Costant time - Electronic Circuits Projects Diagrams Free

marcoo_ · 11th June 2008, 02:51 PM

Hi, i would like to know what are time constants of this circuit.
I mean the time constants of both capacitor.
I know this is an active band pass filter, but i'm not interested about the center frequency of the filter, i would like to know the formula of both cutoff frequencies.
Thanks (sorry for my english...)

Hero999 · 13th June 2008, 05:45 PM

Are you talking about the time or frequency domain?

marcoo_ · 15th June 2008, 04:22 PM

Hi, i'm interested about the cutoff frequiences introduced by capacitors.

Hero999 · 15th June 2008, 06:17 PM

The gain of the amplifier is simply -R3/R1.

C1 and C2 determine the lower and upper cut-off frequencies respectively.
$F_{C_{lower}} = \frac{1}{2\pi R1 C1}$
$F_{C_{upper}} = \frac{1}{2\pi R1 C2}$

I don't know what R2 does, it will reduce the gain but not by much as it's at a vitual earth potential, I'd remove it if I were you. If you want add a resistor to limit the effect of the bias currents then it should be the same value as R3 and be inbetween the +input and 0V.

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11th June 2008, 02:51 PM	(permalink)
marcoo_	Costant time Hi, i would like to know what are time constants of this circuit. I mean the time constants of both capacitor. I know this is an active band pass filter, but i'm not interested about the center frequency of the filter, i would like to know the formula of both cutoff frequencies. Thanks (sorry for my english...)

13th June 2008, 05:45 PM	(permalink)
Hero999	Are you talking about the time or frequency domain? __________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez

15th June 2008, 04:22 PM	(permalink)
marcoo_	Hi, i'm interested about the cutoff frequiences introduced by capacitors.

15th June 2008, 06:17 PM	(permalink)
Hero999	The gain of the amplifier is simply -R3/R1. C1 and C2 determine the lower and upper cut-off frequencies respectively. $F_{C_{lower}} = \frac{1}{2\pi R1 C1}$ $F_{C_{upper}} = \frac{1}{2\pi R1 C2}$ I don't know what R2 does, it will reduce the gain but not by much as it's at a vitual earth potential, I'd remove it if I were you. If you want add a resistor to limit the effect of the bias currents then it should be the same value as R3 and be inbetween the +input and 0V. __________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez