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.
Best Answer
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.
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
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:
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
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
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 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).