ATTENTION: 4:29 In case of car's alternator failure (of its internal voltage regulator circuit) the "V bat" could increase up to 22V (or maybe more), proportional to engine RPM. To avoid a burnt arduino, it's safer to assume maybe 25V to "V bat" at voltage divider circuit calculation. Many thanks for this lesson, I was searching exactly for this topic! Regards from Brazil!
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.
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.
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.
Thank you for a very good and technically convincing explanation of the connections between the knife and an Arduino and the special features. That was a great help to me!
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.
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.
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 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.
National Instruments says a high speed ADC can be more accurate than a high resolution one. You take 4 or more samples and mediate them ! High resolution ADC measures, with precision !, spikes !
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.
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 ?
This is the clearest and shortest explanation about Arduino ADC.
Subscribed!
Glad it was helpful!
ATTENTION: 4:29 In case of car's alternator failure (of its internal voltage regulator circuit) the "V bat" could increase up to 22V (or maybe more), proportional to engine RPM. To avoid a burnt arduino, it's safer to assume maybe 25V to "V bat" at voltage divider circuit calculation.
Many thanks for this lesson, I was searching exactly for this topic!
Regards from Brazil!
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.
Very good video. Thank you for your work.❤
Our pleasure!
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.
Thank you for a very good and technically convincing explanation of the connections between the knife and an Arduino and the special features. That was a great help to me!
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!
thanks for your detailed demo+explanation, now i know what to do with these current sense, again thanks
At 7:25 you measured ONE value (4.834V) but inputted a slightly different value in the sketch (4.863V).
Great video. Thank you for this information.
you are welcome
New subscriber always watching your videos very interesting❤
Awesome! Thank you!
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.
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
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/
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.
Thanks for the information. Can you please advice how to accurately measure mains voltage.
is there possible to read data from 20 batteries/cells from 1 adruino ?
National Instruments says a high speed ADC can be more accurate than a high resolution one. You take 4 or more samples and mediate them !
High resolution ADC measures, with precision !, spikes !
Can you please help me connect the following sensor to arduino: SUI-101a
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.
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.
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 ?
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.
Very nice!
amazing
You have no answer of my question ❓
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.