Introduction to filters Consider v ( t ) = v 1 ( t ) + v 2 ( t ) = V - - PowerPoint PPT Presentation
Introduction to filters Consider v ( t ) = v 1 ( t ) + v 2 ( t ) = V m 1 sin 1 t + V m 2 sin 2 t . 1 v 1 v v 2 0 1 0 5 10 15 20 0 5 10 15 20 t (msec) t (msec) Introduction to filters Consider v ( t ) = v 1 ( t ) + v 2 ( t )
Introduction to filters
Consider v(t) = v1(t) + v2(t) = Vm1 sin ω1t + Vm2 sin ω2t .
−1 1 5 10 15 20 0 5 10 15 20 t (msec) t (msec)
v2 v1 v
Introduction to filters
Consider v(t) = v1(t) + v2(t) = Vm1 sin ω1t + Vm2 sin ω2t .
−1 1 5 10 15 20 0 5 10 15 20 t (msec) t (msec)
v2 v1 v LPF vo = v1 v
A low-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the low-frequency component v1(t) and remove the high-frequency component v2(t).
Introduction to filters
Consider v(t) = v1(t) + v2(t) = Vm1 sin ω1t + Vm2 sin ω2t .
−1 1 5 10 15 20 0 5 10 15 20 t (msec) t (msec)
v2 v1 v LPF vo = v1 v HPF v vo = v2
A low-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the low-frequency component v1(t) and remove the high-frequency component v2(t). A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequency component v2(t) and remove the low-frequency component v1(t).
Introduction to filters
Consider v(t) = v1(t) + v2(t) = Vm1 sin ω1t + Vm2 sin ω2t .
−1 1 5 10 15 20 0 5 10 15 20 t (msec) t (msec)
v2 v1 v LPF vo = v1 v HPF v vo = v2
A low-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the low-frequency component v1(t) and remove the high-frequency component v2(t). A high-pass filter with a cut-off frequency ω1 < ωc < ω2 will pass the high-frequency component v2(t) and remove the low-frequency component v1(t). There are some other types of filters, as we will see.
Ideal low-pass filter
1
ωc ω vo(t) H(jω) H(jω) vi(t)
Vo(jω) = H(jω) Vi(jω) .
Ideal low-pass filter
1
ωc ω vo(t) H(jω) H(jω) vi(t)
LPF
ωc ωc ω ω Vi(jω) Vo(jω)
Vo(jω) = H(jω) Vi(jω) .
Ideal low-pass filter
1
ωc ω vo(t) H(jω) H(jω) vi(t)
LPF
ωc ωc ω ω Vi(jω) Vo(jω)
Vo(jω) = H(jω) Vi(jω) . All components with ω < ωc appear at the output without attenuation. All components with ω > ωc get eliminated. (Note that the ideal low-pass filter has ∠H(jω) = 1, i.e., H(jω) = 1 + j0 .)
Ideal filters
1 Low−pass
ωc ω H(jω)
Ideal filters
1 Low−pass
ωc ω H(jω)
1 High−pass
ωc ω H(jω)
Ideal filters
1 Low−pass
ωc ω H(jω)
1 High−pass
ωc ω H(jω)
1 Band−pass
ωL ωH ω H(jω)
Ideal filters
1 Low−pass
ωc ω H(jω)
1 High−pass
ωc ω H(jω)
1 Band−pass
ωL ωH ω H(jω)
1 Band−reject
ωL ωH ω H(jω)
Ideal low-pass filter: example
1 −1 1.5 −1.5 1 Filter transfer function Filter output 1 −1 5 10 15 20 t (msec) f (kHz) t (msec) 5 10 15 20 0.5 1 1.5 2 2.5
v v3 v2 v1 H(jω)
Ideal high-pass filter: example
1 −1 1.5 −1.5 Filter transfer function Filter output 1 −1 1 5 10 15 20 t (msec) f (kHz) t (msec) 5 10 15 20 0.5 1 1.5 2 2.5
v v3 v2 v1 H(jω)
Ideal band-pass filter: example
1 −1 1 −1 1.5 −1.5 Filter transfer function Filter output 1 5 10 15 20 t (msec) f (kHz) t (msec) 5 10 15 20 0.5 1 1.5 2 2.5
v v3 v2 v1 H(jω)
Ideal band-reject filter: example
1.5 −1.5 1 1 −1 1.5 −1.5 Filter transfer function Filter output 5 10 15 20 t (msec) f (kHz) t (msec) 5 10 15 20 0.5 1 1.5 2 2.5
v H(jω) v3 v2 v1
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitable H(jω) that can be obtained with circuit elements. For example, H(s) = 1 a5s5 + a4s4 + a3s3 + a2s2 + a1s + a0 represents a 5th-order low-pass filter.
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitable H(jω) that can be obtained with circuit elements. For example, H(s) = 1 a5s5 + a4s4 + a3s3 + a2s2 + a1s + a0 represents a 5th-order low-pass filter. * Some commonly used approximations (polynomials) are the Butterworth, Chebyshev, Bessel, and elliptic functions.
Practical filter circuits
* In practical filter circuits, the ideal filter response is approximated with a suitable H(jω) that can be obtained with circuit elements. For example, H(s) = 1 a5s5 + a4s4 + a3s3 + a2s2 + a1s + a0 represents a 5th-order low-pass filter. * Some commonly used approximations (polynomials) are the Butterworth, Chebyshev, Bessel, and elliptic functions. * Coefficients for these filters are listed in filter handbooks. Also, programs for filter design are available on the internet.
Practical filters
High−pass Practical
1 1
Ideal Practical Ideal Amax |H| ωc ω ωs ωc ω Amin |H| ω ωs ωc ω |H| Amax ωc |H| Amin
Practical filters
High−pass Practical
1 1
Ideal Practical Ideal Amax |H| ωc ω ωs ωc ω Amin |H| ω ωs ωc ω |H| Amax ωc |H| Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, e.g., Amax = 1 dB.
Practical filters
High−pass Practical
1 1
Ideal Practical Ideal Amax |H| ωc ω ωs ωc ω Amin |H| ω ωs ωc ω |H| Amax ωc |H| Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, e.g., Amax = 1 dB. * Amin is the minimum attenuation to be provided by the filter, e.g., Amin = 60 dB.
Practical filters
High−pass Practical
1 1
Ideal Practical Ideal Amax |H| ωc ω ωs ωc ω Amin |H| ω ωs ωc ω |H| Amax ωc |H| Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, e.g., Amax = 1 dB. * Amin is the minimum attenuation to be provided by the filter, e.g., Amin = 60 dB. * ωs: edge of the stop band.
Practical filters
High−pass Practical
1 1
Ideal Practical Ideal Amax |H| ωc ω ωs ωc ω Amin |H| ω ωs ωc ω |H| Amax ωc |H| Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, e.g., Amax = 1 dB. * Amin is the minimum attenuation to be provided by the filter, e.g., Amin = 60 dB. * ωs: edge of the stop band. * ωs/ωc (for a low-pass filter): selectivity factor, a measure of the sharpness of the filter.
Practical filters
High−pass Practical
1 1
Ideal Practical Ideal Amax |H| ωc ω ωs ωc ω Amin |H| ω ωs ωc ω |H| Amax ωc |H| Amin
* A practical filter may exhibit a ripple. Amax is called the maximum passband ripple, e.g., Amax = 1 dB. * Amin is the minimum attenuation to be provided by the filter, e.g., Amin = 60 dB. * ωs: edge of the stop band. * ωs/ωc (for a low-pass filter): selectivity factor, a measure of the sharpness of the filter. * ωc < ω < ωs: transition band.
Practical filters
For a low-pass filter, H(s) = 1
n
ai(s/ωc)i . Coefficients (ai) for various types of filters are tabulated in handbooks. We now look at |H(jω)| for two commonly used filters.
Practical filters
For a low-pass filter, H(s) = 1
n
ai(s/ωc)i . Coefficients (ai) for various types of filters are tabulated in handbooks. We now look at |H(jω)| for two commonly used filters. Butterworth filters: |H(jω)| = 1
Practical filters
For a low-pass filter, H(s) = 1
n
ai(s/ωc)i . Coefficients (ai) for various types of filters are tabulated in handbooks. We now look at |H(jω)| for two commonly used filters. Butterworth filters: |H(jω)| = 1
Chebyshev filters: |H(jω)| = 1
n (ω/ωc)
where Cn(x) = cos
Cn(x) = cosh
Practical filters
For a low-pass filter, H(s) = 1
n
ai(s/ωc)i . Coefficients (ai) for various types of filters are tabulated in handbooks. We now look at |H(jω)| for two commonly used filters. Butterworth filters: |H(jω)| = 1
Chebyshev filters: |H(jω)| = 1
n (ω/ωc)
where Cn(x) = cos
Cn(x) = cosh
H(s) for a high-pass filter can be obtained from H(s) of the corresponding low-pass filter by (s/ωc) → (ωc/s) .
Practical filters (low-pass)
Butterworth filters: Chebyshev filters:
1 −100 1 −100 3 4 5 2 n=1 n=1 4 3 2 5 n=1 2 3 4 5 2 3 n=1 4 5 2 3 4 5 1 0.01 0.1 1 10 100 2 3 4 5 1 0.01 0.1 1 10 100
|H| (dB) ω/ωc |H| ω/ωc ω/ωc |H| ω/ωc |H| (dB) ǫ = 0.5 ǫ = 0.5
Practical filters (high-pass)
Butterworth filters: Chebyshev filters:
1 −100 1 −100 n=1 2 4 5 n=1 2 3 4 5 3 n=1 2 3 4 5 n=1 2 3 4 5 0.01 0.1 1 10 100 0.01 0.1 1 10 100 1 2 3 4 1 2 3 4
ω/ωc |H| ω/ωc |H| (dB) ω/ωc |H| ω/ωc |H| (dB) ǫ = 0.5 ǫ = 0.5
Passive filter example
5 µF C 100 Ω Vs Vo R
Passive filter example
5 µF C 100 Ω Vs Vo R
(Low−pass filter)
with ω0 = 1/RC → f0 = ω0/2π = 318 Hz H(s) = (1/sC) R + (1/sC) = 1 1 + (s/ω0) ,
Passive filter example
5 µF C 100 Ω Vs Vo R
(Low−pass filter)
with ω0 = 1/RC → f0 = ω0/2π = 318 Hz H(s) = (1/sC) R + (1/sC) = 1 1 + (s/ω0) ,
f (Hz) −60 −40 −20 20
105 104 103 102 101 |H| (dB)
(SEQUEL file: ee101 rc ac 2.sqproj)
Passive filter example
Vs Vo R 100 Ω C 4 µF 0.1 mF L
Passive filter example
Vs Vo R 100 Ω C 4 µF 0.1 mF L
(Band−pass filter)
with ω0 = 1/ √ LC → f0 = ω0/2π = 7.96 kHz H(s) = (sL) (1/sC) R + (sL) (1/sC) = s(L/R) 1 + s(L/R) + s2LC
Passive filter example
Vs Vo R 100 Ω C 4 µF 0.1 mF L
(Band−pass filter)
with ω0 = 1/ √ LC → f0 = ω0/2π = 7.96 kHz H(s) = (sL) (1/sC) R + (sL) (1/sC) = s(L/R) 1 + s(L/R) + s2LC
−20 −40 −60 −80 f (Hz)
105 104 103 102 |H| (dB) (SEQUEL file: ee101 rlc 3.sqproj)
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.
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 (“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.
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.
Op-amp filters: example
Vs RL C R2 R1 Vo 10 k 10 nF 1 k
Op-amp filters: example
Vs RL C R2 R1 Vo 10 k 10 nF 1 k
Op-amp filters are designed for op-amp operation in the linear region → Our analysis of the inverting amplifier applies, and we get, Vo = − R2 (1/sC) R1 Vs (Vs and Vo are phasors) H(s) = − R2 R1 1 1 + sR2C
Op-amp filters: example
Vs RL C R2 R1 Vo 10 k 10 nF 1 k
Op-amp filters are designed for op-amp operation in the linear region → Our analysis of the inverting amplifier applies, and we get, Vo = − R2 (1/sC) R1 Vs (Vs and Vo are phasors) H(s) = − R2 R1 1 1 + sR2C This is a low-pass filter, with ω0 = 1/R2C (i.e., f0 = ω0/2π = 1.59 kHz).
Op-amp filters: example
Vs RL C R2 R1 Vo 10 k 10 nF 1 k
f (Hz) 20 −20
105 104 103 102 101 |H| (dB)
Op-amp filters are designed for op-amp operation in the linear region → Our analysis of the inverting amplifier applies, and we get, Vo = − R2 (1/sC) R1 Vs (Vs and Vo are phasors) H(s) = − R2 R1 1 1 + sR2C This is a low-pass filter, with ω0 = 1/R2C (i.e., f0 = ω0/2π = 1.59 kHz).
Op-amp filters: example
Vs RL C R2 R1 Vo 10 k 10 nF 1 k
f (Hz) 20 −20
105 104 103 102 101 |H| (dB)
Op-amp filters are designed for op-amp operation in the linear region → Our analysis of the inverting amplifier applies, and we get, Vo = − R2 (1/sC) R1 Vs (Vs and Vo are phasors) H(s) = − R2 R1 1 1 + sR2C This is a low-pass filter, with ω0 = 1/R2C (i.e., f0 = ω0/2π = 1.59 kHz). (SEQUEL file: ee101 op filter 1.sqproj)
Op-amp filters: example
Vs RL C R2 R1 Vo 1 k 100 nF 10 k
Op-amp filters: example
Vs RL C R2 R1 Vo 1 k 100 nF 10 k
H(s) = − R2 R1 + (1/sC) = − sR2C 1 + sR1C .
Op-amp filters: example
Vs RL C R2 R1 Vo 1 k 100 nF 10 k
H(s) = − R2 R1 + (1/sC) = − sR2C 1 + sR1C . This is a high-pass filter, with ω0 = 1/R1C (i.e., f0 = ω0/2π = 1.59 kHz).
Op-amp filters: example
Vs RL C R2 R1 Vo 1 k 100 nF 10 k
f (Hz) 20 −20 −40
105 104 103 102 101 |H| (dB)
H(s) = − R2 R1 + (1/sC) = − sR2C 1 + sR1C . This is a high-pass filter, with ω0 = 1/R1C (i.e., f0 = ω0/2π = 1.59 kHz).
Op-amp filters: example
Vs RL C R2 R1 Vo 1 k 100 nF 10 k
f (Hz) 20 −20 −40
105 104 103 102 101 |H| (dB)
H(s) = − R2 R1 + (1/sC) = − sR2C 1 + sR1C . This is a high-pass filter, with ω0 = 1/R1C (i.e., f0 = ω0/2π = 1.59 kHz). (SEQUEL file: ee101 op filter 2.sqproj)
Op-amp filters: example
Vs RL C1 R2 R1 Vo 100 k 10 k 0.8 µF C2 80 pF
Op-amp filters: example
Vs RL C1 R2 R1 Vo 100 k 10 k 0.8 µF C2 80 pF
H(s) = − R2 (1/sC2) R1 + (1/sC1) = − R2 R1 sR1C1 (1 + sR1C1)(1 + sR2C2) .
Op-amp filters: example
Vs RL C1 R2 R1 Vo 100 k 10 k 0.8 µF C2 80 pF
H(s) = − R2 (1/sC2) R1 + (1/sC1) = − R2 R1 sR1C1 (1 + sR1C1)(1 + sR2C2) . This is a band-pass filter, with ωL = 1/R1C1 and ωH = 1/R2C2 . → fL = 20 Hz, fH = 20 kHz.
Op-amp filters: example
Vs RL C1 R2 R1 Vo 100 k 10 k 0.8 µF C2 80 pF
f (Hz) 20
100 102 104 106 |H| (dB)
H(s) = − R2 (1/sC2) R1 + (1/sC1) = − R2 R1 sR1C1 (1 + sR1C1)(1 + sR2C2) . This is a band-pass filter, with ωL = 1/R1C1 and ωH = 1/R2C2 . → fL = 20 Hz, fH = 20 kHz.
Op-amp filters: example
Vs RL C1 R2 R1 Vo 100 k 10 k 0.8 µF C2 80 pF
f (Hz) 20
100 102 104 106 |H| (dB)
H(s) = − R2 (1/sC2) R1 + (1/sC1) = − R2 R1 sR1C1 (1 + sR1C1)(1 + sR2C2) . This is a band-pass filter, with ωL = 1/R1C1 and ωH = 1/R2C2 . → fL = 20 Hz, fH = 20 kHz. (SEQUEL file: ee101 op filter 3.sqproj)
Graphic equalizer
f (Hz) (Ref.: S. Franco, "Design with Op Amps and analog ICs") 20 −20 R2 C1 a 1−a R3A R3B R1A R1B C2 a=0.9 0.7 0.5 0.3 0.1
Vs RL Vo 105 104 103 102 101 R3A = R3B = 100 kΩ R1A = R1B = 470 Ω R2 = 10 kΩ C1 = 100 nF C2 = 10 nF |H| (dB)
Graphic equalizer
f (Hz) (Ref.: S. Franco, "Design with Op Amps and analog ICs") 20 −20 R2 C1 a 1−a R3A R3B R1A R1B C2 a=0.9 0.7 0.5 0.3 0.1
Vs RL Vo 105 104 103 102 101 R3A = R3B = 100 kΩ R1A = R1B = 470 Ω R2 = 10 kΩ C1 = 100 nF C2 = 10 nF |H| (dB)
* Equalizers are implemented as arrays of narrow-band filters, each with an adjustable gain (attenuation) around a centre frequency.
Graphic equalizer
f (Hz) (Ref.: S. Franco, "Design with Op Amps and analog ICs") 20 −20 R2 C1 a 1−a R3A R3B R1A R1B C2 a=0.9 0.7 0.5 0.3 0.1
Vs RL Vo 105 104 103 102 101 R3A = R3B = 100 kΩ R1A = R1B = 470 Ω R2 = 10 kΩ C1 = 100 nF C2 = 10 nF |H| (dB)
* 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)
Sallen-Key filter example (2nd order, low-pass)
f (Hz) 40 20 −20 −40 −60 C1 RB R1 R2 C2 RA (Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL V1 Vs Vo 105 104 103 102 101 |H| (dB) R1 = R2 = 15.8 kΩ C1 = C2 = 10 nF RA = 10 kΩ, RB = 17.8 kΩ
Sallen-Key filter example (2nd order, low-pass)
f (Hz) 40 20 −20 −40 −60 C1 RB R1 R2 C2 RA (Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL V1 Vs Vo 105 104 103 102 101 |H| (dB) R1 = R2 = 15.8 kΩ C1 = C2 = 10 nF RA = 10 kΩ, RB = 17.8 kΩ V+ = V− = Vo RA RA + RB ≡ Vo/K .
Sallen-Key filter example (2nd order, low-pass)
f (Hz) 40 20 −20 −40 −60 C1 RB R1 R2 C2 RA (Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL V1 Vs Vo 105 104 103 102 101 |H| (dB) R1 = R2 = 15.8 kΩ C1 = C2 = 10 nF RA = 10 kΩ, RB = 17.8 kΩ V+ = V− = Vo RA RA + RB ≡ Vo/K . Also, V+ = (1/sC2) R2 + (1/sC2) V1 = 1 1 + sR2C2 V1 .
Sallen-Key filter example (2nd order, low-pass)
f (Hz) 40 20 −20 −40 −60 C1 RB R1 R2 C2 RA (Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL V1 Vs Vo 105 104 103 102 101 |H| (dB) R1 = R2 = 15.8 kΩ C1 = C2 = 10 nF RA = 10 kΩ, RB = 17.8 kΩ V+ = V− = Vo RA RA + RB ≡ Vo/K . Also, V+ = (1/sC2) R2 + (1/sC2) V1 = 1 1 + sR2C2 V1 . KCL at V1 → 1 R1 (Vs − V1) + sC1(Vo − V1) + 1 R2 (V+ − V1) = 0 .
Sallen-Key filter example (2nd order, low-pass)
f (Hz) 40 20 −20 −40 −60 C1 RB R1 R2 C2 RA (Ref.: S. Franco, "Design with Op Amps and analog ICs")
RL V1 Vs Vo 105 104 103 102 101 |H| (dB) R1 = R2 = 15.8 kΩ C1 = C2 = 10 nF RA = 10 kΩ, RB = 17.8 kΩ V+ = V− = Vo RA RA + RB ≡ Vo/K . Also, V+ = (1/sC2) R2 + (1/sC2) V1 = 1 1 + sR2C2 V1 . KCL at V1 → 1 R1 (Vs − V1) + sC1(Vo − V1) + 1 R2 (V+ − V1) = 0 . Combining the above equations, H(s) = K 1 + s [(R1 + R2)C2 + (1 − K)R1C1] + s2R1C1R2C2 . (SEQUEL file: ee101 op filter 5.sqproj)
Sixth-order Chebyshev low-pass filter (cascade design)
20 −20 −40 −60 −80 f (Hz) (Ref.: S. Franco, "Design with Op Amps and analog ICs") SEQUEL file: ee101_op_filter_6.sqproj 5.1 n 10 n 62 n 2.2 n 510 p 220 p 2.49 k 4.64 k 6.49 k 8.25 k 10.2 k 10.7 k
Vs RL Vo 105 104 103 102 |H| (dB)
Third-order Chebyshev high-pass filter
7.68 k 100 n (Ref.: S. Franco, "Design with Op Amps and analog ICs") SEQUEL file: ee101_op_filter_7.sqproj 100 n 54.9 k 100 n 15.4 k 154 k f (Hz) 20 −20 −40 −60 −80
Vs RL Vo 103 102 101 100 |H| (dB)
Band-pass filter example
f (Hz) 20 −20 −40 40 (Ref.: J. M. Fiore, "Op Amps and linear ICs") SEQUEL file: ee101_op_filter_8.sqproj 5 k 5 k 5 k 5 k 370 k 5 k 5 k 7.4 n 7.4 n
Vs Vo 105 104 103 102 |H| (dB)
Notch filter example
(Ref.: J. M. Fiore, "Op Amps and linear ICs") SEQUEL file: ee101_op_filter_9.sqproj f (Hz) −40 −20 10 k 10 k 10 k 89 k 10 k 1 k 10 k 265 n 265 n 10 k 10 k 10 k
Vs Vo 101 102 |H| (dB)
Half-wave rectifier
Vo slope = 1 Vi Ideal half-wave rectifier Vi Vo
Half-wave rectifier
Vo slope = 1 Vi Ideal half-wave rectifier Vi Vo T/2 T 3T/2 2T
1 Vi Vo
Half-wave rectifier
Vo slope = 1 Vi Ideal half-wave rectifier Vi Vo T/2 T 3T/2 2T
1 Vi Vo Vo Von slope = 1 Vi Vi Vo
Half-wave rectifier
Vo slope = 1 Vi Ideal half-wave rectifier Vi Vo T/2 T 3T/2 2T
1 Vi Vo Vo Von slope = 1 Vi Vi Vo Vo T/2 T 3T/2 2T Vi Von
1
Half-wave rectifier
Vo slope = 1 Vi Ideal half-wave rectifier Vi Vo T/2 T 3T/2 2T
1 Vi Vo Vo Von slope = 1 Vi Vi Vo Vo T/2 T 3T/2 2T Vi Von
1 → need an improved circuit
Half-wave precision rectifier
D
R Vi Vo
Half-wave precision rectifier
D
R Vi Vo R Von Vi Vo1 Vo i− iD iR
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
D
R Vi Vo R Von Vi Vo1 Vo i− iD iR
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, iR = iD. V+ − V− = Vo1 AV = Vo + 0.7 V AV ≈ 0 V → Vo = V− ≈ V+ = Vi.
Half-wave precision rectifier
D
R Vi Vo R Von Vi Vo1 Vo i− iD iR
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, iR = iD. V+ − V− = Vo1 AV = Vo + 0.7 V AV ≈ 0 V → Vo = V− ≈ V+ = Vi. This situation arises only if iD > 0 (since the diode can only conduct in the forward direction), i.e., iR > 0 → Vo = iRR > 0, and therefore Vi = Vo > 0 V .
Half-wave precision rectifier
D
R Vi Vo R Von Vi Vo1 Vo i− iD iR
slope=1
Vo Vi
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, iR = iD. V+ − V− = Vo1 AV = Vo + 0.7 V AV ≈ 0 V → Vo = V− ≈ V+ = Vi. This situation arises only if iD > 0 (since the diode can only conduct in the forward direction), i.e., iR > 0 → Vo = iRR > 0, and therefore Vi = Vo > 0 V .
Half-wave precision rectifier
D
R Vi Vo R Von Vi Vo1 Vo i− iD iR
slope=1
Vo Vi
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, iR = iD. V+ − V− = Vo1 AV = Vo + 0.7 V AV ≈ 0 V → Vo = V− ≈ V+ = Vi. This situation arises only if iD > 0 (since the diode can only conduct in the forward direction), i.e., iR > 0 → Vo = iRR > 0, and therefore Vi = Vo > 0 V . Note: Von does not appear in the graph.
Half-wave precision rectifier
D
R Vi Vo
Half-wave precision rectifier
D
R Vi Vo R Vo1 Vi Vo
(ii) D is not conducting → Vo = 0 V .
Half-wave precision rectifier
D
R Vi Vo R Vo1 Vi Vo
(ii) D is not conducting → Vo = 0 V . What about Vo1? Since the op-amp is now in the open-loop configuration, a very small Vi is enough to drive it to saturation.
Half-wave precision rectifier
D
R Vi Vo R Vo1 Vi Vo
(ii) D is not conducting → Vo = 0 V . What about Vo1? Since the op-amp is now in the open-loop configuration, a very small Vi is enough to drive it to saturation. Note that Case (ii) occurs when Vi < 0 V (we have already looked at Vi > 0). Since V+ − V− = Vi − 0 = Vi is negative, Vo1 is driven to −Vsat.
Half-wave precision rectifier
D
R Vi Vo R Vo1 Vi Vo Vo Vi Vo = 0
(ii) D is not conducting → Vo = 0 V . What about Vo1? Since the op-amp is now in the open-loop configuration, a very small Vi is enough to drive it to saturation. Note that Case (ii) occurs when Vi < 0 V (we have already looked at Vi > 0). Since V+ − V− = Vi − 0 = Vi is negative, Vo1 is driven to −Vsat.
Half-wave precision rectifier
D on D off diode Super Super diode D
R R Vi Vo = Vi Vo = 0 Vo Vi Vo Vi Vo Vo1 iR
Half-wave precision rectifier
D on D off diode Super Super diode D
R R Vi Vo = Vi Vo = 0 Vo Vi Vo Vi Vo Vo1 iR
* The circuit is called “super diode” (an ideal diode with Von = 0 V).
Half-wave precision rectifier
D on D off diode Super Super diode D
R R Vi Vo = Vi Vo = 0 Vo Vi Vo Vi Vo Vo1 iR
* The circuit is called “super diode” (an ideal diode with Von = 0 V). * When D conducts, the op-amp operates in the linear region, and we have V+ ≈ V−.
Half-wave precision rectifier
D on D off diode Super Super diode D
R R Vi Vo = Vi Vo = 0 Vo Vi Vo Vi Vo Vo1 iR
* The circuit is called “super diode” (an ideal diode with Von = 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+ = Vi, and Vo1 = −Vsat.
Half-wave precision rectifier
D on D off diode Super Super diode D
R R Vi Vo = Vi Vo = 0 Vo Vi Vo Vi Vo Vo1 iR
* The circuit is called “super diode” (an ideal diode with Von = 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+ = Vi, and Vo1 = −Vsat. * Where does iR come from?
25 50 75 100 1.5 −1.5 1.5 −1.5 1.5 −1.5
m (t) y (t) c (t) time (µsec) A = 1 M = 0.3 fc = 200 kHz fm = 10 kHz
Application: AM demodulation
25 50 75 100 1.5 −1.5 1.5 −1.5 1.5 −1.5
m (t) y (t) c (t) time (µsec) A = 1 M = 0.3 fc = 200 kHz fm = 10 kHz
Application: AM demodulation
Carrier wave: c(t) = A sin(2πfct)
25 50 75 100 1.5 −1.5 1.5 −1.5 1.5 −1.5
m (t) y (t) c (t) time (µsec) A = 1 M = 0.3 fc = 200 kHz fm = 10 kHz
Application: AM demodulation
Carrier wave: c(t) = A sin(2πfct) Signal (e.g., audio): m(t) = M sin(2πfmt + φ)
25 50 75 100 1.5 −1.5 1.5 −1.5 1.5 −1.5
m (t) y (t) c (t) time (µsec) A = 1 M = 0.3 fc = 200 kHz fm = 10 kHz
Application: AM demodulation
Carrier wave: c(t) = A sin(2πfct) Signal (e.g., audio): m(t) = M sin(2πfmt + φ) AM wave: y(t) = [1 + m(t)] c(t) (Assume M < 1)
25 50 75 100 1.5 −1.5 1.5 −1.5 1.5 −1.5
m (t) y (t) c (t) time (µsec) A = 1 M = 0.3 fc = 200 kHz fm = 10 kHz
Application: AM demodulation
Carrier wave: c(t) = A sin(2πfct) Signal (e.g., audio): m(t) = M sin(2πfmt + φ) AM wave: y(t) = [1 + m(t)] c(t) (Assume M < 1) e.g., Vividh Bharati: fc = 1188 kHz, fm ≃ 10 kHz (audio).
AM demodulation using a peak detector
0.15 −0.15
filter AM input
diode Super t (ms) 0.3 0.4 0.5 0.2 t (ms) 1 2
V1 V1 Vi Vo Vi
AM demodulation using a peak detector
0.15 −0.15
filter AM input
diode Super t (ms) 0.3 0.4 0.5 0.2 t (ms) 1 2
V1 V1 Vi Vo Vi
* charging through super diode, discharging through resistor
AM demodulation using a peak detector
0.15 −0.15
filter AM input
diode Super t (ms) 0.3 0.4 0.5 0.2 t (ms) 1 2
V1 V1 Vi Vo Vi
* charging through super diode, discharging through resistor * The time constant (RC) needs to be carefully selected.
AM demodulation using a peak detector
0.15 −0.15
filter AM input
diode Super t (ms) 0.3 0.4 0.5 0.2 t (ms) 1 2
V1 V1 Vi Vo Vi
* charging through super diode, discharging through resistor * The time constant (RC) needs to be carefully selected.
SEQUEL file: super diode.sqproj
Clipping and clamping
D RL Vo R Vi VR D RL Vo Vi VR C D RL Vo Vi VR C D RL Vo R Vi VR
* What is the function provided by each circuit?
Clipping and clamping
D RL Vo R Vi VR D RL Vo Vi VR C D RL Vo Vi VR C D RL Vo R Vi VR
* What is the function provided by each circuit? * Verify with simulation (and in the lab).
D RL Vo R Vi VR iD When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR.
D RL Vo R Vi VR iD When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD = VR RL + VR − Vi R .
D RL Vo R Vi VR iD When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD = VR RL + VR − Vi R . Since iD > 0, VR 1 RL + 1 R
R → Vi < VR R + RL RL
D RL Vo R Vi VR iD When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD = VR RL + VR − Vi R . Since iD > 0, VR 1 RL + 1 R
R → Vi < VR R + RL RL
For Vi > Vi1, D does not conduct → Vo = RL R + RL Vi.
D RL Vo R Vi VR iD Vo Vi Vi1 Slope = RL R + RL VR When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD = VR RL + VR − Vi R . Since iD > 0, VR 1 RL + 1 R
R → Vi < VR R + RL RL
For Vi > Vi1, D does not conduct → Vo = RL R + RL Vi.
D RL Vo R Vi VR iD Vo Vi Vi1 Slope = RL R + RL VR When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD = VR RL + VR − Vi R . Since iD > 0, VR 1 RL + 1 R
R → Vi < VR R + RL RL
For Vi > Vi1, D does not conduct → Vo = RL R + RL Vi. If RL ≫ R, Vi1 = R, and slope = 1 for Vi > Vi1.
D RL Vo R Vi VR iD When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR.
D RL Vo R Vi VR iD When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD + VR RL + VR − Vi R = 0.
D RL Vo R Vi VR iD When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD + VR RL + VR − Vi R = 0. Since iD > 0, −VR 1 RL + 1 R
R > 0 → Vi > VR R + RL RL
D RL Vo R Vi VR iD When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD + VR RL + VR − Vi R = 0. Since iD > 0, −VR 1 RL + 1 R
R > 0 → Vi > VR R + RL RL
For Vi < Vi1, D does not conduct → Vo = RL R + RL Vi.
D RL Vo R Vi VR iD Vo Vi Vi1 VR Slope = RL R + RL When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD + VR RL + VR − Vi R = 0. Since iD > 0, −VR 1 RL + 1 R
R > 0 → Vi > VR R + RL RL
For Vi < Vi1, D does not conduct → Vo = RL R + RL Vi.
D RL Vo R Vi VR iD Vo Vi Vi1 VR Slope = RL R + RL When D conducts, feedback path is closed → V− ≈ V+ = VR → Vo = VR. KCL: iD + VR RL + VR − Vi R = 0. Since iD > 0, −VR 1 RL + 1 R
R > 0 → Vi > VR R + RL RL
For Vi < Vi1, D does not conduct → Vo = RL R + RL Vi. If RL ≫ R, Vi1 = R, and slope = 1 for Vi < Vi1.
D RL Vo Vi C VR iD VC Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle).
D RL Vo Vi C VR iD VC Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle). When D conducts, V− ≈ VR, and VC (t) = VR − Vm sin ωt. → V max
C
= VR − (−Vm) = VR + Vm.
D RL Vo Vi C VR iD VC Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle). When D conducts, V− ≈ VR, and VC (t) = VR − Vm sin ωt. → V max
C
= VR − (−Vm) = VR + Vm. In steady state, VC remains equal to V max
C
→ Vo(t) = Vi(t) + V max
C
= Vm sin ωt + VR + Vm.
D RL Vo Vi C VR iD VC Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle). When D conducts, V− ≈ VR, and VC (t) = VR − Vm sin ωt. → V max
C
= VR − (−Vm) = VR + Vm. In steady state, VC remains equal to V max
C
→ Vo(t) = Vi(t) + V max
C
= Vm sin ωt + VR + Vm. Note: Von of the diode does not appear in the expression for Vo(t).
D RL Vo Vi C VR iD VC 0.5 1 1.5 2 Time (msec)
2 4 6 8 Vi VR Vo Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle). When D conducts, V− ≈ VR, and VC (t) = VR − Vm sin ωt. → V max
C
= VR − (−Vm) = VR + Vm. In steady state, VC remains equal to V max
C
→ Vo(t) = Vi(t) + V max
C
= Vm sin ωt + VR + Vm. Note: Von of the diode does not appear in the expression for Vo(t).
D RL Vo Vi C VR iD VC Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle).
D RL Vo Vi C VR iD VC Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle). When D conducts, V− ≈ VR, and VC (t) = Vm sin ωt − VR. → V max
C
= Vm − VR.
D RL Vo Vi C VR iD VC Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle). When D conducts, V− ≈ VR, and VC (t) = Vm sin ωt − VR. → V max
C
= Vm − VR. In steady state, VC remains equal to V max
C
→ Vo(t) = Vi(t) − V max
C
= Vm sin ωt + VR − Vm.
D RL Vo Vi C VR iD VC Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle). When D conducts, V− ≈ VR, and VC (t) = Vm sin ωt − VR. → V max
C
= Vm − VR. In steady state, VC remains equal to V max
C
→ Vo(t) = Vi(t) − V max
C
= Vm sin ωt + VR − Vm. Note: Von of the diode does not appear in the expression for Vo(t).
D RL Vo Vi C VR iD VC 0.5 1 1.5 2 4 2
Time (msec) Vo VR Vi Time constant for the discharging process is RLC. Assume RLC ≫ T → VC can only increase (in one cycle). When D conducts, V− ≈ VR, and VC (t) = Vm sin ωt − VR. → V max
C
= Vm − VR. In steady state, VC remains equal to V max
C
→ Vo(t) = Vi(t) − V max
C
= Vm sin ωt + VR − Vm. Note: Von of the diode does not appear in the expression for Vo(t).
Half-wave precision rectifier
10 20 30 40 time (msec) 0.6 −0.6 0.4 0.2 −0.2 −0.4 D on D off 2 −2 −4 −6 −8 −10 −12 −14 Super diode D
R Vi Vo = Vi Vo = 0 Vo Vi Vo Vo1 Vo Vi Vi Vo1
Half-wave precision rectifier
10 20 30 40 time (msec) 0.6 −0.6 0.4 0.2 −0.2 −0.4 D on D off 2 −2 −4 −6 −8 −10 −12 −14 Super diode D
R Vi Vo = Vi Vo = 0 Vo Vi Vo Vo1 Vo Vi Vi Vo1
* When Vi > 0, the op-amp operates in the linear region, and Vo1 = Vo + Von.
Half-wave precision rectifier
10 20 30 40 time (msec) 0.6 −0.6 0.4 0.2 −0.2 −0.4 D on D off 2 −2 −4 −6 −8 −10 −12 −14 Super diode D
R Vi Vo = Vi Vo = 0 Vo Vi Vo Vo1 Vo Vi Vi Vo1
* When Vi > 0, the op-amp operates in the linear region, and Vo1 = Vo + Von. * When Vi < 0, the op-amp operates in the
saturation, and Vo1 = −Vsat.
Half-wave precision rectifier
10 20 30 40 time (msec) 0.6 −0.6 0.4 0.2 −0.2 −0.4 D on D off 2 −2 −4 −6 −8 −10 −12 −14 Super diode D
R Vi Vo = Vi Vo = 0 Vo Vi Vo Vo1 Vo Vi Vi Vo1
* When Vi > 0, the op-amp operates in the linear region, and Vo1 = Vo + Von. * When Vi < 0, the op-amp operates in the
saturation, and Vo1 = −Vsat. * The Vi < 0 to Vi > 0 transition requires the
relatively slow process and is limited by the
Half-wave precision rectifier
10 20 30 40 time (msec) 0.6 −0.6 0.4 0.2 −0.2 −0.4 D on D off 2 −2 −4 −6 −8 −10 −12 −14 Super diode D
R Vi Vo = Vi Vo = 0 Vo Vi Vo Vo1 Vo Vi Vi Vo1
* When Vi > 0, the op-amp operates in the linear region, and Vo1 = Vo + Von. * When Vi < 0, the op-amp operates in the
saturation, and Vo1 = −Vsat. * The Vi < 0 to Vi > 0 transition requires the
relatively slow process and is limited by the
SEQUEL file: ee101 super diode 1.sqproj
Half-wave precision rectifier
D on D off Super diode D
R Vi Vo = Vi Vo = 0 Vo Vo Vo1 Vi
* The time taken by the op-amp to come out of saturation can be neglected at low signal frequencies.
Half-wave precision rectifier
D on D off Super diode D
R Vi Vo = Vi Vo = 0 Vo Vo Vo1 Vi
0.6 −0.6 10 20 30 40 time (msec)
Vo Vi f = 50 Hz
* The time taken by the op-amp to come out of saturation can be neglected at low signal frequencies.
Half-wave precision rectifier
D on D off Super diode D
R Vi Vo = Vi Vo = 0 Vo Vo Vo1 Vi
0.6 −0.6 10 20 30 40 time (msec)
Vo Vi f = 50 Hz
* The time taken by the op-amp to come out of saturation can be neglected at low signal frequencies. * At high signal frequencies, it leads to distortion in the
Half-wave precision rectifier
D on D off Super diode D
R Vi Vo = Vi Vo = 0 Vo Vo Vo1 Vi
0.6 −0.6 10 20 30 40 time (msec)
Vo Vi f = 50 Hz
0.6 −0.6 0.5 1 1.5 2 time (msec)
Vo Vi f = 1 kHz
* The time taken by the op-amp to come out of saturation can be neglected at low signal frequencies. * At high signal frequencies, it leads to distortion in the
Half-wave precision rectifier
D on D off Super diode D
R Vi Vo = Vi Vo = 0 Vo Vo Vo1 Vi
0.6 −0.6 10 20 30 40 time (msec)
Vo Vi f = 50 Hz
0.6 −0.6 0.5 1 1.5 2 time (msec)
Vo Vi f = 1 kHz
* The time taken by the op-amp to come out of saturation can be neglected at low signal frequencies. * At high signal frequencies, it leads to distortion in the
* Hook up the circuit in the lab, and check it out!
Improved half-wave precision rectifier
Vo1 Vi Vo R1 D1 D2 R R2 iD1 iR1 iD2 iR2 iR
Improved half-wave precision rectifier
Vo1 Vi Vo R1 D1 D2 R R2 iD1 iR1 iD2 iR2 iR
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V .
Improved half-wave precision rectifier
Vo1 Vi Vo R1 D1 D2 R R2 iD1 iR1 iD2 iR2 iR
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V . D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo). → iR2 = 0, Vo = V− = 0 V .
Improved half-wave precision rectifier
Vo1 Vi Vo R1 D1 D2 R R2 iD1 iR1 iD2 iR2 iR Vo1 Vi Vo Vi > 0 R1 D1 D2 R R2
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V . D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo). → iR2 = 0, Vo = V− = 0 V .
Improved half-wave precision rectifier
Vo1 Vi Vo R1 D1 D2 R R2 iD1 iR1 iD2 iR2 iR Vo1 Vi Vo Vi > 0 R1 D1 D2 R R2
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V . D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo). → iR2 = 0, Vo = V− = 0 V . iR1 = iD1 which can only be positive ⇒ Vi > 0 V .
Improved half-wave precision rectifier
Vo1 Vi Vo R1 D1 D2 R R2 iD1 iR1 iD2 iR2 iR Vo1 Vi Vo Vi > 0 R1 D1 D2 R R2
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V . D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo). → iR2 = 0, Vo = V− = 0 V . iR1 = iD1 which can only be positive ⇒ Vi > 0 V . (ii) D1 is off; this will happen when Vi < 0 V .
Improved half-wave precision rectifier
Vo1 Vi Vo R1 D1 D2 R R2 iD1 iR1 iD2 iR2 iR Vo1 Vi Vo Vi > 0 R1 D1 D2 R R2 Vo1 Vi Vo Vi < 0 R1 D1 D2 R R2
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V . D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo). → iR2 = 0, Vo = V− = 0 V . iR1 = iD1 which can only be positive ⇒ Vi > 0 V . (ii) D1 is off; this will happen when Vi < 0 V . In this case, D2 conducts and closes the feedback loop through R2.
Improved half-wave precision rectifier
Vo1 Vi Vo R1 D1 D2 R R2 iD1 iR1 iD2 iR2 iR Vo1 Vi Vo Vi > 0 R1 D1 D2 R R2 Vo1 Vi Vo Vi < 0 R1 D1 D2 R R2
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V . D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo). → iR2 = 0, Vo = V− = 0 V . iR1 = iD1 which can only be positive ⇒ Vi > 0 V . (ii) D1 is off; this will happen when Vi < 0 V . In this case, D2 conducts and closes the feedback loop through R2. Vo = V− + iR2R2 = 0 + 0 − Vi R1
R1 Vi .
Improved half-wave precision rectifier
Vo1 Vo1 Vi > 0 Vi Vo Vi < 0 Vi Vo Vo Vi Vo = 0 −R2 R1 Vi R1 D1 D2 R R2 R1 D1 D2 R R2 1 k 1 k 1 k 1 k
Improved half-wave precision rectifier
Vo1 Vo1 Vi > 0 Vi Vo Vi < 0 Vi Vo Vo Vi Vo = 0 −R2 R1 Vi R1 D1 D2 R R2 R1 D1 D2 R R2 1 k 1 k 1 k 1 k
2 1 −1 t (ms) 1 2
Vi Vo
Improved half-wave precision rectifier
Vo1 Vo1 Vi > 0 Vi Vo Vi < 0 Vi Vo Vo Vi Vo = 0 −R2 R1 Vi R1 D1 D2 R R2 R1 D1 D2 R R2 1 k 1 k 1 k 1 k
2 1 −1 t (ms) 1 2
Vi Vo Vo1
Improved half-wave precision rectifier
Vo1 Vo1 Vi > 0 Vi Vo Vi < 0 Vi Vo Vo Vi Vo = 0 −R2 R1 Vi R1 D1 D2 R R2 R1 D1 D2 R R2 1 k 1 k 1 k 1 k
2 1 −1 t (ms) 1 2
Vi Vo Vo1
* Note that the op-amp does not enter saturation since a feedback path is available for Vi > 0 V and Vi < 0 V .
Improved half-wave precision rectifier
Vo1 Vo1 Vi > 0 Vi Vo Vi < 0 Vi Vo Vo Vi Vo = 0 −R2 R1 Vi R1 D1 D2 R R2 R1 D1 D2 R R2 1 k 1 k 1 k 1 k
2 1 −1 t (ms) 1 2
Vi Vo Vo1
* Note that the op-amp does not enter saturation since a feedback path is available for Vi > 0 V and Vi < 0 V .
SEQUEL file: precision half wave.sqproj
Improved half-wave precision rectifier
Vo1 Vo = 0 Vo Vi Vi Vo −R2 R1 Vi R1 D1 D2 R2 R
The diodes are now reversed.
Improved half-wave precision rectifier
Vo1 Vo = 0 Vo Vi Vi Vo −R2 R1 Vi R1 D1 D2 R2 R
The diodes are now reversed. By considering two cases: (i) D1 on, (ii) D1 off, the Vo versus Vi relationship shown in the figure is obtained (show this).
Improved half-wave precision rectifier
Vo1 Vo = 0 Vo Vi Vi Vo −R2 R1 Vi R1 D1 D2 R2 R
The diodes are now reversed. By considering two cases: (i) D1 on, (ii) D1 off, the Vo versus Vi relationship shown in the figure is obtained (show this).
SEQUEL file: precision half wave 2.sqproj
Full-wave precision rectifier
Half−wave rectifier (inverting) x (−1) x (−2)
Vo1 Vo1 Vi Vo VB Vi Vi VA Vi Vi VB Vo VA
Full-wave precision rectifier
Half−wave rectifier (inverting) x (−1) x (−2)
Vo1 Vo1 Vi Vo VB Vi Vi VA Vi Vi VB Vo VA
inverting half−wave rectifier inverting summer
Vo R1 D1 R1 D2 Vo1 R R/2 R Vi (SEQUEL file: precision full wave.sqproj)
Full-wave precision rectifier
Half−wave rectifier (inverting) x (−1) x (−2)
Vo1 Vo1 Vi Vo VB Vi Vi VA Vi Vi VB Vo VA
inverting half−wave rectifier inverting summer
Vo R1 D1 R1 D2 Vo1 R R/2 R Vi (SEQUEL file: precision full wave.sqproj)
1 −1 −2 2 t (ms) 1 2
Vi Vo
Wave shaping with diodes
A D
i R R′ 0 V V −V0 R0
Wave shaping with diodes
A D
i R R′ 0 V V −V0 R0
A D off
i R R′ R0 0 V V −V0
Wave shaping with diodes
A D
i R R′ 0 V V −V0 R0
A D off
i R R′ R0 0 V V −V0 When D is off, i = V R0 , and VA is (by superposition), VA = V R′ R + R′ − V0 R R + R′ .
Wave shaping with diodes
A D
i R R′ 0 V V −V0 R0
A D off
i R R′ R0 0 V V −V0 When D is off, i = V R0 , and VA is (by superposition), VA = V R′ R + R′ − V0 R R + R′ . For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak = R R′ (V0 + Von) + Von .
Wave shaping with diodes
A D
i R R′ 0 V V −V0 R0
A D off
i R R′ R0 0 V V −V0
A D on
i R R′ R0 0 V V −V0 Von When D is off, i = V R0 , and VA is (by superposition), VA = V R′ R + R′ − V0 R R + R′ . For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak = R R′ (V0 + Von) + Von .
Wave shaping with diodes
A D
i R R′ 0 V V −V0 R0
A D off
i R R′ R0 0 V V −V0
A D on
i R R′ R0 0 V V −V0 Von When D is off, i = V R0 , and VA is (by superposition), VA = V R′ R + R′ − V0 R R + R′ . For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak = R R′ (V0 + Von) + Von . When D is on, i = V R0 + V − Von R + −V0 − Von R′ = V 1 R0 + 1 R
Wave shaping with diodes
A D
i R R′ 0 V V −V0 R0
A D off
i R R′ R0 0 V V −V0
A D on
i R R′ R0 0 V V −V0 Von When D is off, i = V R0 , and VA is (by superposition), VA = V R′ R + R′ − V0 R R + R′ . For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak = R R′ (V0 + Von) + Von . When D is on, i = V R0 + V − Von R + −V0 − Von R′ = V 1 R0 + 1 R
i.e., V = (R0 R) i + (constant) .
Wave shaping with diodes
A D
i R R′ 0 V V −V0 R0
A D off
i R R′ R0 0 V V −V0
A D on
i R R′ R0 0 V V −V0 Von i V Vbreak slope = R0 R slope = R0 When D is off, i = V R0 , and VA is (by superposition), VA = V R′ R + R′ − V0 R R + R′ . For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak = R R′ (V0 + Von) + Von . When D is on, i = V R0 + V − Von R + −V0 − Von R′ = V 1 R0 + 1 R
i.e., V = (R0 R) i + (constant) .
Wave shaping with diodes
A D A D off A D on
i R R′ i R R′ i R R′ i 0 V V −V0 R0 R0 0 V V −V0 R0 0 V V −V0 Von V Vbreak slope = R0 R slope = R0
(a) Vbreak = R R′ (V0 + Von) + Von. (b) When D is on, V = (R0 R) i + (constant) .
Wave shaping with diodes
A D A D off A D on
i R R′ i R R′ i R R′ i 0 V V −V0 R0 R0 0 V V −V0 R0 0 V V −V0 Von V Vbreak slope = R0 R slope = R0
(a) Vbreak = R R′ (V0 + Von) + Von. (b) When D is on, V = (R0 R) i + (constant) . * Vbreak depends on the ratio R/R′.
Wave shaping with diodes
A D A D off A D on
i R R′ i R R′ i R R′ i 0 V V −V0 R0 R0 0 V V −V0 R0 0 V V −V0 Von V Vbreak slope = R0 R slope = R0
(a) Vbreak = R R′ (V0 + Von) + Von. (b) When D is on, V = (R0 R) i + (constant) . * Vbreak depends on the ratio R/R′. * The slope R0 R depends on the resistance values.
Wave shaping with diodes
A D A D off A D on
i R R′ i R R′ i R R′ i 0 V V −V0 R0 R0 0 V V −V0 R0 0 V V −V0 Von V Vbreak slope = R0 R slope = R0
(a) Vbreak = R R′ (V0 + Von) + Von. (b) When D is on, V = (R0 R) i + (constant) . * Vbreak depends on the ratio R/R′. * The slope R0 R depends on the resistance values. * Given the break point and the two slopes, the resistance values can be easily determined.
Wave shaping with diodes
i R0 0 V V
Wave shaping with diodes
i R0 0 V V i V
Wave shaping with diodes
i R0 0 V V i V
D1B
R1B′ R1B −V0
Wave shaping with diodes
i R0 0 V V i V
D1B
R1B′ R1B −V0 i V
Wave shaping with diodes
i R0 0 V V i V
D1B
R1B′ R1B −V0 i V
D2B
R2B′ R2B
Wave shaping with diodes
i R0 0 V V i V
D1B
R1B′ R1B −V0 i V
D2B
R2B′ R2B i V
Wave shaping with diodes
i R0 0 V V i V
D1B
R1B′ R1B −V0 i V
D2B
R2B′ R2B i V
D1A
R1A R1A′ V0
Wave shaping with diodes
i R0 0 V V i V
D1B
R1B′ R1B −V0 i V
D2B
R2B′ R2B i V
D1A
R1A R1A′ V0 i V
Wave shaping with diodes
i R0 0 V V i V
D1B
R1B′ R1B −V0 i V
D2B
R2B′ R2B i V
D1A
R1A R1A′ V0 i V
D2A
R2A R2A′
Wave shaping with diodes
i R0 0 V V i V
D1B
R1B′ R1B −V0 i V
D2B
R2B′ R2B i V
D1A
R1A R1A′ V0 i V
D2A
R2A R2A′ i V
Wave shaping with diodes
D2B D1B D1A D2A
R1B′ R2B′ i R2A R1A R2B R1B R1A′ R2A′ i Vo RL Vi −V0 R0 V0 Ra ∼ 0 V Ra = 5 k R0 = 20 k R1A = R1B = 15 k R2A = R2B = 5 k R1A′ = R1B′ = 60 k R2A′ = R2B′ = 10 k V0 = 15 V Vo i
Wave shaping with diodes
D2B D1B D1A D2A
R1B′ R2B′ i R2A R1A R2B R1B R1A′ R2A′ i Vo RL Vi −V0 R0 V0 Ra ∼ 0 V Ra = 5 k R0 = 20 k R1A = R1B = 15 k R2A = R2B = 5 k R1A′ = R1B′ = 60 k R2A′ = R2B′ = 10 k V0 = 15 V Vo i
Since Vi = −Rai, the Vo versus Vi plot is similar to the V versus i plot, except for the (−Ra) factor.
Wave shaping with diodes
D2B D1B D1A D2A
R1B′ R2B′ i R2A R1A R2B R1B R1A′ R2A′ i Vo RL Vi −V0 R0 V0 Ra ∼ 0 V Ra = 5 k R0 = 20 k R1A = R1B = 15 k R2A = R2B = 5 k R1A′ = R1B′ = 60 k R2A′ = R2B′ = 10 k V0 = 15 V Vo i
−10 5 −5 −5 5 10
Vi (V) Vo (V)
Since Vi = −Rai, the Vo versus Vi plot is similar to the V versus i plot, except for the (−Ra) factor.
Wave shaping with diodes
D2B D1B D1A D2A
R1B′ R2B′ i R2A R1A R2B R1B R1A′ R2A′ i Vo RL Vi −V0 R0 V0 Ra ∼ 0 V Ra = 5 k R0 = 20 k R1A = R1B = 15 k R2A = R2B = 5 k R1A′ = R1B′ = 60 k R2A′ = R2B′ = 10 k V0 = 15 V Vo i
−10 5 −5 −5 5 10
Vi (V) Vo (V)
2 4 6 8 time (msec) 10 −10
Vi Vo
Since Vi = −Rai, the Vo versus Vi plot is similar to the V versus i plot, except for the (−Ra) factor.
Wave shaping with diodes
D2B D1B D1A D2A
R1B′ R2B′ i R2A R1A R2B R1B R1A′ R2A′ i Vo RL Vi −V0 R0 V0 Ra ∼ 0 V Ra = 5 k R0 = 20 k R1A = R1B = 15 k R2A = R2B = 5 k R1A′ = R1B′ = 60 k R2A′ = R2B′ = 10 k V0 = 15 V Vo i
−10 5 −5 −5 5 10
Vi (V) Vo (V)
2 4 6 8 time (msec) 10 −10
Vi Vo
Since Vi = −Rai, the Vo versus Vi plot is similar to the V versus i plot, except for the (−Ra) factor. SEQUEL file: ee101 wave shaper.sqproj
Wave shaping with diodes: spectrum
10 −10 5
Vi
Wave shaping with diodes: spectrum
10 −10 5
Vi
10 −10 2 4 6 8 time (msec) N 2 4 6 8 10 10
Vo