Hello ,
I have build an high side differential amplifier using OPA991.
In this schematic i have used an offset about 2.5V.
My expectation was that if the current is 12A (1 Ohm load at 12V)the voltage at output of the opamp (pin 1 ) the voltage should be 12mV x 14(amplif|) + 2.5V = 2.668V
Instead of this I have measured a voltage of 2.660V.
Resistors are smd resistors that have 1% tolerance.
Could be this the cause of difference in measurement ?
Thank you
Hey Marian,
Welcome to e2e! :)
My expectation was that if the current is 12A (1 Ohm load at 12V)the voltage at output of the opamp (pin 1 ) the voltage should be 12mV x 14(amplif|) + 2.5V = 2.668V
This assumption is correct when dealing with an ideal op amp, there are non-idealities of the op amp that contribute to error. We have series on this on ti.com: https://www.ti.com/video/series/precision-labs/ti-precision-labs-op-amps.html
If you account for the offset voltage and the input bias current along with PSRR, CMRR, etc. the output of the op amp will not match exactly what ideal calculations lead you to.
The simulation models available online have all these parameters included with typical characteristics.
I ran a monte carlo sim changing the resistor tolerances to 1% and saw that an expected value can be between 2.44V and 2.9V. Therefore the measured value of 2.66V is accurate.
To make the design more precise you may choose to:
In order to facilitate your design process, I have attached the sim I used. MC_Diff.zip
Please let us know if you have additional questions.
All the best,
Caro
Hey Marian,
The effect of the input voltage offset in my circuit if is negative can lead to a negative output voltage(considering without 2.5V reference , but connected to GND) ?
There are input and output limitations, the output limitations for the OPAx991-Q1 is: (V–) + 0.1 V < VO < (V+) – 0.1 V
Considering a 12V single supply rail: (0) + 0.1 V < VO < (12) – 0.1 V = 0.1 V < VO < 11.9V.
Here is the link for the respective training video.
Therefore the output will never be negative in a single supply configuration, it is more likely the output will saturate around 50mV (considering the 10kohm load).
All the best,
Carolina
Hey Marian,
Yes, the input offset voltage is introduced due to the mismatch of the input transistors and therefore is plus or minus (not including temperature drift, common mode, and power supply).
Therefore:
But if the 2.5V reference is placed like in the schematic the output offset can be 2.5V- Voffset ?
This can occur in the following way:
12mV x 14(amplif|) + 2.5V = 2.668V
Except the equation is more like:
(12mV ± offset) X 14 (gain) + 2.5V (reference voltage) - this doesn't account for resistor variance.
Similarly, the statement below is also possible.
Or if the 2.5V reference is not placed and the resistor R417 is connected to GND, and the supply is V+=12V V-=12V (instead +12V and GND) The output can be Vout _Voffset ?
Here is a slide that summarizes all offset voltage that may be introduced:
There are higher precision op amps that have low offset voltage, temperature drift, good CMRR/PSRR. Please let me know if you would like recommendations for those.
All that being said, the op amp models for general purpose take all these into account at room temperature in typical values. A simulation is a good starting point but in the end for the evaluation to be thorough I recommend statistically significant testing on the final solution.
All the best,
Carolina
