Current sense and diff. amp question

I think RG1 is an essential part of the current sensing. The amplifier forces the transistor connected to its output to carry enough current so the drop across RG1 exactly matches the drop across the external Rsense, making the amplifier differential input voltage equal to zero. That current then follows the current through Rsense in the ratio Rsense / RG1. Then RG2 is probably to match the resistance seen by the two inputs' bias currents to assure high CMRR.

The high-side measurement could have more errors due to the high common mode voltage applied to the inputs, unless the opamp has high CMRR and the resistors are well-matched. (The article talks about this, in the paragraph just after figure 3. It seems to be one of the main selling points for the integrated sensor.) On the other hand, if the DC voltage level at the high side is regulated, maybe you could calibrate out whatever errors you have, depending on just how accurate the measurement has to be.

-- john

right...ow side does not suffer from common mode errors...but sometimes you cannot put the sense resistor in the low side....if you are able to put the sense resistor in the low side, that is the way to go....eliminates common mode rejection errors...

Mark

Why are the op-amp inputs of figures 4 and 5 reversed, for the identical circuit?

They drew figure 5 wrong- the inputs have to be reversed from what they have to get negative feedback.

-- john

-- john

Hi John,

Thanks for your resp> >

Do you have any idea then why RG1 & RG2 do not appear in the Linear Tech circuit (above)?

Thanks for the additional info on high-side and low-side considerations...

Regards. John.

Hi Johnny,

The first thing you should consider is, whether you know the direction of the current and it is ony this one direction or if the currend may flow in both directions.

In the first case take the scematic from Linear, it is quite easier to built. Take a look at Zetex. They have a few devices ready to use... I'm not shure that you may bet the price with the same accuracy.

Why it should? A differential amplifiers CMRR depends on few things, CMRR of OP, but most the tolerances of the resistors and the parasitaric capacitances too. The current mirror designs dont relay on those tolerances. The almost only parameter is the CMRR of the OPAMP. It will be much cheaper than diffamp. Take a OP with low offset voltage and enjoy.

Marte

Have a look at this app note.

Well, it does have RG1, but they call it R1. That is the main scaling resistor. They may have omitted the other resistor because the input bias current is only 10 nA; it might be more important if R1 was higher than 200 ohms.

-- john

Hi Marte,

Thank you for your resp>

For now I am dealing with unidirectional current.

The only issue with the Zetex parts, which I came across in my searches, they have a max gain of 10x, and I need 50x. So I would have to follow the Zetex part with an op-amp gain of 5x.

Thus I figured using an op-amp for the whole task MIGHT be better. Then again, I suspect I would have to worry less about the input voltage offset if I only had a gain of 5x...

Sorry for being unclear in my use of terminology. When I said "differential technique" I was referencing the LTI way of using an op- amp with high CMRR to amplify the difference in sense resistor voltage to drive a current mirror.

Thanks again, John.

Kell,

Thanks for the l>