I was watching a video of the legendary Bob Pease, in which he says that the regular LM324/LM358 is not a low distortion amplifier, however, if you add a 10K resistor from the output of the opamp to the negative supply rail, then, distortion is greatly reduced.

It appears that in the video they are using bipolar power supplies, so my question is: if im using an LM324/LM358 with a single supply, say 9V and ground, will adding a resistor from the output to ground also lower the distortion? I must add that im adding a 4.5V bias to the input of the opamp so the output is idle at 4.5V. The following schematic displays what I'm doing


simulate this circuit – Schematic created using CircuitLab

The video link is the following: Whats All This Distortion Stuff, Anyhow?

  • \$\begingroup\$ It depends on the load (in addition to the resistor you add). \$\endgroup\$
    – Whit3rd
    Nov 26, 2017 at 3:20
  • \$\begingroup\$ With that biasing arrangement and no feedback, shouldn't the output clip to the positive rail all the time (unless the AC input has an extreme amplitude)? \$\endgroup\$ Nov 26, 2017 at 15:12
  • \$\begingroup\$ I forgot to add the feedback resistors, I just edited the schematic \$\endgroup\$
    – S.s.
    Nov 26, 2017 at 16:49

4 Answers 4


The output stage of the LM324 has low output impedance when it sources output current (when the 'push' half of the push-pull stage is active), and when it sinks output current (when the 'pull' half of the push-pull stage is active). It turns off (goes high impedance) at zero output current, and that causes a 'dead spot' in the transfer characteristic.

To reduce distortion, you must never allow zero output-stage current. $$I_{output} = -I_{load} + {V_{output} -V_R \over R} \ne 0$$ That ensures that the load current plus the addon resistor's output current is always nonzero, at all output signal voltage values.

That means a pullup resistor in conjunction with a load that has a limited ability to sink current, or a pulldown resistor in conjunction with a load that has a limited ability to source current. A resistor to 'ground' might meet neither requirement.

  • \$\begingroup\$ VR in your formula is the voltage across the shunt resistor? Will a resistor to ground if the output is at 4.5V idle, wont satisfy the requirement? \$\endgroup\$
    – S.s.
    Nov 26, 2017 at 3:27
  • \$\begingroup\$ @A.J. A resistor to ground is only suitable if it, plus whatever load is on the op amp, draws a NET positive or negative current. And, never zero current. An op amp can drive lots of odd loads and not always another resistor-to-ground. \$\endgroup\$
    – Whit3rd
    Nov 26, 2017 at 5:33
  • \$\begingroup\$ @A.J. - the voltage across the shunt resistor is the difference, as in the formula, of the output and the resistor-connection voltages. The 'signal around 4.5V' and 'ground' and 'negative power supply' terminology is somewhat confusing, so I didn't want to use it... in the example, VR is shown as the ground triangle. \$\endgroup\$
    – Whit3rd
    Nov 26, 2017 at 5:40

The output totem pole stage in a 324 is biassed for low quiescent current, which means both devices go low current when there is no output current demanded. This reduces the gain and leads to crossover distortion. If the load is bipolar, then at times the output current will be zero.

A load that only sources or sinks current will keep one output transistor on, and significantly reduce this source of distortion.

A bipolar load can be biassed into a unipolar one with a sufficiently small pullup or pulldown resistor placed across it.

Should we use pullup, or pulldown? The 324 output transistors will source more current than they can sink. So a heavier bias current, to cope with a larger load, can be used if it's pulldown.


An added load to ground that keeps the upper output transistor conducting under all load and signal conditions will eliminate crossover distortion (but other effects such as limited slew rate and nonlinearity with finite gain will still cause some distortion).

You may not be able to get as much voltage swing as with a resistor to the positive supply rail, but you might be able to get more current. The value of the resistor may have to be lower than 10K, depending on the load.

  • \$\begingroup\$ Does it matter if the resistor is connected from the output to the positive rail VS connecting it from the output to the negative rail?, what is the criteria to determine the value of the resistor depending on the load? \$\endgroup\$
    – S.s.
    Nov 26, 2017 at 1:54
  • \$\begingroup\$ @SpehroPefhany unless I'm misreading the data sheets, the 324 is better at sourcing current than sinking, where 'better' means more current specified, though I've not analysed the voltage swiing \$\endgroup\$
    – Neil_UK
    Nov 26, 2017 at 5:50
  • \$\begingroup\$ @Neil_UK Oops that's what I mean to write. Will replace comment. \$\endgroup\$ Nov 26, 2017 at 7:28
  • \$\begingroup\$ It changes the swing and the LM324 is better at sourcing current than sinking it, so it will make some difference as to what range of resistances will work. The criteria is to make sure the amplifier is always sourcing (or sinking) net current regardless of the load and signal conditions. If you have an inductive or capacitive load you will have to account for that- or if the load is connected to ground, the + rail, somewhere in between etc.. Modelling it might be easiest- just look at the current at the op-amp output and see if it changes sign (and give it some margin) \$\endgroup\$ Nov 26, 2017 at 7:28

Your answer is in the first paragraph of your question.

If you look at the internal schematic of the LM358, you will see why adding the external resistor from o/p pin to the negative supply rail. The op-amp doesn't care whether you are using a single-ended or bipolar power supply.

The distortion is reduced because of the current through that external resistor. In your case, 4.5V across the resistor.

Although the datasheet recommends a value of 6,2k, I normally use 4.7k just to keep my BOM component count down. Either value works well.


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