Im new to arduino and I wanna know something about analogread and reference. I have an input from a shunt that measures amps of a circuit. It outputs a value between 0 and 50 mV. I wanna use the analogreference and the aref pin to lower my upper limit to 50 mV, So then I can read in raw voltage between 0 and 50 mV with 1023 steps because its 10 bit?
Is there a way to set aref to 50 mV using a 5 volt power supply?
Also, do I need a resistor in the analog input? safety? or a diode? im not sure.
There is an internal 1.1V reference voltage you could use. However you can apply a lower external voltage than this but there is little point because then you get into the noise of the system and the values returned in the least significant bits are meaningless. This is true of all A/D converters when you use them with lower than the recommended reference voltage.
If you don't know how to hook up an op-amp, then you need to start a little closer to ground zero. I'd find a tutorial and learn a little.
In case you didn't get what Grumpy_Mike is trying to tell you: The minimum allowable Vref is 1.0 volts. You can't use 100 mV. You'd be better off using the internal reference, Vint, since you don't gain anything by supplying your own.
Normally shunt resistances are chosen so as to not effect the circuit they are measuring. The OP possibly doesn't have the option of raising the shunt resistance enough to help. If he doesn't want to amplify the signal, an alternative would be to use an external ADC with a large enough dynamic range to accurately handle the 0-50 mV signal and read it into the Arduino for further processing and display. A 24-bit ADC would give you about 0.1 microvolt resolution with a 1 volt reference. Ought to do the trick.
I can't agree with your logic. If OP have no knowledge to build simple OPA amplifier, how he/she could follow your advise to implement external 24-bit ADC? It doesn't make any sense.
I just assume, that OP probably doesn't work on something high power or high precision, when increasing shunt resistance wouldn't be desirable. The same time it is the easiest way for someone w/o electronics background. If shunt resistor "build-in" IC, there is always option to add external one.
I wasn't using any logic. I was just suggesting an alternative. If OP is going to have to wire something up based on a tutorial or suggestion rather than from their own knowledge, an external ADC is no more complicated than an op-amp. You didn't say how much higher you would suggest raising the shunt resistance. You have to worry about compliance voltage problems if you raise it to much.
Well the OP didn't state what full scale amps through the shunts creates the 50mv drop, but typically one doesn't like the wasted power and decreased voltage that results when raising the resistance of the shunt. I would think a op-amp set for a gain of 20 and using the internal band-gap reference would get the best results for the least complexity. A good low noise op-amp with rail to rail output capablity is not a hard component to spec, design, and build these days.
Are you sure, when I search for this I get either an LED or voltage regulator.
inverting input to output with 100K resistor, then 1K to ground
No, put the 1K to a signal reference ground. This can be made with two 1K resistors connected like a potential divider giving 2.5V, this is the point where you put the end of your 1K resistor. Also put two 0.1uF caps across each of the two potential divider resistors.
Op amps require a positive and negative power supply. You haven't got one, so to trick it you make a false signal ground at +2.5V with a potential divider and use some capacitors to lower the impedance.
oh sorry it was 741.
Look at the data sheet, it won't work with just a 5V supply. The data sheet says it needs a +/- 12V supply. That's 24V in total. You are only giving it +/- 2.5V.
An op amp has a gain of several thousand, when you say no input do you mean the two inputs connected together, that is the only way to have no input on an op amp.