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Original question:

OPA2192-Q1: current measurement for Moto application

OPA2192-Q1: current measurement for Moto application

Ryan Wang
Expert 5470 points
Part Number: OPA2192-Q1
Other Parts Discussed in Thread: OPA192, OPA388

Dear expert,

https://e2e.ti.com/support/amplifiers-group/amplifiers/f/amplifiers-forum/1054256/opa2192-q1-current-measurement-for-moto-application/3902768#3902768

Follow up for above attached link, Vcc18 and Vee18 mean +/- 18V, right? Can I change it to +3.3V and GND? Thanks

  • 0
    TI__Guru* 93105 points

    Hi Ryan,

    First, the OPA192 is not specified to operate with a supply less than +/-2.25 V (+4.5) so it can't be used in a circuit powered by +3.3 V. A low-voltage op amp such as the OPA388 that Michael mentioned is usable with a single supply of +3.3 V. Second, reducing the supply from +/-18 V down to a single +3.3 V supply will vastly change the output voltage (Vo) swing range. The differential input voltage range, common-mode input voltage and gain would have to be assessed in relation to the much reduced Vo output range.

    Once you decide on the low-voltage op amp the design should use you need to make sure that the circuit still meets the required performance levels. Some changes may be required to assure the op amp operates within its specified linear operating ranges.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • 0 in reply to Thomas Kuehl
    TI__Expert 5470 points

    Thomas,

    Thanks for your feedback.

    For the circuit Michael suggest to me, my understanding is :

    1. +18V is needed , because there is +12V common voltage at input.

    2. -18V is not needed, because there is 1V reference voltage provided which guarantee there is no negative voltage at input. Signal swing is +/-10mV*50 around 1V voltage. So GND is enough to replace -18V

    Is my understanding correct?

  • 0 in reply to Ryan Wang
    TI__Guru* 93105 points

    Hi Ryan,

    The common-mode voltage is the average voltage applied to the two inputs of the op amp, or VCM = (Vin+ + Vin-)/2. Therefore, if you view the TINA circuit I created based on your schematic the VCM is about 19.6 mV, which is much less than the 12 V you mention. Now if the Vin generator in the schematic does provide a voltage that moves up to +12 V, then the op amp's linear VCM range would need to include +12 V voltage for the positive supply (V+ or Vcc) being used. The OPA192 VCM range is (V+) + 0.1 V, but for practical purposes V+ should be higher than the maximum intended VCM.

    Yes, the VCM and differential voltage being applied result in an output voltage that is positive and above the lower Vo limit of the OPA192. Therefore, the circuit can be operated with a single positive supply and GND providing V2 doesn't become too high.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

    OPA192_diff_cir_01.TSC

  • 0 in reply to Thomas Kuehl
    TI__Expert 5470 points

    Thomas,

    great thanks