I want to make system that would measure dc current through the load. I have DC adopter and potentiometer. if I set the potentiometer at any resistance system should be able to measure dc current through the loadSo with all of this do we have it yet? Things making sense?
Ron

because now I don't have power supply that can give 30 A current that's reason just using 5v DC adopter to simple testIf you only have a 2 amp power supply why are you designing for a maximum current range of 30 amps ?
Les.
I want to measure current through at different resistance. so i though potentiometer is one of choiceHello,
Is the potentiometer the only load?
When the potentiometer is turned to almost 0 Ohms, the potentiometer will burn out.
Do you want to "measure" the potentiometers position?
Bertus
I have mcp3008 and ACS712 30 Amp version on my tableHello,
What will be the lowest and highest load resistance values?
When your powersupply can not deliver more than 2 Amps, use a ACS712 5 Amp version.
That will give you more output signal.
Also a higher resolution ADC like the MCP3208 will help.
Bertus
I think I can measure measure 0.2 A - 2A current in the voltage range of 2.5-5V.Hello,
If you want "full" range for 2 Amp, you could amplify the output signal of the ACS.
You would need a rail to rail opamp.
Bertus
It's big progress i really appreciate all of your help and support. I think I can determine the following calculation on paperThe answer to the last line of your post #69 was answered in post #57 (Line 3)
Les.
You may want to give what is covered in there some thought. When you start looking at load currents and load resistances you also need to consider power. 5.0 Volts at 2.0 Amps for example would be a 2.5 Ohm resistance but what else? I also see 10 Watts so any small potentiometer when that resistance gets low will start getting real hot real quick and needs to be able to dissipate the power expressed as watts. Even in your above example:Hello,
Is the potentiometer the only load?
When the potentiometer is turned to almost 0 Ohms, the potentiometer will burn out.
Do you want to "measure" the potentiometers position?
Bertus
Even a low 5.0 Volts * 0.250 Amp = 1.25 Watts so your 20 Ohm resistor needs to be rated well above 1.25 Watts like maybe a 3 to 5 watt resistor. The problem is as heat is generated the resistor gets hot and then the resistance increases and your current decreases.Load : 20 Ohms
ACS712 Voltage : 2.5165 Volts
ADC value : 516
Load Current : 0.25A
Do you have any suggestion regarding circuit to reduce noise ?There are also other tips and tricks which can be used to improve the uncertainty of your measurements like taking multiple sample averages and knowing exactly what your Vref is. That all comes later.
Ron
There have been no less than about a dozen references to the data sheet including the manufacturer's stating the ACS712 should no longer be used for new design. I also suggest you read the Wikipedia Analog-to-digital converter as it is informative. You can't just jump into A/D conversion without any understanding of some of the basic fundamentals. You need to read and more importantly understand data sheets because the question you just asked is well addressed in the data sheet. When you don't understand the data sheet then by all means ask questions.Hello,
Did you read page 13 of the datasheet?
There you will find information on filtering the signal.
Bertus
I have searched http://pinouts.ru/Slots/USB_pinout.shtml but I don't understand vcc and ground wire.Use a hand held multi-meter to make sure you have the correct wires. Also you can ring the cable out measuring resistance.
Ron