If I have two 1k resistors rated at 1% fed by a perfect 10V source then what is the otuput guaranteed to be? 5V ±1% or 5V ±2% or some other value?

I can work out this case easily, assume that R1 is at the minimum, R2 is at the maximum, then the output is 4.95V; for the reverse case it is 5.05V, which is ±1%.

But is there a general rule for differing values? What about differing tolerances - what if one resistor were ±0.1% and one ±1%. While you can work it out by plugging values into the voltage divider formula, I'm looking for some general rule of thumb.

  • \$\begingroup\$ There isn't a rule of thumb that I know of. I think there would have to be a different "rule" for every different type of circuit combination. What circumstance are you needing to actually have a rule that you can't just very easily compute? \$\endgroup\$
    – Kellenjb
    Commented Oct 27, 2010 at 2:06

4 Answers 4


I think you asking for sensitivity analysis of f(x,y) = x/(x+y). Since there are two variables, I do a general analysis first, then look at the dependence on each variable separately. Since you might not care for the algebra, I tried to sum up each case after the bold heading.

The bad news is that the tolerance on the measured voltage can be very far from the predicted voltage, 1000000% or more relative error. The good news is that in reasonable cases it can only be a tad more than twice as bad as the resistor tolerances, and often you can do much better.

If X is supposed to be x, then the signed relative error is (X-x)/x = dx and X = x*(1+dx). If dx is 1% = 0.01, then X is x*101%, and if dx is -1%, then X is x*99%. In other words, we care about X = x*(1+dx).

If X is the resistance of R1, and Y is the resistance of R2, with R1 and R2 in series connected +10V to ground, then the measured voltage with one probe between R1 and R2 and the other probe at ground is f(X,Y) = X/(X+Y), but it was supposed to be f(x,y) = x/(x+y).

If x changes to X=x*(1+dx) and y changes to Y=y*(1+dy) then f(x,y) changes to f(X,Y):

x*(1+dx)/( x*(1+dx) + y*(1+dy) )

The relative error is:

E(x,y,dx,dy) = | f(x,y) - f(X,Y) | / f(x,y) = ( x/(x+y) - x*(1+dx)/( x*(1+dx) + y*(1+dy) ) ) / ( x/(x+y) )

which simplifies to the:

Exact relative error

E(x,y,dx,dy) = y*|dy-dx| / ( X + Y )

This formula is not too bad to plug values into, and it is not too bad to analyze in specific cases.

Quick symmetric bound

Assuming |dx| and |dy| are bounded by 0 ≤ e < 100%, this is bounded by:

E ≤ 2*y*e / ((x+y)*(1-e)) = y/(x+y) * 2 * ( e + e*e + e*e*e + … )

For instance when x = R1 = 1K and y = R2 = 1K and dx = 1% = 0.01 and dy = -1% = -0.01 you get the relative error E = 1% = 0.01. The bound I gave is a little loose since it predicts 1.0101...% but probably this is not too big a deal.

Large R1

When R1 is very large compared to R2, then the relative error goes down considerably.

If x → ∞, then E(x,y,dx,dy) → 0.

It goes to zero about as fast 1/x: x*E(x,y,dx,dy) → y*(dx-dy)/(1+dx).

This is not too surprising: if you have an open circuit with +10V attached to your probe and the other other probe attached to R2 attached to ground, then the current is 0 and both terminals of R2 remain at +0V, so you measure +10V no matter what the value of R2.

Large R2

When R2 is very large compared to R1, then the relative error can be very large, but for reasonable tolerances on R1 and R2, E is only a little more than twice as bad.

If y → ∞, then r(x,y,dx,dy) → |dx-dy|/(1+dx).

If dy = -dx = 0.10 = 10%, then you get 22% error (a little more tan twice as bad).

If dy = -dx = 0.50 = 50%, then r = 2 = 200% relative error (four times as bad).

As dx → 1 = 100%, r → ∞ (infinitely worse).

If |dx| and |dy| are bounded by e < 1 = 100%, then for large y, r is bounded by 2e/(1-e), which is a little bigger than twice as big as e.

If dy = -dx = 0.01 = 1%, then you get E = 2*1%/(99%) = 2.0202…% relative error on the measured voltage (a little more than twice as bad).

If dy = -dx = 0.001 = 0.1%, then you get 2*0.1%/(99.9%) = 0.2002002…% relative error (a very little more than twice as bad).

Asymmetric tolerances

If R2 is very accurate and R1 ≈ R2, then the error is at most a little more than half as bad.

If dy = 0, then E = dx*y / ( X + y ) and if x = y, then E = dx/(2+dx).

If -dx = 0.01%, then E = 0.01 / 1.99 = 0.005025… = 0.5025…% (a little more than half as bad).

  • 1
    \$\begingroup\$ Bringing out the statistics, very nice. \$\endgroup\$
    – Kellenjb
    Commented Oct 27, 2010 at 3:56
  • 1
    \$\begingroup\$ +1 for statistics - love this stuff. (Took a GCSE in it.) \$\endgroup\$
    – Thomas O
    Commented Oct 27, 2010 at 21:33
  • \$\begingroup\$ Very nice! I use this sort of stuff all the time. \$\endgroup\$
    – Jason S
    Commented May 18, 2011 at 1:22

Any two tolerances differ by a factor of at least 5 (as in 5% vs. 1%), so as a rough approximation, you can ignore the effect of the smaller one. This means that at the edge of the tolerance range, the voltage at the middle of the divider will be off by a percentage equal to roughly half of the larger tolerance.

Note that this is only true if the two resistors have similar magnitudes, as in your 1k example. If the divider consists of a 1k and a 100k resistor, error in the larger resistor could dwarf the smaller resistor.

  • \$\begingroup\$ Wouldn't the error of the lower-value resistor have greater effect on current flow? Or am I just confusing myself? \$\endgroup\$
    – Jesse
    Commented Oct 27, 2010 at 3:36
  • \$\begingroup\$ No, I don't think the low-value resistor matters much. If you imagine a pipe with two constrictions, the flow is dominated by the tighter one. Another way to think of it is to imagine the error as a resistor being added in series. 5% of 100k is 5k, while 5% of 1k is only 50 ohms. Imagine adding each of those in series, and think about what would happen. \$\endgroup\$
    – pingswept
    Commented Oct 27, 2010 at 5:12
  • \$\begingroup\$ Good point, well explained. \$\endgroup\$
    – Jesse
    Commented Oct 27, 2010 at 16:40

If you have a bunch of components, all with different tolerances, you do a Monte Carlo analysis to see what kind of risk the worst case is, but also what a probable case is, since it's not likely that every component is at it's worst. This gets as detailed as you want to make it.


no rule of thumb for different tolerances.

Assume R2 between Vout and GND.

Your math above only holds for R1=R2. That is When R1=R2 both with equal tolerance your output has tolerance equal to the resistor tolerance. When R1 != R2 and the tolerances of each resistor are equal the total error logarithmically approaches 200% of the tolerance of an indicidual resistor as the ratio of R1 to R2 increases. If R1 is 100 times bigger than R2 then the output tolerance is roughly twice the tolerance of a single resistor. I'll leave the inverse up to you for calculation.

Now if the output tolerances are not equal, you've added an additional variable into the equation. You'd be looking at a multidimensional equation and attempting to locate maxima and minima to determine Vout tolerance, which is not trivial. So use the same tolerance for both resistors.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.