EDITOR: B. SOMANATHAN NAIR
1. PROBLEMS OF PASSIVE
FILTERS
·
Cascading several stages of filters produces loading effect.
·
To increase the output voltage, external amplifiers have to
be used.
·
Impedance matching becomes a big problem with passive
filters.
·
If inductors are used in filters, they are bound to produce
noise.
2. ACTIVE (OPERATIONAL AMPLIFIER) ANALOG
FILTERS
Active filters employ
active devices such as operational amplifiers (opamps) for obtaining the
desired filter characteristic. The advantages of active filters are:
·
Filters of any order of may be realized easily.
·
Loading effect does not occur when several stages are
cascaded.
·
No external amplifiers are required; opamps used for
constructing the filters themselves will act as amplifiers.
·
No impedance-matching problems arise since opamps have very
high input impedance and very low output impedance.
·
No inductors are used; they are rather simulated. Therefore
noise problems are reduced.
·
Highly complex and complicated filter circuits can be
realized in the form of ICs.
·
Chip area can be saved by using switched-filter capacitors.
·
Highly reliable due to the use of ICs.
·
Low power consumption, especially when CMOS opamps are used.
·
Long life.
·
Low cost.
·
Light weight.
·
Free form shock and vibration.
The following are the
disadvantages of active filters:
·
Maximum operating frequency is limited as opamps are not
normally operated beyond a few MHz.
·
No repair of individual components is possible; once an
active filter fails, it has to be totally replaced.
3.1
FIRST-ORDER KRC LOW-PASS FILTER
It can be seen from Fig. 1 that the
first-order KRC filter is a simple RC low-pass network with a non-inverting amplifier
attached to it. The analysis of RC LPF had already been done in an earlier blog and the
transfer function was obtained as
H(s) = ωc/(s+ ωc)
(1)
where ωc = 1/RC and s = jω. In (1), the filter components are R and C. Since
there is only one equation, we choose either R or C as the component readily available to us
and determine the value of the other component. It is quite common to select C first
and then compute R. The inverting-amplifier is designed as usual.
3.2 ANALYSIS OF THE SECOND-ORDER KRC LOW-PASS FILTER
The analysis of the second-order KRC low-pass
filter is made with reference to the circuit shown in Fig. 2.
We use node-voltage method for this
analysis. First consider node X. With respect to this node, we write the nodal
equation:
(Vs ‒ Vx)/R = (Vx ‒ Vi)/R +
(Vx ‒ Vo)sC
(2)
Multiplying (2) throughout with R yields
Vs ‒ Vx = Vx ‒ Vi + (Vx ‒ Vo)sCR (3)
Rearranging (3), we find
Vs = Vx (2+sCR) ‒ Vi ‒ sCRVo
(4)
At the input node, we find:
(Vx ‒ Vi)/R = VisC (5)
Rearranging (5)
Vx = ‒Vi(1+sCR)
(6)
Substituting for Vx from (6)
into (4), we obtain
Vs = Vi [(2+sCR)( 1+sCR ‒ 1] ‒ sCRVo
(7)
Simplifying (7) yields
Vs = Vi [1+3sCR)+(sCR)2] ‒ sCRVo (8)
We have the expression for the input voltage
of an amplifier
Vi = Vo/A (9)
Using (9) into (8), we get
Vs = (Vo/A) [1+3sCR)+(sCR)2] ‒ sCRVo (10)
Rearranging (10) gives the relation
AVs =Vo [1+3sCR)+(sCR)2‒
AsCR] (11)
From (11), we get the transfer function of the KRC low-pass
filter as
H(s) =Vo/Vs =A/[1+ (3 ‒ A)sCR)+(sCR)2] (12)
Simplifying (12) further yields
H(s) = A/(CR)2/ [s2+ (3 ‒ A)s/CR)+(1/CR)2] (13)
Let CR = 1/ωc resonant frequency of the filter. Then (13) becomes the standard
expression for the second-order active low-pass filter-transfer function:
H(s) = Aωc2/ [s2+ (3 ‒ A)ωcs)+ ωc2] (14)
Equation (14) may now be compared with the expression for the transfer
function of the passive second-order RC low-pass filter
given by the equation
H(s) = ωc2/ [s2+ 2δωcs)+ ωc2] (15)
Now, by comparing (14) and (15), we find that
3 ‒ A = 2δωc (16)
From Table 1, we find for the second-order
filter coefficient s = 1.414. Using this we get
3 ‒ A = 1.414(17)
which yields the gain of the
second-order filter amplifier as
A = 1.586 (18)
The amplifier shown in Fig. 2 is a
non-inverting amplifier. The resistors in the amplifier can therefore be
designed form the relation
A = 1 + (R2/R1) = 1.586
(19)
which yields the relation
R2 = 0.586 R1 (20)
Equation (20) is the equation for the
design of the amplifier used in the LPF.
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