SBOA231 June 2021 OPA325 , TLV316 , TLV9062

Input | Output | Supply | |||
---|---|---|---|---|---|

V_{iMin} |
V_{iMax} |
V_{oMin} |
V_{oMax} |
V_{cc} |
V_{ee} |

–2.45V | +2.45V | 0.05V | 4.95V | 5V | 0V |

Gain | Cutoff Frequency (f_{c}) |
V_{ref} |
---|---|---|

–1V/V | 10kHz | 1.25V |

**Design Description**

The multiple-feedback (MFB)
low-pass filter (LP filter) is a second-order active filter. V_{ref}
provides a DC offset to accommodate for single-supply applications. This LP filter
inverts the signal (Gain = –1V/V) for frequencies in the pass band. An MFB filter is
preferable when the gain is high or when the Q-factor is large (for example, 3 or
greater).

**Design Notes**

- Select an op amp with sufficient input common-mode range and output voltage swing.
- Add V
_{ref}to bias the input signal to meet the input common-mode range and output voltage swing. - Select the capacitor values first
since standard capacitor values are more coarsely subdivided than the resistor
values. Use high-precision, low-drift capacitor values to avoid errors in
f
_{c}. - To minimize the amount of slew-induced distortion, select an op amp with sufficient slew rate (SR).

**Design Steps**

The first step in design is to find component values for the normalized cutoff frequency of 1 radian/second. In the second step the cutoff frequency is scaled to the desired cutoff frequency with scaled component values.

The transfer function for a second-order MFB low-pass filter is given by:

$\text{H(s)}=\frac{\frac{1}{{R}_{2}\times {R}_{3}\times {C}_{1}\times {C}_{2}}}{{s}^{2}+s\times \frac{1}{{C}_{1}}\left(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\right)+\frac{1}{{R}_{1}\times {R}_{2}\times {C}_{1}\times {C}_{2}}}$

$\text{H(s) =}\frac{{b}_{0}}{{s}^{2}+{a}_{1}\times s+{a}_{0}}$

${Here,a}_{1}=\frac{1}{{C}_{1}}\left(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\right){,a}_{0}=\frac{1}{{R}_{1}\times {R}_{2}\times {C}_{1}\times {C}_{2}}$

- Set normalized values of
R
_{1}and R_{2}(R_{1n}and R_{2n}) and calculate normalized values of C_{1}and C_{2}(C_{1n}and C_{2n}) by setting w_{c}to 1 radian/sec (or fc = 1 / (2 × π) Hz). For a 2nd-order Butterworth filter, (see the*Butterworth Filter Table*in the*Active Low-Pass Filter Design Application Report*).${\text{\omega}}_{\text{c}}\text{= 1}\frac{\text{radian}}{\text{second}}\text{\u2192}{a}_{0}\text{= 1,}{\text{a}}_{1}\text{=}\sqrt{2}{\text{, letR}}_{1n}{\text{= R}}_{2n}{\text{= R}}_{3n}\text{= 1}$$\text{Then}{\text{C}}_{\text{1n}}{\text{\xd7 C}}_{\text{2n}}{\text{= 1orC}}_{\text{2n}}=\frac{1}{{C}_{\mathrm{1n}}},{\text{a}}_{\text{1}}\text{=}\frac{3}{{C}_{1n}}\text{=}\sqrt{2}$$\therefore {C}_{1n}=\frac{3}{\sqrt{2}}=2.1213\text{F},{C}_{2n}=\frac{1}{{C}_{1n}}=0.4714\text{F}$ - Scale the component values and
cutoff frequency. The resistor values are very small and capacitors values are
unrealistic, hence these must be scaled. The cutoff frequency is scaled from 1
radian/second to w
_{0}. If*m*is assumed to be the scaling factor, increase the resistors by*m*times, then the capacitor values have to decrease by 1/*m*times to keep the same cutoff frequency of 1 radian/second. If the cutoff frequency is scaled to be w_{0}, then the capacitor values have to be decreased by 1/w_{o}. The component values for the design goals are calculated in steps 3 and 4.${R}_{1}={R}_{1n}\times m,{R}_{2}={R}_{2n}\times m,{R}_{3}={R}_{3n}\times m$${C}_{1}=\frac{{C}_{1n}}{m\times {\omega}_{0}}=\frac{2.1213}{m\times {\omega}_{0}}F$${C}_{2}=\frac{{C}_{2n}}{m\times {\omega}_{0}}=\frac{0.4714}{m\times {\omega}_{0}}F$ - Set R
_{1}, R_{2}, and R_{3}to 10kΩ.${R}_{1}={R}_{1n}\times m=10k\Omega ,{R}_{2}={R}_{2n}\times m=10k\Omega ,{R}_{3}={R}_{3n}\times m=10k\Omega $$\text{Therefore,}m=10000$ - Calculate
C
_{1}and C_{2}based on*m*and w_{0}.${C}_{1}=\frac{2.1213}{m\times {\omega}_{0}}F=\frac{2.1213}{\mathrm{10k}\times 2\times \pi \times \mathrm{10kHz}}=\mathrm{3.376nF}\approx \mathrm{3.3nF}\text{(StandardValue)}$${C}_{2}=\frac{0.4714}{m\times {\omega}_{0}}F=\frac{0.4714}{\mathrm{10k}\times 2\times \pi \times \mathrm{10kHz}}=\mathrm{0.75nF}\approx \mathrm{0.75nF}\text{(StandardValue)}$ - Calculate the minimum required GBW
and SR for f
_{c}. Be sure to use the noise gain for GBW calculations. Do not use the signal gain of –1V/V.${\text{GBW=100 \xd7 Noise Gain \xd7 f}}_{\text{c}}\text{=100 \xd7 2 \xd7 10kHz=2MHz}$${\text{SR = 2 \xd7 \pi \xd7 f}}_{c}{\text{\xd7 V}}_{\mathrm{iMax}}\text{=2 \xd7 \pi \xd7 10kHz\xd7 2.45V=0.154}\frac{V}{\mathrm{\mu s}}$

The TLV9062 device has GBW of 10MHz and SR of 6.5 V/µs, so the requirements are met.

**Design
Simulations**

**AC Simulation Results**

**Transient Simulation Results**

The following image shows the filter
output in response to a 5-V_{pp}, 100-Hz input signal (gain = –1V/V).

The following image shows the filter
output in response to a 5-V_{pp}, 10-kHz input signal (gain = –0.01V/V).

**Design References**

- See Analog Engineer's Circuit Cookbooks for TI's comprehensive circuit library.
- SPICE Simulation File SBOC597
- TI Precision Labs.
*Active Low-Pass Filter Design Application Report*

**Design Featured Op Amp**

TLV9062 | |
---|---|

V
_{ss}_{} |
1.8V to 5.5V |

V
_{inCM}_{} |
Rail-to-Rail |

Vout |
Rail-to-Rail |

V
_{os}_{} |
0.3mV |

Iq |
538µA |

Ib |
0.5pA |

UGBW |
10MHz |

SR |
6.5V/µs |

#Channels |
1, 2, 4 |

www.ti.com/product/TLV9062 |

**Design Alternate Op Amp**

TLV316 | OPA325 | |
---|---|---|

V_{ss} |
1.8V to 5.5V | 2.2V to 5.5V |

V_{inCM} |
Rail-to-Rail | Rail-to-Rail |

Vout |
Rail-to-Rail | Rail-to-Rail |

V
_{os}_{} |
0.75mV | 0.150mV |

Iq |
400µA | 650µA |

Ib |
10pA | 0.2pA |

UGBW |
10MHz | 10MHz |

SR |
6V/µs | 5V/µs |

#Channels |
1, 2, 4 | 1, 2, 4 |

www.ti.com/product/TLV316 | www.ti.com/product/OPA325 |