Electronic – High resistance voltage divider

The question is a little ambiguous, so I will try my best to interpret your question correctly.

The term "optimal" needs a bit more clarification. Given that you've made three readings of an \$R_x\$ value that does not change between readings, the question is now how to deduce what the value of \$R_x\$ is. Though you could pick one of the three that is closest to the true value, this is not the only way.

Doing an ADC conversion cuts off the lower portions of the digits from an analog read. As I'm sure you already know, the equation to find \$R_x\$ is as follows:

$$ A = \lfloor 2^n \frac{R_L}{R_x + R_L} \rfloor $$

Where \$A\$ is the \$n\$ bit analog read (10 bits, say) and the \$\lfloor \cdot \rfloor\$ is the floor function (that is, rounding down). The problem arises with the rounding down. Since the analog read throws away the lower order bits, any value of \$R_x\$ is admissible that will produce the same rounded down value of \$A\$.

The range of values for which \$R_x\$ will produce the same \$A\$ value is as (by a simple rearrangment of terms):

$$ R_{x \text{ lower}} = ( \frac{2^n}{A+1} - 1 ) R_L $$ $$ R_{x \text{ upper}} = ( \frac{2^n}{A} - 1 ) R_L $$

where the true value of \$R_x\$ can be anywhere within the range of \$R_{x \text{ lower}}\$ to \$R_{x \text{ upper}}\$. If you were doing one analog read of \$n\$ bits and assumed a uniform distribution on \$R_x\$ within the range of \$R_{x \text{ lower}}\$ to \$R_{x \text{ upper}}\$, you would just pick the midpoint and be done.

But we don't have one read, we have three. In this case, each analog read gives us a range of values for \$R_x\$. The intersection of all three of these gives a restricted range for an admissible value of \$R_x\$. If we assume again a uniform distribution of \$R_x\$ values within this range of values, the best pick would then just be the midpoint of the restricted range.

To me, this is the definition of optimal: Pick the midpoint value of \$R_x\$ from the restricted range. This will minimize error assuming the \$R_x\$ value is uniform within the restricted range.

As an example, consider doing three read values on a true value of \$R_x = 4424\$. This would result in (assuming the number of bits \$n = 10\$):

$$ A_{1k} = 188 $$ $$ A_{11k} = 730 $$ $$ A_{111k} = 984 $$

Which corresponds to calculated ranges of \$R_x\$ for each lower resistance value:

$$ R_{x \text{ 1k lower}} = 4417.99\dots $$ $$ R_{x \text{ 1k upper}} = 4446.81\dots $$

$$ R_{x \text{ 11k lower}} = 4409.03\dots $$ $$ R_{x \text{ 11k upper}} = 4430.13\dots $$

$$ R_{x \text{ 111k lower}} = 4394.92\dots $$ $$ R_{x \text{ 111k upper}} = 4512.20\dots $$

Picking the lower bound and upper bounds to find the smallest range of values gives us a choice of \$R_{x \text{ 1k lower}}\$ and \$R_{x \text{ 11k upper}}\$ respectively. Choosing the midpoint from this restricted range gives us:

$$ \frac{ R_{x \text{ 11k upper}} - R_{x \text{ 1k lower}} } { 2 } + R_{x \text{ 1k lower }} = 4424.06 $$

Picking the midpoints for the individual ranges would give \$R_{x \text{ 1k}} = 4461.37\dots\$, \$R_{x \text{ 11k}} = 4453.56\dots\$ and \$R_{x \text{ 111k}} = 4419.58\dots\$ respectively. Please note that I cherry picked the 4424 value to illustrate this example.

In the end, you're only getting 1.5 more bits of information and maybe doing this type of fine grained analysis isn't worth it, especially if this calculation is required to be done on an embedded system with limited resources.

One alternative, which is perhaps more along the lines of what you wanted in the first place, is to choose one of ADC readings based on what the average absolute error is.

If we take \$R_{x \text{ lower}}\$ and \$R_{x \text{ upper}}\$ as above, then one way to measure the error for a measured value relative to the true value is:

$$ M = \frac{ R_{x \text{ upper}} - R_{x \text{ lower}} }{2} + R_{x \text{ lower}} $$ $$ \text{Error} = \int_{R_{x \text{ lower}}}^{ M } (M - x) dx + \int_{ M }^{R_{x \text{ upper}}} (x - M) dx $$

That is, measure the absolute (and unscaled) error from the midpoint of the range.

Using this, you can get a closed form solution for each range of values and calculate absolute error for each resistor combination. I have done this and used GNUPLOT to generate a pretty picture (ADC conversion assumed to be 10 bit):

Note that this is plotted on a log-log scale. Looking at the graph, it looks like picking 3316 and 34943 are suitable cross-over points.

I picked absolute unscaled error as the definition of "optimal" in this case and this may not be appropriate. Once you define what optimal is for your needs, then you should be able to pick suitable values of cross over points if you wanted use this method.

Please let me know if I've done any of the analysis in error.

Since the bulb draws \$0.5\$A when \$1.5\$V is across it, its equivalent resistance by Ohm's Law is

$$R_{B} = \frac{1.5\text{V}}{0.5\text{A}} = 3\Omega$$

As you've stated in your question the solution is to use a series resistor \$R_{1}\$ and parallel (to the light bulb) resistor \$R_{2}\$. When the bulb is connected this creates a voltage divider composed of \$R_1\$ and \$R_2||R_B\$:

$$v_{s} = 1.5\text{V} = \frac{R_2 \parallel R_B}{R_2 \parallel R_B + R_1}6\text{V}$$

Also recall that

$$R_2 \parallel R_B = \frac{R_{2}R_{B}}{R_{2} + R_{B}}$$

When the bulb is disconnected the path through the bulb becomes an open circuit, which means that \$R_{B} \to \infty\$ (the actual bulb resistance is the same of course, but \$R_{B}\$ appears to be infinite to the circuit since the path is open). You have the same equation as above, except that \$v_{s} = 2\$V and \$R_2 \parallel R_B = R_2\$ (the equivalent resistance for a resistor in parallel with an infinite resistance is just the resistor value):

$$v_{s} = 2\text{V} = \frac{R_2}{R_2 + R_1}6\text{V}$$

You've got two equations (the two equations for \$v_{s}\$) and two unknowns (since \$R_{B}\$ is known) so you can solve the system for \$R_1\$ and \$R_2\$.