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 = 1/2RC = (6.28)(10k)(0.001uf) = 15.9khz
Capacitive Reactance
Below cutoff frequency at 1khz
Xc = 1/2fC
= 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_ =__ 2vpp159.235k1.99v
Z
159.548k
At cutoff frequency
Xc
= 1/(6.28)(15.9khz)(0.001uf) = 10k
Z = √ R2+Xc2
= √10k2(10k2) = 14.142k
Vout = _VinXc_ =__ 2vpp10k1.414v
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_ =__ 2vpp1.592k0.3144v
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 = 2fc = __1__
2RC
R =
__1__
2fcC
C =
__1__
2fcR
Design procedure 40db Butterworth low pass filter:
Identify/choose the desired cutoff frequency 1.125khz
fc = _0.707_
2RC
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_
2fcC1
=
______0.707________
6.281.125khz)(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 = 2fc = __1__
2RC3
Choose C3
=
0.01uf
C1 =0.5C3
=
0.005uf
C2 = 2C3
=
0.02uf
Use the formula below to find resistor R
R = __1__
=
_________1_________
2fcC3
=
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 = 1/2RC = (6.28)(10k)(0.001uf) = 15.9khz
Capacitive Reactance
Below cutoff frequency at 1khz
Xc = 1/2fC
= 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_ =__ 2vpp10k0.1253v
Z
159.548k
At cutoff frequency
Xc
= 1/(6.28)(15.9khz)(0.001uf) = 10k
Z = √ R2+Xc2
= √10k2(10k2) = 14.142k
Vout = _VinXc_ =__ 2vpp10k1.414v
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_ =__ 2vpp10k1.975v
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 = 2fc = __1__
RC
Chose C
Find R
R = __1__ = _______1________
2fcC
6.28(15.9khz).001uf
C =
= 0.001uf
= 10k
__1__
2fcR
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
2fcC1 (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 = 2fc = __1__
R3C
C1 = C2 = C3
=
0.01uf
Find R3
R3 = __1__
= _______1________
= 15.923k
2fcC3
(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