Passive lowpass filter circuit. R2 Vin 10kΩ Vout C1 0.001µF Fig.1 Passive lowpass filter circuit The Low Pass Filter – the low pass filter only allows low frequency signals from 0Hz to its cut-off frequency, ƒc point to pass while blocking those any higher How do the circuit works? We already know in fundamentals of electricity that the capacitor has an AC resistance Xc that is a function of frequency, and Xc is inversely proportional to the frequency ( f) of the AC voltage that flows through it. If the AC input voltage has 1hz fed to the circuit Xc will be very large and as frequency f of the AC input voltage reaches fc (cutoff frequency) Xc starts to decease and as f continually increases it will come to a point where Xc reaches 0 resistance. If the input voltage for the circuit is 2vpp Cutoff frequency (fc) fc = 1/2RC = (6.28)(10k)(0.001uf) = 15.9khz Capacitive Reactance Below cutoff frequency at 1khz Xc = 1/2fC = 1/(6.28)(1khz)(0.001uf) = 159.235k R2 Vin 10kΩ Vout Xc Fig.2 Lowpass filter equivalent circuit 1 And the AC total resistance of the circuit is termed as impedance (Z) and it is the square root of the summation of R2 & Xc2 Z = √ R2+Xc2 At cutoff frequency = √10k2(159.235k = 159.548k If the input voltage for the circuit is 2vpp using voltage divider formula Vout = _VinXc_ =__ 2vpp159.235k1.99v Z 159.548k At cutoff frequency Xc = 1/(6.28)(15.9khz)(0.001uf) = 10k Z = √ R2+Xc2 = √10k2(10k2) = 14.142k Vout = _VinXc_ =__ 2vpp10k1.414v Z 14.142k Above cutoff frequency Xc = 1/(6.28)(100khz)(0.001uf) = 1.592k Z = √ R2+Xc2 = √10k2(1.59k2) = 10.125k Vout = _VinXc_ =__ 2vpp1.592k0.3144v Z 10.125k Vin = 2vpp 1hz 100hz 1khz 15.9kz 50khz 2v 1.414v 0.314v Vout 100khz fig.3 frequency response of a 20db low pass filter 2 R1 70kΩ VCC 15V 7 R2 1 5 U1 3 70kΩ 6 741 2 4 VEE -15V Fig.4 Non-inverting buffer/unity gain amplifier Fig.1 is a non-inverting unity gain or buffer amplifier R2 is the input resistance of the circuit and R1 is the feedback resistor. Since the differential voltage between the inverting and non-inverting input is virtually shorted, thus it can be seen that R2 is virtually connected to R1, giving us a non-inverting amplifier with a closed loop gain (ACL) = Rf/Rin = R1/R2 since R1 = R2 = 1 Active Low pass filter R1 10kΩ VCC 15V 7 R2 Vin 1 5 U1 3 10kΩ 6 2 741 Vout C1 0.001µF 4 VEE -15V Fig.5 20db active low pass filter Integrating a non-inverting unity gain amplifier will limit the lost due to the passive devices the response to the input voltage is basically the same. With the desired frequency in mind, designing a low pass filter will require you to arbitrarily choose the value of the capacitor that you are going to use in the range of 0.001uf to less than 1uf since it is much easier to find the resistors commercially than commercial value of capacitors. 3 Design procedure 20db Low pass filter 1. Identify/choose the desired cutoff frequency fc 2. Choose the value of your capacitor 3. Use the formula below to find resistor R fc = 2fc = __1__ 2RC R = __1__ 2fcC C = __1__ 2fcR Design procedure 40db Butterworth low pass filter: Identify/choose the desired cutoff frequency 1.125khz fc = _0.707_ 2RC Choose the value of your capacitor C1 = 0.01uf Make C2 = 2C1 = 0.02uf Use the formula below to find resistor R R = _0.707_ 2fcC1 = ______0.707________ 6.281.125khz)(0.01uf) Chose Rf =2R = 20k = 10k C2 0.02µF Rf 20kΩ VCC 7 Vin R1 R2 10kΩ 10kΩ 15V 1 5 U1 3 6 741 2 C1 0.01µF Vout 4 VEE -15V Fig.6 40db lowpass active filter circuit 4 Design procedure 60db low pass filter: Identify/choose the cutoff frequency = 995.23hz fc = 2fc = __1__ 2RC3 Choose C3 = 0.01uf C1 =0.5C3 = 0.005uf C2 = 2C3 = 0.02uf Use the formula below to find resistor R R = __1__ = _________1_________ 2fcC3 = 6.28)(995.23hz)(0.01uf) R1 = R2 = R3 = R = 16k Rf1 = 2R = 32k Rf2 = R = 16k = 16k C2 0.02µF Rf1 Rf2 32kΩ 16kΩ VCC 15V VCC 15V 7 R1 16kΩ Vin R2 1 5 U1 7 3 16kΩ 6 741 1 5 6 2 741 Vout 4 VEE -15V U2 3 16kΩ 2 C1 0.005µF R4 C3 0.01µF 4 VEE -15V Fig.7 60db active highpass filter circuit 5 Passive highpass filter circuit. C1 0.001µF Vin R2 10kΩ Vout Fig.8 Passive highpass filter circuit The Low Pass Filter is a filter allows low frequency signals above its cut-off frequency, ƒc point to pass while blocking those below it. How do the circuit works? If the AC input voltage has 1hz fed to the circuit Xc will be very large and as frequency f of the AC input voltage reaches fc (cutoff frequency) Xc starts to decease and as f continually increases it will come to a point where Xc reaches 0 resistance. If the input voltage for the circuit is 2vpp Cutoff frequency (fc) fc = 1/2RC = (6.28)(10k)(0.001uf) = 15.9khz Capacitive Reactance Below cutoff frequency at 1khz Xc = 1/2fC = 1/(6.28)(1khz)(0.001uf) = 159.235k Xc Vin R2 10kΩ Vout Fig.9 High pass filter equivalent circuit 6 Impedance (Z) is the total AC resistance of the RC filter circuit and it is the square root of the summation of R2 & Xc2 Z = √ R2+Xc2 At cutoff frequency = √10k2(159.235k = 159.548k If the input voltage for the circuit is 2vpp using voltage divider formula Vout = _VinXc_ =__ 2vpp10k0.1253v Z 159.548k At cutoff frequency Xc = 1/(6.28)(15.9khz)(0.001uf) = 10k Z = √ R2+Xc2 = √10k2(10k2) = 14.142k Vout = _VinXc_ =__ 2vpp10k1.414v Z 14.142k Above cutoff frequency Xc = 1/(6.28)(100khz)(0.001uf) = 1.592k Z = √ R2+Xc2 = √10k2(1.59k2) = 10.125k Vout = _VinXc_ =__ 2vpp10k1.975v Z 10.125k 2v 1.414v Vin = 2vpp Vout 1hz 100hz 1khz 15.9kz 50khz 100khz fig.10 frequency response of active high pass filter circuit 7 Design procedure for 20db Active High pass filter Choose/identify the cutoff frequency fc = 15.9khz Fc = 2fc = __1__ RC Chose C Find R R = __1__ = _______1________ 2fcC 6.28(15.9khz).001uf C = = 0.001uf = 10k __1__ 2fcR Rf 10kΩ VCC 15V 7 C 1 5 U1 3 Vin 0.001µF 6 R 10kΩ 2 741 Vout 4 VEE -15V Fig. 11 20db active high pass filter circuit 8 Design procedure for 40db active high pass filter circuit Choose the cutoff frequency = 1khz fc = _1.414_ RC Pick C1, = 0.01uf Make C1 = C2 = 0.01uf R1 = _1.414_ = ____1.414_______ = 22.5k 2fcC1 (6.28)(1khz)(0.01uf) R2 = ( ½)R1 or = R1 = 22.5k Chose Rf =R2 = 22.5k R2 11.3kΩ Rf 22.5kΩ VCC C1 7 C2 15V 1 5 U1 3 0.01µF Vin 6 741 0.01µF 2 R1 22.5kΩ Vout 4 VEE -15V Fig.12 40db active filter circuit 9 Design procedure for 60dbActive filter circuit Choose /identify the cutoff frequency fc = 1khz fc = 2fc = __1__ R3C C1 = C2 = C3 = 0.01uf Find R3 R3 = __1__ = _______1________ = 15.923k 2fcC3 (6.28)(1khz)(0.01uf) R1 = 2R3 = R2 = (½)R3 = Rf1 = R1 = Rf2 = R3 = 31.846k 7.9615k 31.846k 15.923k R2 7.965kΩ Rf1 Rf2 31.846kΩ VCC 15.923kΩ 15V VCC 15V C1 7 C2 1 5 U1 3 0.01µF Vin 6 741 0.01µF 2 R1 31.846kΩ 7 C3 5 U2 3 0.01µF 4 1 6 R3 15.923kΩ VEE 2 741 Vout 4 -15V VEE -15V Fig. 13 60db active high pass filter circuit 10