Voltage selector design problem

Question

I designed a 5/24Vout voltage selector circuit with the schematic design that I drew below:

By bridging either Pin 1-2 or Pin 2-3 of J6, I could have two different voltage dividers leading to two different output voltages.

After producing the PCB for this, I tested the output voltage and found that if I selected 5V, the output voltage was about 9.35V and if I selected 24V the output voltage was 48V, about twice the intended output voltage of 5V/24V.

Knowing that the boost converter chip can only support up to 28Vout, I assumed that I had somehow drawn a series circuit on the PCB.

Would greatly appreciate it if somebody could point out the problem in the schematic/PCB, thanks!

PCB:

Note: I selected either voltage by bridging Pin 1-2 or Pin 2-3 of J6 using this:

Link to boost converter:
https://www.olimex.com/Products/Breadboarding/BB-PWR-3608/resources/MT3608.pdf

Answer 1

If I rearrange resistors R32, R33, R34 and R35, you can see immediately what the problem is:

^{simulate this circuit – Schematic created using CircuitLab}

Check for yourself that this is equivalent to what you have built. Clearly R33 and R35 are connected in parallel, and their combined resistance is:

$$ \begin{aligned} R_P &= \frac{1}{\frac{1}{R_{33}} + \frac{1}{R_{35}}} \\ \\ &= \frac{1}{\frac{1}{2.20k\Omega} + \frac{1}{2.32k\Omega}} \\ \\ &= 1.13k\Omega \end{aligned} $$

Now with the jumper positioned to select the left path, the output voltage will be:

$$ \begin{aligned} V_{OUT} &= 0.6\left(1+\frac{16.5k\Omega}{1.13k\Omega}\right) \\ \\ &= 9.36V \end{aligned} $$

Fix this by completely removing R35, so that only \$R_P = R_{33} = 2.2k\Omega\$ remains. Then the left path will work as expected, and \$V_{OUT} = 5V\$.

You then have to change the value of R34. When the right path is selected by the jumper (for 24V), now you have two resistances R34 and R33, where R33 is now 2.2kΩ:

$$ \begin{aligned} V_{OUT} &= 0.6\left(1+\frac{R_{34}}{R_{33}}\right) \\ \\ R_{34} &= R_{33}\left(\frac{V_{OUT}}{0.6}-1\right) \\ \\ &= 2.2k\Omega\left(\frac{24}{0.6}-1\right) \\ \\ &= 85.8k\Omega \end{aligned} $$

So your final design looks like this:

^{simulate this circuit}

Answer 2

R33 and R35 are connected in parallel here resulting in a equivalent resistance of 1.13K. No wonder therefore that you are getting your quoted output voltages.

To fix the problem remove R35. You'll then get your selected 5v output. You'll have to slightly tweak R34 to 86K to get 24V with the bottom 2.2K resistance.