• State
  • Locked Locked
  • Replies 1 reply
  • Subscribers 48 subscribers
  • Views 381 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

INA149: Clipping input on purpose

Josh Deng1
Prodigy 50 points

Part Number: INA149

I have a 5VAC input signal that I’d like to bring in via a diff amp, to protect against ground noise. Downstream from the diff amp, the signal gets rectified by a single diode (only the positive part of the sine wave gets kept). It then it goes through a filter and then to the ADC of a microcontroller.

I currently don’t have a negative VCC to amplify the negative side of the sine wave, but I also don’t need the negative side. I would love to avoid a double-supply amp if possible (it would require a new DC-DC converter).

I was wondering if I can use a single-supply diff amp and simply let the negative input get clipped by the amplifier. Is it ok to put -7V on the input of a single-supply amp?

Here is a simulation to show my plan. (Note that I used a generic LTspice diff amp IC for the sim.)

Would love any thoughts you have on whether this overstresses the inputs of a diff amp. Thanks!

  • 0
    TI__Guru* 77501 points

    Hi Josh,

    I am going to use INA149 (per your interest) to address your question. I am not going to comment on the competitor's part LTC6363. 

    Your requirements:

    1. Single power supply rail

    2. Input 5Vrms, 400Hz with Gain =1 --> Vout is going to swing 2*7.07Vp = 14.14Vp-p

    The datasheet is listed V+ = 5V and V- = 0 or GND, see the captured image above. Therefore 12Vdc rail (from your schematic above) is not going to meet the Vocm requirements. With Vout swinging 14.14Vp-p, you will need minimum 14.14Vdc + 1.5V = 15.64Vdc. So I selected 18Vdc. 

    Since V+ is selected at 18Vdc, Vcm = REFA = REFB = V+/2 = 9Vdc. This is why you only see half of the voltage swing, because your Vcm = 0Vdc. Please see the simulation above and simulated file below.

    /cfs-file/__key/communityserver-discussions-components-files/14/INA149-Diff-Amp-11232020.TSC

    If you are going to intentionally clip the negative part of sine wave, you still have to meet the input Vcm requirements. 

    If you have additional questions, please let us know. 

    Best,

    Raymond