ECE307-10 Active Filter Circuits Z. Aliyazicioglu Electrical and Computer Engineering Department Cal Poly Pomona Active Filter Circuits Introduction Filter circuits with RLC are passive filter circuit Use op amp to have active filter circuit Active filter can produce band-pass and band-reject filter without using inductor. Passive filter incapable of amplification. Max gain is 1 Active filter capable of amplification The cutoff frequency and band-pass magnitude of passive filter can change with additional load resistance This is not a case for active filters We look at few active filter with op amps. We look at that basic op amp filter circuits can be combined to active specific frequency response and to attain close to ideal filter response ECE 307-10 2 1 Active Filter= Circuits First-Order Low-pass Filters C Zf R2 R1 Zi - - Vi OUT Vo Vi OUT + + Vo + + −Zf 1 R2 Transfer function of the circuit Hs()= −R || − Z 2 sR C + 1 i Hs()==SC 2 RR −R ω 11 Hs()= 2 Hs()=− K c RsRC12(1)+ ()s + ωc Transfer function in jω The Gain Cutoff frequency 1 R2 1 Hj()ω =− K K = ω = ω R c RC (1+ j ) 1 2 ω ECE 307-10c 3 Active Filter Circuits Example • Find R2 and C values in the following active Low-pass filter for gain of 1 C and cutoff frequency of 1 rad/s. 1F R12 1 R1 From the gain R2 - K = = 1 RR21= =Ω1 1 R Vi OUT 1 + Vo + From the cutoff frequency 1 1 ωc = = 1 CF==1 RC2 R2 1 Hj()ω = ω (1+ j ) 1 ECE 307-10 4 2 Active Filter Circuits Example >> w=0.1:.1:10; >> h=20*log10(abs(1./(1+j*w))) ; >> semilogx(w,h) >> grid on >> xlabel('\omega(rad/s)') >> ylabel('|H(j\omega)| dB') >> ECE 307-10 5 Active Filter Circuits A first order high-pass filter R2 R1 C - Vi OUT + Vo + −Zf −RRsC− Transfer function of the circuit Hs()= Hs()==22 Z 1 RsC+ 1 i R + 1 −Rs 1 Hs()= 2 s sC 1 Hs()=− K Rs1()+ ()s + ω RC c 1 Transfer function in jω The Gain Cutoff frequency jω R K = 2 1 ω ωc = c R1 RC Hj()ω =− K 1 ω (1+ j ) ωc ECE 307-10 6 3 Active Filter Circuits Example R2 200K R1 C - 20 K Vi 0.1 uF OUT + Vo + • Find R2 and R1 values in the above active High-pass filter for gain of 10 and cutoff frequency of 500 rad/s. From the cutoff frequency 1 1 ωc ==500 RK1 = =Ω20 RC12 500C R From the gain K ==2 10 RR21= 10=Ω 200 K R1 jω 500 Transfer function in jω Hj()ω =− 10 ω (1+ j ) 500 ECE 307-10 7 Active Filter Circuits Example >> w=1:10000; >> h=20*log10(10*(abs((j*w/500 )./(1+j*w/500)))); >> semilogx(w,h) >> grid on >> xlabel('\omega(rad/s)') >> ylabel('|H(j\omega)| dB') >> ECE 307-10 8 4 Active Filter Circuits Scaling • In filter design, we can transform RLC values in to realistic values, this process is called scaling • Two types of scaling, magnitude and frequency scaling •In magnitude scaling, we multiply all L and R by scaling factor km, multiplying all C by 1/km C RkR' = m LkL' = m C ' = km •km,is positive real number ECE 307-10 9 Active Filter Circuits Scaling • frequency scaling, we multiply all L, C by 1/kf where kf is scaling factor. L C RR' = L' = C ' = kf kf • A circuit can be scaled in both magnitude and frequency in simultanously km C RkR' = m LL' = C ' = kf kkmf ECE 307-10 10 5 Active Filter Circuits Example • Example 1 , Find R2 and R1 values in the active Low-pass filter for gain of 5 C and cutoff frequency of 1Khz and 1F R12 c=0.01 µF R1 1 - 1 Vi OUT + Vo ωc ' 21000π + kf == =6283.185 ωc 1 11C k == =15915.5 m −8 kCf ' 6283.185(10 ) RkR22'==m 15915.5(1) = 15.9 K Ω • For gain specification, we need to change R1 R 15.9K RK= 2 ==Ω3.18 1 K 5 ECE 307-10 11 Active Filter Circuits Example >> f=1:10000; >> w=2*pi*f; >> h=20*log10(5*abs(1./(1 +j*w/(2*pi*1000)))); >> semilogx(f,h) >> grid on >> xlabel(‘f(Hz)') >> ylabel('|H(jf)| dB') ECE 307-10 12 6 Active Filter Circuits Op Amp Band-Pass Filters • Three components • A unity gain low-pass filter, cutoff frequency is ωc2 • A unity gain high-pass filter , cutoff frequency ωc1 • A gain component to provide the desired level ω c2 ≥ 2 ωc1 Vi Low-pass filter High-pass filter Inverting amp. Vo ECE 307-10 13 Active Filter Circuits Op Amp Band-Pass Filters CL RL RH RL Rf - RH CH Vi OUT - Rf + OUT - + OUT + Vo + −ωc2 −s Rf −Ksω Hs()=−Hs()= c2 ss++ωωcci21 R ()()ss++ωωcc21 −Ksω Hs()= c2 βs 2 Hs()= ss++()ωωcc12 + ωω cc 12 22 ss++β ω0 R 1 1 Hj()ω =− K =− f ωcc21 ω ωc2 = ωc1 = 0 RC RCH max Ri LL H ECE 307-10 14 7 Active Filter Circuits Example: • Design a band-pass filter for a graphical equalizer that has gain 2 within the frequency between 100 and 10,000 Hz. Use 0.1 µF capacitors • For upper cutoff frequency from LP filter 1 11 ω = R = ==Ω80 c2 L −6 RCLL ωcL2C 2π 10000(0.1)10 • For Lower cutoff frequency from HP filter 1 11 ω = R = ==Ω7958 c1 H −6 RCHH ωcL1C 2100(0.1)10π • For gain, choose Ri=1KΩ R K = f RRK= ==Ω1000(2) 2 K Ri fi ECE 307-10 15 Active Filter Circuits From transfer function −−2πω 1000j 2000 Hj()ω =− AHj= 20log | (ω ) | jjωπ++2 1000 ωπ 2 100 1000 dB 10 >> f=10:80000; >> w=2*pi*f; >> H=((- 2*pi*10000)./(j*w+2*pi* 10000)).*((- j*w)./(j*w+2*pi*100))*( -2); >> A=20*log10(abs(H)); >> semilogx(f,A) >> grid on; >> ylabel ('A_{dB}') >> xlabel ('F (Hz)') ECE 307-10 16 8 Active Filter Circuits Op Amp Band-Reject Filters • Three components • A unity gain low-pass filter, cutoff frequency is ωc1 • A unity gain high-pass filter , cutoff frequency ωc2 • A gain component to provide the desired level Low-pass filter Vi Inverting amp. Vo High-pass filter ECE 307-10 17 Active Filter Circuits Op Amp Band-Reject Filters CL RL −ωc1 −s Rf Hs()=+ − ss++ωω R RL cci12 - Rf OUT + Rf Vi RH - OUT + Vo RH CH + - Rf For ω >> ω OUT cc21 + 1 1 ωc1 = ωc2 = RCLL RCHH R ss2 ++2ωωω Hs()= f ccc112 Rs()()++ωω s R icc12 Hj()ω == K f max Ri ECE 307-10 18 9 Active Filter Circuits Example: • Design an active band-reject filter that has gain 5 and the stop frequency between 100 and 2000 Hz. Use 0.5 µF capacitors Fcc12==100 Hz and F 2000 Hz For ωcc21>> ω 11 1 RK= ==Ω3.18 ωc1 = L ω C −6 RCLL cL1 2100(0.5)10π 1 11 ω = R = ==Ω159 c2 H −6 RCHH ωcH2C 2π 2000(0.5)10 • For gain, choose Ri=1KΩ R K = f RRK= ==Ω1000(5) 5 K Ri fi ECE 307-10 19 Active Filter Circuits −ω − jω R Hj()ω =+c1 −f AHjdB = 20log10 | (ω ) | jωω++cci12jR ωω >> f=10:80000; >> w=2*pi*f; >> H=(((- 2*pi*100)./(j*w+2*pi*100)) +((- j*w)./(j*w+2*pi*2000)))*(- 5); >> A=20*log10(abs(H)); >> semilogx(f,A) >> grid on; >> xlabel ('F (Hz)') >> ylabel ('A_{dB}') ECE 307-10 20 10.