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According to LM358 datasheet from TI it should be capable to reach GND when operated from a single power supply in linear mode. Thus I have decided to use it as input buffer to an ADC as shown:

enter image description here

with 1.65 V implemented as:

enter image description here

However it is not doing what I expect. In the following image CH1 is showing the + pin of C3 and CH2 is showing "ADC_INPUT":

enter image description here

Any idea?

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    \$\begingroup\$ You mention the LM358, but the diagram shows the RC4558. The two are very different. The RC4558 is a dual version of the ancient and grotty 741, which cannot get its output close to the rails. \$\endgroup\$
    – JRE
    Commented Apr 15, 2018 at 19:25
  • \$\begingroup\$ Yes, but I am using an LM358. I forgot to change the label in the picture. \$\endgroup\$
    – user171780
    Commented Apr 15, 2018 at 19:28
  • \$\begingroup\$ The circuit is AC coupled, so the input to the opamp can go below ground. That would cut off that part if the signal. \$\endgroup\$
    – JRE
    Commented Apr 15, 2018 at 19:33
  • \$\begingroup\$ @JRE: Yes, but obsolete a741 is just a previous version of LM358. Next was OP07 and even further, OP27. The RC4558 is just something inbetween. PS: Even LM358 is a double op.amp. inside single case. \$\endgroup\$
    – Jakey
    Commented Apr 15, 2018 at 19:33
  • \$\begingroup\$ @Jakey: The LM358 has rail to rail output. The RC4558 doesn't. The LM358 can operate on a single rail down to 3V. The RC4558 can't. In all respects, the RC4558 is more like a 741 than it is an LM358. The name of the RC4558 is even a call back to the original dual 741 - that was the 1458. \$\endgroup\$
    – JRE
    Commented Apr 15, 2018 at 19:43

3 Answers 3

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LM358 datasheet says it is only able to sink a couple tens of µA when the output is close to 0V (see page 6, output current/sink).

If you check the internal schematics (page 13) this is quite obvious (Q13 can't pull the output below 1 Vbe). This isn't a true rail to rail output. It can only go to 0V if the load is something like a resistor connected to ground, in this case the opamp does not have to sink any current, so it works.

In your schematic, when the output is at 0V, the opamp output will have to sink 1.65V/10kOhm = 165µA which is too much.

You need a true rail to rail opamp, or larger feedback resistor values (which will increase noise due to bias current).

Note: LM358 was introduced in 1972. It is still produced because it works fine in its application domain, and it is very cheap. It's a timeless classic. However, real rail-to-rail opamps that actually work well are a much more recent development. Don't expect it to compare to a 50c modern RRIO opamp... also 50 cents is very expensive compared to the price of LM358...

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    \$\begingroup\$ Or a pull-down resistor on the output \$\endgroup\$
    – Harry Svensson
    Commented Apr 15, 2018 at 19:59
  • \$\begingroup\$ @HarrySvensson yeah, a 1k pulldown would bring the output down to less than 0.2V. I wonder about the output impedance of the opamp though, it will switch from an emitter follower (quite low Z) to the value of the pulldown when the output transistor turns off, this will be in series with the feedback network, could influence frequency response depending on output voltage. \$\endgroup\$
    – bobflux
    Commented Apr 15, 2018 at 23:27
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The LM358 is NOT a rail to rail op amp.
Reference to the schematic shows that there is a limited current source that allows you to get close to the negative rail, but the datasheet shows that you can only get as close as about 200 mV.

To get rail-to-rail you need a FET op-amp.
Read this for more information, it shows the current limitations as you approach the positive and negative rail.

Select a rail-to-rail op-amp such as the MCP6411 which can get you to within a couple of mV of the rails.

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I would try grounding the output probe to see where the actual ground level is. You might be at actual 0V there in the bottom ,but have clipped the signal by too much gain.

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