Hi All
I seem to be having a bad day. I am trying to calc to freq. resp. (-3db) of the following circuit.
Vin =A6 -- =A6 =A6 =A6 =A6 2K R1 =A6 =A6 -- =A6 =A6--------------- Vout -- =A6 =A6 =A6 ___=A6__ =A6 =A6 20K R2 C1 100nF =A6 =A6 ______ -- =A6 =A6 =A6 ---------------- =A6 ___ _
Cheers
WayneL
Low frequency gain is 20/22.
That drops off at high freqs. The 3dB corner freq is
F = 1/(2*pi*r*c)
where r = 2k || 20k
which is a tad under 1 KHz.
John
WayneL:
The resistance that the Capacitor "sees" is R1 || R2 = ` R1R2/(R1+R2) = 2X20/(2+20) = 1.818181......K ` Time constant (Tau) = RC = 1.8181818...E3 X 1E-7 = ` 1.81818E-4. f3db = 1/(2 X Pi X Tau) = 1/(2 X Pi X 1.81818E-4) = ' 8.7535E2 = 875.35 Hz
Unless your problem is a difference between the above prediction and an actual real live measument -- in which the first thing I would look at would be the impedance of the source, and the second would be the actual (not nominal) value of the components, particularly the cap.
You're not kidding it's a bad day, it's just the AC version of Ohm's Law:
Calculate R2||Xc as though Xc were a resistance - hint Xc = 1/wC. Call it Z.
Calculate the output of Vo = Vi Z/(R+Z).
Sort out to get the form G* 1/(1+wT) where G is some resistive combination and T is some sort of RC. Set wT = 1, solve for w hence f0.
You'll learn more if you DIY.
Paul Burke
2k || 20k isn't much different from 2k || infinity.
John
Yes, but if the source impedance is 2k then the 3dB frequency is nearly halved. Granted, that's a pretty high source impedance - unless it's some sort of transducer to be amplified...
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