reducing amplitude of sine wave

[aashishsharma]

Hey I have sine wave whose p-p voltage is 4 [3-(-1)] common mode voltage is 1. I need to feed this sine wave into an ADC but problem is my ADC's input range is -1.75 to 1.75 V in bipolar mode and 0 - 1.75 in uni polar mode. Since I have -ve voltage also i was hoping to use bipolar mode
But don't know how to reduce sine wave's amplitude proportionally
Thanks in advance

 

[crutschow]

The simplest way is to use a resistive voltage divider.

What is the signal frequency and the impedance of your voltage source?

 

[aashishsharma]

basically its IC AD8347

https://www.analog.com/static/imported-files/data_sheets/AD8347.pdf
last stage as you can see in block diagram is amplifier. It is i think ADC driver input impedance given in datasheet you can check is 1||3 Mohms||pF. Output imedance is not mentioned but i guess i can take that as 0 idelly.
As far as frequency is concerned it wont be more than 1kHz

 

[Mithun_K_Das]

Yes! simple way is using a voltage divider with proper capacitor in parallel with the the resistor. calculate the maximum and minimum resistor required and then choose one.

First calculate taking the maximum voltage and which will not cross 1.75V. and then calculate the resistors for suitable range that is available.

 

[crutschow]

For a 4Vpp voltage with an offset, you can run the signal through a capacitor and then a voltage divider to get the desired ±1.75V. The attenuation from the voltage divider should be 3.5V / 4V = 0.875 The rated output load for the AD8347 is 10kΩ. A 1% resistor pair to meet those two requirements is 1330Ω in series with 9310Ω to ground.

The capacitor is in series with the 1330Ω resistor with the junction of the 1330Ω resistor and the 9310Ω resistor going to the A/D input. The value of the capacitor determines the low frequency rolloff. A 2µF cap will give a corner frequency below 10Hz.