Hi,everybody. The voltage of my signal is of mv level.this signal is the input of ADCs.I want to raise the input impdedance of the signal,then the signal will be isolated from the ADCs.
Simplely,An op-amp(OPA602) is used.
the signal---PIN3(IN+) PIN2(IN-)--PIN5(OUT)
but I found the result is so bad.Such as,when the signal is about
Who know the reason and better ways?
Best regards.
Thank you.John.
I used a voltage meter to measure the signal.
best regards.
How do you know the signal is 2.5 mV when it is connected to the opamp?
The OPA602 has a typical input offset voltage of 1mV
Your result is typical therefore.
You need an op-amp with better DC accuracy or you need to trim the OPA602 ( see pin 5 ).
Graham
1) You do not want to "raise the impedance of the *signal*. 2) You want the signal to drive a rather high impedance *load*. 3) The output pin of a standard opamp is pin 6; usually pin 5 is reserved for one end of a trimming pot. 4) At least you picked a JFET input opamp, which gives the highest input impedance. BUT the offset voltage can be as much as 0.25mV, and add that to (about) 1pA times the input resistance, and one may get a total offset near that 1mV you see. 1Kmeg times 1pA gives 1mV; warm the opamp up by 10C and that doubles. If the opamp draws a fair amount of power, it can warm itself up by quite a bit. So, a 200Meg input resistance and a fairly warm opamp could give that result. Add a trimpot...
My reference gives 0.25mV max at 25C.
He used his 1K/V VOM...
Almost all DVMs have a 10Meg input resistance, so the signal would have to have an equivalent internal resistance of less than 100K for a one percent error, less than 10K for a 10 percent error... Know your equipment...
Ooops ! I missed that. maybe the OP made a typo ?
That's the 'headline' figure on the data sheet.
Look at the detailed spec and it's 1mV typical for the A spec part.
3mV max too.
Indeed.
Graham
How do you know that the input voltage did not load down from 3.3 mV to 2.5 mV when you attached your meter? Did you measure the opamp output voltage while the meter was attached to the input?
Thank you.John.
No matter 3.3mV or 2.5mV,The two results are difference is the fact. I want to know the reason and how to solve this problem.
best regards.
My point is that if the input is actually at 3.3 mV (but is not being measured) when you are measuring the output at 3.3 mV but is loaded down to 2.5 mV by the meter only when you are measuring the input (and at that moment, the output also follows the loaded input down to 2.5 mV, but is not being measured at that instant), then there may not actually be any difference between input and output at any given instant.
Have you ruled out this possibility? Are you measuring both voltages simultaneously (or applying a dummy load to the input that matches the meter impedance, while you measure the output)?
If you have, then your opamp is not a very good example of the type. If your result is taking place in a simulation, then the simulated opamp is a worst case example, not a typical one.
If you really need a small fraction of a millivolt input accuracy under wide temperature range, you may have to change the opamp to a self zeroing (chopper) type. These can have DC errors down to less than .1 mV.
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