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Topic: Measuring Small Currents (<= 100mA) (Read 505 times) previous topic - next topic

Coding Badly


I've decided to try measuring the current consumption of a battery powered gadget.  The measured current is always less then 100mA.  The voltage varies between 3.1 V and 1.8 V.  I'm currently using an Uno to measure the voltage and I want to also measure the current.  The Uno is powered from a USB hub (5.05 V).

I can use an instrumentation amplifier on another more permanent project so I've decided to try using one for this test.

Browsing Mouser, I settled on this instrumentation amplifier...
http://www.mouser.com/ProductDetail/Texas-Instruments/INA126PAG4/?qs=sGAEpiMZZMuvJlO0BgfYbBOz9neZdvOf
http://www.ti.com/lit/ds/sbos062a/sbos062a.pdf

Some things I like about it: The gain is in range (I believe I need 100 or 500 gain).  The supply voltage is very versatile (1.35V to 18V).  It appears to be easy to use (one external resistor to set the gain and two bypass capacitors).  It's bread board friendly.

My plan is to power the amplifier from the Uno (5.05V).  The input will be from a low-side 0.1 ohm shunt.  Absolute accuracy is not important.  I'm primarily interested in relative differences (e.g. if I change X what is the percent change in current consumption).

My problem is that I don't have a clue how to read the datasheet.  I don't know which parameters might be a cause for concern, which parameters I can ignore, and which parameters indicate this is a good choice for amplifying a 0 to 10 mV signal to be fed into an Uno's analog input.

Any guidance about this (or any other appropriate) amplifier is appreciated.

MarkT

That device's common-mode input range doesn't extend to either supply rail - so you can't put the shunt at 0V.

However since your circuit under test is from 1.8 to 3.1V it is probably possible to place the shunt high-side, which would
I think be in the common mode range OK.

Its plenty accurate enough, just be sure to run the sense wires from directly on the shunt resistor (so it becomes a
4-terminal resistor in effect), for best accuracy.

pin 12 will need to go to a virtual ground (about 2V from a voltage divider would do), and the output voltage
would be limited to about 2V to 3.5V I think.

It would actually be a lot easier with a rail-to-rail opamp - you don't need an instrumentation grade amp by
the sound of things, and they usually aren't rail-to-rail as its not the most important aspect of an instrumentation
amp.  With a rail-to-rail you just set up a standard differential amplifier circuit  - and the shunt can be at
any voltage from 0 to 5V....
[ I won't respond to messages, use the forum please ]

oric_dan

As Mark basically indicates, an instrumentation amp is good for differential measurements, and
especially useful for high-side measurements where you need to measure 2 voltages for comparison.

For the case of a resistor with 1 side grounded, you can use a simple non-inverting opAmp ckt
with a gain of 10-25. As he says, just be sure to get an 0pAmp with ground in its "input common
mode voltage range". This one will run off 5V supply. Note the output voltage does not go all the
way to Vcc=5V.

http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_23966_-1

Here is the sort of ckt to use, non-inverting variety,

http://www.elexp.com/t_gain.htm




Krodal

http://www.maximintegrated.com/app-notes/index.mvp/id/746

Coding Badly


Thank you for the reply.

That device's common-mode input range doesn't extend to either supply rail...


Did you determine that from the INPUT COMMON-MODE RANGE section?

Quote
...- so you can't put the shunt at 0V.

Quote
The common mode range is limited on the negative side...


If I did put the shunt at 0V, would I be giving up the low range?  In other words, would the circuit work correctly until the voltage across the shunt approached zero?

oric_dan


If I did put the shunt at 0V, would I be giving up the low range?  In other words, would the circuit work correctly until the voltage across the shunt approached zero?

See reply #2.

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