Practical filters Low−pass High−pass | H | | H | | H | | H | Amax Amax 1 1 ������������ ������������ ������������ ������������ Ideal Practical Ideal ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ Amin Amin ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ Practical ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ 0 0 0 0 0 0 ω 0 0 ω ω ω ω c ω c ω s ω c ω s ω c * A practical filter may exhibit a ripple. A max is called the maximum passband ripple, e.g., A max = 1 dB. * A min is the minimum attenuation to be provided by the filter, e.g., A min = 60 dB. M. B. Patil, IIT Bombay
Practical filters Low−pass High−pass | H | | H | | H | | H | Amax Amax 1 1 ������������ ������������ ������������ ������������ Ideal Practical Ideal ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ Amin Amin ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ Practical ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ 0 0 0 0 0 0 ω 0 0 ω ω ω ω c ω c ω s ω c ω s ω c * A practical filter may exhibit a ripple. A max is called the maximum passband ripple, e.g., A max = 1 dB. * A min is the minimum attenuation to be provided by the filter, e.g., A min = 60 dB. * ω s : edge of the stop band. M. B. Patil, IIT Bombay
Practical filters Low−pass High−pass | H | | H | | H | | H | Amax Amax 1 1 ������������ ������������ ������������ ������������ Ideal Practical Ideal ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ Amin Amin ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ Practical ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ 0 0 0 0 0 0 ω 0 0 ω ω ω ω c ω c ω s ω c ω s ω c * A practical filter may exhibit a ripple. A max is called the maximum passband ripple, e.g., A max = 1 dB. * A min is the minimum attenuation to be provided by the filter, e.g., A min = 60 dB. * ω s : edge of the stop band. * ω s /ω c (for a low-pass filter): selectivity factor, a measure of the sharpness of the filter. M. B. Patil, IIT Bombay
Practical filters Low−pass High−pass | H | | H | | H | | H | Amax Amax 1 1 ������������ ������������ ������������ ������������ Ideal Practical Ideal ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ Amin Amin ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ Practical ������������ ������������ ������������ ������������ ������������ ������������ ������������ ������������ 0 0 0 0 0 0 ω 0 0 ω ω ω ω c ω c ω s ω c ω s ω c * A practical filter may exhibit a ripple. A max is called the maximum passband ripple, e.g., A max = 1 dB. * A min is the minimum attenuation to be provided by the filter, e.g., A min = 60 dB. * ω s : edge of the stop band. * ω s /ω c (for a low-pass filter): selectivity factor, a measure of the sharpness of the filter. * ω c < ω < ω s : transition band. M. B. Patil, IIT Bombay
Practical filters 1 For a low-pass filter, H ( s ) = . n � a i ( s /ω c ) i i =0 Coefficients ( a i ) for various types of filters are tabulated in handbooks. We now look at | H ( j ω ) | for two commonly used filters. M. B. Patil, IIT Bombay
Practical filters 1 For a low-pass filter, H ( s ) = . n � a i ( s /ω c ) i i =0 Coefficients ( a i ) for various types of filters are tabulated in handbooks. We now look at | H ( j ω ) | for two commonly used filters. Butterworth filters: 1 | H ( j ω ) | = 1 + ǫ 2 ( ω/ω c ) 2 n . � M. B. Patil, IIT Bombay
Practical filters 1 For a low-pass filter, H ( s ) = . n � a i ( s /ω c ) i i =0 Coefficients ( a i ) for various types of filters are tabulated in handbooks. We now look at | H ( j ω ) | for two commonly used filters. Butterworth filters: 1 | H ( j ω ) | = 1 + ǫ 2 ( ω/ω c ) 2 n . � Chebyshev filters: 1 | H ( j ω ) | = where � 1 + ǫ 2 C 2 n ( ω/ω c ) n cos − 1 ( x ) C n ( x ) = cos � � for x ≤ 1, n cosh − 1 ( x ) � � C n ( x ) = cosh for x ≥ 1, M. B. Patil, IIT Bombay
Practical filters 1 For a low-pass filter, H ( s ) = . n � a i ( s /ω c ) i i =0 Coefficients ( a i ) for various types of filters are tabulated in handbooks. We now look at | H ( j ω ) | for two commonly used filters. Butterworth filters: 1 | H ( j ω ) | = 1 + ǫ 2 ( ω/ω c ) 2 n . � Chebyshev filters: 1 | H ( j ω ) | = where � 1 + ǫ 2 C 2 n ( ω/ω c ) n cos − 1 ( x ) C n ( x ) = cos � � for x ≤ 1, n cosh − 1 ( x ) � � C n ( x ) = cosh for x ≥ 1, H ( s ) for a high-pass filter can be obtained from H ( s ) of the corresponding low-pass filter by ( s /ω c ) → ( ω c / s ) . M. B. Patil, IIT Bombay
Practical filters (low-pass) Butterworth filters: ǫ = 0.5 0 1 n=1 | H | (dB) n=1 2 | H | 2 3 3 4 4 5 −100 5 0 0 1 2 3 4 5 0.01 0.1 1 10 100 ω/ω c ω/ω c Chebyshev filters: 0 ǫ = 0.5 1 n=1 n=1 | H | (dB) 2 | H | 2 3 3 4 −100 4 5 5 0 0 1 2 3 4 5 0.01 0.1 1 10 100 ω/ω c ω/ω c M. B. Patil, IIT Bombay
Practical filters (high-pass) Butterworth filters: 0 1 n=1 n=1 | H | (dB) 2 | H | 2 3 3 4 4 5 ǫ = 0.5 5 −100 0 0 1 2 3 4 0.01 0.1 1 10 100 ω/ω c ω/ω c Chebyshev filters: 0 1 n=1 n=1 | H | (dB) 2 | H | 3 2 4 5 3 ǫ = 0.5 −100 4 5 0 0 1 2 3 4 0.01 0.1 1 10 100 ω/ω c ω/ω c M. B. Patil, IIT Bombay
Passive filter example R Vs Vo 100 Ω C 5 µ F
Passive filter example R Vs Vo (1/sC) 1 100 Ω H(s) = R + (1/sC) = 1 + (s /ω 0 ) , C 5 µ F with ω 0 = 1/RC → f 0 = ω 0 / 2 π = 318 Hz (Low−pass filter)
Passive filter example R Vs Vo (1/sC) 1 100 Ω H(s) = R + (1/sC) = 1 + (s /ω 0 ) , C 5 µ F with ω 0 = 1/RC → f 0 = ω 0 / 2 π = 318 Hz (Low−pass filter) 20 0 | H | (dB) −20 −40 −60 101 102 103 104 105 f (Hz) (SEQUEL file: ee101 rc ac 2.sqproj ) M. B. Patil, IIT Bombay
Passive filter example R Vs Vo 100 Ω L C 0.1 mF 4 µ F
Passive filter example R Vs Vo ( sL ) � ( 1/sC ) s(L/R) H(s) = R + (sL) � (1/sC) = 100 Ω 1 + s(L/R) + s 2 LC √ L C with ω 0 = 1 / LC → f 0 = ω 0 / 2 π = 7.96 kHz 0.1 mF 4 µ F (Band−pass filter)
Passive filter example R Vs Vo ( sL ) � ( 1/sC ) s(L/R) H(s) = R + (sL) � (1/sC) = 100 Ω 1 + s(L/R) + s 2 LC √ L C with ω 0 = 1 / LC → f 0 = ω 0 / 2 π = 7.96 kHz 0.1 mF 4 µ F (Band−pass filter) 0 −20 | H | (dB) −40 −60 −80 102 103 104 105 f (Hz) (SEQUEL file: ee101 rlc 3.sqproj ) M. B. Patil, IIT Bombay
Op-amp filters (“Active” filters) * Op-amp filters can be designed without using inductors. This is a significant advantage since inductors are bulky and expensive. Inductors also exhibit nonlinear behaviour (arising from the core properties) which is undesirable in a filter circuit. M. B. Patil, IIT Bombay
Op-amp filters (“Active” filters) * Op-amp filters can be designed without using inductors. This is a significant advantage since inductors are bulky and expensive. Inductors also exhibit nonlinear behaviour (arising from the core properties) which is undesirable in a filter circuit. * With op-amps, a filter circuit can be designed with a pass-band gain. M. B. Patil, IIT Bombay
Op-amp filters (“Active” filters) * Op-amp filters can be designed without using inductors. This is a significant advantage since inductors are bulky and expensive. Inductors also exhibit nonlinear behaviour (arising from the core properties) which is undesirable in a filter circuit. * With op-amps, a filter circuit can be designed with a pass-band gain. * Op-amp filters can be easily incorporated in an integrated circuit. M. B. Patil, IIT Bombay
Op-amp filters (“Active” filters) * Op-amp filters can be designed without using inductors. This is a significant advantage since inductors are bulky and expensive. Inductors also exhibit nonlinear behaviour (arising from the core properties) which is undesirable in a filter circuit. * With op-amps, a filter circuit can be designed with a pass-band gain. * Op-amp filters can be easily incorporated in an integrated circuit. * However, there are situations in which passive filters are still used. - high frequencies at which op-amps do not have sufficient gain - high power which op-amps cannot handle M. B. Patil, IIT Bombay
Op-amp filters: example R 2 10 k C R 1 10 nF V s 1 k V o R L
Op-amp filters: example R 2 10 k C R 1 10 nF V s 1 k V o R L Op-amp filters are designed for op-amp operation in the linear region → Our analysis of the inverting amplifier applies, and we get, V o = − R 2 � (1 / sC ) V s ( V s and V o are phasors) R 1 H ( s ) = − R 2 1 1 + sR 2 C R 1
Op-amp filters: example R 2 10 k C R 1 10 nF V s 1 k V o R L Op-amp filters are designed for op-amp operation in the linear region → Our analysis of the inverting amplifier applies, and we get, V o = − R 2 � (1 / sC ) V s ( V s and V o are phasors) R 1 H ( s ) = − R 2 1 1 + sR 2 C R 1 This is a low-pass filter, with ω 0 = 1 / R 2 C (i.e., f 0 = ω 0 / 2 π = 1 . 59 kHz).
Op-amp filters: example R 2 20 10 k C R 1 10 nF | H | (dB) V s 1 k 0 V o R L −20 10 1 10 2 10 3 10 4 10 5 f (Hz) Op-amp filters are designed for op-amp operation in the linear region → Our analysis of the inverting amplifier applies, and we get, V o = − R 2 � (1 / sC ) V s ( V s and V o are phasors) R 1 H ( s ) = − R 2 1 1 + sR 2 C R 1 This is a low-pass filter, with ω 0 = 1 / R 2 C (i.e., f 0 = ω 0 / 2 π = 1 . 59 kHz).
Op-amp filters: example R 2 20 10 k C R 1 10 nF | H | (dB) V s 1 k 0 V o R L −20 10 1 10 2 10 3 10 4 10 5 f (Hz) Op-amp filters are designed for op-amp operation in the linear region → Our analysis of the inverting amplifier applies, and we get, V o = − R 2 � (1 / sC ) V s ( V s and V o are phasors) R 1 H ( s ) = − R 2 1 1 + sR 2 C R 1 This is a low-pass filter, with ω 0 = 1 / R 2 C (i.e., f 0 = ω 0 / 2 π = 1 . 59 kHz). (SEQUEL file: ee101 op filter 1.sqproj ) M. B. Patil, IIT Bombay
Op-amp filters: example R 2 10 k C R 1 V s 1 k 100 nF V o R L
Op-amp filters: example R 2 10 k C R 1 V s 1 k 100 nF V o R L R 2 sR 2 C H ( s ) = − R 1 + (1 / sC ) = − 1 + sR 1 C .
Op-amp filters: example R 2 10 k C R 1 V s 1 k 100 nF V o R L R 2 sR 2 C H ( s ) = − R 1 + (1 / sC ) = − 1 + sR 1 C . This is a high-pass filter, with ω 0 = 1 / R 1 C (i.e., f 0 = ω 0 / 2 π = 1 . 59 kHz).
Op-amp filters: example 20 R 2 10 k C R 1 0 V s | H | (dB) 1 k 100 nF V o R L −20 −40 10 1 10 2 10 3 10 4 10 5 f (Hz) R 2 sR 2 C H ( s ) = − R 1 + (1 / sC ) = − 1 + sR 1 C . This is a high-pass filter, with ω 0 = 1 / R 1 C (i.e., f 0 = ω 0 / 2 π = 1 . 59 kHz).
Op-amp filters: example 20 R 2 10 k C R 1 0 V s | H | (dB) 1 k 100 nF V o R L −20 −40 10 1 10 2 10 3 10 4 10 5 f (Hz) R 2 sR 2 C H ( s ) = − R 1 + (1 / sC ) = − 1 + sR 1 C . This is a high-pass filter, with ω 0 = 1 / R 1 C (i.e., f 0 = ω 0 / 2 π = 1 . 59 kHz). (SEQUEL file: ee101 op filter 2.sqproj ) M. B. Patil, IIT Bombay
Op-amp filters: example R 2 100 k C 2 C 1 80 pF R 1 V s 10 k 0.8 µ F V o R L
Op-amp filters: example R 2 100 k C 2 C 1 80 pF R 1 V s 10 k 0.8 µ F V o R L H ( s ) = − R 2 � (1 / sC 2 ) R 1 + (1 / sC 1 ) = − R 2 sR 1 C 1 (1 + sR 1 C 1 )(1 + sR 2 C 2 ) . R 1
Op-amp filters: example R 2 100 k C 2 C 1 80 pF R 1 V s 10 k 0.8 µ F V o R L H ( s ) = − R 2 � (1 / sC 2 ) R 1 + (1 / sC 1 ) = − R 2 sR 1 C 1 (1 + sR 1 C 1 )(1 + sR 2 C 2 ) . R 1 This is a band-pass filter, with ω L = 1 / R 1 C 1 and ω H = 1 / R 2 C 2 . → f L = 20 Hz, f H = 20 kHz.
Op-amp filters: example R 2 20 100 k C 2 | H | (dB) C 1 80 pF R 1 V s 10 k 0.8 µ F V o R L 0 10 0 10 2 10 4 10 6 f (Hz) H ( s ) = − R 2 � (1 / sC 2 ) R 1 + (1 / sC 1 ) = − R 2 sR 1 C 1 (1 + sR 1 C 1 )(1 + sR 2 C 2 ) . R 1 This is a band-pass filter, with ω L = 1 / R 1 C 1 and ω H = 1 / R 2 C 2 . → f L = 20 Hz, f H = 20 kHz.
Op-amp filters: example R 2 20 100 k C 2 | H | (dB) C 1 80 pF R 1 V s 10 k 0.8 µ F V o R L 0 10 0 10 2 10 4 10 6 f (Hz) H ( s ) = − R 2 � (1 / sC 2 ) R 1 + (1 / sC 1 ) = − R 2 sR 1 C 1 (1 + sR 1 C 1 )(1 + sR 2 C 2 ) . R 1 This is a band-pass filter, with ω L = 1 / R 1 C 1 and ω H = 1 / R 2 C 2 . → f L = 20 Hz, f H = 20 kHz. (SEQUEL file: ee101 op filter 3.sqproj ) M. B. Patil, IIT Bombay
Graphic equalizer C1 20 a 1−a R1A R2 R1B V s a=0.9 C2 0.7 | H | (dB) 0.5 0 R3A R3B 0.3 R1A = R1B = 470 Ω R3A = R3B = 100 k Ω V o 0.1 R2 = 10 k Ω R L C1 = 100 nF C2 = 10 nF −20 10 1 10 2 10 3 10 4 10 5 f (Hz) (Ref.: S. Franco, "Design with Op Amps and analog ICs") M. B. Patil, IIT Bombay
Graphic equalizer C1 20 a 1−a R1A R2 R1B V s a=0.9 C2 0.7 | H | (dB) 0.5 0 R3A R3B 0.3 R1A = R1B = 470 Ω R3A = R3B = 100 k Ω V o 0.1 R2 = 10 k Ω R L C1 = 100 nF C2 = 10 nF −20 10 1 10 2 10 3 10 4 10 5 f (Hz) (Ref.: S. Franco, "Design with Op Amps and analog ICs") * Equalizers are implemented as arrays of narrow-band filters, each with an adjustable gain (attenuation) around a centre frequency. M. B. Patil, IIT Bombay
Graphic equalizer C1 20 a 1−a R1A R2 R1B V s a=0.9 C2 0.7 | H | (dB) 0.5 0 R3A R3B 0.3 R1A = R1B = 470 Ω R3A = R3B = 100 k Ω V o 0.1 R2 = 10 k Ω R L C1 = 100 nF C2 = 10 nF −20 10 1 10 2 10 3 10 4 10 5 f (Hz) (Ref.: S. Franco, "Design with Op Amps and analog ICs") * Equalizers are implemented as arrays of narrow-band filters, each with an adjustable gain (attenuation) around a centre frequency. * The circuit shown above represents one of the equalizer sections. (SEQUEL file: ee101 op filter 4.sqproj ) M. B. Patil, IIT Bombay
M. B. Patil, IIT Bombay
Sallen-Key filter example (2 nd order, low-pass) 40 C1 20 R1 R2 V s V 1 V o 0 | H | (dB) C2 R L −20 R1 = R2 = 15.8 k Ω RB C1 = C2 = 10 nF RA −40 RA = 10 k Ω , RB = 17.8 k Ω −60 (Ref.: S. Franco, "Design with Op Amps and analog ICs") 10 1 10 2 10 3 10 4 10 5 f (Hz) M. B. Patil, IIT Bombay
Sallen-Key filter example (2 nd order, low-pass) 40 C1 20 R1 R2 V s V 1 V o 0 | H | (dB) C2 R L −20 R1 = R2 = 15.8 k Ω RB C1 = C2 = 10 nF RA −40 RA = 10 k Ω , RB = 17.8 k Ω −60 (Ref.: S. Franco, "Design with Op Amps and analog ICs") 10 1 10 2 10 3 10 4 10 5 f (Hz) R A V + = V − = V o ≡ V o / K . R A + R B M. B. Patil, IIT Bombay
Sallen-Key filter example (2 nd order, low-pass) 40 C1 20 R1 R2 V s V 1 V o 0 | H | (dB) C2 R L −20 R1 = R2 = 15.8 k Ω RB C1 = C2 = 10 nF RA −40 RA = 10 k Ω , RB = 17.8 k Ω −60 (Ref.: S. Franco, "Design with Op Amps and analog ICs") 10 1 10 2 10 3 10 4 10 5 f (Hz) R A V + = V − = V o ≡ V o / K . R A + R B (1 / sC 2 ) 1 Also, V + = R 2 + (1 / sC 2 ) V 1 = V 1 . 1 + sR 2 C 2 M. B. Patil, IIT Bombay
Sallen-Key filter example (2 nd order, low-pass) 40 C1 20 R1 R2 V s V 1 V o 0 | H | (dB) C2 R L −20 R1 = R2 = 15.8 k Ω RB C1 = C2 = 10 nF RA −40 RA = 10 k Ω , RB = 17.8 k Ω −60 (Ref.: S. Franco, "Design with Op Amps and analog ICs") 10 1 10 2 10 3 10 4 10 5 f (Hz) R A V + = V − = V o ≡ V o / K . R A + R B (1 / sC 2 ) 1 Also, V + = R 2 + (1 / sC 2 ) V 1 = V 1 . 1 + sR 2 C 2 1 ( V s − V 1 ) + sC 1 ( V o − V 1 ) + 1 KCL at V 1 → ( V + − V 1 ) = 0 . R 1 R 2 M. B. Patil, IIT Bombay
Sallen-Key filter example (2 nd order, low-pass) 40 C1 20 R1 R2 V s V 1 V o 0 | H | (dB) C2 R L −20 R1 = R2 = 15.8 k Ω RB C1 = C2 = 10 nF RA −40 RA = 10 k Ω , RB = 17.8 k Ω −60 (Ref.: S. Franco, "Design with Op Amps and analog ICs") 10 1 10 2 10 3 10 4 10 5 f (Hz) R A V + = V − = V o ≡ V o / K . R A + R B (1 / sC 2 ) 1 Also, V + = R 2 + (1 / sC 2 ) V 1 = V 1 . 1 + sR 2 C 2 1 ( V s − V 1 ) + sC 1 ( V o − V 1 ) + 1 KCL at V 1 → ( V + − V 1 ) = 0 . R 1 R 2 K Combining the above equations, H ( s ) = . 1 + s [( R 1 + R 2 ) C 2 + (1 − K ) R 1 C 1 ] + s 2 R 1 C 1 R 2 C 2 (SEQUEL file: ee101 op filter 5.sqproj ) M. B. Patil, IIT Bombay
Sixth-order Chebyshev low-pass filter (cascade design) 5.1 n 10 n 62 n V o V s 10.7 k 10.2 k 8.25 k 6.49 k 4.64 k 2.49 k RL 2.2 n 510 p 220 p 20 (Ref.: S. Franco, "Design with Op Amps and analog ICs") SEQUEL file: ee101_op_filter_6.sqproj 0 −20 | H | (dB) −40 −60 −80 10 2 10 3 10 4 10 5 f (Hz) M. B. Patil, IIT Bombay
Third-order Chebyshev high-pass filter 20 15.4 k 154 k 100 n 7.68 k 0 V o V s −20 | H | (dB) R L 100 n 100 n 54.9 k −40 −60 (Ref.: S. Franco, "Design with Op Amps and analog ICs") SEQUEL file: ee101_op_filter_7.sqproj −80 10 0 10 1 10 2 10 3 f (Hz) M. B. Patil, IIT Bombay
Band-pass filter example 40 5 k 5 k 7.4 n 20 5 k 7.4 n V s 5 k | H | (dB) 5 k 0 −20 370 k V o 5 k −40 10 2 10 3 10 4 10 5 (Ref.: J. M. Fiore, "Op Amps and linear ICs") f (Hz) SEQUEL file: ee101_op_filter_8.sqproj M. B. Patil, IIT Bombay
10 k Notch filter example 10 k 265 n 10 k 265 n V s 10 k 10 k 10 k 10 k V o 89 k 10 k 1 k 0 (Ref.: J. M. Fiore, "Op Amps and linear ICs") SEQUEL file: ee101_op_filter_9.sqproj | H | (dB) −20 −40 10 1 10 2 f (Hz) M. B. Patil, IIT Bombay
Half-wave rectifier V o slope = 1 Ideal half-wave V i V o V i rectifier
Half-wave rectifier V o 1 slope = 1 Ideal V o 0 half-wave V i V o V i rectifier V i -1 0 T/2 T 3T/2 2T
Half-wave rectifier V o 1 slope = 1 Ideal V o 0 half-wave V i V o V i rectifier V i -1 0 T/2 T 3T/2 2T V o V i V o slope = 1 V i V on
Half-wave rectifier V o 1 slope = 1 Ideal V o 0 half-wave V i V o V i rectifier V i -1 0 T/2 T 3T/2 2T V o 1 V on V i V o V i slope = 1 V o 0 V i V on -1 0 T/2 T 3T/2 2T
Half-wave rectifier V o 1 slope = 1 Ideal V o 0 half-wave V i V o V i rectifier V i -1 0 T/2 T 3T/2 2T V o 1 V on V i V o V i slope = 1 V o 0 V i V on -1 → need an improved circuit 0 T/2 T 3T/2 2T M. B. Patil, IIT Bombay
Half-wave precision rectifier V o D V i R
Half-wave precision rectifier i D i − V o V o V o1 D V on V i V i i R R R Consider two cases: (i) D is conducting: The feedback loop is closed, and the circuit looks like (except for the diode drop) the buffer we have seen earlier.
Half-wave precision rectifier i D i − V o V o V o1 D V on V i V i i R R R Consider two cases: (i) D is conducting: The feedback loop is closed, and the circuit looks like (except for the diode drop) the buffer we have seen earlier. Since the input current i − ≈ 0, i R = i D . V + − V − = V o 1 = V o + 0 . 7 V ≈ 0 V → V o = V − ≈ V + = V i . A V A V
Half-wave precision rectifier i D i − V o V o V o1 D V on V i V i i R R R Consider two cases: (i) D is conducting: The feedback loop is closed, and the circuit looks like (except for the diode drop) the buffer we have seen earlier. Since the input current i − ≈ 0, i R = i D . V + − V − = V o 1 = V o + 0 . 7 V ≈ 0 V → V o = V − ≈ V + = V i . A V A V This situation arises only if i D > 0 (since the diode can only conduct in the forward direction), i.e., i R > 0 → V o = i R R > 0, and therefore V i = V o > 0 V .
Half-wave precision rectifier V o i D i − V o V o V o1 D V on V i V i slope=1 i R R R V i Consider two cases: (i) D is conducting: The feedback loop is closed, and the circuit looks like (except for the diode drop) the buffer we have seen earlier. Since the input current i − ≈ 0, i R = i D . V + − V − = V o 1 = V o + 0 . 7 V ≈ 0 V → V o = V − ≈ V + = V i . A V A V This situation arises only if i D > 0 (since the diode can only conduct in the forward direction), i.e., i R > 0 → V o = i R R > 0, and therefore V i = V o > 0 V . M. B. Patil, IIT Bombay
Half-wave precision rectifier V o i D i − V o V o V o1 D V on V i V i slope=1 i R R R V i Consider two cases: (i) D is conducting: The feedback loop is closed, and the circuit looks like (except for the diode drop) the buffer we have seen earlier. Since the input current i − ≈ 0, i R = i D . V + − V − = V o 1 = V o + 0 . 7 V ≈ 0 V → V o = V − ≈ V + = V i . A V A V This situation arises only if i D > 0 (since the diode can only conduct in the forward direction), i.e., i R > 0 → V o = i R R > 0, and therefore V i = V o > 0 V . Note: V on does not appear in the graph. M. B. Patil, IIT Bombay
Half-wave precision rectifier V o D V i R
Half-wave precision rectifier V o V o V o1 D V i V i R R (ii) D is not conducting → V o = 0 V .
Half-wave precision rectifier V o V o V o1 D V i V i R R (ii) D is not conducting → V o = 0 V . What about V o 1 ? Since the op-amp is now in the open-loop configuration, a very small V i is enough to drive it to saturation.
Half-wave precision rectifier V o V o V o1 D V i V i R R (ii) D is not conducting → V o = 0 V . What about V o 1 ? Since the op-amp is now in the open-loop configuration, a very small V i is enough to drive it to saturation. Note that Case ( ii ) occurs when V i < 0 V (we have already looked at V i > 0). Since V + − V − = V i − 0 = V i is negative, V o 1 is driven to − V sat .
Half-wave precision rectifier V o V o V o V o1 D V i V i R R V o = 0 V i (ii) D is not conducting → V o = 0 V . What about V o 1 ? Since the op-amp is now in the open-loop configuration, a very small V i is enough to drive it to saturation. Note that Case ( ii ) occurs when V i < 0 V (we have already looked at V i > 0). Since V + − V − = V i − 0 = V i is negative, V o 1 is driven to − V sat . M. B. Patil, IIT Bombay
Half-wave precision rectifier V o Super D on diode D off D V o V o V i V o = V i V o1 V i R i R R V o = 0 Super diode V i M. B. Patil, IIT Bombay
Half-wave precision rectifier V o Super D on diode D off D V o V o V i V o = V i V o1 V i R i R R V o = 0 Super diode V i * The circuit is called “super diode” (an ideal diode with V on = 0 V). M. B. Patil, IIT Bombay
Half-wave precision rectifier V o Super D on diode D off D V o V o V i V o = V i V o1 V i R i R R V o = 0 Super diode V i * The circuit is called “super diode” (an ideal diode with V on = 0 V). * When D conducts, the op-amp operates in the linear region, and we have V + ≈ V − . M. B. Patil, IIT Bombay
Half-wave precision rectifier V o Super D on diode D off D V o V o V i V o = V i V o1 V i R i R R V o = 0 Super diode V i * The circuit is called “super diode” (an ideal diode with V on = 0 V). * When D conducts, the op-amp operates in the linear region, and we have V + ≈ V − . * When D is off, the op-amp operates in the saturation region, V − = 0, V + = V i , and V o 1 = − V sat . M. B. Patil, IIT Bombay
Half-wave precision rectifier V o Super D on diode D off D V o V o V i V o = V i V o1 V i R i R R V o = 0 Super diode V i * The circuit is called “super diode” (an ideal diode with V on = 0 V). * When D conducts, the op-amp operates in the linear region, and we have V + ≈ V − . * When D is off, the op-amp operates in the saturation region, V − = 0, V + = V i , and V o 1 = − V sat . * Where does i R come from? M. B. Patil, IIT Bombay
1.5 A = 1 M = 0.3 m (t) f c = 200 kHz 0 f m = 10 kHz Application: AM demodulation −1.5 1.5 c (t) 0 −1.5 1.5 y (t) 0 −1.5 0 25 50 75 100 time ( µ sec ) M. B. Patil, IIT Bombay
1.5 A = 1 M = 0.3 m (t) f c = 200 kHz 0 f m = 10 kHz Application: AM demodulation −1.5 1.5 Carrier wave: c ( t ) = A sin(2 π f c t ) c (t) 0 −1.5 1.5 y (t) 0 −1.5 0 25 50 75 100 time ( µ sec ) M. B. Patil, IIT Bombay
1.5 A = 1 M = 0.3 m (t) f c = 200 kHz 0 f m = 10 kHz Application: AM demodulation −1.5 1.5 Carrier wave: c ( t ) = A sin(2 π f c t ) c (t) Signal (e.g., audio): 0 m ( t ) = M sin(2 π f m t + φ ) −1.5 1.5 y (t) 0 −1.5 0 25 50 75 100 time ( µ sec ) M. B. Patil, IIT Bombay
1.5 A = 1 M = 0.3 m (t) f c = 200 kHz 0 f m = 10 kHz Application: AM demodulation −1.5 1.5 Carrier wave: c ( t ) = A sin(2 π f c t ) c (t) Signal (e.g., audio): 0 m ( t ) = M sin(2 π f m t + φ ) AM wave: −1.5 y ( t ) = [1 + m ( t )] c ( t ) 1.5 (Assume M < 1) y (t) 0 −1.5 0 25 50 75 100 time ( µ sec ) M. B. Patil, IIT Bombay
1.5 A = 1 M = 0.3 m (t) f c = 200 kHz 0 f m = 10 kHz Application: AM demodulation −1.5 1.5 Carrier wave: c ( t ) = A sin(2 π f c t ) c (t) Signal (e.g., audio): 0 m ( t ) = M sin(2 π f m t + φ ) AM wave: −1.5 y ( t ) = [1 + m ( t )] c ( t ) 1.5 (Assume M < 1) e.g., Vividh Bharati: y (t) 0 f c = 1188 kHz, f m ≃ 10 kHz (audio). −1.5 0 25 50 75 100 time ( µ sec ) M. B. Patil, IIT Bombay
AM demodulation using a peak detector 0.15 V i V 1 Super diode V i V 1 filter V o AM 0 input −0.15 0 1 2 0.2 0.3 0.4 0.5 t (ms) t (ms) M. B. Patil, IIT Bombay
AM demodulation using a peak detector 0.15 V i V 1 Super diode V i V 1 filter V o AM 0 input −0.15 0 1 2 0.2 0.3 0.4 0.5 t (ms) t (ms) * charging through super diode, discharging through resistor M. B. Patil, IIT Bombay
AM demodulation using a peak detector 0.15 V i V 1 Super diode V i V 1 filter V o AM 0 input −0.15 0 1 2 0.2 0.3 0.4 0.5 t (ms) t (ms) * charging through super diode, discharging through resistor * The time constant ( RC ) needs to be carefully selected. M. B. Patil, IIT Bombay
AM demodulation using a peak detector 0.15 V i V 1 Super diode V i V 1 filter V o AM 0 input −0.15 0 1 2 0.2 0.3 0.4 0.5 t (ms) t (ms) * charging through super diode, discharging through resistor * The time constant ( RC ) needs to be carefully selected. SEQUEL file: super diode.sqproj M. B. Patil, IIT Bombay
Clipping and clamping C R V i V i D D V o V o V R V R R L R L C R V i D V i D V o V o V R V R R L R L * What is the function provided by each circuit? M. B. Patil, IIT Bombay
Clipping and clamping C R V i V i D D V o V o V R V R R L R L C R V i D V i D V o V o V R V R R L R L * What is the function provided by each circuit? * Verify with simulation (and in the lab). M. B. Patil, IIT Bombay
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