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Why not load your voltage sensing rectifier with a resistor, and then use an active low pass filter to filter out all the ripple ?
That should respond much faster.
In your circuit, the cap C35 is being charged by R64. They will determine the charging time constant. R65 will discharge the capacitor beyond the potential set by R64/(R64+R65). Roughly speaking, your output voltage is being integrated over a time period of 1 msec; This is excellent in my opinion.
To understand (rather crude, in the mathematical sense) the operation of this part of the circuit:
1. Consider an arbitrary waveform Vin(t)- it may have some spikes or noises...
2. Consider another square pulse of width 1 ms and unity amplitude...
3. Multiply the two voltages - point by point and add them... (you are taking an average within the window)...
4. The result will be Vout(t)
5. Move the square pulse by one point to the right and repeat the calculation...
6. You get the output waveform. It will not have any spikes that are much shorter than 1ms.
In reality the window of 1ms is not exactly flat-top because the cap is both charging and discharging at the same time but this the idea...
If your mains frequency is 50Hz, ripple will be at 100Hz, or one ripple cycle every 10mS.
How can it possibly have a time constant of 1mS and still get rid of the ripple ?
The best you will ever be able to do is average over several cycles.
What you need is a very sharp low pass filter set to something like 20Hz.
That will get rid of all the spikes and ripple, and the dc coming out will follow the average ac amplitude reasonably quickly in either direction.
The sharper the cut off frequency, the higher you can make it, and if your low pass filter has a good phase response without excessive ringing or overshoot it should do everything expected of it fairly well.
The formula is simple: R*C; if R is in Ohms and C in farads, the product will be in secs.
You can increase R or C or both and get a higher time constant. It will be better to filter the power supply ripple at the input side. If you try to use a rather long time constant on the output side, your transient response (at the output) will suffer.
It will be better to filter the power supply ripple at the input side. If you try to use a rather long time constant on the output side, your transient response (at the output) will suffer.
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