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Johnny electronics forum beginner
Joined: 21 Jul 2006
Posts: 1
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Posted: Fri Jul 21, 2006 8:53 am Post subject:
Current sense and diff. amp question
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Hi,
I am considering both low-side and high-side approach to DC current
measurement, 100 mA to 6 A range. Accuracy is important and I am
considering discrete implementations (op-amps, etc) over integrated
parts (i.e. MAX4372) for cost reasons.
- Can someone explain what the resistors at the input stages of the op-
amp in this page does, figure 8, Rg1, Rg2:
http://www.maxim-ic.com/appnotes.cfm/appnote_number/746/
I am considering a similar circuit and want to know if I use a generic
op-amp if I need those resistors:
http://www.linear.com/pc/productDetail.do?
navId=H0,C1,C1154,C1009,C1099,P1102
It seems to me the resistors serve to isolate the op-amp circuitry from
the sense resistor and limit the input current to the op-amp terminals.
- For DC accuracy is high-side any better or worse than low-side
assuming either implementation is with an op-amp that has low input
voltage offset? My gain will be 50x , current sense will be supplied
from a switcher, resistors 1% tolerance.
I would think the differential (high-side) technique would yield a
slightly better result since it would elimate any common mode noise.
John. |
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John O'Flaherty electronics forum addict
Joined: 08 Mar 2005
Posts: 68
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Posted: Fri Jul 21, 2006 11:38 am Post subject:
Re: Current sense and diff. amp question
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Johnny wrote:
| Quote: | Hi,
I am considering both low-side and high-side approach to DC current
measurement, 100 mA to 6 A range. Accuracy is important and I am
considering discrete implementations (op-amps, etc) over integrated
parts (i.e. MAX4372) for cost reasons.
- Can someone explain what the resistors at the input stages of the op-
amp in this page does, figure 8, Rg1, Rg2:
http://www.maxim-ic.com/appnotes.cfm/appnote_number/746/
I am considering a similar circuit and want to know if I use a generic
op-amp if I need those resistors:
http://www.linear.com/pc/productDetail.do?
navId=H0,C1,C1154,C1009,C1099,P1102
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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.
| Quote: | It seems to me the resistors serve to isolate the op-amp circuitry from
the sense resistor and limit the input current to the op-amp terminals.
- For DC accuracy is high-side any better or worse than low-side
assuming either implementation is with an op-amp that has low input
voltage offset? My gain will be 50x , current sense will be supplied
from a switcher, resistors 1% tolerance.
I would think the differential (high-side) technique would yield a
slightly better result since it would elimate any common mode noise.
|
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 |
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Mark electronics forum Guru
Joined: 03 May 2005
Posts: 329
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Posted: Fri Jul 21, 2006 2:05 pm Post subject:
Re: Current sense and diff. amp question
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John O'Flaherty wrote:
| Quote: | Johnny wrote:
Hi,
I am considering both low-side and high-side approach to DC current
measurement, 100 mA to 6 A range. Accuracy is important and I am
considering discrete implementations (op-amps, etc) over integrated
parts (i.e. MAX4372) for cost reasons.
- Can someone explain what the resistors at the input stages of the op-
amp in this page does, figure 8, Rg1, Rg2:
http://www.maxim-ic.com/appnotes.cfm/appnote_number/746/
I am considering a similar circuit and want to know if I use a generic
op-amp if I need those resistors:
http://www.linear.com/pc/productDetail.do?
navId=H0,C1,C1154,C1009,C1099,P1102
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.
It seems to me the resistors serve to isolate the op-amp circuitry from
the sense resistor and limit the input current to the op-amp terminals.
- For DC accuracy is high-side any better or worse than low-side
assuming either implementation is with an op-amp that has low input
voltage offset? My gain will be 50x , current sense will be supplied
from a switcher, resistors 1% tolerance.
I would think the differential (high-side) technique would yield a
slightly better result since it would elimate any common mode noise.
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
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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 |
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