I am using ads8320 ADC IC which have IN+ and IN- analig inputs. I need to operate this IC with 3.3V VCC and 3.3V Vref. My analog sensor gives 0V-12V range signal which is required to set 0V-3.3V by using an Inverting amplifier. As far now I found the load of the inverting amplifier can be Resistive load or Current output. I don't know what to use in my application.The circuit i have added was generated using Texas Instruments Web-bench. And I have used resistive load.Can I directly connect upper end of the load resistor to my IN+ of the ADC IC (ads8320)

The datasheet for the ADS8320 gives an example circuit. See Figure 22 of the datasheet.

For reference figure 22 from the ADS8320 datasheet:

The given circuit doesn't quite match what you are doing, but it and the text give you a couple of pointers.

1. You MUST have that 1nF capacitor
2. You (almost certainly) need that 100 Ohm resistor. If you use the LMP7716 from the example, then you MUST have that 100 Ohm resistor.

The 1nF capacitor is needed by the ADC input. Section 8.2.1.2 explains the reasoning.

From a practical standpoint, if you don't have it the input voltage will scitter around as it is being sampled and result in incorrect values.

The 100 Ohm resistor is needed because op-amps in general don't like driving capacitors. Putting that 100 Ohms in there prevents the op-amp from oscillating.

The 100 Ohm resistor and the 1nF also form a low pass filter. The given values have a cutoff of about 1.5MHz. That's not really low enough considering that the ADS8320 sample rate only goes up to 100kHz. If you can guarantee that the signal will not contain any frequencies above half of your sampling rate then this won't be a problem.

For your task, you need a rail to rail op-amp, a voltage divider, and the parts mentioned above.

That would look like this:

simulate this circuit – Schematic created using CircuitLab

The voltage divider (R1 and R2) brings the 12V sensor voltage down to something less than 3.3V because it uses standard resistor values. You should be able to figure out what ratio the divider really uses.

The two diodes are schottky diodes. They clamp the input to the op-amp to the limits that the datasheet for the LMP7716 gives. That's less than 0.3V above or below the power rails.

C1 is the 1nF required by the ADS8320 datasheet to keep the input from scittering around during the sampling.

R3 is the series resistor needed so that the LMP7716 can drive the 1nF without oscillating.

This is non-inverting unity gain buffer amplifier. I do not know why you think you need an inverting amplifier. That would be an unusual thing for the situation you describe.

You can add a capacitor in parallel with R2 to form a simple low pass filter. This will reduce the noise being sampled, and help avoid aliasing.

10nF should be sufficient (calculating the corner frequency is a task for you.)

Once you have it sampled, you can apply a filter in the digital domain and get better results than an analog filter would give you.

You need to get acquainted with Shannon's theorem and the Nyquist rate. This will help you avoid problems when sampling analog signals.

• thank you by the way what do we call this opamp configuration? – vassidefuk Apr 29 '17 at 12:30
• As mentioned above, it is a non-inverting unity gain buffer. – JRE Apr 29 '17 at 12:36
• can you explain the purpose of unity gain buffer in above circuit? – vassidefuk May 14 '17 at 9:20
• The sensor can only supply a (very) small current. The ADC draws more current than the sensor can supply. The buffer delivers the same voltage, but can deliver enough current for the ADC. – JRE May 14 '17 at 10:00

You don't need an inverting amplifier; just connect the -input to 0 volts and attenuate the input with a resistor potential divider. Make sure the divider is enough to reduce the biggest signal from your sensor to below 3.3 volts. Also add a capacitor across the input to ground as per the data sheet recommends.

Data sheet note about input: -

Inverting input: Connect to ground or to remote ground sense point.

• The input signal frequency is around 1000Hz bell shaped analog signal. Which way is better resister divider or inverting amplifier – vassidefuk Apr 28 '17 at 18:47
• Considering that an inverting amplifier uses two resistors as a potential divider it seems a no brainer to use the simplest solution i.e. a simple potential divider. Also, given that an inverting amplifier turns a positive signal into a negative signal, you would need an extra amplifier to restore it to being a positive signal. Still a no brainer. – Andy aka Apr 28 '17 at 18:52
• Thank you for the advise. My supervisor asked to use an inverting amplifier stage. So I need to know the theoretical definitions that I have asked in above quastions – vassidefuk Apr 28 '17 at 18:54
• Understand this; I don't give advice that I believe encourages anyone to follow a path that is dubious or is problematic. If you feel you must somehow use an inverting amplifier then all I can say is that your supervisor is ignorant (can never be ruled out) or, you didn't listen to what he said very well. Are you sure he didn't tell you to use a NON inverting amplifier because there is some degree of sense in that but, you will still need a resistor divider. – Andy aka Apr 28 '17 at 19:02
• My supervisor said to use amplifier stage to regulate 0-12V to 0-3.3V and while reducing noise at higher frequencies. sensor output changes at 170Hz which attached to a engine combustion chamber. engine is 10000rpm/60 = 166.66rps approx 170Hz – vassidefuk Apr 28 '17 at 19:15

That circuit almost works however...

• Your reference on the + pin should be $3.3/2 = 1.65V$ not 2.603V.

• You should add a resistor and two germanium diodes at the output. One from ground, the other to the 3.3V rail, to protect the micro in case something odd happens. As follows.

simulate this circuit – Schematic created using CircuitLab

• Know the signal you receive will be upside down from the sensor output.

• Why your boss told you to do it that way makes little sense to me... This would be sufficient if you need the high impedance input.

simulate this circuit