pp 129-138
Active Filters
129
2 pi 2 pi x f2C 300 x 100 x 10-9
= 5.3 kO L R = 5.6 kO (E12 Series)
and R2 = 4 x 10 kO = 40 kO L R2 = 39 kO (Series E12)
The Sch.5.14 the response curve of the filter computers
with (as revealed by the simulation program mCap III).
Figure 5.14. PHY with K = 5 and f2 = 300 Hz: Response Curve
5.8 FTT 2nd order
The Sch.5.15 shows an active kykioma FTT 2nd order, type
Sallen-Key. It turns out that the transfer function is given by
the relation:
Figure 5.15. PHY second order
130
Analogue Electronics
(5.8.1)
where a is a coefficient whose value is found by an expert panel - (rates Butterworth) - and f is the frequency cutoff
filter, which is given by:
f = 1 (5.8.2)
Strengthening K again given by Eq. (5.6.2). This filter is second order, because the denominator of the transfer function, as we see from Eq. (5.8.1) is a polynomial of second degree. (This is because the circuit has two energy storage elements, ie the two capacitors). The slope of the filter is -40 dB / Oct. or -12 dB / oct, see Eq. (5.5.1).
It turns out that the Handbook of calculating this filter is given by the equations:
(5.8.3)
(5.8.4)
where:
(K-1) (5.8.5)
Example 5-3
We compute FTT Sallen-Key second order, Butterworth-type support with 5 and cutoff frequency f1 = 700 Hz.
Solution
From the table for special second order Butterworth filters in a = 1.414. We accept that R = 10 kQ.
Therefore,
l = a + ^ / a2 + 8 (K-1) = 1.414 W1.4142 + 8 x 4 = 7245,
l 7.245 = 4.12 nF LC1 = 39nF, (E12 series)
Active Filters
131
4o1 R 4 x 2 pi x 500 x 104
C = 4 12.6 nF LC2 = 12nF, (E12 series)
lo1 ch2p CH500CH R 7.245 104 2 k
Then we R1 = 10 KW and calculate the
R = (K - 1) R = 4 x 10 kn = 40 KW • R = 39 kO (line F12)
Fig 5.16 shows the filter and the calculated response curve.
Figure 5.16. PHY second degree Butterworth with f1 = 500 Hz, K = 10 and the response curve of
5.9 PHY second degree Sallen-Key
The Sch.5.17 shows the circuit of an active PHY 2nd order, type Sallen-Key. It turns out that the transfer function of this filter is:
where f2 is the frequency cutoff filter and a filter coefficient. The Handbook of calculating the filter is given by the equations:
132
Analogue Electronics
Figure 5.17. PHY second order
(5.9.1)
(5.9.2)
(5.9.3)
(5.9.4)
where,
l = a ^ a2 + 8K-1
(5.9.5)
The dc strengthen K still given by Eq. (5.6.2).
39 K
Active Filters
133
Example 5-4
Asked to calculate PHY Butterworth with f2 = 100 Hz and K = 10.
Solution
From the special table we find that Butterworth filters, filters for 2 nd order, a = 1.414.
Besides, we accept C = 100 nF
Working as in Example 5-3, we arrive at:
l = a + \ / a2 + 8 (K-1) = 1.414 + \ / 1.4142 + 8 x 9 = 10 016
R1 = - ^ = 4 = 6.36 kO k R1 = 6.8kO (E12)
'LoC 10,016 x 2 pi x 100 x 102 x 10-9'
L = 39.9 kO R9 = 39kO (E12)
Then we R3 = 10 kO. Thus, according to Eq. (5.6.2),
compute the
R4 = (K-1) R3 = 9 x 10 kO = 90 kO O R4 = 82 kO (E12)
The Sch.5.18 shows the filter curve calculated and hide my-
ing it.
Figure 5.18. PHY second order, f2 = 100Hz, K = 10 and the response curve of
134
Analogue Electronics
5.10 Narrowband Filter transit
To achieve the narrow band a FSZD, prosforoteros way is to supply the circuit with many anasyzefxeis. This technique is called Multi-anasyzefxis (PA) from the entrance to the exit of the filter. In particular, feeds the inverting (-) input of TE Sch.5.19, which generates 180 ° phase difference between the input and therefore the transfer function of this filter has a negative prosimo.
Figure 5.19. FSZD. Multi anasyzefxeon
The filter transfer function is given by:
(5.10.1)
where K is the strengthening of the filter center frequency fo (¬ frequent coordination quantity) and Q is the quality factor or growth factor (known by us from the selectivity coefficient of resonant circuits).
We should say that with this filter can have up to 10 Q, if we want to work properly.
The resonant frequency of the filter is by:
(5.10.2)
R2 = (5.10.4)
Active Filters
135
The Handbook of calculation of this filter is the following:
R1 = k ^ O (5 * 10 • 3)
R3 = (5.10.5)
o0 C (2Q2 - K)
where o0 = 2 pi / 0.
The filter aid is calculated by:
R2
k = 2 | R1 (5 • 1a6)
Example 5-5
We compute narrowband pass filter with a PA f0 = 500 Hz,
support 5 and Q = 10.
Solution
We choose C = 100 nF and we have:
R1 = Q, = 10 5 = 31.8kO L R1 = 33kO (E12)
'C o0 K 2 pi x 500 x 100 x 10-9 x 5'
R2 63.7kO L R2 = 68kO (E12)
2 o ^ C 2 pi x 500 x 100 x 10-9 2
R3 = Q = 10 = 163 R3 = O O 180O
3 ff> 0C (2Q2-K) 2T x 500 x 100 x 10-9 (200-5) 3
The filter shows the calculated Sch.5.20 with the response curve of the mCap III.
136
Analogue Electronics
Figure 5.20. FSZD PA with f0 = 500 Hz, K = 5, Q = 10 and the response curve of
5.11 narrowband filter cutoff
To Sch.5.21 shows the filter narrowband filter cut the tooth.
At the stopband frequencies faded side. Eg sometimes it is necessary to weaken the 50 Hz <to 400 Hz from the network <in the second case of an engine.
Figure 5.21. FZA (close) or tooth
The transfer function of a narrowband filter cutoff is given by:
(5.11.1)
Active Filters
137
where Q is the quality factor of the filter cutoff frequency f0 of the filter. The frequency f0, the filter is given by:
(5.11.2)
The formulas for calculating the filter apodeknyetai that given by
types:
R1 = c (5.11.3)
12Q othC
R2 = (5.11.4)
2 othC
R4 = 2Q2 R3 (5.11.5)
assuming appropriate values of C and R3.
Example 5-6
We compute filter tooth that will cut the frequency of 50 Hz and has Q = 10.
Solution
We choose C = 1 Mf and R3 = 1 kQ
R1 = ^ 1 ^ = 1 - = 159.2 l O R1 = 150O (E12)
1 2Q oo C 2 H 2 pi x 10 x 50 x 10-6 '
R4 = 2Q2R3 = 2 x 102 x 1 kO = 200 kO O R4 = 180 kO (E12)
The Sch.5.22 shows the filter and the calculated response curve.
Figure 5.22. Filter tooth with f0 = 50 Hz and Q = 10
