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OPA2348: Output voltage with no input applied

Guru 16770 points

Replies: 7

Views: 257

Part Number: OPA2348

Hi

Our customer has question about OPA2348.

I attach the circuit below with simplified.

2476.e2e.xlsx

Two OPA2348 are tested in same circuit.

Device A : 100mV at Probe_X and 100mV at Probe_Y.

Device B : Nearly 0V in both Probe point.

1.

OPA2348 offers voltage output swing from fail in electrical characteristics.

(25mV when RL=100K, A >94dB and 125mV when RL=5K, A > 90dB)

Should this voltage appear at output when the input is not applied?

2.

In this time, Device A and B show different result.

Could it be possible result?

The customer uses OPA2348 for current sensing, so they want to avoid such variation because it leads to measurement error.

BestRegards

  • Hi Na Na78,

    The circuit has a gain of 44 and the maximum offset voltage of the op-amp is 5mV. If non-inverting input to the first stage is 0V, the output of the op-amp can be as high as 220mV from the offset voltage alone. Each copy of the amplifier will have a different offset voltage so you might see different results when doing an A-B-A swap.

    Let me know if I misunderstood the question, I hope that my answer is clear. I would recommend watching the TI Precision Labs videos on offset voltage for some further explanation:
    training.ti.com/ti-precision-labs-op-amps

    And consider taking a look at this cookbook circuit:
    www.ti.com/.../sboa215a.pdf

    Thanks,
    Paul
  • Guru 16770 points

    In reply to Paul Goedeke:

    Hi Paul

    Thank you for your reply.

    I will check the customer circuit as you advised.

    Let me confirm the meaning of voltage output swing from fail.

    If voltage output swing from rail is 25mV, would the output swing 0V+25mV ti VCC-25mV?

    BestRegards
  • In reply to na na78:

    I'm not sure that I understand your question, but I think you're trying to understand the output swing specification.

    The specification says that if the op-amp is put in a circuit where it is trying to output a voltage equal to the positive or negative supply, that it will be within 25mV if the load is 100kOhms or 100mV if the load is 5kOhms. This also relates to the claw curves (Figure 6) because the load changes the output current of the op-amp.

    This means that the output voltage for a 100kΩ load can swing from VEE+25mV to VCC-25mV.

    -Paul
  • Guru 16770 points

    In reply to Paul Goedeke:

    Hi Paul

    Thank you for your reply.

    So, in the following two cases, I added my understanding. Could you please review?

    Device A : 100mV at Probe_X and 100mV at Probe_Y.
    -> If the offset voltage of two opamp are called Vos1 and Vos2 conveniently,
    Vos1 at first stage opamp is increase with gain of 44 and it appears at second stage opamp.
    Basically, we can not say that Vos1 and Vos2 are equal.

    Device B : Nearly 0V in both Probe point.
    -> In this case, it can be considered that Vos1 x 44 is smaller than output swing limit (VEE+25mV).
    It can be considered that ~25mV (or output swing limit) appear at output.

    Is my understanding correct?

    BestRegards

  • In reply to na na78:

    I think your understanding is correct for your point on Device A. I don't quite understand what you're asking for Device B, but if the desired output voltage is greater than VEE+25mV, then you'll get the desired output voltage. If it is less, you'll hit the output swing limit.

    Thanks,
    Paul
  • Guru 16770 points

    In reply to Paul Goedeke:

    Hi Paul

    Thank you for your reply.

    Regarding to Device B, I wanted to understand why the output of Device B is almost 0V regardless under same condition as Device A.

    However, actually, I think that the output of Device B would not be just 0V but a slight voltage should appear by output limit.

    Please tell me if you have additional information.
    If not, I will close this thread with my gratitude to your help.

    BestRegards
  • In reply to na na78:

    Hi Na Na78,

    It is difficult to say exactly what is going on without better measurements and data but one device could have a much smaller offset than the other, causing the different results.

    -Paul

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