I'm doing a project to get one water pump running in case of power shortage. I'm using a bank with 3 x 12v batteries and an inverter to keep running this pump. I chose to use a couple of chip multimeter ($4.99) to monitor voltage of batteries and also the power line. Building a voltemeter with the ICL7106 plus an LCD display was more expensive and also more work. I also have a curent Sensor CSLA that can be used until 100 Amps with a sensitivity of 33 mV/A. I would like to use it to monitor the inverter consumption. In the same line of work(the lazy way) could it be possible to adapt one multimeter to display the Amps in in a scale of 0-100?. I mean convert in some way the voltage of the CSLA and display it in one of the multimeter scale in a way that , for instance, when the CSLA indicate 3.3V the display show 100.
Thank you for your help

 

Sounds to me that you need to program some type of pic micro-controller or something along those lines. And for cheap: What about an analogue meter? You can change the back ground to your own design. https://www.sparkfun.com/products/10285 something like that.

 

It's good some points are not clear about your project please share some more details or pics of your project.I think you need to try latest design of sensor which are more reliable and provide you more accurate results.

 

Here's a simple solution. At 100A the output from the CSLA would be 3.3V so you could use a voltage divider with an attenuation of 0.0303 to reduce that to 100mV. This could then be read by your multimeter on the 200mV scale to read 100.0mV on a 3 1/2 digit multimeter corresponding to 100.0A. 115kΩ in series with 3.57kΩ to ground will generate a nominal attenuation of .301. This is within 0.64% of the desired attenuation ±1% if you use 1% resistors.

If you want better accuracy then you could add a 10k pot in series with a 100kΩ resistor for the top of the attenuator with 3.24kΩ going to ground, and adjust for exactly 100.0mV from the attenuator with a 3.3V input.

 

Thanks crutschow!!. I went to try your solution and I realized that the voltage of CSLA has a base voltage of 4.0V so the range is from 4.0V - 7.33V. It was my mistake because I didn't read carefully the Datasheet , if I power the CSLA with a 8.0V the measurable range is half this voltage plus 0.033mV/A. How can overcome this?. There is a way to shift the reading 4.0V?
Thanks a lot for your help!!

 

You can use another resistor divider to generate 4.0V offset from the power supply. Just use two equal value 750Ω ±1% resistors connected in series between the power supply and ground. Use the junction of those two resistors as the common (-) connection for your voltmeter.

Then use that common point to connect a 35.7kΩ resistor in series with a 1.15MegΩ resistor to the CSLA output. Their junction goes to the meter plus input. That should give you 0 to 100mV input to the meter for 0-100A from the CSLA. (I increased this second set of divider resistors by a factor of ten to minimize loading effects.)

Note: I believe you mean 33mV/A not 0.033mV/A.

 

Thank again Crutschow. Now is working!!!

 

Sorry, You were right, it was a typo is 33mV/A

 

Thanks crutschow!!. I went to try your solution and I realized that the voltage of CSLA has a base voltage of 4.0V so the range is from 4.0V - 7.33V. It was my mistake because I didn't read carefully the Datasheet , if I power the CSLA with a 8.0V the measurable range is half this voltage plus 0.033mV/A. How can overcome this?. There is a way to shift the reading 4.0V?
Thanks a lot for your help!!

Me totally agree with that person about sensors info.He had good idea and knowledge about the usage of sensors.