I have a sensor that outputs a signal change between -4 to 4 V.
I want to use a bipolar ADC with a 3.3 V reference. I know that I should use a voltage divider, but what is the impact on resolution? I will use a 12-bit ADC (though even 8 bits will be enough).
Will an op-amp with 0.5 V/V gain do the job ? Or maybe a voltage divider with resistors?
I know that I must be using some voltage divider but what is the impact on resolution?
Well, if you don't use the voltage divider either your ADC will be destroyed or your signal range above 3.3 volts and below -3.3 volts is lost.
So, you need a voltage divider irrespective of how the resolution might changes. And anyway, if you are using a 12 bit ADC (but would be OK with an 8 bit ADC), halving the signal input range (voltage divider) is only going to reduce resolution by 1 least-significant-bit to 11 bits.
It's a bit of a no-brainer really.
Op amp with 0.5v/v gain will do the job ? or maybe a voltage divider with resistors?
An op-amp with a gain of 0.5 still has to use a resistive voltage divider.
It depends a bit on how the ADC is designed and specified.
If it's a 12-bit including polarity and it reads +/-3.3V then you can use a voltage divider to reduce the input by the ratio 3.3/4.0 = 0.825.
If it's a 3.3V single-supply ADC with differential inputs then you'd have to bias one input at 3.3V/2 (maybe with a couple resistors) and divide and offset the signal (maybe with three resistors) so it remains within the 0~3.3V range for -4~+4 input.
In both cases, you'd get full 12-bit resolution.
Whether you need or want an op-amp depends on the details of the source and required output impedances (and the available power supply rails may affect the complexity).
Since you're saying that 8 bits of resolution is enough (though you don't say what that's based on), this means that you can sample signals 16 times smaller than your full range. Since you're cutting your full range down by between a factor of 2 and a factor of 4, a 12-bit ADC should do the job.
A more precise answer would probably rely on you telling us more about why you're sure 8 bits is enough.
Your 12-bit ADC with a reference voltage of 3.3V can measure in steps of 3.3V/4096 which is 0.0008V.
If you have some circuit in front of the ADC, which converts -4V - 4V into 0V - 3.3V, then the ADC will measure the output of this circuit in 0.0008V steps. Because this circuit will squish the voltages together, a 0.0008V on the output will mean a 0.002V step on the input. You could also calculate this as 8V/4096.
Note that you didn't ask how to build this circuit, just how it would impact the ADC resolution.
