Please, check my calculations (Diff. amp errors referred to input)

[Kirill V.]

Hi every one!
Some time ago, on other forum was asked question:

In datasheet of diff. amplifier LT1997 says about output offset formula (page 6, Note 9). In this formula uses R_p and R_n terms. What this terms means?
https://www.analog.com/media/en/technical-documentation/data-sheets/lt1997-3.pdf

I decided to make my own calculations to get a similar expression. Now I am interested, are my results correct? Please, check it!
Please, note that I uses IEC symbols and much mathematics:)
 Diff. amp calculation.pdf (840.52 kB - downloaded 56 times.)

[Kirill V.]

Can't anyone help me? After all, this is a simple algebraic analysis of the circuit

[iMo]

My guess would be the Rn = R1||R2 and Rp = R3||R4 in your doc.

[RandallMcRee]

In the datasheet:

Note 9: Offset voltage, offset voltage drift and PSRR are defined as
referred to the internal op amp. You can calculate output offset as follows.
In the case of balanced source resistance, VOS,OUT = (VOS • NOISEGAIN)
+ (IOS • 22.5k) + (IB • 22.5k • (1– RP/RN)) where RP and RN are the total
resistance at the op amp positive and negative terminal, respectively.

In cases like this, where doubts arise, you might want to create a simple LTSPICE model and verify your calculations.

[Kirill V.]

Thanks!
The question was what did the phrase mean: total resistance at the op amp positive and negative terminal, respectively. In other words, this is addition of resistances or parallel value of resistances. I think I've been able to prove mathematically that it's a parallel connection R1||R2 and R3||R4 respectively. In datasheet not says about this details directly.
LT1997 provides the user with the ability to select different internal resistors which can cause output offset. Standard differential amplifier has unchangeable internal resistor networks, therefore R1||R2=R3||R4 in this case and this calculation do not need.
I would like you to appreciate the accuracy and consistency of my theoretical calculations. I've already tested it with the simulator and that's right, seem to me.

[Kirill V.]

If you don't say anything about it I can conclude that my theory is correct.
In any case, thank you all for any help:)

[T3sl4co1l]

Your math looks good.  Cool to see the input resistors drop out leaving bias currents depending on R2 only.  Input offset voltage is multiplied only by noise gain, basically by definition; and note that goes for DC offset as well as input noise e_n -- same thing.

I would only suggest, in the last steps, splitting the bias current into a common and differential mode, Ib+ = Ibc + Ibd, Ib- = Ibc - Ibd, and working that out.  Ibc should cancel out then, and you can be sure only the difference matters.  (Actually, as drawn, since the currents are in and out of the amp, it should be the other way around.  Alternately, it's enough to prove this separately, and continue with Ibd --> Ib+ = Ib- as shown.)

And likewise, the noise currents of Ibd and Ibc may vary; in conventional bipolar amps, Ibc is dominant (base bias current), while in compensated amps (like LT1115) the majority of base current is supplied internally and Ibc is smaller than, or comparable to, Ibd.  In FET amps, they are comparable.

Also, I guess I should write Ib and Ios instead of Ibc and Ibd.

Tim

[Kirill V.]

Great thanks, Tim!
If I understand you correctly, you're suggesting that I consider all the details in this analysis. This is interesting but initially I wanted to show the derivation of the formula from the datasheet.
I based it on the classic op-amp with bipolar input stage and which input currents direction into inputs. Not any internal bias cancellation.
Later I will try to edit all this and add analysis with unequal internal resistances of signal sources.