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OPA4348: OPA4348 Current Sense Application Question

Part Number: OPA4348
Other Parts Discussed in Thread: OPA348

Hi Sir,


Currently I use OPA4348 implement current sense application Motor, but have some question need you help.

May I know as below FIG. have error? or if you have better solution, please advise me, thanks.

Another question Tina simulation output is always clamp 6 mV, why?

sboc201b - autosave 17-01-22 09_24.TSC

  • Wen,

    First, the overall gain of your application is G=(2.55k+2.55k)/1k*(10k/1k)=51 instead of G=35.5 you show.

    Even though the input common-mode voltage of OPA348 is truely rail-to-rail, the output voltage range may NEVER reach either rail. With the output load of RL=100k, Vout is specified to be within 25mV of either rail (see output voltage range) BUT with RL=5k load (close to 10k you use) it may only get within 125mV above the negative rail - see below.  Please watch the TI Precision Labs video explaining the topic:

    Additionally, considering that the maximum input voltage offset of OPA348 may be as high as +/-6mV, with zero shunt current, IS1, the output may be driven to Vout=G*Vos_max =51*(-6mV)= -306mV and therefore 125mV+306mV = 431mV below its specified linear output voltage swing limit, thus resulting in significant error. Therefore, in order to assure the linear operation of the output stage, the minimum required input voltage across Rs shunt resistor would have to be: Vshunt = 431mV/51 = 8.4mV (thus Is>4.2A).  Of course, all of the above calculations assume perefectly match gain resistors and any mismatch between them would also have to be accounted for.

    To get around the output voltage swing limitation, you may use the fourth channel of the quad in a buffer configuration with 1k/9k supply voltage divider to reference the output voltage to 0.5V above the negative rail instead of Vss. This will assure that even under Ishunt=0, the output voltage will stay within the OPA348 linear output volatge range.

  • Hello,

    In addition to Marek's comments keep in mind that the input common-mode and output range of a discrete instrumentation amplifier will not have the same swing to rail limits as the individual amplifiers it is created out of.  Take a look at the following FAQ page which discusses instrumentation amplifier VCM vs. Vout plots:

  • Hi Sir,

    I'll study it. Thanks for your help.

    Wen