Design Goals

Design Description

This circuit amplifies an AC signal and shifts the output signal so that it is centered at half the power supply voltage. Note that the input signal has zero DC offset so it swings above and below ground. The key benefit of this circuit is that it accepts signals which swing below ground even though the amplifier does not have a negative power supply.

Design Notes

1. R₁ sets the AC input impedance. R₄ loads the op amp output.
2. Use low feedback resistances to reduce noise and minimize stability concerns.
3. Set the output range based on linear output swing (see A_ol specification).
4. The cutoff frequency of the circuit is dependent on the gain bandwidth product (GBP) of the amplifier. Additional filtering can be accomplished by adding a capacitor in parallel to R₄. Adding a capacitor in parallel with R₄ will also improve stability of the circuit if high-value resistors are used.

Design Steps

1. Select R₁ and R₄ to set the AC voltage gain.
   ${\mathrm{R}}_1=1 \mathrm{k\Omega }\text{ (Standard Value)}$
   ${\mathrm{R}}_4={\mathrm{R}}_1\times \left|{\mathrm{G}}_{\mathrm{ac}}\right|=1 \mathrm{k\Omega }\times \left|-10\frac{\mathrm{V}}{\mathrm{V}}\right|=10\mathrm{k\Omega }\text{ (Standard Value)}$
2. Select R₂ and R₃ to set the DC output voltage to 2.5 V.
   ${\mathrm{R}}_3=4.99\mathrm{k\Omega }\text{ (Standard Value)}$
   ${\mathrm{R}}_2=\frac{{\mathrm{R}}_3\times {\mathrm{V}}_{\mathrm{ref}}}{{\mathrm{V}}_{\mathrm{DC}}}-{\mathrm{R}}_3=\frac{4.99\mathrm{k\Omega }\times 5\mathrm{V}}{2.5\mathrm{V}}-4.99\mathrm{k\Omega }=4.99\mathrm{k\Omega }$
3. Choose a value for the lower cutoff frequency, f_l, then calculate C₁.
   ${\mathrm{f}}_{\mathrm{l}}=16\mathrm{Hz}$
   ${\mathrm{C}}_1=\frac{1}{2\times \mathrm{\pi }\times {\mathrm{R}}_1\times {\mathrm{f}}_{\mathrm{l}}}=\frac{1}{2\times \mathrm{\pi }\times 1 \mathrm{k\Omega }\times 16\mathrm{Hz}}=9.94\mathrm{\mu F}\approx 10\mathrm{\mu F}\text{ (Standard Value)}$
4. Choose a value for f_div, then calculate C₂.
   ${\mathrm{f}}_{\mathrm{div}}=6.4\mathrm{Hz}$
   ${\mathrm{R}}_{\mathrm{div}}=\frac{{\mathrm{R}}_2\times {\mathrm{R}}_3}{{\mathrm{R}}_2+{\mathrm{R}}_3}=\frac{4.99\mathrm{k\Omega }\times 4.99\mathrm{k\Omega }}{4.99\mathrm{k\Omega }+4.99\mathrm{k\Omega }}=2.495\mathrm{k\Omega }$
   ${\mathrm{C}}_2=\frac{1}{2\times \mathrm{\pi }\times {\mathrm{R}}_{\mathrm{div}}\times {\mathrm{f}}_{\mathrm{div}}}=\frac{1}{2\times \mathrm{\pi }\times 2.495\mathrm{k\Omega }\times 6.4\mathrm{Hz}}=9.96\mathrm{\mu F}\approx 10\mathrm{\mu F}\text{ (Standard Value)}$
5. The upper cutoff frequency, f_h, is set by the noise gain of this circuit and the gain bandwidth (GBW) of the device (LMV981).
   $\mathrm{GBW}=1.5\mathrm{MHz}$

Design Simulations

AC Simulation Results

Transient Simulation Results

Design References

See Analog Engineer's Circuit Cookbooks for TI's comprehensive circuit library.

See circuit SPICE simulation file SBOC504.

See TIPD185.

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