I am trying to estimate the exact tolerance of a simple circuit, consisting of an op-amp based, inverting comparator with hysteresis.

This is the schematic (electronic-tutorials.ws)

enter image description here

For which I have a given Vref derived from a buffered voltage divider powered from the same +Vcc of the op-amp and all the resistors are +-1% of tolerance. Now, I want to calculate the tolerances on the threshold voltages.

Let's say the voltage divider is made by Rup and Rdown, my Vref is: $$ V_{ref} = V_{cc} * \frac{R_{down}}{R_{up}+R_{down}} $$

The positive and negative threshold will be:

$$ V_{highToLow} = V_{oh}*\frac{R_1}{R_1+R_2} + V_{ref}*\frac{R_2}{R_1+R_2} $$ $$ V_{LowToHigh} = V_{ref}*\frac{R_2}{R_1+R_2} $$

Assuming that the equations are correct, I thought to use the worst cases. For example, taking the Vref at its higher value, with R2 at higher value and R1 at lower one, so to increase the ratio and the multiplication. Viceversa when estimating the lower bound of the overall equation I will do the same with the lower values or the combination that lowers the ratios. The strange (or cool?) thing is that I have two ratios, one with R1 on the top and the other with R2, and they seems to compensate a bit each other.

Assuming that the flow until now is correct, now the last (or maybe the first) thing to do is how to estimate the Voh. I think to add in this estimation, also the 1% of the Vcc (which is nominal 5V). So, trying to explain better, from the following picture:

enter image description here

The Voh would be ((4.965-4.875)/2)+4.875 = 4.92V with tolerance of [+-%]: $$\frac{\frac{4.965-4.875}{2}}{4.92}*100 = 0.9$$

But since the power supply is already at 1% of tolerance, I would consider a total error on the output saturation voltage of around +-(0.9+1) = +-1.9 => +- 2%.

And in this case, just reiterate what I have already done with the previous equations, but also considering the worst case of the Voh.

I'd like to have an opinion on this approach and if there are some things that I can neglect.

  • \$\begingroup\$ You do realize that the circuit has a positive feedback, not negative feedback, right? \$\endgroup\$ – dannyf Jul 27 '17 at 12:09
  • \$\begingroup\$ I don't get the point, it is a comparator, it needs a positive feedback. \$\endgroup\$ – thexeno Jul 27 '17 at 13:22
  • \$\begingroup\$ Yes, the circuit is correct. For your worst-case analysis - you want to know the extremes of the comparison voltages - you need to calculate the sensitivity of the threshold equations to each of the variables. To do that, you must run partial differentiations with respect to the considered term, \$V_{ref}\$, \$V_{out}\$, \$R_1\$ and \$R_2\$. Once you have all these coefficients, they will tell you by their sign and weight how they affect the threshold. Now "push" all components to their limits according to their data-sheet and plug them in the equation featuring the partial coefficients. \$\endgroup\$ – Verbal Kint Jul 27 '17 at 16:02
  • \$\begingroup\$ Yes, that makes sense. Just to satisfy all my curiosity: is there a more "ingegneristic" fast way of knowing a priori the order of magnitude? Like "Vcc 1%, resistors 1%, total 2%"? \$\endgroup\$ – thexeno Jul 27 '17 at 18:52

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