# How to bias a voltage for an ADC

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: 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: How to design the bias?

• Is a negative voltage rail available this time? – Transistor Feb 10 '18 at 0:37
• yes there is a negative rail – rur2641 Feb 10 '18 at 0:44
• Are you going to share the voltage? – Transistor Feb 10 '18 at 0:45
• the dual supply source is XP Power IZ2415S, it can supply +/- 100 mA at +/- 15 V. There are 2 LEM LAH 100-P sensors running of the source. They each draw 10 mA in addition to 1/2000 of the current sensed. They are each expected to sense an average of + 70 A which amounts to 90 mA total. I don't know what is meant in the power source datasheet by +/- 100 A. – rur2641 Feb 10 '18 at 1:12
• Is your power system DC or AC??? I would expect AC from your $\pm 11\:\text{mA}$ on the schematic. But I expect DC from your statement that $10\:\text{V}$ should map to $0\:\text{V}$ (because, otherwise, what should $-10\:\text{V}$ and $-38\:\text{V}$ map to??) So what is the deal? Is this an AC system or a DC system. I'm confused. Also, it appears you have two questions -- (1) mapping the system voltage to within the ADC range; and (2) mapping the Hall Effect sensor range to with a similar ADC range. .... I admit, I'm confused right now. What exactly are you doing/asking? – jonk Feb 10 '18 at 6:42

Rearranging:

• 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.

How:

• 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.

"Mapping" is a good description, but level shifting is mostly what is needed. If you source voltage is 12 volts to 36 volts (n), then use n/10 to get 1.2 volts to 3.6 volts. Subtract 1.2 volts to get 0 - 2.4 volts.

Use an op-amp with a gain of 1.375 to get a range of 0 to 3.3 volts. An op-amp with a negative supply of -3.3 volts will allow an output as low as zero volts simulate this circuit – Schematic created using CircuitLab