Level shifter using LM324 Op Amp for ADC reading

Question

I'm want to digitize an +/- 2.5 V signal using an ADS1256. Since the IC can't take negative voltages I designed a level shifter with a LM324 to translate the +/-2.5 V signal to 0 - 5 V but I was only able to translate up to 3.4 V since the LM324 isn't rail-to-rail. I also realized that the input signal (+/-2.5V) and output signal (0-3V) have different phases. I would like to know if I'm losing resolution or accuracy reading designing the level shifter with a LM324 instead of a rail-to-rail Op-Amp. The frequency range of the signal is 1 to 10 Hz.

The circuit I used is from this website: https://vidnyankendra.org/2017/02/17/level-shifter-using-op-amp/

Answer 1

You don't need an op amp. You can offset the signal with just a resistor network. You wouldn't use the full input range of the ADC but you don't care: it's a 24 bit ADC.

Answer 2

At -2.5V in, the voltage at pin 3 should indeed be 0V, at +2.5V in, it will be 2.5 + (2.5/3) = 3.3V in. As the op amp is a follower, I would expect the same out. The circuit is doing what it should. You could use a rail to rail opamp, but note that you cannot get the input to +5V with this method of level shifting.

Also note with your circuit that you try to bring the input very close to 0V. Due to resistor tolerances, you might well be clipping the input slightly. Whether this is a problem depends really on the application.

Your circuit is non-inverting, input and output have the same phase.

If you really want to get the largest swing, another way might be to generate an artificial mid-rail and sum the input with a DC offset. Schema below:

^{simulate this circuit – Schematic created using CircuitLab}

The circuit is now inverting, and has lower input impedance. However you are still DC coupled, and you use the full voltage. You can use any opamp that gives rail to rail swing at 5V. You also have a bit more possibility to adjust gain, using the feedback resistor. C1/R2 are sized to provide a stable mid rail at 1Hz.

Answer 3

If you are using the same circuit and same input range, the maximum output voltage you will get is \$\frac {2 \times V_{in} +5}{3}\$ (use superposition theorem.) For \$V_{in}\$ =2.5 V (peak), the \$V_o\$ = 3.34V (peak) and 0V (minimum.)

Answer 4

Preferred ratios for 5V Rail to Rail In/Out

^{simulate this circuit – Schematic created using CircuitLab}

The non-inverting input sees 0 to 2.5 and is then amplified by 1+1=2