hi,
You will have to get some resistors of known value and with a suitable wattage rating if are going to try to do a calibration check of the project.

E

P = V * I = 20 * 4 = 80 Watt

4 A current will flow through 5 ohms
2 A current will flow through 10 ohms
1 A current will flow through 20 ohms

What would be power rating for 10 ohms, 5 ohms and 20 ohms

 

Hello,

Have a look at Ohms Law Formula Wheel:



Bertus

 

I connected 5 ohms 10 ohms and 20 ohms resistor one by one with 5V at 2 A supply But the three resistor were burnt after few second. I think its because power rating of resistor

All of this has been covered. Eric provided some pictures of various wattage resistors. Many post back it was covered what happens as resistors get hot and the result on current.

Your load resistor needs to easily, with a margin, dissipate the power. 5 ohms 10 ohms and 20 ohms with 5 volts applied need to dissipate 5 Watts, 2.5 Watts and 1.250 Watt respectively. Allowing a margin I get 10.0 Watts, 5.0 Watt and 2.5 Watt respectively. All of this has been covered previously. Go back and look at the image Eric posted.

<EDIT> Fixed my lousy math. </EDIT>

Ron

 

Hello,

Your load resistor needs to easily, with a margin, dissipate the power. 5 ohms 10 ohms and 20 ohms with 5 volts applied need to dissipate 1 Watt, 0.5 Watt and 0.250 Watt respectively

Are you sure?
I get 5 watts for the 5 Ohm, 2.5 Watts for the 10 Ohm and 1.25 Watt for the 20 Ohm.
P = (V * V) / R

Bertus

 

A clear photo of the resistors would help.

When i used this resistor It doesn't burn but I don't see any changes at output of sensor in voltage

 

Morning Fanfire,
What is the resistance value of that resistor in the photo ?
E
Do you have an Ohm meter.?

 

Morning Fanfire,
What is the resistance value of that resistor in the photo ?
E
Do you have an Ohm meter.?

Good Morning ericgibbs

Sorry I forgot to mention resistance Its 5 ohms resistor
Can you guess What would be the power rating of this resistor ?

 

hi,
I would guess that is a 0.5W resistor.
What voltage did you measure across the resistor. during test on post #145.?

 

hi,
I would guess that is a 0.5W resistor.
What voltage did you measure across the resistor. during test on post #145.?

I connected 5 ohms resistor with 2V supply so When I measured voltage across resistor it show 0 voltage

 

hi,
You have got to get to grips with the maths...

I= V/R
I= 2v/5R = 0.4Amps.
W= V * I = 2 * 0.4A = 0.8Watts

What is the resolution of your current measuring project.??
What is it's lower limit of measurement.?

E

 

hi,
What is the resolution of your current measuring project.??
What is it's lower limit of measurement.?

E

System should be measure current between 0.5 A to 30 A DC

This is final specification

At 30 A using 60 Volts the load resistance will be 60 Volts / 30 Amps = 2 Ohms.
P = V * I = 60 * 30 = 180 W

At 0.5 A using 60 Volts the load resistance will be 60 Volts / 0.5 Amps = 120 Ohms.
P = V * I = 60 * 0.5 = 30 W

I don't have 60 V supply that's the reason I am doing experiment with 5V and 20 V supply

 

Hello,


Are you sure?
I get 5 watts for the 5 Ohm, 2.5 Watts for the 10 Ohm and 1.25 Watt for the 20 Ohm.
P = (V * V) / R

Bertus

I don't believe I did that. I just took the 5 Volts / Resistance and got Current then for reasons I will never get, I stopped as if it were the power. Then too, after 150 or so post in this thread I believe I achieved burn out more than the resistors are burned up. Maybe I shopuld return to the scene of the crime and fix it.

Thank You a Bunch
Ron

 

System should be measure current between 0.5 A to 30 A DC

This is final specification

At 30 A using 60 Volts the load resistance will be 60 Volts / 30 Amps = 2 Ohms.
P = V * I = 60 * 30 = 180 W

At 0.5 A using 60 Volts the load resistance will be 60 Volts / 0.5 Amps = 120 Ohms.
P = V * I = 60 * 0.5 = 30 W

I don't have 60 V supply that's the reason I am doing experiment with 5V and 20 V supply

You are going to need some power resistors with the power you are showing. Also whet exactly is this project as it sounds like a school assignment? What in an actual application will be drawing this power and where did the 60 volts come from and yes, I understand you do not have a 60 Volt supply but where did the 60 Volts come from?

Ron

 

I don't believe I did that. I just took the 5 Volts / Resistance and got Current then for reasons I will never get, I stopped as if it were the power. Then too, after 150 or so post in this thread I believe I achieved burn out more than the resistors are burned up. Maybe I shopuld return to the scene of the crime and fix it.

Thank You a Bunch
Ron

But I burned all, And now it's turn to my mobile
Can I connect in this way

 

Your mobile is likely not going to draw any current to speak of. Likely so low you will not notice it. Also while the ACS 712 sensor is isolated I would place it on the low side (the negative path for current flow) rather then the high side as drawn. If you want to pursue this you really need some decent loads and one more time is this a school project?

Ron

 

I would place it on the low side (the negative path for current flow) rather then the high side as drawn. If you want to pursue this you really need some decent loads and

I didn't understand can you show me connection in my diagram

one more time is this a school project?

I have to give one topics for project The project has not been assign yet but I am thinking to make meter to measure DC current If my professor likes Idea and research work, he will assign the project.

 

Never mind as it is on the low side. Low side / High Side just means where the sensor is placed. Placing the sensor on the source negative as you have it would be called low side current measuring.

Ron

 

Never mind as it is on the low side. Low side / High Side just means where the sensor is placed. Placing the sensor on the source negative as you have it would be called low side current measuring.

Ron

I connected mobile as we connect any load with acs712

At no load output voltage is 2.56 V

If I connect mobile as shown in picture so output voltage of sensor should be change but There is no change happen

 

hi,
Do you have a multimeter that can measure low levels of current, say 100mA to 500mA.?
If yes, measure the current into the mobile using the meter.? What is the value.?
E

 

If your mobile device has a charged battery, as drawn in your image, the current draw of the mobile device will be close to zero current. My phone, a Samsung Galaxy S7 when charging off a 5.0 volt 2,000 mA charger draws about 2 to 6 watts starting at 6 Watts which is about 1.2 amps maximum on a really low battery. That quickly changes to 2 watts or roughly 400 mA. When my battery has a full charge or close to 100% charge that trickles to several mA.

Start as Eric suggested:

hi,
Do you have a multimeter that can measure low levels of current, say 100mA to 500mA.?
If yes, measure the current into the mobile using the meter.? What is the value.?
E

Remove the ACS712 current sensor and use a meter to actually measure the current. Till you do that you have no idea what you are looking for. A charged or close to charged mobile device is going to draw little to no current.

Ron