Why this opamp has offset at input? - Page 1
Forum section: electronics / beginners

author: zulunation
author_rank: Contributor
timestamp: June 26, 2023, 06:49:02 pm
content: Hi ALL,I have decided to study operational amplifiers.I have AD8031 connected as a non inverted amplifier.The schematics is attached.I found that when 1K and 1 uF are disconnected I see a voltage offset at non inverting input.And amplifier has a constant voltage at the output which is logical.Why free non inverted pin has a voltage offset (about 500mV)?When I connect it through the 1M resistor to the ground it still has offset.But when I connect it through 1K resistor the input has zero.I measure the voltage with oscilloscope (1Mohm 15pf).

author: cristoper
author_rank: Contributor
timestamp: June 26, 2023, 06:56:58 pm
content: Looking at its datasheet, the AD8031 has a high input bias current (.45uA - 2uA). If it is 0.5uA with a 1M resistor at the input, that would produce an offset of 500mV.

author: wraper
author_rank: Unknown
timestamp: June 26, 2023, 07:11:22 pm
content: Also you need to be aware that if you order parts from dodgy sources like ebay and Ali, they may be not what they seem. Although in this case it's a high speed opamp with its own downsides such as high bias and offset current, and lack of such would be a big suspect.

author: magic
author_rank: Super Contributor
timestamp: June 26, 2023, 10:18:51 pm
content: Offset voltage is the difference between IN+ and IN-. I'm not sure if this is what you meant.You probably found that IN+ is simply at +500mV when 1k is disconnected. The reason is 500nA bias current flowing out of the input pin and through the oscilloscope input (1MΩ) to ground. Connecting 1MΩ resistor and the scope should decrease this voltage by half (250mV). With 1kΩ resistor, it falls down to fairly negligible 0.5mV.If you want to know the true open circuit voltage at this pin, setup the opamp for unity gain and measure its output. With this chip it should be ~1V below the positive supply, or +2.7V.

author: Terry Bites
author_rank: Super Contributor
timestamp: June 27, 2023, 03:37:32 am
content: Without a bias path to ground the voltage at the ninv pin is indeterminate. Technically, we say the input node its flappin around in the breeze. The output voltage of the amp cannot be defined from the circuit parameters.All opamp input must have a dc path to ground. Opamp forgotten golden Rule 0.0

author: magic
author_rank: Super Contributor
timestamp: June 27, 2023, 07:13:33 am
content: Quote from: Terry Bites on June 27, 2023, 03:37:32 am"Indeterminate" is a concept which only exists in theory.In real world, where the OP appears to be working, everything determines itselfsomehowand it's only a question of knowing how exactly.

author: Terry Bites
author_rank: Super Contributor
timestamp: June 27, 2023, 07:53:51 am
content: Saved from hopeless ignorance - thank you.

author: Vovk_Z
author_rank: Super Contributor
timestamp: June 28, 2023, 03:48:04 pm
content: So, what is a right term: 'somehow determinated'?

author: TimFox
author_rank: Super Contributor
timestamp: June 28, 2023, 03:53:43 pm
content: Quote from: magic on June 27, 2023, 07:13:33 am"Indeterminate" is the mathematical term applied, for example, to 0/0.In plain English, it just means that the value cannot be determined in advance, just like leaving an unknown bias current into a very high resistance giving you some voltage that you can't predict.

author: zulunation
author_rank: Contributor
timestamp: June 28, 2023, 06:53:16 pm
content: Thanks to all for replies.I have played with opamp yesterday.This bias current is a significant thing.If we connect something to opamp the resistance of it should be as low as possible.Otherwise the bias current will create a voltage drop on the input pin and this voltage will be amplifiedand we will have a bit offset at the output of the amplifier.That will reduce the amplitude of the output signal.

author: magic
author_rank: Super Contributor
timestamp: June 29, 2023, 12:13:53 am
content: Quote from: TimFox on June 28, 2023, 03:53:43 pmThat's why I said the concept only exists in theory.Regarding this IC, we can absolutely predict its open circuit input voltage in advance, using information about bias current and its variation with common mode input voltage provided by the datasheet. Namely, it's the exact voltage where input bias current is zero, which happens to be ~1V below the positive railQuote from: zulunation on June 28, 2023, 06:53:16 pmAnother option is to exploit the fact that bias currents of both inputs are approximately equal and ensure that both inputs see equal external series resistance. Then the two voltage errors are equal and cancel out.A small difference between bias currents exists, though, which is specified as "input offset current" in datasheets. So cancellation isn't 100% accurate, but it is often better than nothing.

author: TimFox
author_rank: Super Contributor
timestamp: June 29, 2023, 02:49:57 am
content: The difference between the two bias currents is the "offset current", as you said.For bipolar opamps, it is typically 10% of the bias current, and is indeterminate within that range.However, for FET opamps, it is often a higher fraction of the bias current, since the bias current starts out low.