SBOA226 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 | 2.5V |

**Design Description**

The Butterworth Sallen-Key low-pass
filter is a second-order active filter. V_{ref} provides a DC offset to
accommodate for single-supply applications. A Sallen-Key filter is usually preferred
when small Q factor is desired, noise rejection is prioritized, and when a
non-inverting gain of the filter stage is required. The Butterworth topology
provides a maximally flat gain in the pass band.

**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 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 second order Sallen-Key low-pass filter is given by:

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

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

Here,

${a}_{1}=\frac{1}{{R}_{1}\times {C}_{1}}+\frac{1}{{R}_{2}\times {C}_{1}},{a}_{0}=\frac{1}{{R}_{1}\times {R}_{2}\times {C}_{1}\times {C}_{2}}$

- 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 f_{c}= 1 / (2 × π) Hz). For the second-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}=\sqrt{2}{\text{, letR}}_{1n}={\text{R}}_{2n}=\text{1,then}{\text{C}}_{\text{1n}}\times {\text{C}}_{\text{2n}}={\text{1orC}}_{\text{2n}}=\frac{1}{{C}_{\mathrm{1n}}},{\text{a}}_{\text{1}}\text{=}\frac{2}{{C}_{1n}}=\sqrt{2}$${\text{\u2234C}}_{1n}=\sqrt{2}{\text{= 1.414F,C}}_{\mathrm{2n}}=\frac{1}{{C}_{1n}}\text{= 0.707F}$ - Scale the component values and
cutoff frequency. The resistor values are very small and capacitors values
are unrealistic, hence these have to be scaled. The cutoff frequency is
scaled from 1 radian/sec to w
_{0}. If*m*is assumed to be the scaling factor, increase the resistors by*m*times, then the capacitor values have to decrease by 1/*m*times to keep the same cutoff frequency of 1 radian/sec. If the cutoff frequency is scaled to be w_{0}, then the capacitor values have to be decreased by 1 / w_{o}. The component values for the design goals are calculated in steps 3 and 4.Equation 1. ${R}_{1}={R}_{1n}\times m,{R}_{2}={R}_{2n}\times m$Equation 2. ${C}_{1}=\frac{{C}_{1n}}{m\times {\omega}_{0}}=\frac{1.414}{m\times {\omega}_{0}}F$Equation 3. ${C}_{2}=\frac{{C}_{2n}}{m\times {\omega}_{0}}=\frac{0.707}{m\times {\omega}_{0}}F$ - Set R1 and R2
values:$\text{m=10000}$Equation 4. ${R}_{1}=\left({R}_{1n}\times m\right)=10k\Omega $Equation 5. ${R}_{2}=\left({R}_{2n}\times m\right)=10k\Omega $
- Calculate C
_{1}and C_{2}based on*m*and w_{0}.${\text{Given\omega}}_{0}{\text{= 2 \xd7 \pi \xd7 f}}_{c}{\text{, wheref}}_{c}\text{= 10kHzandm = 10000 = 10k}$${\text{C}}_{1}\text{=}\frac{\text{1.414}}{{\text{m \xd7 \omega}}_{0}}\text{F =}\frac{\text{1.414}}{\text{10k \xd7 2 \xd7 \pi \xd7 10kHz}}\text{= 2.25nF \u2248 2.2nF(StandardValue)}$${C}_{2}=\frac{\text{0.707}}{{\text{m \xd7 \omega}}_{0}}\text{F =}\frac{\text{0.707}}{\text{10k \xd7 2 \xd7 \pi \xd7 10kHz}}\text{= 1.125nF \u2248 1.1nF(StandardValue)}$ - Calculate the minimum required
GBW and SR for f
_{c}.${\text{GBW=100 \xd7 Gain \xd7 f}}_{\text{c}}\text{=100 \xd7 1 \xd7 10kHz=1MHz}$${\text{SR = 2 \xd7 \pi \xd7 f}}_{c}{\text{\xd7 V}}_{\mathrm{ipeak}}\text{=2 \xd7 \pi \xd7 10kHz\xd7 2.45V=0.154}\frac{V}{\mathrm{\mu s}}$The TLV9062 device has a GBW of 10MHz and SR of 6.5V/µs, so the requirements are met.

**Design
Simulations**

**AC Simulation Results**

**Transient Simulation Results**

The following image shows the filter output in response to 5-Vpp, 1-kHz input signal (gain = 1V / V).

The following image shows the filter output in response to 5-Vpp, 100-kHz input signal (gain = 0.01 V/V).

**Design References**

- See Analog Engineer's Circuit Cookbooks for TI's comprehensive circuit library.
- SPICE Simulation File SBOC598.
- TI Precision Labs.
*Active Low-Pass Filter Design Application Report*

**Design Featured Op Amp**

TLV9062 | |
---|---|

Vss |
1.8V to 5.5V |

VinCM |
Rail-to-Rail |

Vout |
Rail-to-Rail |

Vos |
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 | |
---|---|---|

Vss |
1.8V to 5.5V | 2.2V to 5.5V |

VinCM |
Rail-to-Rail | Rail-to-Rail |

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

Vos |
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 |