Often, newer systems use a single power supply to improve efficiency and reduce the cost of the power supply. The OPA836 and OPA2836 devices are designed for use with a single supply with no change in performance compared to a split supply, as long as the input and output are biased within the linear operation of the device.

To change the circuit from split supply to single supply, level shift of all voltages by half the difference between the power supply rails. For example, changing from ±2.5-V split supply to 5-V single supply is shown in Figure 8-7.

Figure 8-7 Single-Supply Concept

A practical circuit will have an amplifier or other circuit providing the bias voltage for the input, and the output of this amplifier stage provides the bias for the next stage.

Figure 8-8 shows a typical noninverting amplifiercircuit. With 5-V single-supply, a mid-supply reference generator is needed to bias the negative side through R_G. To cancel the voltage offset that would otherwise be caused by the input bias currents, R₁ is selected to be equal to R_F in parallel with R_G. For example, if gain of 2 is required and R_F = 1 kΩ, select R_G = 1 kΩ to set the gain and R₁ = 499 Ω for bias-current cancellation. The value for C depends on the reference; TI recommends a value of at least 0.1 µF to limit noise.

Figure 8-8 Noninverting Single Supply With Reference

Figure 8-9 shows a similar noninverting single-supply scenario with the reference generator replaced by the Thevenin equivalent using resistors and the positive supply. R_G’ and R_G” form a resistor divider from the 5-V supply and are used to bias the negative side with their parallel sum equal to the equivalent R_G to set the gain. To cancel the voltage offset that would otherwise be caused by the input bias currents, R₁ is selected to be equal to R_F in parallel with R_G’ in parallel with R_G” (R₁= R_F || R_G’ || R_G”). For example, if gain of 2 is required and R_F = 1 kΩ, selecting R_G’ = R_G” = 2 kΩ gives equivalent parallel sum of 1 kΩ, sets the gain to 2, and references the input to mid supply (2.5 V). R₁ is then set to 499 Ω for bias-current cancellation. The resistor divider costs less than the 2.5 V reference in Figure 8-8 but may increase the current from the 5-V supply.

Figure 8-9 Noninverting Single Supply With Resistors

Figure 8-10 shows a typical inverting amplifier situation. With 5-V single supply, a mid-supply reference generator is needed to bias the positive side through R₁. To cancel the voltage offset that would otherwise be caused by the input bias currents, R₁ is selected to be equal to R_F in parallel with R_G. For example if gain of –2 is required and R_F = 1 kΩ, select R_G = 499 Ω to set the gain and R₁ = 332 Ω for bias-current cancellation. The value for C is dependent on the reference, but TI recommends a value of at least 0.1 µF to limit noise into the operational amplifier.

Figure 8-10 Inverting Single Supply With Reference

Figure 8-11 shows a similar inverting single-supply scenario with the reference generator replaced by the Thevenin equivalent using resistors and the positive supply. R₁ and R₂ form a resistor divider from the 5-V supply and are used to bias the positive side. To cancel the voltage offset that would otherwise be caused by the input bias currents, set the parallel sum of R₁ and R₂ equal to the parallel sum of R_F and R_G. C must be added to limit coupling of noise into the positive input. For example if gain of –2 is required and R_F = 1 kΩ, select R_G = 499 Ω to set the gain. R₁ = R₂ = 665 Ω for mid-supply voltage bias and for operational amplifier input bias-current cancellation. A good value for C is 0.1 µF. The resistor divider costs less than the 2.5-V reference in Figure 8-10 but may increase the current from the 5-V supply.

Figure 8-11 Inverting Single Supply With Resistors