I have breadboarded a simple differential amplifier:

schematic http://www.ti.com/lit/ds/symlink/lm158-n.pdf

what I expect is that OUT=B-A but that only works when the voltage of B is larger than A+0.6, i:e there seems to be an offset of about 0.6V on the output. Since the output LM358 is supposed to be able to go to ground I can't figure out where this offset is comming from. Any ideas?

Edit: VDD is 16V in my tests and if I tie A to ground, then OUT=B without the offset.

  • \$\begingroup\$ Do you have any load on OUT that isn't shown in the schematic? \$\endgroup\$
    – The Photon
    Jan 11, 2014 at 19:48
  • 1
    \$\begingroup\$ You should use an op-amp with a bipolar supply to do these kinds of experiments. rail-to-rail op-amps will get close but you'll always have some weirdness at the supply rails. A bipolar supply means that there is nothing really special about 0V, it's in fact in the middle of the supply range. \$\endgroup\$
    – akohlsmith
    Jan 11, 2014 at 20:28
  • \$\begingroup\$ @ThePhoton: At first there was only my multimeter, when I added a 1k load as suggested I was able to get down to about 0.2V, at 100R load I got down to about 0.07V. \$\endgroup\$
    – Johan
    Jan 11, 2014 at 23:51

1 Answer 1


Here are some: -

Scenario 1 - A is at 3 V and B is at 0 V - the output will be trying to make -3V but limits at 0V.

Scenario 2 - A is at 3 V and B is at 3 V - +Vin will be 1.5 volts and the output will be trying to make -1.5 volt + 1.5 volt = 0V = GND.

Scenario 3 - B is at 6 volt, A is at 3 V - +Vin will be 3V and the output won't be trying to force anything into the feedback loop so the output will be +3 volts.


Following a look at the circuit diagram of the LM358 it is apparent that it will have problems in the circuit getting down to 0V: -

enter image description here

Without the 50uA current sink the output would only get down to about 0.6V. The current sink trys to pull it the rest of the way but, with R2 and R3 pulling the output up to some value (maybe half rail at 8V) then there will be a problem that can only be overcome with a slight negative rail or a good load on the output to ground.

8V across 2x 18k is a current of 222uA so it might help if the 18k resistors were made 180k.

  • \$\begingroup\$ Thanks for the comment, in scenario A the output is 0.6, in scenario B the output is also 0.6, in scenario 3 the output is +3V just as you suggest. \$\endgroup\$
    – Johan
    Jan 11, 2014 at 19:58
  • \$\begingroup\$ I understand that the output is clamped when it wants to go below 0V, but in my case it seems to clamp already at 0.6V. \$\endgroup\$
    – Johan
    Jan 11, 2014 at 20:02
  • \$\begingroup\$ @Johan OK I see what the problem is. To get the output to swing down to 0V means you cannot have any load that is lifting the output and taking more than 50 uA. Take a look at page 21 of the brochure. There is a 50uA current sink to help pull the output down to 0V but you are pulling it up with R3 and R2. Try loading the output with 1k ohm. The natural bottom limit for the amp is realistically 0.6V unloaded. I'll add this to my answer. \$\endgroup\$
    – Andy aka
    Jan 11, 2014 at 20:07
  • \$\begingroup\$ The LM358 should go nearly to 0V much closer then 0.6. Can you confirm that your meter reads 0V when both leads are at ground? \$\endgroup\$ Jan 11, 2014 at 20:10
  • \$\begingroup\$ @dfowler7437 it's not that dude - see the edit in my answer but thanks for the thoughts. \$\endgroup\$
    – Andy aka
    Jan 11, 2014 at 20:18

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