crutschow
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Note that this circuit detects the peak AC voltage.
To detect the true RMS would require the addition of a true RMS conversion IC.
View attachment 152815
I don't think it's peak because of the filter capacitor after the diode bridge. This capacitor tends to average the rectified dc pulses.Hi,
This means peak value measurement.
It is even less precise than your previous method, because it takes only a very short time of the fullwave into account. All the rest of the time the signal is ignored.
Yes, I guess I may have mixed up precision and accuracy but I can't affect accuracy very much in this case since that would depend on how much the actual mains waveform differs from the ideal sine wave and how much harmonics is present on the power line at any particular time.It seems you mix accuracy and precision.
If you don't need accuracy but precision to be within +/-2% for a small area around the setpoint...then this eases design.
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Use excel to do a simple simulation:
Let's use a sine with 10% third overtone.
V(t)= sin(t) + 0.1 × sin(3 × t + phi)
Phi is phase shift.
RMS calculation gives a constant value independent of phi.
Every other method will vary the output value depending on phi...even if the amplitude of both sine is constant...
Klaus
Hi,
You say "some"... Must ask: Is wide-ranging hysteresis (between VthL and VthH) possible with the part or does it seem to have a somewhat limited/restrictive range?
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The larger the capacitance, the closer all points on the filtered waveform gets to ((Vac*1.414) - Vd)*2/pi
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I don't think it's peak because of the filter capacitor after the diode bridge. This capacitor tends to average the rectified dc pulses.
The larger the capacitance, the closer all points on the filtered waveform gets to ((Vac*1.414)-Vd)*2/pi
That's the purpose of the pot.....I hope that would be able to do the job. I'll just have to tune the TL431 reference voltage divider resistors to obtain my setpoint.
The current transfer ratio may drop, but at that low current it's unlikely to degrade significantly.I noticed that the current through the optocoupler LED is less than 2mA. Please, over time, how will CTR degradation affect the performance of the opto?
I just ran simulation of crutschow's circuit on LTspice. After the bridge rectifier, the average and the RMS values of the voltage equals the peak.
I was wrong with averaging it as if I was averaging a sine wave.
Now that LTspice is reporting Vpk = Vrms = Vavg, we can just use the Vdc = ((Vac*1.414) - 2*Vd) relationship.
I hope that would be able to do the job. I'll just have to tune the TL431 reference voltage divider resistors to obtain my setpoint.
I still don´t know which value you want to have. Thus I assume you want RMS.but I can't affect accuracy very much in this case since that would depend on how much the actual mains waveform differs from the ideal sine wave and how much harmonics is present on the power line at any particular time.
Hi,
...
We talk about distortion (overtones) and noise ... hopefully you are aware that your simulation is with unrealistic perfect sinewave.
I recommend to do some simulation runs which include noise and distortion to find out your expectable precision.
(Otherwise I see the simulation rather useless...)
I still don´t know which value you want to have. Thus I assume you want RMS.
Maybe I was not clear with my previous posts:
* You can have accurate RMS values if you do true RMS measurement.
* You can have accurate peak values if you do peak value measurement.
* You can have rectified accurate average values if you do rectified average measurment.
* But you can´t have accurate RMS values when you do a peak measurement....but this is what you do.
According accuracy (and precision)
* Peak value method is the worst (Thus I don´t recommend it)
* rectified average may be good or not: It depends on your requirements and the signal wavefrom.
* RMS method is perfect.
Klaus
Below is the updated schematic to include a 4N32 optocoupler with a Darlington output for a higher CTR, giving a more conservative design.
View attachment 152864
Nobody can help without your input.Okay. I think I need some help here.
I understand that noise can ride on the peak of the mains voltage and can probably cause false triggering. Assuming the mains voltage is very far away from the true sine wave, how can I achieve true RMS or rectified average in this case that I need a single bit digital signal from the monitor?
The PC817 has a minimum CTR of 50% at a 5mA input, but that should be sufficient to work reliably in the circuit at >1mA with a 47k ohm output load.Okay. I think I need some help here.
Actually, I'm feeding the output into a CMOS. With that I thought I could use PC817 optocoupler and increase the emitter resistor to say 47kohm in value so I can hopefully achieve saturation for a long time but I'm not sure whether the idea of using PC817 here is a good one.
I am really interested with cost saving although I'm not going to do it at the expense of a long lasting device.
Below is the circuit modified to respond to the average (filtered) value of the rectified input, which is much less sensitive to noise and distortion, as compared to the peak voltage.
View attachment 152904
R7 and C2 act as a low-pass filter/running integrator, which gives an average value of the rectified sine-wave (actually with the load added by R2, U3, and R3 the voltage becomes about 1/2 the average value).
Note that R7 is connected to the unfiltered, rectified output.
This is peak filtered by D6 and C1 to generate DC power for the rest of the circuit.
I think you meant that Vmains_rectfd = 0.5*((Vac*1.414) - 2*Vd)*2/pi...
(...the voltage becomes about 1/2 the average value).
...
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