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Recently, I encountered a unexpected problem when used the operation amplifier AD8397.Before described the problem, please see the attach file that is circuit schematic.

the range of the output is 0-10V(4*Vin)

When the Vin was less than 0.25V, the output was always low, and not equaled to 4*Vin. That means the output can not reach to less than 1V.

As soon as the negative power(Pin.4 of AD8397) changed to -3V or used another IC(TI: LM7322), the output can be able to reach to less than 1V. the circuit can achieve design requirement(output:0-10V).

My problem is that,

1, the negative power is GND, and why output cannot reach to less than 1V. From the datasheet, the output of swing is -11.83V at the condition of negative power -12V.I think that the output is about 0.2V at the condition of GND. so the output was 0.2-10V, but the actual result is 1-10V.

2,Anyway, as I described that, Can ADI engineer point out how the problem is or give some comments

• Hi,

AD8397 is "rail-to-rail" output but not rail-to-rail input. The amplifier's input voltage range is ~1V from the supply rails. Means that the amplifier can only accept voltage levels within this range. Beyond this, the amplifier output would be limited by the IVR (Vout = Gain * IVR) until of course, the Output Swing limitation takes over.

Regards,

Jino

• Hi Webber,

IVR stands for Input Voltage Range. It specifies the voltage range where the amplifier's operation is linear. For the case of the AD8397 datasheet, it appears to be left out. I will discuss this with the product owner regarding this concern.

Regards,

Jino

• Hi Webber,

There are actually two solutions that you could implement to have the amplifier operate linearly:

1. Use dual supplies (Vs+=11, Vs-=-1) instead of single. Vs = + 6V (or Vs+=11and Vs-=-1) instead of Vs = 12V. or

2. Provide sufficient bias voltage on the inputs to make sure that the signal of interest is within the IVR and OVR of the amplifier.

I hope this helps.

Regards,

Jino

• Hi Jino,

I cannot get that the input voltage range from the datasheet. Can you point out that in the datasheet?

In additional, what is the IVR?

Cheers,

Webber

• Hi Jino,

Best Regards

Webber

• Hi Jino,

Thank you very much.

Current soltution,I used the dual supply(V+=12V, V-=-3V) to implement the function.

and backup solution is that TI LM7322 took the place of AD8397.Because LM7322 is cheaper than AD8397.

if used LM7322, use single supply(V+=12V, V-=GND) to work normally. That can save cost of power IC(-3V). The total cost is lower than current solution.

Cheers

Webber

• Hello Jino,

i stumbled over almost the same problem which is described by the thread starter. Obviously the IVR is still not included in the AD8397 datasheet...

One thing is very strange: we wanted to use the AD8397 as an output buffer for a resolver excitation circuit. This is described in your circuit note CN-0276 "High Performance, 10-Bit to 16-Bit Resolver-to-Digital Converter". The schematic we used could be found in figure 2. We did a spice simulation of this circuit (with the AD8397 spice model) and found this thing with the limited input voltage range of the AD8397: input voltages approx. lower than 1.4 V will result in a clipped output signal (as you described in a previous post). For me it seems quite clear, that the proposed circuit in CN-0276 could not work as suggested. But if you have a look at figure 5, 6 and 7 in CN-0276, there are measurements from the input and output of the non-inverting configuration of the AD8397. It is clearly stated that the input signal is below 1 V and the output signal is not a bit distorted or clipped. Unfortunatelly i cant get this together. The simulation validates your statement but CN-0276 says something completely different.

Maybe someone could bring light into the dark concerning this issue..?