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Part Number: INA180
I have not used a sensing amp before and would like to have some advice.
I am planning to use the INA180 for two applications.
How would i calculate the sense resistor to use and how to determine the current consumption from the voltage output?
1. A simulation load of 20 ohms will kind of work, but you must be careful to understand that as you change the voltage from 0V to 5V to create varied load currents, you will also be changing the common mode of the device. This will affect the error performance of the part. In terms of offset, it could likely be in your favor as with a common mode of 0V, the offset will be under 150uV guaranteed, but at larger values, it could be up to 500uV.
Another thing to note is the swing to ground spec. If you are trying to measure currents down near 0mA, there will be some point where the amplifier goes non-linear. For the INA180, this is around 5mV.
For shunt resistors, A 1 ohm shunt resistor with 240mA through it creates 240mV of input voltage, and using the A1 version of the device, is x20, so that would produce 4.8V on the output of the shunt resistor. So overall, that would be about the largest value I would expect, but lower values offer lower power consumption and smaller voltage drop. If you have this device in a high side configuration, know that the bus rail at full 240mA is no longer 5V but 4.76V. Current consumption through the shunt is just I=V/R. For reference, the output of the current sense amp is Ishunt * Rshunt * Gain. So, if you had a 20mOhm shunt, a gain of 200 (A4 device), and your current of 240mA, then you would get .24 * .020 * 200 = 0.96V on the output. The output must be lower than the Vs supply of the device, and just a little lower than that to account for swing-to-supply-rail specs.
The error as you get lower and lower in current will increase. There is a sidebar widget in the product page that can help you see what the error curve for your application will look like. I put in a few values and here are the results.
To better understand implementations, the data sheet specs and all the sources of error in current sense amplifiers, there is a training video series we made that can help.
2. Depending on how you implement the current sense amp in the circuit, there are things to consider. The INA180 may not be the best choice for a solenoid application if it's not a low-side or high-side only implementation. For that, I would strongly recommend you look at the INA240, which excels in PWM and in-line solenoid and motor control applications. Also, we produce tech notes that are like very short, 2 page app notes that can help you understand implementations and get product recommendations. There are 5 tech notes already on solenoid and motor control. Please take a look and see if they match your desired implementation.
Jason Bridgmon, TI Sensing Products Applications Support
Current Shunt Monitor Video Training Series
Hall Effect Sensor Video Training Series
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In reply to Jason Bridgmon:
In reply to Dharmesh Joshi:
Jason is out of the office so I will jump in to provide you feedback.
1. If you are using the solar panel as the common mode voltage of our device, that is fine. The INA180 common mode voltage is: –0.2 V to +26 V. As long as you stay within this range it should be fine. Do have in mind, changes in VCM will affect the accuracy of the device. Also, I wanted to mention that the INA180 operate from a single 2.7-V to 5.5-V power supply. In case you decide to power the device with the solar panel, the voltage can't go lower than 2.7-V. Seems like you are using the same 3.3-V rail as the micro controller, so that's great.
2. The INA180 will suffice for the application you are describing.
3. About the value of the shunt resistor. The 1 ohm resistor will not work if you have the 240mA current and a Vs= 3.3V. As Jason mentioned, a 1 ohm shunt resistor with 240mA through creates 240mV of input voltage, and using the A1 version of the device, is x20, so that should produce 4.8V on the output of the shunt resistor. In your application, the output of the device will saturate at 3.3V due to your supply voltage.
Please refer to our video section 4: How to chose and appropriate shunt resistor. This video will help you calculate the value needed for your application.
TI Current Shunt Monitors Applications Support
In reply to Mayrim Verdejo:
Just preforming some calculation and after watching the video.
As my DC-DC step will output 9V 1A, but I have kept ILoad max to 1.5A for my calculation so I have some flexibility.
Presuming Vout is 3.3V and not 9V,
My shunt calculator will be 0.11R for my DC_DC application, which is outputting 9V 1.5A
And for solar application
My shunt calculator will be 0.66R for my DC_DC application, which is outputting 5V 0.240mA.
Does this seem correct
regarding 1 - yes, you should size you shunt appropriately so that the maximum current * rshunt * gain < adc max input voltage.
Also remember to account for a little headroom - if you power the INA180 with 3.3 and expect 3.3V output, it won't reach 3.3 because of the amplifier headroom needed. See this article for more on that. However, if you only need to measure 1A and scale for 1.5A, then you're probably just fine.
for your other application, I'm assuming you meant 5V 0.240A and not 5V 0.240mA. With a 660mΩ shunt and a gain = 20, then that would produce 3.168V at 240mA, which is reasonable.
Thank you for your reply. Yes you are correct it is 240mA.
I will be placing a sample order forthe INA180 and testing them out on a test PCB.
The aim of measuring the current on the 5V 240mA application is also determine the LUX of the light, i am not sure if this is possible to determine this from solar current output or is this a matter of building some correlation table ?
For the application of 9V 1.5A, we need to determine if external device is connected correctly, which i guess will be straightforward.
I have attached my schematic diagram of the INA180, can you advice me if this circuit will be good enough or do i need to have additional filters?
The INA180 does not have a sleep mode enabled and when its not measuring any current i would like to conserve energy, would it be advice-able to power the device directly from GPIO pin of my micro-controller as there current requirement of the device is only 300uA? This way i can do away with any load switch?
From the datasheet it mentions there are two pin configurations , how would i determine which configuration is being ordered as i could not find information the part number indication?
The schematic looks ok to me, and the only thing I usually recommend is adding a spot for a low-pass pi filter to the front end. You can always not populate the cap and install 0 ohm resistors if you don't need it, but it's great to have if you do, and it's good for debug as well.
It looks something like this:
Next, about pin configurations, they are labeled A and B - see below:
You can tell them apart in the ordering table at the end of the data sheet because it says INA180A1 or INA180B1 and so on. The package markings are on the ordering table as well, so you can see which one you have if all you have is the physical part in your hand.
You can power the device from a GPIO switch, but make sure to have a bypass cap on it for the power supply still.
Another thing to think about is that you are driving an ADC, which means you're going to be pulling some current from the INA180 output drive circuitry, which is pretty weak. Adding an RC filter to the output, and possibly even a buffer IC, is something to consider. There is a parameter called "closed loop output impedance" which you can use to determine if a filter will be stable. See this article and this app note for more.
One final note - there is a lux light sensor that TI makes (OPT3001) and it measures darkness up to bright sunlight with good IR rejection and has an i2c interface. Not sure if that helps, but I thought I would throw it out there.
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