Electrical – How to bias a voltage for an ADC

biasingcircuit analysis

I am part of a project that is implementing a power storage system. The storage device voltage must be monitored in order to direct power. The storage device voltage should remain between 12 and 36 V. The project uses a TI TMS320F28027 MCU which operates at 3.3 V. How can the 12 to 36 V be mapped onto 0 to 3.3 V for the MCU.

I posted a similar question here but it specified that the voltage would range from 0 to 60 V. The difference for this question is how to bias the voltage appropriately.

I also posted another question about biasing the output of a current sensor for an ADC, but it involved biasing a balanced voltage around 0 V. I am having trouble adapting the answer to this problem. Diagram from answer:

diagram from answer
Figure 1. Diagram from previous answer

An attempt to emulate the design methodology would be:

24 V maps to 1.65 V
38 V maps to 3.3 V
10 V maps to 0 V

The most basic circuit model:

enter image description here

How to design the bias?

Best Answer


  • 38 V maps to 3.3 V
  • 24 V maps to 1.65 V
  • 10 V maps to 0 V


simulate this circuit – Schematic created using CircuitLab

Figure 1. An 11.5:1 potential divider.

The simplest solution is to use a potential divider with a ratio of 38:3.3 or 11.5:1. This would result in:

  • 38 V maps to 3.3 V
  • 24 V maps to 2.08 V
  • 10 V maps to 0.868 V

The 0.868 V offset can be removed in software. Again, you lose a little resolution with this approach.

If a negative voltage supply is available then the offset can be removed.


simulate this circuit

Figure 2. With a negative rail available the offset at minimum input voltage can be removed.


  • The span is 38 - 10 = 28 V.
  • This has to be scaled to 3.3 V so a divider ratio of 28 / 3.3 = 8.5:1. Let's use 7.5k and 1k to give us the required ratio.

Now we need to figure out the negative reference voltage.

  • At 10 V in Vout will be 0 V. With the 8.5:1 ratio we will need to hold Vref at \$ - \frac {1}{8.5} 10 = -1.18 V \$.

So, R1 = 7.5k, R2 = 1k, Vref = -1.18 V should do the trick.

I'll leave it to you to work out how to create the reference voltage.

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