# Multiple voltage dividers : finding resistances in a circuit

I have to set the battery voltage to 14.5V from the first divider, then from that I have to attenuate it to 1V to feed to the input of an op amp. Now this answer is inspiring but doesn't solve my purpose. Maximum I can assume the R2 & R4. R3 & R5 still unknown.

• One thing you should note about voltage dividers is that the divider current must dwarf the output current. You should look up the input current of the op amp and size r4 and r5 to pass no less than 10 times that(100x is better). Then figure out the ratio of r2 to r3 and size the resistors to pass another 10 to 100x. Note that cascading voltage dividers will not increase accuracy and that is why I recommend a regulator.
– K H
Sep 14 '20 at 4:22
• You should be able to solve for R5 right way: R4 and R5 form a voltage divider that divides 14.5V down to 1V. This is independent of R2 and R3. Now that you know R5, you know that R4+R5 (a now known value) in parallel with R3 is one half of your voltage divider (with R2) that takes 16V down to 14.5V. Sep 14 '20 at 4:26
• This can't be done with the initial values you give for R2 and R4. You have 1.5 volts across the 10K R2. That current can't possibly produce 13.5 volts across 1 K R4, and you don't say how much of the 150 uA through R2 is going to the 14.5 volt load. Sep 14 '20 at 4:29
• what is the point of having R3 in the circuit? Sep 14 '20 at 4:38
• @KH: A regulator isn't the right choice, here. This circuit apppears to be intended to read the battery voltage with the ADC of a microcontroller. A regulator isn't appropriate.
– JRE
Sep 14 '20 at 5:57

Well, we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag2$$

Substitute $$\(2)\$$ into $$\(1)\$$, in order to get:

$$\begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag3$$

Now, we know the values of $$\\text{V}_\text{i}\$$, $$\\text{R}_1\$$, $$\\text{R}_3\$$, $$\\text{V}_1\$$, and $$\\text{V}_2\$$. So coloring the things we know red, we see:

$$\begin{cases} \frac{\color{red}{\text{V}_\text{i}}-\color{red}{\text{V}_1}}{\color{red}{\text{R}_1}}=\frac{\color{red}{\text{V}_1}}{\text{R}_2}+\frac{\color{red}{\text{V}_1}-\color{red}{\text{V}_2}}{\color{red}{\text{R}_3}}\\ \\ \frac{\color{red}{\text{V}_\text{i}}-\color{red}{\text{V}_1}}{\color{red}{\text{R}_1}}=\frac{\color{red}{\text{V}_1}}{\text{R}_2}+\frac{\color{red}{\text{V}_2}}{\text{R}_4} \end{cases}\tag4$$

So, we have two equations and two unknowns, namely $$\\text{R}_2\$$ and $$\\text{R}_4\$$. Using your values, we see that:

$$\begin{cases} \frac{16-\frac{29}{2}}{10000}=\frac{\frac{29}{2}}{\text{R}_2}+\frac{\frac{29}{2}-1}{1000}\\ \\ \frac{16-\frac{29}{2}}{10000}=\frac{\frac{29}{2}}{\text{R}_2}+\frac{1}{\text{R}_4} \end{cases}\tag5$$

It is not hard to solve this (so I let you do that). I used a simple Mathematica code to solve this:

In:=Vi = 16;
V1 = 29/2;
V2 = 1;
R1 = 10000;
R3 = 1000;
FullSimplify[
Solve[{I1 == I2 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2,
I3 == (V1 - V2)/R3, I3 == V2/R4}, {I1, I2, I3, R2, R4}]]

Out={{I1 -> 3/20000, I2 -> -(267/20000), I3 -> 27/2000,
R2 -> -(290000/267), R4 -> 2000/27}}


So, we get:

$$\text{R}_2=-\frac{290000}{267}\approx-1086.14\space\Omega\space\space\space\wedge\space\space\space\text{R}_4=\frac{2000}{27}\approx74.0741\space\Omega\tag6$$

Which is not possible because the resistance must be bigger than zero.