??? 02/07/11 11:00 Read: times |
#181010 - Re: Waveform comparison Responding to: ???'s previous message |
Actually, in domestic inverters, you compare the feedback with the drive, and not with a sine table. Not sure whether this is correct, but this is what seems to be working I think it is mainly because of the cost pressure on the BoM. Presently what is "working" for me in this design is: I am taking 32 measurements and averaging them simply. Comparing this average value with two different thresholds (High and low) and controlling the multiplier accordingly. This works even when the sinewave is clipped since more clipping gives more values at peak level thus increasing the average. The output voltage is not constant at 230V. It varies between 226V to 235V. I think this should be an 'acceptable' solution. Applying point by point waveform correction seems quite quite difficult with 16F72, which is a "cost-driven" choice for this design. |
Topic | Author | Date |
How to measure a clipped and non-clipped sinewave accurately | 01/01/70 00:00 | |
peak voltages does not represent RMS if distorted curve | 01/01/70 00:00 | |
RE: How to measure a clipped and non-clipped sinewave | 01/01/70 00:00 | |
Still can't look at a single point in time | 01/01/70 00:00 | |
RE: Still can't look at a single point in time | 01/01/70 00:00 | |
peak-to-RMS relation In general | 01/01/70 00:00 | |
I have posted this thread on some other forum | 01/01/70 00:00 | |
Sine feedback for inverter | 01/01/70 00:00 | |
Absolutely | 01/01/70 00:00 | |
Re: Sine feedback for inverter | 01/01/70 00:00 | |
hints | 01/01/70 00:00 | |
Waveform comparison | 01/01/70 00:00 | |
What if? | 01/01/70 00:00 | |
Drive and table | 01/01/70 00:00 | |
Re: Waveform comparison | 01/01/70 00:00 |