channel(pin=pin2, attn=ADC.ATTN_11DB) self.samples = time // self.sample_period # Test code for creating the artificial signal for phase noise # self.pi_f = (2 * math.pi * self.freq) / 1000000 # angle step per us # self.start_us = ticks_us() self.count = 0 self.busy = True self.q1_1 = 0.0 self.q2_1 = 0.0 self.q1_2 = 0.0 self.q2_2 = 0.0 self.alarm = Timer.Alarm(self.read_adc, 0, us=(self.sample_period * 1000 - 3), periodic=True) def stop(self): self.alarm.cancel() def read_adc(self, alarm): state = disable_irq() v1 = self.pin_adc1() v2 = self.pin_adc2() # Test code using an artificial Signal # v1 = math.sin(ticks_diff(ticks_us(), self.start_us) * self.pi_f) # v2 = math.sin(ticks_diff(ticks_us(), self.start_us) * self.pi_f) enable_irq(state) self.q0_1 = v1 + self.coeff * self.q1_1 - self.q2_1 self.q2_1 = self.q1_1 self.q1_1 = self.q0_1 self.q0_2 = v2 + self.coeff * self.q1_2 - self.q2_2 self.q2_2 = self.q1_2 self.q1_2 = self.q0_2 self.count += 1 if self.count >= self.samples: self.alarm.cancel() self.busy = False def result(self): while self.busy == True: sleep_ms(self.sample_period) real = self.q1_1 - self.q2_1 * math.cos(self.omega) imag = self.q2_1 * math.sin(self.omega) amplitude_1 = 2 * math.sqrt(real * real + imag * imag) / self.count phase_1 = math.atan2(real, imag) real = self.q1_2 - self.q2_2 * math.cos(self.omega) imag = self.q2_2 * math.sin(self.omega) amplitude_2 = 2 * math.sqrt(real * real + imag * imag) / self.count phase_2 = math.atan2(real, imag) # diff = 40e-6 * 2 * self.freq * math.pi # phase diff caused by 40µs sampling delay phase = phase_1 - phase_2 - diff # if phase > math.pi: phase -= 2 * math.pi elif phase < -math.pi: phase += 2 * math.pi return amplitude_1, amplitude_2, phase def reading(self, pin1, pin2, time): self.start(pin1, pin2, time) return self.result() # def run(period = 500, n=10): acq = Acquire(50.0, sample_period=1) # 50 Hz l=[] for _ in range(n): value = acq.reading("P13", "P14", period) l.append(value[0]) print (value, math.cos(value[2])) avg = sum(l)/n sqsum = 0 for _ in l: sqsum += (_ - avg) * (_ - avg) print ("average: ", avg, "SDev:", math.sqrt(sqsum/n)) -
@robert-hh
Hello,
I built my electronic circuit, now how should i use the result of the value [2] to get the correct voltage?value = acq.reading("P13", "P14", 500)
print(value[2])Thank a lot
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@Stefano-Floriani The values returned by acq.reading are the two values from P13 and P14 and the phase difference phi between the signals at P13 and P14. So if P13 is connected to the SCT013 sensor, and P14 to the voltage sensor, you will get the current I from value[0], and the voltage U from value[1]. The number in value[2] can then be used to calculate the effective power consumed by the load, which is
Peff = U * I * cos(phi)
with I derived from value[0], U derived from value[1] and phi being value[2], which already has the proper scaling. The number cos(phi) is called the power factor. It is 1 for pure resistive loads, and 0 for pure capacitive or inductive loads. Values close to 1 are common, values close to 0 are rare.
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@robert-hh
Dear Robert,
The reading voltage with external voltmeter it's:
At 230 v and U from value[1] it's from 4
At 220 v and U from value[1] it's from 2How can I find the right formula?
Thanks a lot
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