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OP-AMP ACTIVE FILTERS

8.35 The circuit shown in Figure P8.35 is an active filter with C = 1 F, R = 10 k andRL = 1 k. Determine (a) the gain (in decibels) in the passband; (b) the cutofffrequency; (c) whether this is a low or high pass filter.

1. First, identify the type of filter: LP or HP ormixed configuration i.e. BP or BR

2. If it doesnt resemble directly to the abovestandard type (as is the case here), derive anexpression for Gain (transfer-function)

3. Since it is in inverting configura

tion, use gain =ZF / Z S (see eqn 8.45, pg 413).

4. Compare your expression with the high-passfilter (see eqn 8.56, pg 415 for HP filter).

Any relationship?

(hint: value of R s=0)

RC j jv

v

S

o =)(

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8.35 continued

5. The above expression gives you gain! Convert itin decibel units (logarithmic computation).

6. The Gain at cut-off frequency is found byequating to 0 in the Gain expression.

7. By varying the frequency from 0 to a largenumber, and observing the Gain magnitude, youwill be able to tell if it is a LP (high gain at low

freq. and 0 gain at high freq.s) or, a HP (0 gainat low freq. and high gain at high freq.s)

8. More details of active filters can be found frompg 413 to pg 418.

)log(.20|)(| RC jvv

dBS

o =

)log(.20|)(| RC jA dBHP =

dBRC jA dBHP 3)log(.20|)(| 0=

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OP-AMP ACTIVE FILTERS

8.36 The op-amp circuit shown in P8.36, is used as a filter. C =0.1 F, RL =333 , R1 =1.8 k and R 2 = 8.2k. Determine (a) whether the circuit is a low- or high-pass filter; (b) thegain Vo / Vs in decibels in the pass-band, that is at the frequencies being passed bythe filter; (c) the cutoff frequency.

1. First, identify the type of filter: LP or HP ormixed configuration i.e. BP or BR

2. Use similar argument as in the previousexample (this is a standard filterconfiguration! so you would be able to usethe expression for gain directly).

(see pg 414-415 for HP filter)

20

120

)/(1

)/)(/(|)(|

+

=RR

jAHP

Now at =0; the num. is 0, hence, 0)( = jAHP

)1()(

1

2

CR jCR j

jAHP

+=

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1

2|)(|R

R jAHP = Lim

The pass-and in HP is from 0 to

= 0. freqoff cutAt2

8.1/2.82/|)(| 120 == RR jAHP

dBRR jAscaledecibelinGainbandPass

dBHP 17.13)/log(.20|)(|:

120 ==

dBtoreducewillitfreq.off cutAt 17.10317.13 =

The cut-off frequency 0 = 1/R 1.C = 5555rad/s = 884.1Hz