Help Me understand Input/Output offset voltage. - Page 1
Forum section: electronics / beginners

author: sansan
author_rank: Contributor
timestamp: June 06, 2022, 12:19:13 pm
content: ++++ First of all, I've no electronics degree, just a hobbyist with curiosity. ++++Guys, help me understand some of the term below.• Which number is "input offset voltage"?• Which number is "input bias current"?I believe each voltage (blue text) is referred to Gnd.

author: Berni
author_rank: Super Contributor
timestamp: June 06, 2022, 12:33:19 pm
content: The output offset voltage is just the input offset voltage multiplied by the gain set by the two resistors.So the term output offset voltage is rarely used. Any offset in the output would have been compensated for by the opamps feedback. So in general we only talk about input offset voltage with opamps.The reason the output offset voltage is useful is when you are trying to measure the input offset voltage. Say for example you want to measure a really good opamp that has only 1uV of input offset voltage. Measuring such a low voltage accurately can be difficult even if you have a really nice multimeter. But you can construct a amplifier with a gain 1000 using this opamp (so the opamp amplifies its own offset voltage). This makes the output offset 1000x larger than the input offset, so you can then for example measure 1.34mV of output offset (easy to measure) then divide it by 1000 to get the 1.34uV of input offset.This is not always fully accurate tho since there are other effects such as input bias currents, open loop gain...etcEDIT: For example in your case you can help reduce the error from input bias currents by making the 18K resistor to ground be the same as the impedance on the other leg of the opamp (18K parallel with 2K so 1.8K)

author: sansan
author_rank: Contributor
timestamp: June 06, 2022, 01:00:17 pm
content: Quote from: Berni on June 06, 2022, 12:33:19 pmthanks for your input Bernie, but I don't think it has the answer to my question (& regarding to the number in the circuit).I'm more interested in the process, not only in the measurement.Is my understanding correct:"to measure the input offset voltage, put the DMM probe across opamp input pin?"I more like:"What is happening there?Why is the DCo is -27.5mV, and not -13.2mV?that is like 100% more than I expect"

author: TimFox
author_rank: Super Contributor
timestamp: June 06, 2022, 01:26:49 pm
content: The best way to think about input offset voltage for a high-gain op-amp is this:The input offset voltage is the voltage, when applied between the input pins, that results in zero output voltage.Since op-amp DC gains can be 105or more, the practical meaning is that the output voltage in this test is within its useful range (not saturated near either supply voltage).

author: Berni
author_rank: Super Contributor
timestamp: June 06, 2022, 05:30:27 pm
content: Quote from: sansan on June 06, 2022, 01:00:17 pmThose equations work for ideal opamps (Also not sure about the way that equation attempts to compensate for the resistor values).Real opamps have more complex behavior than that. The input bias currents can vary depending on operating conditions, the open loop gain can vary depending on how loaded the output is etc... The LT Spice models of opamps can be very sophisticated and model a lot of this. If you are just exploring opamp theory i would recommend instead using the purely parametric opamp model (found at the end of the list under opamps in LTspice) that does not try to mimic any of the real life imperfections of actual production partnumbers.In your particular case you are likely getting mostly thrown off by feeding your opamp a fairly high mismatches impedance on the input(As mentioned in my previous post). The positive input is driven by 18K while the negative input is driven by a 1.8K impedance. At the 100s of nanoamp bias currents this causes extra voltage drops in the milivolt ranges, those extra milivolts in the resistors end up amplified on the output too. This is the reason why it is a good idea to feed both inputs of a opamp with the same impedance (generally opamps are designed to have a reasonably closely matched bias current on both inputs).

author: Peabody
author_rank: Super Contributor
timestamp: June 07, 2022, 04:43:03 am
content: My assumption would be that a Spice model would by default have zero input offset voltage. If you wanted to simulate a specific input offset voltage, you would insert a voltage source in series into the non-inverting input.

author: sansan
author_rank: Contributor
timestamp: June 07, 2022, 11:02:13 am
content: Quote from: Berni on June 06, 2022, 05:30:27 pmSo I add R4 to make the impedance at both input appear "similar".DCO decreased by a lot.I think I understand what's going on in the 1st circuit (DC wise),may be.What do you call for the -4mv at +in and -3mV at -in?DC bias voltage?? that voltage appear to be relatively the same (compared to the 2nd Circuit).On the previous circuit, it seem like R3 is sitting at the inverting bias voltage (-3.07mV). In which making it pulling current of-3.07mV/2k = -1.537µA.(close enough)since -in bias current is relatively the same (-178nA), the rest(current) must came from the output through R2.-1.537µA - -178nA = -1.359 µA.(should it anadditionorsubtraction, I don't quite understand why/how).As current through a resistor(R2), it generate voltage.-1.359µA * 18k = -24.462mVThen, added the -in bias voltage of -3.07mV,it become -27.532mV. (close enough).I don't quite know where that small difference should came from (-27.532mV vs DCo -27.5394mV).If I change R4 to 15k49(R1 vs R4+(R2|R3) = 18k vs 17k29), the DCo become only 0.75µV.I don't understand why (yet).I mean, 18k and 17k29 is not quite "matched". Is it because of the difference between +in & -in bias current (-226nA vs -178nA)?

author: Peabody
author_rank: Super Contributor
timestamp: June 07, 2022, 02:08:39 pm
content: Could the series resistance in the supply rails be having any effect? Do you get different results if you change them to zero?

author: sansan
author_rank: Contributor
timestamp: June 10, 2022, 12:33:11 pm
content: Quote from: Peabody on June 07, 2022, 02:08:39 pmit,s not affect much, only change in tenth of femto volt DCO.

author: sansan
author_rank: Contributor
timestamp: June 10, 2022, 01:02:04 pm
content: So IdoubleR1, R2 & R3 size.Input Bias Current remain the "same", 226nA & 178nA. So the input DC bias voltage goes double (-8.14mV & -7.14mV)?Although the input DC bias voltage doubling, input offset voltage is remain the same, about 1mV (-8.14mV - -7.14mV).even if I change R4 to 1k, input offset voltage only shift about 2µV (but DCo goes wacky,, -63mV).It seems like the input bias current is more of "dominant role", and the resistor should follow (tuned) accordingly to get lowest DCo.

author: Berni
author_rank: Super Contributor
timestamp: June 11, 2022, 06:49:03 am
content: Quote from: sansan on June 10, 2022, 01:02:04 pmYes these are the non ideal behaviors of a LM1875 versus a fictional ideal opamp (that most of the math formulas are made for). LTSpice tries to emulate the crappy aspects of a real opamp.All of this is why being able to properly read a datasheet becomes particularly important when it comes to getting high performance out of opamps. What you have there is a "LM1875 20W Audio Power Amplifier" and it makes a pretty shitty opamp for precision analog applications since acording to the datasheet it has a lot of input offset, the input offset is not stable, there is a lot of input bias current, and the bias current is not tightly matched between the inputs. It doesn't need any of this because it is designed to be a power amplifier.If you do need the powerful output drive capability of that opamp and have precision at the same time, then you should use this opamp as simply a voltage follower power output stage, then drive this using an actual precision oapmp with good DC specs in the datasheet.

author: sansan
author_rank: Contributor
timestamp: June 11, 2022, 09:20:54 am
content: Quote from: Berni on June 11, 2022, 06:49:03 amSometimes I wondering using this IC for headphone amp, but DC offset is....must use DC blocking cap at R3 I suppose.Quote from: Berni on June 11, 2022, 06:49:03 amDatasheet says "minimum gain is 10x for it to stay stable".voltage follower is a buffer(no voltage gain) right?

author: Berni
author_rank: Super Contributor
timestamp: June 11, 2022, 10:30:06 am
content: Quote from: sansan on June 11, 2022, 09:20:54 amYep that is a common way to AC couple audio amplifiers without needing a large DC blocking capacitor on the output. (It does not remove all offset tho, just makes it much smaller)Why do you even care about DC offset in a audio amplifier anyway? You can't hear DC, the offset is too small to hurt voice coils in any way and can easily be filtered away using a DC blocking capacitor. This opamp is designed for outputting lots of power at low harmonic distortion. It does a superb job at that according to the datasheet. Perfect for making a headphone amp.This is the reason why many thousands of different opamps are available on the market. Each one is designed to hit a different set of performance targets. Some are powerful, some are precise, some are fast, some are low power, some are low cost...etc So you can choose the kind of opamp that fits your design best. Sometimes the kind of performance you want is not available out of the box or the opamps that offer it don't fit the budget constraints. In those cases you have to design a way around the chips limitations using other tricks.Quote from: sansan on June 11, 2022, 09:20:54 amSorry i did miss that detail.Yes in this case you have to just give it some fake gain to keep it happy or add in some compensation circuitry around the feedback.

author: TimFox
author_rank: Super Contributor
timestamp: June 11, 2022, 01:41:06 pm
content: The "fake gain" Berni mentioned above to get stable feedback is usually referred to as "noise gain", which you should look up in order to get a stable unity-gain follower with an opamp not stable at unity gain.