Electronic – the best way to convert a 10V power supply into a 5V power supply

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).

Here is one way of understanding the problem and thus arriving at the solutions you seek:

1. You have a voltage V applied across a "black box", consisting of a series of resistors R1, R2 and R3 in this case. The resistances are in series so they add up, thus the Black Box has a cumulative resistance of R = R1 + R2 + R3.
2. A voltage applied across a resistance causes a current I to flow, thus: I = V / R.
3. Since the constituent resistors are in series, the SAME amount of current must flow through each of them. There is no alternative path for current to flow from V+ to ground.
4. A current across a resistance implies a voltage across said resistance, by the same formula as above, thus: V(r1) = I * R1. That is the potential difference between the two ends of resistor R1.
5. Similarly, V(r2) = I * R2, and so on.
6. Evidently, one of these resistors, R3, has one end at ground potential, i.e. 0 volts. Thus, the voltage from there to the other end of that resistor is V(r3). The voltage at the next higher measurement point is V(r3) + V(r2), since the voltages add up, and as stated above, reference to ground.

By following this process, the voltages at each of the points of any series resistance network can be computed if either the applied voltage V (15 volts in this case) or the flowing current due to it, is known.

Now, how does one decide what resistances to use? Well, make the total resistance too small, and the current will be high, potentially burning out the resistors or the power supply, or causing the supplied voltage to droop, depending on how ideal we are assuming things to be. Similarly, use too high a resistance, and too little current will flow, thus the readings will be swamped by other noise effects that exist in practical electronics from various causes.

So pick a number that you like, and divide it in the ratio you want the test-point voltages to be. The resistances need not be equal, just as the voltages need not be at 33% each - calculate for any ratio you want.

I hope this helped.