Need help calibrating ADC input using a voltage divider

Question

I have been using Arduino Nano analog input to measure voltages, in the range between 22-30 Volts. The Nano's ADC expects a 0-5 volt range and has 0-1023 as its output.

Following the instructions found here:

Read Analog Voltage (www.arduino.cc)

Which basically states that all I need is this code:

int sensorValue = analogRead(A0);
float voltage= sensorValue * (5.0 / 1023.0);

5.0/1023 = 0.004888 volts per increment in analog readings. Turning that around, voltage input divided by this value should give me the sensor value.

My voltage divider is 1 Mohm (R1) between the voltage input and A2, then 200K (R2) from A2 to ground

30 Volts measures as 4.2 Volts. 22 Volts measures as 3.1 Volts.

I think the range between 22 volts and 30 Volts doesn't present enough variation to the ADC - every single-digit change in the analogRead really make a big difference.

The result is that it is giving me some crazy readings. It is definitely nonlinear, so I am having real trouble trying to get it to be accurate at both ends of my measurement range. Each slight ADC increment makes too much of a difference in the measured voltage because I am not using the full 0-5V input range.

So I have been taking 10 readings in a row, at 500 ms intervals, then averaging.

And yet it is not working well enough.

One thing that seems to really matter is the power supply. When I put 4.97 into the Vin it gets different readings than if I rely on getting power from the USB connection to the Raspberry Pi 3B.

Just yesterday I put it on the power supply for good - had been relying on the Pi all along. But it still is not giving accurate readings.

--EDIT from Answers (Thank you for showing me where to look)

Input impedance: I'm trying 147/23.5 for the next attempt. Getting 3.0-4.1 for the 22-30 range. Same as the 1M/100K, basically, but lower resistances.

The reference voltage issue: It measures 4.36 Volts on the 5V output from the Nano. Vin is 5V exactly.

Here is an article I found that says the 5/1023 formula should use that actual value:

Measuring DC Voltage using Arduino

And another that confirms it:

Analog to Digital Conversion

SO NOW the immediate problem is how can I map 22-30 volts to 0-4 or so, thus allowing the utilization of more than just a few % of the allowable values?

Cross-posted at How can I seriously calibrate ADC voltage readings with Arduino Nano?

Answer 1

The best way is offset and less attenuation.

Spec

• yes convert 20~30 to 0 to 5V with 0.1% accuracy if poss.

Rev A

• 22~30 converted to 0~4 becomes Vin * 0.5000 -11.00V = Vout
  • 11.000V = k* 4.380V using k= 2.5114
  • with a non-inv OpAmp of gain 1+Rf/Rin
• one way: V+/2 - 11.00 V using precision Vref or programmable zener.
• using R array with matched R to <0.1%
• This will be 3x better resolution using 1/2 instead of 1/6 attenuation
• ensure your connections are twisted pair and filtered with low ESR RF cap

What supplies are avail?

p.s. I grew up in awe of Aurora , chasing smoke trails that appeared to be at ground level, watching sky waves like the ocean for hours at night and the fiery colours in Churchill Mb at the NRC rocket research range for Plasma Physics.

Answer 2

Aside from reducing the voltage divider resistances, which is necessary, but not sufficient, you need to consider the reference if you want accurate readings. For the divider, 10K is a good maximum, so if you use a 49.9K 1% resistor and a 10.0K 1% resistor you'll be pretty close o the desired ratio and the impedance from the ADC's input will look like 8.3K. A 10n-100n ceramic cap across the 10K won't hurt.

The Arduino does not have an ADC voltage reference so the reading is going to be ratiometric to the supply voltage. A 5% difference in supply voltage means a 5% difference in reading. There is a bandgap reference on the ATmega chip, however it is loosely specified (+/-10%), is relatively low voltage (1.1V nominal) and can only be used as an input to the ADC. If you can measure an accurate voltage even with the ratiometric reading you can correct for supply voltage with calculations if the supply voltage stays constant enough.

Or you could attach a precision external ADC and reference to the Arduino. `

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If you want to do an offset zero you can use a circuit like this one:

^{simulate this circuit – Schematic created using CircuitLab}

Use a RRIO op-amp such as Microchip MCP6001.

If you don't like my choice of input/output voltages, you can recalculate R1-R5 to get what you want, it's just simple algebra that anyone can do with a bit of patience.

R1/R5 are two resistors rather than one to get the 9:1 ratio with standard E96 series resistor values. R6 is not critical- it just ensures enough current gets to D1 to maintain regulation. The others should be precision types.

Since a fairly accurate and very stable voltage is subtracted from the input voltage you may find using the supply voltage to be acceptable as a reference, depending on what your actual requirements are. The reduction in sensitivity is 3:1 in this case.

Answer 3

As JRE says in the comments, Atmel (now Microchip :-( ), who makes the processors in Arduinos, generally suggest to not use any circuits with an output impedance higher than 100kOhm on most of its processors.

In fact with an impedance of 100kOhm they often suggest adding a capacitor as well, as follows:

^{simulate this circuit – Schematic created using CircuitLab}

Because the processor has an input impedance, which will pull the read voltage down (or possibly up in some cases?), which is likely more than 10Mohm. But it also has a small capacitance inside which it uses to "hold" the voltage while it does a conversion.

When it connects that capacitance a small amount of extra current will flow, which is likely the culprit for your loss of voltage. If you use 10k and 2k that current isn't large enough to create more error than is already in the ADC, but for larger resistances you will start noticing it. With 1MOhm on top I'm not sure if you can fully fix it with a capacitor (the internal leakage probably already becomes noticeable), so maybe the golden middle is 100k and a small capacitor.

The extra capacitor in the case of 100k impedance buffers the voltage all the time that the ADC is not connected to your network, and then when the ADC shortly samples your voltage with its hold capacitor, the capacitor you added will provide a lot of the energy to charge that internal hold capacitor, so the error will be much lower.

Answer 4

Firstly, don't power your arduino through 5 volts into the VIN pin. Connect 5V to the 5V pin (If you are sure it is 5V) or power it though a higher voltage though the VIN pin. Second, I would recommend changing the divider to work with the 1.1 volt band-gap voltage reference (aim for 0-1V). It is probably the most accurate reference (on similar products it has a 90 ppm temperature coefficient) at hand and you can calibrate out the 10% uncertainty with a multi-meter. Third, mapping the 0 to 1024 value to 0V to 30V can be done using the builtin map function: map(value, 0, 1024, 0, 30000)/1000.0

Answer 5

Thank you for your answers - there is a lot of collective knowledge there. It helped me find a way to stabilize and calibrate this system with what I had on hand. The Arduino people helped me with some concepts about the particular device but it is here that I am seeing the answers I really need.

(The biggest problem was the reference voltage and resolution were assumed to be 5/1024 which is completely incorrect. Unless, as someone pointed out, I fed the Nanos directly at the 5V pin - which is still on the table - then it would be 5/1023)

I changed my resistor network to be $$ 147K/23.5K$$

(Parts list: 1x 100K, 3x 47K)

$$ R1 = \frac{47K}{47K} = 23.5K $$

$$ R2 = 100K + 47K = 147K $$

Also added a 100uF 50V electrolytic across the input from the battery bank I am monitoring.

I am feeding 5V to Vin. And have figured out that the reference is measurable on the 5V pin, after the regulator, as 4.6V.

Knowing that

$$ \frac{Resolution}{Vref} = \frac{ADC}{Vin} $$

I turned that around to be

$$ Vin = \frac{(ADC * Vref)}{Resolution} $$

Or, $$ Vin = A2 * \frac{4.6}{1023} $$

Using a laboratory power supply I was able to confirm that with those settings 22 Volts came in at 3 Volts, and 29.98 came in right at 4 Volts.

So

$$ Vbat = A2 * (4.6/1023) / (R2/(R1+R2)) $$

$$ Vbat = A2 * \frac{(4.6/1023.0)}{(147000/170500)} $$

It is now operating reasonably well.

--UPDATE-- Changed resistors to 100K/20K for the divider and brought power to the 5V pin, bypassing the regulator for a reference. It measures at 5.02 and is stable.

It is tracking pretty well with the reading on the Charge Controller's screen.

(Yellow means the inverter is on)

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Similarly, I went to 147K/47K for the 11-15 Volt circuit, and moved a higher-quality P/S over to the 5V pin, where it measures 5.03V; the results are better there, too.

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So anyway, I'll put this to rest until I can figure out an op-amp circuit.

Thank you all for your help