Each time I power up the Arduino, the max reference voltage which is represented by 1023 changes. I observed this by measuring the voltage from the Aref pin each time. Is there any internal function or way to find this max voltage? Because, the voltage reading from the analog input pins of a constant voltage source keeps changing every time as a result of varying max reference voltage.
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\$\begingroup\$ Look in the datasheet and also the Arduino documentation. The analog reference voltage can be the supply voltage (VCC) or an internal reference voltage, see: arduino.cc/reference/en/language/functions/analog-io/… The maximum is always the supply voltage VCC. No voltage inside the IC can (should) exceed VCC. \$\endgroup\$– BimpelrekkieJun 24, 2019 at 13:50
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\$\begingroup\$ I have seen this, and have called analogReference(DEFAULT) in the setup function. However, sometimes Aref is 4.8V and when I reconnect the board to the laptop I get 4.6 V. It seems to vary around this value. It's neither 5V nor a constant value. \$\endgroup\$– Nick RogersJun 24, 2019 at 13:57
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\$\begingroup\$ The "obviously" your supply isn't a stable 5 V. The easiest solution is to use on of the (much more stable) internal reference voltages. If that limits the input voltage range of your ADC input, use a voltage divider (and maybe a buffer) to make the input voltage within the proper range. \$\endgroup\$– BimpelrekkieJun 24, 2019 at 14:14
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\$\begingroup\$ I'm using the internal default reference voltage. On measuring Aref using a multimeter, I get variations of 0.1-0.2V. I got 4.7,4.6V ... \$\endgroup\$– Nick RogersJun 24, 2019 at 14:18
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\$\begingroup\$ I'm using the internal default reference voltage I got 4.7,4.6V Both sentences cannot be true at the same time, if you measure 4.6 or 4.7 V then you're NOT using the internal reference voltage, then AREF is connected to VCC. When using the default internal reference voltage you should measure much less: 1.1 V or 2.56 V. Look at the diagram in CrossRoads' answer. \$\endgroup\$– BimpelrekkieJun 24, 2019 at 14:22
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The default reference should be 5V, fed from Vcc to Aref by way of an internal reference select multiplexer. If might be less than 5V if the board is powered from USB.
Measure your 5V source - is that varying as well?
answered Jun 24, 2019 at 14:04
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\$\begingroup\$ I have powered the board via usb, because I need to serially transmit back the readings to the computer. So how can I give a 5V supply as well as read serial data on a computer? \$\endgroup\$ Jun 24, 2019 at 14:07
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\$\begingroup\$ You can't get 5V from USB, there will always be some loss thru the USB protection fuse. You can add a 4.5V precision reference and connect that to Aref. Or use the internal reference and scale back the analog inputs with a resistor divider. This one is nice, only needs 4.75V in to make 4.5V out digikey.com/product-detail/en/analog-devices-inc/REF194GSZ/… \$\endgroup\$ Jun 24, 2019 at 14:18
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\$\begingroup\$ Isn't, analogReference (DEFAULT) function using the internal reference voltage of 5V? \$\endgroup\$ Jun 24, 2019 at 14:20
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\$\begingroup\$ Yes, if Avcc = 5V. But powered from USB, AVCC will not reach 5V due to the USBVCC protection fuse, there is some voltage drop across it, which will increase with current load (it is a low value resistor that heats up and melts with too much current flow). Measure the two sides of the fuse (big green or gold colored part next to the USB connector) and see what you get. \$\endgroup\$ Jun 24, 2019 at 14:25
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\$\begingroup\$ Isn't, analogReference (DEFAULT) function using the internal reference voltage of 5V? No, as that would be very confusing to many! Imagine using VCC = 5 V vs VCC = 3.3 V and getting completely different ADC values for the same input voltage. That's why an internal reference voltage is used, to keep it simple. \$\endgroup\$ Jun 24, 2019 at 14:26