How-to: Accurate Voltage Measurements with Arduino
ฝัง
- เผยแพร่เมื่อ 8 มิ.ย. 2024
- In this video, we show how to configure an Arduino UNO board for precise voltage measurements. A good-quality power supply is required for this. Also, you must know the exact value of the reference voltage the AD converter is using. Do not simply assume that it is 5 V or so, measure it. If you follow the instructions carefully then you too can take accurate measurements with an Arduino board.
#Arduino @Arduino
Contents
========
0:00 intro
0:15 Analog-to-digital converters
1:12 analogRead()
1:57 Convert to volts
2:49 Floating-point calculations
3:36 A practical application
5:15 Better resistors
6:38 Measure the reference
7:41 Use a good power supply
8:52 analogReference()
10:00 Interference
10:59 Summarizing
Resources
=========
* Arduino UNO Rev3: www.elektor.com/arduino-uno-rev3
* A good multimeter: www.elektor.com/hioki-dt4256-...
* A good power supply: www.elektor.com/peaktech-6225...
Check out our TH-cam offers: www.elektor.com/youtube
Subscribe to our Newsletter: www.elektormagazine.com/elekt...
Join this channel to get access to perks:
/ @elektortv - วิทยาศาสตร์และเทคโนโลยี
This is the clearest and shortest explanation about Arduino ADC.
Subscribed!
Glad it was helpful!
highly appreciated. logical progressive and methodical teaching. 👌
Glad it was helpful!
perfect!
I didn't know about the resolution of the ADC, and I also hadn't thought about the reference voltage. Thank you very much! The video was very useful.
Glad it was helpful!
Very helpful. Clear and, as far as can tell, all the relevant information needed to get the job done. Also happy to see the comment below that filtering the external voltage supply will improve stability.
Glad you enjoyed it!
Excellent start! As a continuation, you can show how to amplify a weaker signal, and how to adapt a voltage of maybe 2-4V to use all the bits in the AD converter.
I would add that the reason why we divide by 1023 instead of 1024 is because the 0 (zero) is included as one of the values, so we still have 1024 values but zero is one of them; in other words, we're counting from 0 instead of counting from 1, so that's why it ends at 1023.
We don't divide by 1023, because the datasheet says that you must divide by 1024 (section "ADC Conversion Result"). 1023 equals to Vref minus one LSB. So, if Vref=5V, then 1024 = 0x400 = 5V; 1023 = 0x3ff = 4.995V. Unfortunately, 1024 doesn't fit in 12 bits.
Very good video. Thank you for your work.❤
Our pleasure!
New subscriber always watching your videos very interesting❤
Awesome! Thank you!
thanks for your detailed demo+explanation, now i know what to do with these current sense, again thanks
external interference performance can be improved by adding a capacitor between ground and the analog input (as close to the ADC pin as possible). 1nF should be a good compromise between speed and filtering quality.
Less accurate power supplies can be used if a large value low ESR capacitor is used across the 5V rail and the 1.1V reference is selected.
Thanks for the detailed explanation 🙏
Glad it was helpful!
At 7:25 you measured ONE value (4.834V) but inputted a slightly different value in the sketch (4.863V).
Very nice!
The bandgap reference on the Atmega328 can vary a bit from unit to unit. In order to have the highest accuracy, you can measure such value and make your calculations with the real value, not the one mentioned on the datasheet. I got great results doing that.
@2:49 additionally to floating-point calculation, isn't it more appropriate to multiply by 5 first, then do the dividing by 1024 as the last step ? Or does the compiler already do this optimization automatically ?
The compiler does not do calculation quality optimizations, only code optimizations. Multiplying by 5 works for integer maths but it remains integer so the output will be between 0 and 4. Floating point has enough precision that multiplying by 5 first doesn't improve much. larger factors like 100 or 1000, yes.
amazing
is there possible to read data from 20 batteries/cells from 1 adruino ?
Very informative, Now how can we measure voltage and current and the phase angle (power factor) with an Arduino?
Search the internet, e.g. solarduino.com/how-to-measure-power-factor-and-phase-angle-with-arduino/
Can you please help me connect the following sensor to arduino: SUI-101a
Sir I will try to make dc volt meter using ESP32 and I2C 16*2 LCD.
Am facing some critical condition
Actual voltage 12.7V
LCD display 12.7V
But
Actual voltage 7.4V
LCD display 6.9V
This problem only for ESP32, Arduino UNO is perfect work in range.
How to solve this problem sir ?
Since the max value of the ADC is 1023, I would divide by 1023 instead of 1024.
other than that, very interesting video
Yes, but no. The ADC's maximum theoretical value is 1024. 1024/1024=1. But 1024 = 0x400 which requires 13 bits whereas the registers are only 12 bit. Therefore, the maximum *register* value is 0x3ff, which is 1023. So the maximum value in practice is 1023/1024 = 0.999. See also the MCU's datasheet.
@@ElektorTV Indeed, there is a catch... from the datasheet : "and the maximum value represents the voltage on the AREF pin minus 1 LSB."
@@ElektorTV Yes, the MCU is not able to measure 5V. Now it depends on you how to overcome this problem. You can use 1024 to show the error only on 5V or use 1023 to spread out the error over the full range.
@@ElektorTV PS.: The real problem is not the question 1023/1024, it is the multiplication with 5. It should be the 5-1 LSB instead.
What about a higher voltage inputs 0 to 500 volts maybe
The same principles apply, but the voltage divider needs more attention to support high voltages (use several resistors in series for instance).
I did it up to 600v. No problem. You will have a step of about 0.6v with the 1024 bit of th arduino uno.
@@programmer1111x great
can you make Ac voltage meter 0-400v with 0-20mA output or Rs485 Modbus RTU
Of course, you can. Just be careful with the high voltages.
can use esp32 12bit mcu is more prisesion?
In theory, yes, but keep in mind that more bits doesn't always mean more precision. The input circuitry must be up to the job too.
I need to make an arduino voltmeter to read from -20V to 20V
Why you divide by 1024 when you have between 0 and 1023 only 1023 intervals of length 1.
Because there are 1024 intervals. Please refer to the section entitled "ADC Conversion Result" of the ATmega328 datasheet. 1023 corresponds to VCC minus one LSB.
Good job, but 1.1/16 = (R1+R2)/R1 is never true.
You are right, well spotted. It should read 1.1/16 = R1/(R1+R2). Sorry for any confusion caused.