# Output low voltage of op amp?

Discussion in 'General Electronics Discussion' started by eem2am, Nov 15, 2012.

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1. ### eem2am

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Aug 3, 2009
Hello,

I am using the following differential amplifier to measure current in the sense resistor.

I dont need great accuracy, but just need to know if its 1A, 2A, 3A, ...10A
schematic:

(the opamp used will be OPA335 .....-> http://www.ti.com/lit/ds/symlink/opa335.pdf )

....when the current is 1A, the output of the opamp will be just 100* 1mV = 100mV.

...unfortunately, page 3 of the OPA335 datasheet says that the VOL of the OPA335 is 100mV.....so does this mean that i wont be able to measure this 1A current?

Also, regarding the VOL value on page 3 of the OPA335 opamp datasheet, i dont understand how VOL is lower when the load is 100K, compared to when its 10K.......i would have thought that the 10K load would make the VOL lower than when there was a 100K load?

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2. ### Harald KappModeratorModerator

11,522
2,654
Nov 17, 2011
That is a common problem with rail-to-rail amplifiers. Since you have to take into account the output circuit's impedance you can't go all the way to rail. There will always be a (small) voltage left.

Your circuit is supposed to operate as differential amplifier - but it doesn't really.
R2, R5 and C1 are in parallel and effectively the "+" input of the OpAmp is at GND. This leaves you with an inverting amplifier.
For this application I suggest you use an instrumentation amplifier. You can build one from Standard operational amplifiers or use an integrated circuit, e.g. INA333 (no, I'm not associated with TI). I recommend the integrated solution as this is internally trimmed. By referring the REF pin not to GND but some other voltage (e.g. from a reference source) you can have the amplifier's output at this reference voltage for a 0V input differential. A simple, not too precise means would be a biased diode (0.7V) at the ref pin. If your microcontroler has 2 analog inputs, you can measure the output voltage of the amplifier and the reference volate and subtract the latter from the first. Thus you have only the input voltage differential for further processing.

Harald

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Jan 21, 2010