Offset correction for a simple inverting amplifier

[anushaas]

I am simulating a simple inverting amplifier configuration using LF444A IC in TINA.

The circuit diagram is :


The input VG1 is a square wave of 75Hz frequency and 22mV peak amplitude.
Both R1 and R2 are 10k.
Theoretically, output VF1 should be -VG1 but I am getting the positive peak of VF1 at 32mV and negative peak at -12mV.

The wave forms are shown below:


Why is there an offset at the output and how can it be corrected?

 

[FvM]

Apparently the TINA LF444 model comes with 10 mV built-in offset. According to datasheet it's rather a worst case than a typical value, but not impossible with this cheap general purpose amplifier.

A straightforward solution is to use a precision (low offset) amplifier instead of asking for "offset correction". In any case: read the datasheet.

 

[crutschow]

Note that the offset is amplified by the non-inverting gain of the amplifier (the input offset appears to be in series with the differential inputs), which in this case is 2, so the TINA model offset is apparently 5mV.

So you can add a pot at the plus input to adjust the offset, or use a lower offset op amp as FvM suggested.

 

[anushaas]

Note that the offset is amplified by the non-inverting gain of the amplifier (the input offset appears to be in series with the differential inputs), which in this case is 2, so the TINA model offset is apparently 5mV.

So you can add a pot at the plus input to adjust the offset, or use a lower offset op amp as FvM suggested.

I am sorry but I did not get you.The circuit has an inverting configuration.Then how is "offset amplified by the non-inverting gain of the amplifier "?

 

[dick_freebird]

Unequal input impedances looking outward from the part
are generally a bad idea. Using Iib as a trim "handle" may
be OK for an application where temperature does not
vary and device leakages do not drift.

If input is taken as a "signal" and referred to the IN+,
then the feedback loop does amplify it at 1+Rf/Ri. Same
as if you put the offset there with a voltage source. The
A=-Rf/Ri inverting gain only applies to the inverting input.

 

[KlausST]

Hi,

If the offset (in a true circuit) is caused by the bias current of the OPAMP, then simply installing a 5k resistor from +In to Gnd helps.

Klaus

 

[crutschow]

I am sorry but I did not get you.The circuit has an inverting configuration.Then how is "offset amplified by the non-inverting gain of the amplifier "?

I know it's not obvious why that is so.
But note that the inverting gain is only for voltages at the input to R1, not for voltages directly at the the op amp inputs.

The input offset voltage appears as a voltage in series with the inputs.
A series voltage can be placed anywhere in a series chain with out affecting the circuit operation, so lets put it between the (+) input and ground to make it more obvious what is happening.
So, since the gain from the plus input to the output is 1+ R2/R1 = 2, that means the input offset voltage is also multiplied by a gain of 2.

Another way to look at it is that the op amp can't tell the difference between the input being applied to R1 (where the gain is -1) or the input applied to the plus input (where the gain is +2). The op amp configuration is the same for either case.
Since this is true, then the offset voltage in both cases must also be the same.

This fact is also often missed when calculating the op amp input noise and the frequency response from the op amp gain-bandwidth.
Those both must be calculated using the non-inverting gain as well.

Does that all make sense to you?