This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

on high precision operation amplifier

Other Parts Discussed in Thread: LM358, LMC6482, INA198

Hi guys

I am presently using LM358 N( off set voltage of 2mv and offset current of 30nA )which used as non inverting amplifier(with gain of 10) to measure the very low voltage. but  I am not  getting what I  expect from the  output of lm358IC

this was the result :

firstly ...

result of small circuit  done it in bread board ..

case 1> when input voltage is 0V . and getting output voltage  more than >30MV.

case 2> when input voltage is 0.2v and getting output voltage more than  2.5 V.

I really don't know why  LM358 showing this much of inaccuracy ...
please help on this  ,what may be the root cause

secondly :

please suggest me good  operational amplifier with high precision off-set voltage,current .

Thanks&Regards

Sunil

  • Hi Sunil!

    The LM358 isn't a good choice any more. You can do basic stuff with it, but there are thousands of better opamps.
    Anyway - what is your supply voltage for the opamp? Do you use a single supply - +5V/GND for example? Note that the output cannot go down to zero volts when the input is zero volts. The output voltage swing is limited to a few mV above the negative supply. And the offset voltage is multiplied by your gain, too - this could lead to your higher output voltage. The offset is something between 2 to 7mV. Maybe you could provide a schematic of your application? Especially when reading about your case 2, there seems to be something wrong. An input of 200mV should not lead to an output of 2.5V (at least if everything is connected correctly). So please draw a schematic for that.

    Here you can look for better opamps:
    www.ti.com/.../precision-amplifier-products.page

    But only you know about the specifications you need in your application.

    Dennis
  • Thanks for your answer .. I design simple circuit from which lm358 used as non -inverting amplifier with a gain of 10 ( using two resistor rf =18k ohm and rs =2k ohm , gain =( 1+rf/rs) ). supply used is a single supply (+3.6V and GND) . what is the exact formula to calculate output voltage of op-amp in practical condition (consider off -set voltage and bias current other op-amp parameter) with respect to input voltage and gain


    thanks in advance
  • Your gain calculation is - in general - correct. But what kind of resistors are you using? Are they exactly 18k and 2k? If they were normal precision types then they would fit into the E24 series which has a tolerance of 5%. Let's start from that:

    The worst case is when your 18k has a maximum positive tolerance and your 2k has a maximum negative tolerance, so

    • 18k + 5% = 18k + 900R = 18k9
    • 2k - 5% = 2k - 100R = 1k9

    This would lead to a gain of

    • A = [(18k9 / 1k9) + 1] = 10.95

    If your input is 0.2V (is it exactly that?), then this would lead to an output of

    • 0.2V * A = 0.2V * 10.95 = 2.19V

    Plus the worst case offset of your opamp multiplied by A

    • 2.19V + (10.95 * 7mV) = 2.19V + 0.077V = 2.267V

    So your 2.5V is still a little bit to high, at least from the mathematical point of view. And in almost no case you will have all components at their maximum tolerances - possible, but unusual. You could add bias currents now, but...no, not willing to do that.

    Of course there are models on how to calculate the output including all possible errors, but this is quite complex. Don't care for that if not really needed. I have no example for that now.

    Keep in mind that this opamp is normally specified for an output of (Vcc - 1.5V) @ 2k output load. So when powering it from 3.6V, it's maximum output is 2.1V (minimum). You are operating it in a region that is already critical for expecting linear results.

  • Thanks for your answer ..To calculate the out put voltage of OP -AMP .your considering only off -set voltage of op-amp .. what about off-set current and bias current ..LM358(off -set current is 75nA and bias current is 200nA) .. calculating out put voltage of op -amp ..its not necessary to considering the bias and off set current ? please help me on this ...
  • Yes of course it is, but it gets more complicated than necessary :-) Do you need that precise calculation? Considering all aspects that affect the output voltage isn't that trivial. What is the goal you want to achieve?
  • Another reason for the output of about 2.5V could be a wrong connection making the opamp act as a comparator setting it's output to it's maximum positive output voltage. Just a suggestion. In that case there would be an issue with your feedback.
  • Thanks a lot your help ..I am measuring the current consumed by the particular device .. I am measuring the voltage drop across the small resistor(0.1OHM) around(36micro volts) and I want amplify that voltage which is necessary for my application ... precision is very important for me ..if I get what factor which affect the when I am doing amplify with lm358 so that I can do calibration ..precision is very important for me ...my main question is what and all factor which is consider when amplify the input out voltage from op-amp ...what is formula to calculate the output op-amp under practical condition including all parameter (like off-set voltage and current ,bias current ) .
  • sunil shetty said:
    d(36micro volts) and I want amplify that voltage which is necessary for my application ... precision is very important for me ..if I get what factor which affect the when I am doing amplify with lm358 so that I can do calibration .

     Hi Lm350 is a good old day poplar cheap opamp can do a lot of jobs than precision or rail to rail operation, your application is at almost rail to rail input due to resistor on power rail and output too due to near zero value.

     I suggest an operational like LMC6482 its a dual but you need amplify a lot so don't use one channel but two.

     Another solution are INA198 current shunt amplifier solve a lot of problem involved with differential stage.

    sunil shetty said:
    ..precision is very important for me ...m

     these word sound worst than a non sense, what is precision when no specification is present? This phrase is itself imprecise !!!!

    sunil shetty said:
    my main question is what and all factor which is consider when amplify the input out voltage from op-amp ...what is formula to calculate the output op-amp under practical condition including all parameter (like off-set voltage and current ,bias current ) .

     A good basic opamp theory course can address all these problem, try ordinary or EDX one, basic knowledge is necessary to understand math and electronic basics.

     Also if frontend of, this is OT from MSP430 too.

**Attention** This is a public forum