Arduino Due does not have an analog output voltage from 0 V to Vref, but from 1/6 to 5/6 of the
reference voltage, that is, 0.55 V and 2.75V with Vref = 3.3 V. This is also
confirmed by the Atmel (see bibliography).
The output voltage range of the DAC is only 2.75-0.55
= 2.2 V, with a resolution of 2.2/4095
= 0.5372 mV.
A simple differential amplifier, realized with
an op amp, is used to remove the 0.55 V offset and to amplify the
output signal in order to reach the desired value. The following figure shows
the scheme of this circuit:
To
generate the voltage Vp = 0.55 V, that is, the minimum value of the DAC, is used
the divider formed by R3, R4 and Rp. The trimmer pot Rp is used to adjust Vp in
order to have an output value close to 0 V in correspondence with the number 0
sent to the DAC. With a single power supply you can never get exactly 0V but
you approach some mV, about 8 in my prototype.
The differential amplifier has the following gain:
G = Vo/(Vi-Vp) = R2 / R1
Just change the gain, for a different output voltage,
but less than 4 V,.
With a gain G = 3.3 /2.2 = 1.5, the output of
this circuit has the desired range from
0 to 3.3V.
It is recommended to use resistors with a
tolerance of 1% or less, particularly for the four resistors of the amplifier.
References
1)
“Atmel
ARM Cortex-M3 Product Family (SAM3)”, Atmel application note 42187A−SAM−10/2013
This should be placed somewhere in Arduino's site. If not in a official page, in Playground at least. Thanks for your job
ReplyDeleteYes, I did it yesterday, in Megatopic (in Italian) and replying to some questions in the Arduino Due Forum, inserting this link, Thanks for your interest
ReplyDeletemy problem input voltage 0.7V-4.3V output voltage 0-5V how to analysis circuit,
ReplyDeleteIt's the same principle.
Deletecould you draw out the circuit
DeleteInteresting article!, I've worked with the LM741 Very well, i didn't understand the ''single-supply'' thing you said.
ReplyDeleteHello,
ReplyDeleteWhat is the formula for a voltage divider and a potentiometer ?
Because the resistor from the potentiometer should appear in the formula ..
Thanks
hi,
ReplyDeletei am working on a simple circuit in which i use ADC's output as DAc's input. since the voltage range is different for both i wanted to use this concept to modify the offset voltage so that the range of DAC is same as ADC's (0v - 3.3V ). but i used the differtial amplifier without capacitor's and pot (instead used 0.55V input as suggested Vp = 0.55V and Vi(DACout) = 0.55V), which gave output V0 = 0.03V. when DACout =2.75 then V0 = 2.62V but the reuired V0 = 3.3V. is this because of not using capacitors? i am ataaching the circuit diagram link i used with its values please let me know the changes to be done and if i have to use capacitors will 100nF work with these resistors?
https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwiD5_SkyIDnAhUMyqQKHW4hAqMQMwhHKAEwAQ&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FDifferential_amplifier&psig=AOvVaw2Vzny3KSTdFe45HmwpH7YH&ust=1579004502746851&ictx=3&uact=3
Op Amp used is LM324N
V1 =Vp =0.55V ; V2 = DACout(0.55V - 2.75V)
R1 =22K ohms ; R2 = 10K ohms ; Rf = 33K ohms ; Rg = 10K ohms
have given Vcc and ground connections too (+12V).
This is a useful circuit. I built it with the LM358 op amp. After setting Rp the best Vout minimum was ~30mV. I swapped for a MCP6022 op amp, which is pin compatible and in-expensive. The frequency response is better and the minimum voltage dropped to 6mV.
ReplyDeleteGreetings,
DeleteI knew it: using rail-to-rail op-amps is much better, but many can't easily find them in shops around town or at home.
Best regards,
Hi thanks for posting thhis
ReplyDelete