On the issue of accuracy

But there is a nuance…

Since my dear readers have overwhelmingly expressed a desire to read about the accuracy of resistive circuits, I have no choice and I proceed to the presentation. It should be noted that accuracy issues have historically been developed in the context of measurements, and we will consider them in reverse, although this will in no way affect the results.

First, let's introduce basic concepts – let's assume that we have some true (or exact) value of an arbitrary parameter (Po) (for example, the resistance of a resistor) and a practically measured value of this parameter (P). Then we can introduce the concept of “error,” which is numerically equal to the difference between these two values. If we take the absolute value of the obtained value, we get the “absolute error”:∆P=|P‑Po|.

The practical value of this indicator is somewhat questionable, since an error of 100 Ohms for a 10 MOhm resistor is clearly insignificant, but for a 1 kOhm resistor it is very significant. Therefore, an additional indicator, “relative error”, is introduced: δP=∆P/Po, which is more informative and easier to use due to the lack of dimension. Usually the relative error is indicated as a percentage (%), but parts per mile (%%) and even parts per million (ppm) and parts per billion (ppb) are also found. We will further deal with the first representation – as a percentage, although this is unimportant.
Note in the margin (Pnp): an attentive reader will ask what to do when the true value is equal to zero, because “you cannot divide by zero.” Well, first of all, you can… a necessary clarification for the stunned readers – you can divide, you cannot divide, but these are all jokes. But essentially, they believe that the true value in the denominator (and only in it) is equal to the measured one, and then we get δP=|P-0|/P=100% is a completely logical result.

So, there is an indicator that characterizes the accuracy of maintaining any parameter, in this case the resistance of the resistor. Typically we will be dealing with a value of 5% – the standard E48 series, but in practical applications 2% – the E96 series and even 0.1% – the E240 series are used.
Pnp: the number in the index determines the number of denominations in a row per decade, from where it is easy to obtain the relationship between adjacent denominations 10^(1/48)=1.049~1.05. I read somewhere that the rows of denominations were created in such a way that a resistor of any denomination could be placed in some cell of the cash register, this hypothesis looks plausible.

Since we will rarely use a resistor alone and will more often than not build a system from imprecise resistive components, it is necessary to know how the accuracy of a component will affect the accuracy of the system as a whole. Let's consider various functions from one argument and determine the accuracy of the function's calculation depending on the accuracy of the argument.

Let's first consider linear functions – adding a constant and multiplying by a constant.
Let's start with multiplication and solve the problem head-on. Let f(x)=kx, Then f(x+dx)=k*(x+dx) and relative error of the function value
q(k*x) =|k*xk*dx-k*x|/(k*x)=k*|dx|/(k*x)=|dx|/x=qx.
Pnp: hereinafter the symbol d will replace ∆, a symbol q – δ, nothing personal, it’s just more convenient for me. Conclusion – when multiplying a parameter by a constant (specified absolutely precisely), the relative accuracy is maintained.

Now the addition: f(x)=x+kThen
q(x+k)=|f(x+dx) – f(x)| /f(x) = |x+dx+k-(x+k)| / (x+k) =
|dx|/(x+k) = |dx|/x – |dx|/x*k/(x+k)=qx*(1-k/(x+k)).
Assuming that x and k are always of the same sign (for example, positive), the relative error will not increase, and sometimes even decrease. Pnp: yes, we are lucky, resistances are never negative, otherwise the addition operation would be prohibited from the point of view of assessing accuracy.

Let's move on to nonlinear functions and first consider the quadratic f(x)=x*x. Then q(x*x)=((x+dx)*(x+dx)-x*x)/(x*x) = (2*x*dx+dx*dx)/(x*x) =
2*dx/x+dx*dx/(x*x) = 2*qx+qx*qx.
Neglecting terms of the second order of smallness, we obtain q(x*x)=2*qx, that is, the error when squared increases by 2 times. We can carry out a similar solution for an arbitrary exponent k, but first we will simplify the process.

To do this, we use Newton’s formula and vice versaenim f(x+dx) to f(x)+f'(x)*dx, Then
qf(x) = (f(x)+f'(x)*dx-f(x))/f(x) = f'(x)*dx/f(x)=f'(x)*dx/x *x/f(x)=qx*f'(x)*x/f(x)
and we got a generalized formula. Let's check it on already solved problems
q(k*x) = qx*k*x/(k*x) = qx (+),
q(x*x) = qx*2*x*x/(x*x) = 2*qx (+),
q(x) = qx*1*x/x = qx (+).

Then q(x^k) = qx*k*x^(k-1)*x/(x^n) = k*qx.
Note that q(1/x) = |-1|*qx = qx.

Now it's easy to move on to two argument functions. Let's start with addition.
f(x+y) = qx*1*x/(x+y)+qy*1*y/(x+y)=qx*x/(x+y)+qy*y/(x+y),
that is, when adding, the relative errors are added with different coefficients, each of which is less than one. In an important special case, when qx=qywe get q(x+y) = qxthat is, the relative error of the sum is equal to the error of each term.

Next we will consider multiplication
f(x*y) = qx*y*x/(x*y)+qy*x*y/(x*y) = qx+qy,
which is consistent with the formula for a square.

The formula for division can be obtained in two ways – considering division as multiplication by the inverse of the denominator and then we get
q(x/y)=q(x*1/y)=q(x)+q(1/y)=qx+qy.
But you can use a generalized formula for training, then
q(x/y)=qx*1/y*x/(x/y)+qy*(x/y*y)*y/(x/y)=qx+qyetc.

And now slides You can move on to resistor circuits. Connecting resistors in series leads to a sum formula, which means the error remains unchanged (with the same error of the terms). Parallel connection is also not a problem if we remember that in this case the conductances, or units, are the reciprocal of the resistance. q(1/(1/x+1/y))=qx. This means that in this case the error remains unchanged. Pnp: proof of error for a function x*y/(x+y) We'll leave it to the inquisitive reader.

Let us now consider a very commonly used circuit, namely a resistive divider, which is described by the formula x/(x+y). From a formal point of view, we should get an error 2 times greater than the error of the resistors, since q is at the top and q is at the bottom. But let’s carry out the analysis using a generalized formula, then:
qf = q(x/(x+y)) =
qx*(1*((x+y)-1*x)/((x+y)*(x+y)))*x/(x/(x+y)) + qy*((0-1x)/((x+y)*(x+y)))y/((x+y)*(x+y)) = . qx*y/(x+y)^2*x/x*(x+y)+qy*-x/(x+y)^2*y/x*(x+y) =
qx*y/(x+y)-qy*y/(x+y)=(qx+qy)*y/(x+y).
And for a special case qx=qy we get qf = 2q*y/(x+y).
Since x and y are positive and not zero, the resulting division factor error will always be less than twice the resistor error. If you enter the parameter “resistor ratio” k, equal to x/y, the transfer coefficient of the divider will be equal to 1/(1+k) with an error 2*qx*k/(k+1) and, for example, with identical resistors (k=1), the error will be equal to the error of each resistor. It is interesting that for k<1 we get an error in the transmission coefficient less than the error of one resistor, for example for k=2/3 we have qf=2*qx*2/3/(1+2/3)=qx*2*2 /3*3/5=qx*4/5, which is clearly less than qx. Personally, I am slightly surprised when the result of a function is more accurate than its arguments, but this picture is quite real. For example, the error of the function f(x)=sqrt(x) is two times less than the error of its argument.

Unfortunately, with an increase in the ratio of resistors and, accordingly, a decrease in the transmission coefficient, the error in setting this parameter tends to the sum of the resistor errors and already at k=9 practically does not differ from this sum qf=0.9*2*qx=1.8*qx. The ingenious solution to make a partitioned divider does not work, because the final error will be the sum of the errors of each section and the gain becomes ephemeral, and, most likely, will turn into a loss, although the last statement requires proof.