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Original question:

OPA187: OPA187 is a rail-to-rail amplifier but the output voltage swing are +/-3V

OPA187: output voltage swing in high-voltage operation

haras nosregor
Prodigy 90 points

Part Number: OPA187
Other Parts Discussed in Thread: OPA205

Tool/software:

My application uses the OPA187 with (V-) = 0V, (V+) = 12V. It is configured as a non-inverting amplifier. I would like to know if an output voltage of 0.5V at 5mA load current is within the linear output swing range of the device?

The TI precision labs video on op-amp output swing (https://www.ti.com/content/dam/videos/external-videos/en-us/1/3816841626001/6279999026001.mp4/subassets/opamps-input-output-limitations-output-swing-presentation-quiz.pdf) suggests that conditions for the open-loop voltage range specification can be used to determine the linear output swing range. The open-loop voltage gain is specified from (V-)+0.3 to (V+)-0.3. However, this for a 10kOhm load, which is a far lighter load than my application. 

Figure 10 in the datasheet show the output voltage swing at 5mA output current. But based on the response to Danilo's question, this for a much lower supply voltage (Vs = +/-2.75V) than my application. Is there a similar plot of output voltage swing vs output current for the OPA187 in high-voltage operation?

  • +1
    TI__Guru 70981 points

    The linearity of the output mostly depends on the output current; thus, based on the spec table you showed above, the worst case occurs for +/-18V supply with the 10k load connected to mid-supply (GND).  Under this condition, the output current is Iout=~18V/10k = 1.8mA and it requires 0.3V headroom from either rail.  Therefore, everything else equal, for sinking 5mA with high degree of linearity, where AOL>132dB, you would need to be 5mA/1.8mA*0.3V = 0.83V above ground.  However, if you are sourcing 5mA as you drive the output to ground (most likely the case in your single supply application where the load is connected to ground), the output may swing within 5mV of ground (the same as unloaded condition).  This is because during the sourcing current only the upper output stage transistor delivers the current while lower transistor carries only quiescent current (the same as in unloaded case). 

  • +1
    TI__Mastermind 27418 points

    Haras,

    Your understanding is correct.  I will summarize it from my perspective here:

    1. The Aol specification is a good way to see what the output swing is for low distortion (i.e. high Aol).  If you can adhere to this range you will get the best performance.  However, the specification is limited in that it is only given for one load condition.  In your case it really doesn't help much as your load is much heaver that the the specified condition.
    2. You can look at Figure 10 in the data sheet to get an idea of the typical limits for saturation.  Typical means that some devices will have worse swing to the rail than what is displayed.  Saturation means that the device is non-linear on or beyond the curve.  As you approach the curve the device will begin to become non-linear, so for best performance you should avoid the curve by at least 0.1V.  This curve is also dependent on temperature so you can get better or worse swing depending on the ambient temperature.  In your example you can see a degradation of between 0.25V and 0.5V at 5mA.
    3. You may consider a rail-to-rail output bipolar device if you want to drive a lot of current but still get good output swing.  CMOS devices like OPA187 get very good output swing for light loads.  Bipolar devices, on the other hand, get about 0.2V swing-to-the-rail independent of the load.  Thus, when the load current is low CMOS is much better from an output swing perspective (millivolts from the rail for OPA187).  Bipolar devices like OPA205 will get about 0.2V to 0.35V swing limit with good linearity across a wide load range.  Op Amp Input and Output Swing Limitations covers this in section 11.

    Best regards, Art Kay