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op-amp nV input offset voltage

Discussion in 'Electronic Components' started by Paul, Jun 21, 2008.

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  1. Paul

    Paul Guest

    Hi,

    As you know, the *input* offset voltage is the voltage required across
    the op-amp's input terminals to drive the output voltage to zero.
    Although it has been my experience that for most op-amps the input
    offset voltage is due to the "-" input pin for the *most* part. For
    example, according to Spice the input offset voltage on the "+" input
    pin on a LMC660A op-amp for a non-inverting amp circuit is a few
    nanovolts, disregarding thermoelectric effects mind you, but a few
    millivolts on the "-" input pin. Although as you know the input signal
    is not applied to the "-" input pin for a non-inverting amp circuit,
    which means there's just a few nanovolts on the input of such a
    circuit if we disregard thermoelectric effects.

    I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
    25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
    0.5mV on the input of this Instrumentation op-amp circuit with 3fA
    bias current that the DUT input impedance would have to be 0.50mV /
    3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
    say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
    = 0.6nV, disregarding thermoelectric effects?

    INA116PA datasheet:
    http://focus.ti.com/lit/ds/symlink/ina116.pdf

    Regards,
    Paul
     
  2. Paul

    Paul Guest



    Here's my main concern. If I build the INA116PA for DC application,
    which is an internal Instrumentation op-amp chip (3 op-amps), and the
    impedance of my DUT is 200 Kohms, then what bias currents could a good
    EE such as yourself expect? I mean, for a 200K ohm DUT input source we
    cannot have both 0.5mV offset and 3fA bias on the DUT. I think V=I*R
    applies, so if the bias current is 3fA then V = 3fA * 200Kohms = 0.6
    nV.

    Thanks,
    Paul
     
  3. Paul

    Paul Guest


    The LMC660A has a typical voltage offset of 1mV and bias current of
    2fA, but that depends what type of op-amp circuit. According to Spice
    the input voltage offset for an inverting or differential circuit is
    about what the Vos spec says, but for a non-inverting circuit it's a
    few nanovolts on the "+" input pin. I'm wondering if the Vos in
    datasheets is referring to a certain type of op-amp circuit such as
    the inverting type (http://hyperphysics.phy-astr.gsu.edu/Hbase/
    Electronic/opampvar.html#c2).

    Regards,
    Paul
     
  4. Jamie

    Jamie Guest

    The last time I checked, the offset voltage would be the difference
    between the (-) and (+) input with the Op-amp in (-) loop back mode.

    So, if you were to put an op-amp in (-) loop back and lets say 5 volts
    on the (+) input, the (-) input should be offset no more than what the
    spec's state.

    Or, I guess if you were using a +/- to common supply, you can simply
    tie the (+) to common with op-amp in (-) loop back. You should be seeing
    that offset factor at the (-)/output..

    Maybe things have changed but that is what I go by..

    http://webpages.charter.net/jamie_5"
     

  5. Hello Paul,
    Maybe it helps if you think about the transistor circuit
    of an opamp.

    The first stage of an opamp consists of a differential
    amplifier made by a pair of two well matched transistors.
    The difference of the Vgs(Mosfet opamp) or Vbe(bipolar opamp)
    of these two transistors in the input stage is the main
    contributor for the offset voltage.

    Offset voltage is always measured between the + and - input.
    What you have measured at the +input is the bias(leakage)
    current multiplied by the value of the resistor connected
    to the +pin.

    Best regards,
    Helmut
     
  6. Paul

    Paul Guest



    I appreciate all of the replies! All of these years I've had this
    false idea about the datasheets Vos burnt into my head. I've always
    assumed that if the datasheet said the op-amps Vos was say 50uV then
    that's the lowest input voltage (by my def: the voltage applied on the
    input device due to the op-amp) one can expect with a typical op-amp
    circuit such as an inverter or non-inverter.

    So it's true that one could achieve input voltages in the nanovolt
    region on a 200K ohm DUT from an Instrumentation op-amp chip such as
    INA116PA even though the datasheet Vos spec is 2mV?

    Thanks,
    Paul

    INA116PA datasheet:
    http://focus.ti.com/lit/ds/symlink/ina116.pdf
     

  7. Hello Paul,

    Yes you can apply voltages as small as you like.
    they will be still amplified by the gain G, set with
    the feedback resistors. The drawback of any Vos
    is that you will have an output voltage of (Vos+Vin)*G .
    This menas you have to either adjust the offset voltage
    already at the input or you have to subtract Vos*G at
    the output.

    Best regards,
    Helmut
     
  8. Paul

    Paul Guest


    Thanks! As you said the output offset can always be corrected, but
    it's great to know that a 2mV op-amp chip such as the INA116PA can
    apply DC voltages as low as a few nanovolts on the input device
    without adding shunt resistors. Of course one can always add a shunt
    resistor to lower the input voltage across the DUT, something I knew
    about, but of course that has obvious effects of decreasing the DUT's
    effective input voltage to the op-amp.

    I'm wondering if there are any op-amps or perhaps a BiFET amp circuit
    that could achieve a few nanovolts across say a 200K ohm device while
    consuming no more than a few microwatts. The idea is that such a
    microwatt amp would have considerably less input thermoelectric
    effects. Thermoelectric effects can generate a half dozen or more
    microvolts on the DUT unless carefully balanced with dummy resistors.
    I believe Linear Tech has some microwatt op-amps, but nothing near
    25fA bias current.

    Thanks,
    Paul
     
  9. Paul

    Paul Guest


    Hi,

    I'll try to clarify:

    I am referring to the input voltage on the *DUT* caused by the op-amp,
    and therefore if the bias current through the DUT is decreased then
    the offset voltage on the DUT will be less-- ohms law.

    The op-amps I am working with have offsets around 0.5uV to a few uV.
    Therefore thermoelectric effects should be considered. As far as I
    know instrumentation op-amp appear to have to least thermoelectric
    effects since both input pins go to the same polarity on both op-amps,
    the + pin, but there are still thermoelectric effects since both op-
    amps are not 100% identical. Other circuits such as the inverter
    require dummy resistors and such to help reduce the thermoelectric
    voltages on the DUT.

    My interest in BiFET's is to design a low power amp circuit with low
    bias current.

    Thanks,
    Paul
     
  10. neon

    neon

    1,325
    0
    Oct 21, 2006
    the input offset voltage is never required actualy is an un-wanted situation. and either pin input just the [-] input
     
  11. Pieter

    Pieter Guest

    To prevent thermoelecric voltages, keep all pins at the same
    temparature. But also all surrounding resistors etc. A cooling airflow
    gives temperature differences. And resistors and opamps that get warm
    may give some effects.

    Pieter
     
  12. Tom Bruhns

    Tom Bruhns Guest

    My apologies if this has been covered in the other branch of this
    thread, which I don't have time to fully read...

    As others have pointed out, the input bias current and input offset
    voltage of the part are characteristics of the part. HOWEVER, the
    external circuit strongly influences how well you can take advantage
    of those characteristics. That is, the external circuit can
    completely wipe out the potential benefits of either a low bias
    current or a low offset voltage or both, even. For example, I used a
    chopper-stabilized op amp to amplify the output of a diode RF
    detector. The op amp has typically a pA of input bias current and
    about a uV of input offset voltage. First, I had to be very careful
    to guard the detector traces against currents leaking in from outside,
    and then I had to be careful that the resistance between the guard
    trace and the detector output trace (feeding the op amp input) was
    high enough that a 1uV offset would not result in a current as large
    or larger than the pA op amp bias current. For this part, that's only
    a megohm or so, fairly easy to do, but in an earlier design using a
    non-chopper amp where the offset voltage was up to a millivolt or so,
    it was a killer. The RF detector diode is shunt between the guard and
    the amplifier input, with RF fed in through a capacitor, and the zero
    bias RF detector diode shows considerable current at a millivolt.

    Seems like there should be some good references on applying low
    offset, low bias amplifiers. I know that Bob Pease has had some good
    articles on the trials and tribulations of testing amplifiers down in
    the fA region--not trivial! His articles can be found with a search
    on the web...

    Cheers,
    Tom
     
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