In the circuit attached R17 is the burden resistor for a CT which measures 50/60 Hz currents. The switch is a CMOS analog one, not mechanical. When the switch is closed, the gain is about R16/R14 = 49.9k/832 = 60. When the switch is open, the gain is R16/(R14+R15) = 49.9k/(832+49k) = 1.
The cutoff frequency of the low pass filter is 1/(2pi*R16*C17) = 678 Hz
Now my question is about the AC coupling cap which forms a high pass filter.
When the switch is closed, the cutoff frequency is 1/(2pi*R14*C18) = 4 Hz and time constant RC = 39 ms
When the switch is open, the cutoff frequency is 1/(2pi*(R14+R15)*C18) = 0.068 Hz and time constant RC = 2.34 s
What are the effects of this time constants on my 60 Hz signal (16.67 ms period)?
The cutoff frequency of the low pass filter is 1/(2pi*R16*C17) = 678 Hz
Now my question is about the AC coupling cap which forms a high pass filter.
When the switch is closed, the cutoff frequency is 1/(2pi*R14*C18) = 4 Hz and time constant RC = 39 ms
When the switch is open, the cutoff frequency is 1/(2pi*(R14+R15)*C18) = 0.068 Hz and time constant RC = 2.34 s
What are the effects of this time constants on my 60 Hz signal (16.67 ms period)?
