Why both your oscilloscope are set to DC? Vin should be AC?
can i use 0.1uF?
Hi there,
One way to look at this is to calculate the capacitance required to get the resistive divider center tap to respond up to a percentage of the input.
For example, if we have a 1v peak input and we want the center tap to be within 90 percent of that value, it would reach 0.9v peak relative to the center tap dc value and everything else could be scaled accordingly.
The formula to calculate the capacitance is this:
C=(p*(R2+R1))/(sqrt(1-p^2)*w*R1*R2)
where
C is the capacitance in Farads,
p is the percentage,
R1 is the upper resistor (normally),
R2 is the lower resistor,
w is the angular frequency 2*pi*f where f is the frequency in Hertz.
For example, we have an ADC which reads the center tap, and its max input resistance for decent accuracy is 5k ohms. If we choose R1=10k and R2=10k the input resistance presented to the ADC is 5k ohms, so we satisfied that requirement. Next, we need to calculate the capacitance required to couple the signal to the resistive divider, so we use the equation above. Let's say we want 99 percent of the input to appear at the center tap (a reasonable amount). That means we make 'p' equal to 0.99. Since we are operating at 60 Hz, we make f=60, and since R1 and R2 are both 10k we use those values also, so we get:
C=(p*(R2+R1))/(sqrt(1-p^2)*w*R1*R2)
C=(0.99*(10000+10000))/(sqrt(1-0.99^2)*2*pi*60*10000*10000)
and all we did there was replace the variables with their values, and we get:
C=3.72e-6
which is about 3.7uf.
Now if we apply 1v peak to the input we will see 0.99v peak at the center tap, relative to the DC value there, so if we supply 5v DC to the resistive divider we would see 2.5+0.99=3.49 volts peak at the center of the resistive divider. The minimum value we would see at the tap is 2.50-0.99=1.51 volts. We would then make the algorithm to match this by scaling by the appropriate amount.
Note that if you need attenuation in addition to the coupling itself, you may need to insert a resistor in series with the cap to prevent the input from beaning the input to the ADC when the system is first switched on. This would require a few other calculations or some trial and error. Perhaps a better idea would be to use a clamp with some series resistance, or some series resistance and rely on the ADC input pin protection circuit to clamp the signal.