Passive Filter Design

[minimileman12]

Ok, so I've been working on getting simple first-order passive filters working in a simple circuit. Say you have something like this:
**broken link removed**

I have a 555 circuit in astable mode with a pot to control the frequency as the input into the filter circuit. I'm then measuring the output voltage of the filter circuit. When I change the frequency of the 555 circuit, the filter output hardly changes at all. I know that this first order circuit doesn't have a very steep transfer function so the change will not be very quick, but you would think that as you approach the extremes, the voltage would drop a good amount but it doesn't. Any suggestions?

 

[audioguru]

The extremely simple filter will not change its output voltage much when its input frequency changes.
The load for the filter must have a high resistance.

 

[misterT]

Changing the frequency does not change the average voltage of the signal. You'll have to change the duty cycle of the signal to get different dc-voltage levels.

 

[crutschow]

Changing the frequency does not change the average voltage of the signal. You'll have to change the duty cycle of the signal to get different dc-voltage levels.

You are referring to the DC average of a PWM type signal. The AC output of the filter for a 50% duty-cycle signal will drop when the frequency is above it's cutoff point.

 

[MrAl]

Ok, so I've been working on getting simple first-order passive filters working in a simple circuit. Say you have something like this:
**broken link removed**

I have a 555 circuit in astable mode with a pot to control the frequency as the input into the filter circuit. I'm then measuring the output voltage of the filter circuit. When I change the frequency of the 555 circuit, the filter output hardly changes at all. I know that this first order circuit doesn't have a very steep transfer function so the change will not be very quick, but you would think that as you approach the extremes, the voltage would drop a good amount but it doesn't. Any suggestions?

Hello there,

The kind of output you are looking for can only be seen when you use a sine wave input, not a square wave. A square wave contains lots of sine waves of higher and higher frequencies called harmonics, and if you were to include those in the calculation you would find that the output of your filter for a square wave would be an averaged type wave that is like smooth dc with ripple.

If you want to explore this kind of filter like that you really need to use a sine wave input.
If you dont have a sine wave generator then you'll either have to purchase one or ask here about how to build one. Once you have the sine wave generator you will be able to experiment with this filter a lot better and see exactly what you are looking for as far as amplitude, as long as your volt meter can respond to the kind of frequencies you are going to experiment with. You'll have to be careful there too as many meters are made for 60Hz or around there so they may not work very well at higher frequencies like 1kHz or 10kHz.

With a sine wave input you will see the output drop according to 1/(2*pi*R*C) as you expect it to.
The output amplitude is:
Vo=Vi/sqrt(w^2*R^2*C^2+1) and w=2*pi*F

which gives us an output of about 0.7071 when F=1/(2*pi*R*C)

 

[Mikebits]

In agreement with MrAl, see RC integrator in this tutorial.
**broken link removed**

 

[MrAl]

Hi again,


Just to note, if we include a load resistor across the capacitor (R2) and call the original resistor R1, the amplitude is:
Vo=Vin*R2/sqrt((R2+R1)^2+4*pi^2*C^2*F^2*R1^2*R2^2)

and now the -3db point is at:
Fo=sqrt(R2^2-2*R1*R2-R1^2)/(2*pi*C*R1*R2)

if it exists. Fo does not exist if (R2^2-2*R1*R2-R1^2)<0 however.

If the measuring device on the output has somewhat low impedance then we have to include that in parallel with R2 first before we do any of these calculations.

 

[RCinFLA]

The 555 operates with two comparitors set to 1/3 and 2/3 of Vcc supply.

The frequency is set by the time it takes to transition between these two comparitor trip points.

High R means it takes longer to transition between the comparitor points therefore a lower frequency. The waveform voltage always travels between the two comparitor points regardless of frequency.

 

[MrAl]

Hi again,


Just to note, if we include a load resistor across the capacitor (R2) and call the original resistor R1, the amplitude is:
Vo=Vin*R2/sqrt((R2+R1)^2+4*pi^2*C^2*F^2*R1^2*R2^2)

and now the -3db point is at:
Fo=sqrt(R2^2-2*R1*R2-R1^2)/(2*pi*C*R1*R2)

if it exists. Fo does not exist if (R2^2-2*R1*R2-R1^2)<0 however.

If the measuring device on the output has somewhat low impedance then we have to include that in parallel with R2 first before we do any of these calculations.

Hi again,


Another interesting frequency i forgot to mention is:
Fx=(R2+R1)/(2*pi*C*R1*R2)

which is the frequency at which the amplitude is down 3db from the zero frequency amplitude.
That is, the amplitude will be down by 3db from where it is in the pass band. The amplitude in the pass band is:
VoPB=Vin*R2/(R1+R2)
which is simply the voltage divider output voltage caused by R1 and R2.

When F=Fx the output voltage will be equal to 0.7071*VoPB, which is 3db down from 'normal'.