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OPA333: Output voltage swing limitation

Part Number: OPA333
Other Parts Discussed in Thread: LM7705

Hello,

I have a generic question about opamp functionality, but for my application I am using a OPA333. I intend to read the voltage across a shunt resistor and I am wondering about the limitation of the output voltage swing, namely the lower value threshold (i.e. VOL in datasheets). I am supplying the OPA333 between 5V and GND. According to datasheet the VCM range is -0.1<Vcm<5.1 and the output swing interval is 0.05<Vo<4.95. The shunt resistor has a value of 0.03Ohm and teh minimum load current is 0.35A resulting in a minimum sensed voltage of 0,0105V (no errors considered). At this input voltage I am violating the output voltage swing requirement (considering a voltage follower configuration) and probably my output will be limited around the VOL limit of 0.05V (worst case), which does not reflect the reality. My question is how the lower output voltage threshold will afect my output if I set a gain to my voltage follower stage (i.e. change it to a non-inverting amplifier; e.g. Av=15). Any explanation or literature suggestion is very wellcome.

Also in the datasheets I observed the output voltage swing is given for a device supplied between Vs and GND with the output Vo = Vs/2, 10kOhm load between Vo and Vs/2 (wich technically means no voltage across load resistor). Could you please explain how this is defined and what practically this means!?

Many thanks for your support,

Florin

  • Hi Florin,

    having a dead zone of only 0.05V is a big advantage to the old dino OPAmps we had to use in the 70ies and 80ies, which had dead zones of 2...3V. :-) A minimum dead zone is unavoidable when using "real world" transistors. Not only the output voltage is nonideal in this dead zone but also many other parameters like open loop gain, etc. The dead zone should be shunned in a standard circuit because the OPAmp can behave strange. One remedy is to use a very small negative supply voltage, for instance by the help of LM7705.

    Referencing the load at Vs/2 makes sense because one suited methode to avoid the dead zone is to center the signal arround a reference voltage optimally set at Vs/2.

    Kai
  • Hi Kai,

    Thank you for your feedback, but unfortunately it does not answer my question. I understand that if I apply an input signal below the VOL threshold I can get any value between 0 and VOL at the output, if my opamp is configured as a voltage follower, but I am interested what happens to the output voltage if I set a different gain. E.g. for OPA333 VOL=0.05V (worst case) and when Av=1 if my sensed voltage is 0.01V than for sure I will not get Vo=0.01V. However if I change my gain to 20 than I would expect a Vo=0.2V which is above VOL. I am wondering if VOL does affect Vo by any means or once your output should be higher than VOL (considering the gain), the last one is ousted for good and you only have to consider other error sources.

    For the second question I do not understand why R load is placed between Vs/2 and Vo wich is Vs/2 as well (see the image below).

    Thank you,

    Florin

  • Hi Florin,

    in a first order approximation the dead zone has nothing to do with the gain. It's only a limitation of output transistors. So, if you have a gain of 20 and an input signal of 0.01V, the output voltage of 0.2V is correctly produced by the OPamp. But it still cannot correctly output a voltage in the dead zone between 0...50mV.

    To your second question: The output voltage isn't Vout=Vs/2 for this parameter specification, of course. You saw the phrase "unless otherwise noted"? The output voltage can swing from 30mV to Vs-30mV, typically, with Rs=10k connected to Vs/2.

    Kai
  • Hi Kai,

    Thank you, now I got it.
    One last question regarding the second part. Following your explanation I guess that the reason for testing it in such a way is simetry and in the particular case in which R load is connected to ground for the same 10kOhm I sould consider a typical VOH of more or less Vs-60mV (so 100mV in the worst case). Following the same reasoning I have to consider a typical value of 60mV for VOL when R load connected to Vs. I am asking you because I do not want to overengineer my design, but still I do want to take in consideration all limitation.

    Thanks again for your answers and wish you a nice weekend,
    Florin
  • Hi Florin,

    figure 6 of datasheet can help to find out the maximum output voltage swing versus load current.

    Kai
  • Florin,

    Also note that offset is considered as either positive or negative within the limit (±10uV max in the case of the OPA333). Not all datasheets show the ± sign.

    While a positive offset will set the output to +200mV, a negative offset will set the output to a *theoretical* -200mV.

    This hard-pegs the output against VOL at +50mV as the output tries to go negative.

    So your input signal must now overcome a theoretical 150mV before the output will start moving off of 50mV.

    This creates an even larger 'dead-zone' due to the need to overcome negative offset.

    If you do not want a dead zone, you would need to pick a "zero" baseline voltage that also takes into account the maximum negative offset. You may loose a few lower codes, but you will not have a dead-zone. Servo and feedback systems really hate dead-zones.

  • Hi Paul,

    I have some questions to your remarks:
    - are you referring the input offset voltage (VOS)? In this case the datasheet shows a positive only value of 10uV. Shall I consider the - as well?
    - from where did you get the 200mV value?
    - could you explain how you get the 150mV to be overcame?

    Lots of thanks for your support,
    Florin
  • Hi Florin,

    It is a common misconception, and the point I was trying to get bring up...

    Not all datasheets show or mention the "±". Unless stated otherwise, the offset values should be considered as positive and negative.

    Very few amplifiers have a systematic offset in one direction. If they do, you will usually see something like "+3mV" or "-7mV" in the datasheet table.

    If there is no sign, you should assume both positive AND negative values (±) to be safe. Yes, this usually means two sets of error calculations...

    You had mentioned 200mV above..that is why I used it. The 150mV was a typo...should be 250mV....ignore that..better to think of input referred.

    Simple example...We'll use larger a larger offset than the OPA333 to make numbers easier.

    Assume a non-inverting gain of 200, and the amplifier output low saturation (VOL) is +50mV using a single supply (V+ and GND).

    If the amplifier has a an actual offset voltage of +1mV, and you have a gain of +200, the resulting output should be +200mV. The output is greater than 50mV, so the output is linear and moving, but a little higher than you want. You just loose a few codes around zero.

    But if the actual offset is -1mV, the output w*should* be -200mV.

    On a single supply, the output cannot go lower than 0V.

    As Kai pointed out, the output also cannot go to 0.000V due to the particular minimum VO (VOL) saturation of the amplifier output.

    So the output will be "stuck" at the minimum VOL, or +50mV.

    +50mV on the output is equivalent to +250uV back on the input. So you will not be able to resolve input signals between 0 and +250uV because the output is "stuck" at 50mV. So you have a 0-250uV "dead zone" on the input.

    With a -1mV offset, the output is *supposed* to be at -200mV.

    But the output would still be "stuck" at 50mV...

    You would now need to apply an input signal of +1mV just to get the output back up to theoretical 0V.

    Then you need to overcome the output saturation by applying an *additional* +250uV to the input to get the output up to +50mV.

    The result is the output is "stuck" at +50mV until you apply greater than 1.25mV to the input, and *then* the output will start moving up linearly. So now there is a 1.25mV "dead zone" on the input due to the negative offset.

    With a positive offset, you only need to worry about the minimum saturation voltage. With a negative offset, you need to worry about both the offset "suppression" and saturation.

    The minuscule offsets and near-rail outputs of the OPA chopper family alleviate much of this problem, unless you crank up the gain to the point where the offset (Vos*Av) becomes an issue again....
  • Hi Paul,

    Thank you for the very nice explanation! It gives me a better understanding about the topic.

    I am a bit on the safe side with my application regarding this aspect as the OPA333 has an ofset of ± 10uV (only 10mV mentioned in the datasheet but I will consider bipolar) and the minimum sensed voltage in my application is 10mV unipolar (low side current sensing). Therefore I presume that the offset voltage is affecting my output a s an error relative to the 10mV reference for the minimum load (i.e. Av*(10mV-10uV)<Vo<Av*(10mV+10uV)). Please correct me if wrong!

    Thanks again,

    Florin

  • Florin

    We haven't heard back from you so we assume this answered your questions. If you need more help just post another reply below.

    Thanks
    Dennis
  • Hi Denis,

    Yes, everything is clear now.

    Thanks,

    Florin