Impedance Effect into Filter

Started by jcknowles89, December 10, 2019, 01:35:46 PM

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jcknowles89

Hello all,

I've been struggling with this for a while and cannot seem to figure this out. Can anyone tell me the effect of an impedance value going INTO a filter? so for example a high pass filter (Cap in series with Resistor to Ground) has a certain cut off frequency by itself but I'm convinced it has a different effect with higher impedance.

Using 1/2*Pi*Resistor*Capacitor I can get the cut off frequency under Ideal circumstances.

However if say there is a 39K impedance coming from the output of a 12AX7 or 10K coming from a guitar pickup how does that effect that calculation?

anyone have a formula update? or perhaps a different formula to use?

Thanks,

PRR

So, a coupling network?

At infinite frequency the cap is zero impedance. The 39k and the to-ground resistor makes a voltage divider. So for say 78k resistor, your max gain drops from 50 to 33. Your total resistance is 117k. Cap reactance works against that to define your corner.

It is usually expedient to pick your to-ground resistor "large", like 390k, so the small error vanishes in overall tolerance and slop. That's another great thing about amplifiers-- we usually do not have to "match" for best power transfer, but can mis-match for "voltage transfer" and make-up Power losses in amplification.
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antonis

#2
I'm too lazy to solve equations so you may proceed to calculate cut-off frequency by the following:

R=0.707(Zin+Xc+R), where Zin = Input impedance (set in series with cap), Xc=1/(2πfC), f=frequency of interest, 0.707=Vout/Vin ratio (for -3dB attenuation..)

As Paul said, just consider Zin + Xc as the upper equivalent resistor of a voltage divider.. :icon_wink:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

jcknowles89

Thanks Guys! time to brush up on my Algebra lol. continuing on that topic of impedance going into a filter. Do you know if there is a way to change the Impedance of an AC signal? if seen this done with the primary of a transformer, so theoretically this could be done with an Inductor?  anyone have a way of knowing what size inductor would create what impedance if it is possible?

thanks,

Colton

PRR

> time to brush up on my Algebra lol.
> a way of knowing what size inductor would create what impedance if it is possible?


Reboot your math. Then read up on fundamental AC theory.

The basic impedance of an inductance is 2*Pi*F*L. 1 Henry at 400Hz is 2*3.14*400Hz*1H is about 2,400 Ohms.
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