Capacitive Filters

Thread Starter

Jaziek

Joined Apr 21, 2011
4
I have a high pass and low pass filter, which look like this,



R1 = 1kΩ R2 = 10kΩ , and C1 = 10 nF

I need to find the breakpoint values for each of these filters. When I put them into LTSpice, it gives me 14.1kHz for the low pass filter, and 1.8kHz for the high pass filter.

But when I use the equation 1/(2∏RC) to work out what the values should be ideally, it gives me something noticeably different.

I know that this is because I'm not properly taking into account the second resistor, but I don't know how I to do this.

How do I work out cutoff frequencies for filters in circuits with more than 1 resistor?

thanks :p
 
Last edited:

t_n_k

Joined Mar 6, 2009
5,455
If you add a load resistor Rload, then for the purpose of analysis R2 is replaced with Rload||R2. [ || meaning "in parallel with" ]
 

Thread Starter

Jaziek

Joined Apr 21, 2011
4
If you add a load resistor Rload, then for the purpose of analysis R2 is replaced with Rload||R2. [ || meaning "in parallel with" ]
Sorry I guess I phrased my question badly, used terminology wrongly and whatnot, I'm very new to electronics in general.

What I mean is, I don't know how to properly account for R2, because I've only ever seen examples of filters with 1 resistor.
 

Heavydoody

Joined Jul 31, 2009
140
R1 = 1kΩ R2 = 10kΩ , and C1 = 10 nF

I need to find the breakpoint values for each of these filters. When I put them into LTSpice, it gives me 14.1kHz for the low pass filter, and 1.8kHz for the high pass filter.

But when I use the equation 1/(2∏RC) to work out what the values should be ideally, it gives me something noticeably different.

I know that this is because I'm not properly taking into account the second resistor, but I don't know how I to do this.

How do I work out cutoff frequencies for filters in circuits with more than 1 resistor?

thanks :p
Your question demonstrates the inadequacy of merely memorizing "formulas." Looking at case B, notice that we are seeking the voltage across R2, which is part of a basic voltage divider. Therefore:

\(v_o=\frac{R_2}{R_2+R_1+\frac{1}{j\omega C_1}}v_i\)

Dividing through by R2+R1, we get:

\(v_o=(\frac{R_2}{R_2+R_1})\frac{1}{1-j\frac{1}{\omega C_1(R_2+R_1)}}v_i\)

Now we have separated the resistive component of the divider, which attenuates the signal regardless of frequency, from the reactive component, which IS dependent on frequency (note the presence of ω). Half power (cutoff) occurs when:

\(\frac{1}{sqrt{1-j\frac{1}{\omega C_1(R_2+R_1)}^2}}=\frac{1}{\sqrt{2}}\)

Therefore, at cutoff:

\(\omega C_1(R_2+R_1)=1\)

Solving for ω:

\(\omega=\frac{1}{C_1(R_2+R_1)}\)

Convert angular frequency to cyclical and you have your answer. This technique (NOT this formula) can be applied to part A as well.
 
Last edited:

Thread Starter

Jaziek

Joined Apr 21, 2011
4
Your question demonstrates the inadequacy of merely memorizing "formulas."
It does indeed, but everybody has to start somewhere :p

Looking at case B, notice that we are seeking the voltage across R2, which is part of a basic voltage divider. Therefore:

\(v_o=\frac{R_2}{R_2+R_1+\frac{1}{j\omega C_1}}v_i\)

Dividing through by R2+R1, we get:

\(v_o=(\frac{R_2}{R_2+R_1})\frac{1}{1-j\frac{1}{\omega C_1(R_2+R_1)}}v_i\)

Now we have separated the resistive component of the divider, which attenuates the signal regardless of frequency, from the reactive component, which IS dependent on frequency (note the presence of ω). Half power (cutoff) occurs when:

\(\frac{1}{sqrt{1-j\frac{1}{\omega C_1(R_2+R_1)}^2}}=\frac{1}{\sqrt{2}}\)

Therefore, at cutoff:

\(\omega C_1(R_2+R_1)=1\)

Solving for ω:

\(\omega=\frac{1}{C_1(R_2+R_1)}\)

Convert angular frequency to cyclical and you have your answer. This technique (NOT this formula) can be applied to part A as well.
Thanks for helping. I'll give this a try.
 

Thread Starter

Jaziek

Joined Apr 21, 2011
4
so if I understand this right, which I probably dont,

For the low pass filter, I need to find the parallel impedance of R2 and C1, like this...

\( Z_p = \frac {Z_1 Z_2} {Z_1 + Z_2} \)

where z1 and z2 are r2 and c1

right?

then put that into a voltage divider with R1, which gives

\( \frac {V_o}{V_i} = \frac {Z_p}{R_1 + Z_p} \)

I'm having trouble simplifying what I get at this point to anything meaningful though.
 

Heavydoody

Joined Jul 31, 2009
140
so if I understand this right, which I probably dont,
Actually, yes you do. Good work.

\( \frac {V_o}{V_i} = \frac {Z_p}{R_1 + Z_p} \)

I'm having trouble simplifying what I get at this point to anything meaningful though.
Yes, it gets a bit tricky here. Try dividing through with Zp. Then expand Zp and continue simplifying from there. You should eventually end up with:

\(\frac{v_o}{v_i}=(1+\frac{R_1}{R_2})\frac{1}{1+j \frac{\omega R_1R_2C_1}{R_2+R_1}}\)
 
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