Author Topic: High/Low Pass Filter Questions  (Read 782 times)

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High/Low Pass Filter Questions
« on: April 11, 2007, 07:47:38 PM »
Lately I've been playing around with the idea of designing my own booster to get a better understanding of how some of these effects work. Now I've read Cook Your Own Distortion and an article on Geofex and have a pretty good understanding of high and low pass filters and such.

Ok, say I have an input section like this:

C1 & R1 form a high pass filter, R2 & C3 (which I should've labelled C2 :P) form a low pass filter and the input impedance is the parallel value of R3 and R1. Right?

Okay, now my question is, when I'm working out the value of say, C1 (1/2*pi*R*Freq), does R3 play a role at all? Is R the parallel value of R1 and R3, or just R1? What about when I'm working out C3?

Rob Strand

Re: High/Low Pass Filter Questions
« Reply #1 on: April 11, 2007, 09:48:32 PM »
You are on the right track,

C1: R = R1 // (R2 + R3)

C3: R = R2 // R3

This assumes the centre of the two unmarked resistors have a large cap to ground, as it usually does.  If not then replace R3 above with R3'=(R3+Rx1//Rx2).

If the high-pass frequency is too close to the low pass frequency the above equations are not strictly correct because the two filters interract.
"If it turns out it’s like an onion with millions of layers… then that’s the way it is." — Richard Feynman


Re: High/Low Pass Filter Questions
« Reply #2 on: April 12, 2007, 03:13:09 AM »
This can't be right. :icon_eek:

Okay, so I want to have a large input impedance, say 5M. So I used these resistor values:

R1 = 10e6
R2 = 10e6 + 1e3 = 1.0001e7

RT = 1 / ([1/10e6] + [1/1.10001e7])
     = 5.00025e6

Basically 5meg, so that's all good, but now comes the weird part - calculating C1.

C1 = 1 / (2 * pi * 5.00025e6 * 20,000)
     = 1.59e-12

1.59pF! They don't even make caps that small do they?!


Re: High/Low Pass Filter Questions
« Reply #3 on: April 12, 2007, 04:10:30 AM »
You should move R1 to the input side of C1.  Right now it will set up a voltage divider with R3 dropping the bias voltage that you set up with the two bias supply resistors.  Also you should have a filter cap to ground at the junction of the two bias resistors.

The value of C1 will then set the low end roll off when figured in with R3 and the input imp of the device used.  C3 still sets the high end roll off with R2, but it is effected by the input imp to a degree.

Later, PaulC
Heritage amps/Tim & timmy pedals
I like ham, and jam, and spam alot


Re: High/Low Pass Filter Questions
« Reply #4 on: April 12, 2007, 04:20:13 AM »
C1 = 1 / (2 * pi * 5.00025e6 * 20,000)
     = 1.59e-12
1.59pF! They don't even make caps that small do they?!
This equation would be correct - if you wanted to exclude all frequencies below 20khz.

Actually, I think you want to exclude all frequencies below maybe 40Hz or 75Hz and pass the rest through.

So if you wanted to pass bass guitar at 40Hz, you might calculate

C1 = 1/ (2*pi*5E6*40) = 796E-12 = 796pF; might as well use 1000pf.

If you want to exclude everything over 20kHz, you do that with C2. In this case, the 1k resistor predominates and
C2 = 1/(2*pi*1E3*20E3) = 7nF = 7000pF.

An interesting point on interactions.  With the C1/R1/R3 stuff mixed up with the R3/C3 filter, this will not work as expected because the C1/1 network is loaded down by R2/C2. The impedance level of R2/C2 is too low.

If you rearrange the network so the input signal hits R2/1K first, is shunted to ground by C2 for high pass removal, then connects to R1 to ground and C1 in series to R3, then it works as intended. R2/C2 removes the highs, and is a low impedance source for the very high impedance C1/R1/R3 high pass section.

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