Need your help offset buffer - Page 1
Forum section: electronics / projects

author: corso
author_rank: Newbie
timestamp: July 08, 2016, 06:59:19 pm
content: Hi all,I have a strange problem with an opamp in buffer mode. At the input I have 1.9889V and the output is 1.9969V, so 8mV offset.In the datasheet the MCP6V64 have +/-8uV offset?...I have try with MCP6004 and a AD8630, but same problem.I have try to remove the tension divisor at the input of the opamp (resistor 200k and 50k), the offset was just 150uV, this is perfect for what I need to do, but I need the tension divisor.Some one have a solution or something that I can try to resolve my problem.Thank you,

author: Andy Watson
author_rank: Super Contributor
timestamp: July 08, 2016, 07:05:58 pm
content: Are you using a meter with 10M ohm input impedance? Consider what effect that has on the resistive input divider.

author: corso
author_rank: Newbie
timestamp: July 08, 2016, 07:11:57 pm
content: Yes, I use a Fluke 45, the impedance is 10M ohm.

author: StillTrying
author_rank: Super Contributor
timestamp: July 08, 2016, 07:59:28 pm
content: Very roughly.You're measuring the same voltage at 2 very different impedances, ~0.5R at the opamp output and ~50K at the +Ve input.The Fluke's 10M means the measurement at the output will be low by 0.5/10M = 1 part in 20 million, so you can forget that error.But at the +Ve input the error is 50K/10M = low by 1 part in 200. As you're measuring ~2V, low by 1 part in 200 = 10mV, which is close enough to your -8mV for me!Measure directly between the +Ve input and the output instead of Gnd and the measured offset between them 2 points should be much less than 8mV.

author: corso
author_rank: Newbie
timestamp: July 08, 2016, 08:35:16 pm
content: StillTrying, you are right, I have 152 uV with the +Ve input and the output. Not exactly +/-8uV but it okay for my job...If I put the input to the ground (0V), I get 300uV at output of the opamp, so why not 8uV?

author: RandallMcRee
author_rank: Frequent Contributor
timestamp: July 08, 2016, 08:48:20 pm
content: Hmmm, seems like this is an error due to the input bias (offset) current of the opamp. You can compensate for it...http://www.daenotes.com/electronics/devices-circuits/operational-amplifier-bias-current-compensation-methodRandy

author: StillTrying
author_rank: Super Contributor
timestamp: July 08, 2016, 09:05:44 pm
content: Quote from: corso on July 08, 2016, 08:35:16 pmThe output saturation voltages come into it then. I can't see what it is at OV from the datasheet, possibly 3.5mV with a 20K load, but at 0.3mV your winning on that one. You could try a 20K from output to Gnd to keep it low, it'll depend on whether it has to sink any current from the ADC.

author: Someone
author_rank: Super Contributor
timestamp: July 08, 2016, 11:23:27 pm
content: Quote from: RandallMcRee on July 08, 2016, 08:48:20 pmNope, the max input bias current of 50pA wouldn't be causing enough offset for these measurements (similar in magnitude to the 8uV input offset). Measuring between the input and output should have mostly eliminated the impedance of the multimeter, as its now at almost the same potential as the sensitive input node just like controlling leakage with guard rings.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: July 09, 2016, 08:06:46 am
content: Hi group,I agree with members Andy Watson and StillTrying.Here is some analysis performed with LTspice. The SPICE primitive, E, is a voltage controlled voltage source, essentially an amplifier.First I will show the effect of having a large, but limited gain, in an op-amp.The general formula for a servo with feedback is:Gain = A / 1 + A*BWhere A is the open loop gain of the amplifierand B is the gain of the feedback pathIn the case of a buffer, B=1A= Open loop gain of the amplifier, normally considered very large, but does introduce an error in ppm accuracy circuits.If you assume the A=120dBA = 1E6Gain = 1E6 / (1+1E6) = 0.999999This introduces 1ppm errorIf the gain is 100dB, there is a 10ppm errorMeter LoadingThe 10M input resistance of the DMM will load the 200k, 50k divider:Thevenin EquivalentThe 10V source can be replace with a 2V and resistors of 200K and 50K in parallel (40k):This results in 8mV of error, exactly what the OP was measuring.I have attached the LTspice model.Regards,Jay_Diddy_B

author: StillTrying
author_rank: Super Contributor
timestamp: July 10, 2016, 12:35:07 am
content: I think I prefer the "Guesstimate it as you go" method.