Please tell me about the hysteresis of the AD7785.

Hi miky , 

We usually recommend newer products as a replacement for AD7785 such as the AD7124, AD4170, and AD4130. It has all the building blocks of the above ADC but more advance features and better performance. To add, it also has a more complete datasheet of values (device error before and after calibrations) that you would find useful for your application. 

You may select a reference signal, other than 1.6V, that will have a hysteresis region which can range from 0.1Vmin to 0.17Vmax which is depending on your requirement. Then, you may adjust the Vt(+) and Vt(-) depending on the hysteresis voltage that you will set, However, the Vt(+) and Vt(-) should be bounded by its respective Vmin and Vmax.

Thanks and regards,

Rod

Hi,RAdante 

Thank you for your reply.

I have more question about that.

Does it mean that Vt(+)=2V and Vt(-)=0.8V is never the case?

Does it mean that Vt(+) and Vt(-) are determined by the hysteresis value?

Best regards.

miky.

Hi miky ,

For AD7785, yes, you are correct. 

Thanks and regards,

Rod

Hi, RAdante .

Thank you very much for your answering.

I'm so sorry. I would like to ask you more question about that.

For example, with a hysteresis of 017V, Vt(-)=1.83V when Vt(+)=2V, which violates the maximum value of Vt(-), which is 1.7V.

Does this mean that Vt(+)=2V is impossible?

Similarly, with hysteresis 017V, when Vt(-) = 0.8V, Vt(+) = 0.97V, which violates the 1.4V minimum for Vt(+).
Does this mean that Vt(-)=0.8V is also impossible?

Why is the maximum value of Vt(+) 2V and the minimum value of Vt(-) 0.8V?

Thank you and regards.

This question has been answered on this thread: In the datasheet, why is the maximum value of Vt(+) 2V and the minimum value of Vt(-) 0.8V? - Q&A - Precision ADCs - EngineerZone. I will now close this thread. If you have further question, feel free to ask us and create a new thread for your question/s. 

Thanks and regards,

Rod