Opamp current measurements

[eem2am]

Hello,

I am required to measure current which can be up to 10 Amps.
Which is the most accurate of the following three methods?….

…one is a non-inverting amplifier, and the other is a differential amplifier.

Non inverting amplifier (Gain 101)
http://i44.tinypic.com/2z9d2f7.jpg

Differential amplifier (inverting) (Gain 100)
http://i42.tinypic.com/e25fo.jpg

Differential amplifier (non-inverting) (Gain 100)
http://i40.tinypic.com/24me3vm.jpg



The opamps feed into microcontroller ADC inputs

Here are the schematics in pdf

 

[Harald Kapp]

Any of these circuits is basically suitable. Precision depends mainly on the selection of components. However, all these circuits require the OpAmp to include GND as allowed input and output voltage ranges, since the input signal is near 0V. Note that the LT1006 in single supply operation can not achieve 0V output, but only very close to that. This is an issue with all OpAmps in single supply operation.

I suggest you use a dual supply, possibly generating a small negative supply using a charge pump IC so 0V at the OpAmp's input is still at least 1-2V away from ground. This way handling a signal of 0V is no problem.

You may also look for dedicated current sense amplifiers, e.g. the LT6102 or a similar product.

 

[KrisBlueNZ]

#1 and #3 are the same apart from the smoothing network (R5 and C2) in #3 which will help with filtering if the current being measured is noisy.

Diagram #3 has a gain of exactly 100 mV per amp measured. Diagram #1 has a gain of 101 mV per amp measured; this could be changed to 100 mV per amp by changing R2 to 99k (or making any change that gives a 99:1 ratio between R2 and R3).

Diagram #2 has three unnecessary components: R3, C2 and R4. These should be replaced with a simple connection from the op-amp's non-inverting input to its negative power connection, either directly or via a 1k resistor (if input bias current is an issue, which I doubt it is for a Linear Technology op-amp). The circuit's gain is 101 mV per amp measured; change R5 to 99k for 100 mV per amp as per diagram #1.

As Harald mentioned, all of these designs may have a problem with the op-amp output going all the way to zero volts. Reducing the output load resistor will help; a value of 1k would be low enough to get "near as dammit" to zero.

An alternative is to connect a signal diode such as a 1N914 in series with the op-amp output, with its anode to the op-amp and its cathode to the output load resistor and the feedback resistor. The output load resistor should also be reduced to say 1k. The feedback capacitor should be connected to the op-amp output, before the diode, for stability.

 

[eem2am]

#1 and #3 are the same apart from the smoothing network (R5 and C2) in #3 which will help with filtering if the current being measured is noisy.

...in that case, i prefer #3. Is there any way of giving more filtering to #1?

Diagram #2 has three unnecessary components: R3, C2 and R4

May i ask is the basis for this statement the fact that the high quality LT1006 has minimal bias current?

 

[KrisBlueNZ]

...in that case, i prefer #3. Is there any way of giving more filtering to #1?

Yes. Change it to match schematic #3 The addition of the filtering is the only difference. The capacitor in the negative feedback path will provide a lot of filtering as well - more than the extra capacitor. So there will be very little difference between the two. Run a simulation using white noise or a frequency sweep and see how similar they are.

(re the three unnecessary components in circuit #2) May i ask is the basis for this statement the fact that the high quality LT1006 has minimal bias current?

Not exactly. The circuit formed by the two resistors and the capacitor has no effect as a filter. The fact that the LT1006 has very low bias current means that you don't need to match source impedances, so you don't need a 1k resistor to replace that network; you can just connect the non-inverting input straight to the negative rail.

 

[BobK]

Am I missing something? If the - input is connected to the negative rail, the feedback network is no longer setting the gain, no? You have open loop gain.

Bob

 

[KrisBlueNZ]

Yes. It's the non-inverting input that's connected to the negative rail. That's the "+" input. There is normal negative feedback around the op-amp.