Hello,

I have said it before and I will say it again: I'm not an analog guy, but now I'm trying to get into op-amps so here we go.

I have a circuit and I want to determine it's accuracy. It is a current limiter, and I will do a simplified schematic. What I want to do is show you how to get the total accuracy based on the worst case scenario analysis. I'm learning about this too, so if I got something wrong, feel free to point it out.

**At a glance:**

The uses a negative feedback loop, so the only critical thing there is a single feedback resistor used for current sensing. I have a voltage reference, with two low tolerance resistors for dividing the voltage and the op amp, of course. I don't know if I should get in the expense of getting a chopper amplifier, which has almost no input offset voltage.

Every title in bold is a calculation. Also it might have a small explanation of what it is. As you are doing the worst case scenario, you have to calculate both the lowest possible value and the highest one, and from there calculate the lowest and highest value of the next block, taking in count the values you go on the first block. Then get the worst out of the results and calculate the next one the same way.

**Error in the input offset voltage:**

Input offset voltage is an error any real op amp has. When they are used in a feedback loop, op amps will adjust their output until the two inputs have what they believe is the same voltage. But there is an error.

If the offset is 10mV (a lot), one of the pins could be 10mV higher than the other. You can confirm this by tying the output to the inverting input and the non-inverting input to a voltage in the middle of your supply voltage. If you have a dual supply, that is GND; if you have a single supply, get two resistors of the same value and connect the non-inverting input in the middle, and both of the ends to Vcc and Gnd.

I did a quick approximation and having a 1mV offset with a 1V reference gives me 0.1% of accuracy in the output of the circuit.

`Offset = 1mV`

Vref = 1V <----This has 2% of accuracy in the reference itself and also we have to take in count the resistor tolerance, but this is just a quick approximation to get the idea of how much offset contributes to the lack of accuracy.

Percent of error=?

Error/100=Offset/Voltage

Error/100=0.001/1

Error=0.001*100

Error=0.1

Error is exactly one thousandth of the voltage reference, just like a millivolt is one thousandth of a volt. If the reference was 2.5, then the offset would be even less significant. I am using a volt because I'm working on a 5 volt system and if I increase the input voltage I would have to change the feedback loop in a way that is not convenient for my circuit or I would have to add some gain to compensate the change. That means even more calculations and possibly a trim pot (I won't be adjusting 5 pots per board in a 100 board batch and I won't pay anyone to do it).

Now I just have to figure out the rest. I will consult a book I have on the subject.

*Working on it, please don't give me the answers, I want to figure it out myself*

*Ivan*