Design Goals

Design Description

This circuit uses two op amps to build a discrete, single-supply diff-in diff-out amplifier. The circuit converts a differential signal to a differential output signal.

Design Notes

1. Verify that R₁ and R₂ are well matched with high accuracy resistors to maintain high DC common-mode rejection performance.
2. Increase R₄ and R₅ to match the necessary input impedance at the expense of thermal noise performance.
3. Bias for single-supply operation can also be created by a voltage divider from V_cc to ground.
4. V_ref sets the output voltage of the instrumentation amplifier bias at mid-supply to allow the output to swing to both supply rails.
5. Choose C₁ and C₂ to select the lower cutoff frequency.
6. Linear operation is contingent upon the input common-mode and the output swing ranges of the discrete op amps used. The linear output swing ranges are specified under the Aol test conditions in the op amps data sheets

Design Steps

1. The transfer function of the circuit is shown below.
   $$\begin{gathered} {\mathrm{V}}_{\mathrm{oDiff}}={\mathrm{V}}_{\mathrm{i}}\times \mathrm{G}+{\mathrm{V}}_{\mathrm{ref}} \\ \mathrm{where} {\mathrm{V}}_{\mathrm{i}}=\mathrm{the} \mathrm{differential} \mathrm{input} \mathrm{voltage} \\ {\mathrm{V}}_{\mathrm{ref}}=\mathrm{the} \mathrm{reference} \mathrm{voltage} \mathrm{provided} \mathrm{to} \mathrm{the} \mathrm{amplifier} \\ \mathrm{G}=1+2\times \left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_3}\right) \end{gathered}$$
2. Choose resistors R₁ = R₂ to maintain common-mode rejection performance.
   $\mathrm{Choose} {\mathrm{R}}_1={\mathrm{R}}_2=\text{20 }\mathrm{k\Omega } (\mathrm{Standard} \mathrm{value})$
3. Choose resistors R₄ and R₅ to meet the desired input impedance.
   $\mathrm{Choose} {\mathrm{R}}_4={\mathrm{R}}_5=\text{10 }\mathrm{k\Omega } (\mathrm{Standard} \mathrm{value})$
4. Calculate R₃ to set the differential gain.
   $$\begin{gathered} \mathrm{Gain}=1+\left(\frac{2\times {\mathrm{R}}_1}{{\mathrm{R}}_3}\right)=\text{5 }\frac{\mathrm{V}}{\mathrm{V}} \\ {\mathrm{R}}_1 ={\mathrm{R}}_2= \text{20 }\mathrm{k}\Omega \\ \mathrm{G}=1+\frac{{\displaystyle 2\times \text{20 }\mathrm{k\Omega }}}{{\displaystyle {\mathrm{R}}_3}}=\text{5 }\frac{{\displaystyle \mathrm{V}}}{{\displaystyle \mathrm{V}}}\to \text{5 }\frac{\mathrm{V}}{\mathrm{V}}-1=\frac{{\displaystyle \text{40 }\mathrm{k\Omega }}}{{\displaystyle {\mathrm{R}}_3}}=4\to {\mathrm{R}}_3=\frac{{\displaystyle \text{40 }\mathrm{k\Omega }}}{{\displaystyle 4}}=\text{10 }\mathrm{k\Omega } (\mathrm{Standard} \mathrm{value}) \end{gathered}$$
5. Set the reference voltage V_ref at mid-supply.
   ${\text{V}}_{\mathrm{ref}}=\frac{{\mathrm{V}}_{\mathrm{cc}}}{2}=\frac{\text{5 }\mathrm{V}}{2}\to {\mathrm{V}}_{\mathrm{ref}}=2.5\mathrm{V}$
6. Calculate C₁ and C₂ to set the lower cutoff frequency.
   $$\begin{gathered} {\text{f}}_{\mathrm{c}}=\frac{1}{2\times \mathrm{\pi }\times {\mathrm{R}}_4\times {\mathrm{C}}_1}=\text{16 }\mathrm{Hz} \\ {\mathrm{R}}_4={\mathrm{R}}_5=\text{10 }\mathrm{k\Omega } \\ {\mathrm{f}}_{\mathrm{c}}=\frac{1}{2\times \mathrm{\pi }\times \text{10 }\mathrm{k\Omega }\times {\mathrm{C}}_1}=\text{16 }\mathrm{Hz}\to {\mathrm{C}}_1=\frac{1}{2\times \mathrm{\pi }\times \text{10 }\mathrm{k\Omega }\times \text{16 }\mathrm{Hz}}=0.99\mathrm{\mu F}\to {\mathrm{C}}_1={\mathrm{C}}_2=1\mathrm{\mu F} (\mathrm{Standard} \mathrm{value}) \end{gathered}$$

Design Simulations

AC Simulation Results

In the following figure, notice the lower –3-dB cutoff frequency is approximately 16Hz and the upper cutoff frequency is > 1MHz as required for this design.

Transient Simulation Results

References

Texas Instruments, SBOMAU5 SPICE Simulation, file download

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