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'[EE]:Shunts versus ceramic resistors'
2003\02\17@104246 by Ian McLean

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I am building a 0 to 8A current sensor as part of a PIC16F877 project.  The
voltage supply is 12V.

I am wondering whether I should really be using proper shunt resistors to
create the voltage drop, or is it OK to use normal, easily obtained 5W or
10W ceramic wirewound resistors ?

I was figuring on using 2 x 5W 0.1R wirewound resistors in parallel for
0.05R at 10W, with the following calculations...

From my calculations, the power dissipation is I2R = 64*0.05 = 3.2W - so 2 x
5W resistors better than one to avoid overheating.

Now, the voltage developed across the resistors is = IR = 8 x 0.05 = 400mV.
I am planning to op-amp this to get about 0 to 3V out for the PIC AD input.
This is easily achieved using a cheap op-amp such as an LM358, esp. if a
negative rail is supplied to get the output down to zero.  If I forego the
negative rail, I cannot quite get down to zero (not that that would be any
great hassle anyway) - I figured I would need a more expensive rail-to-rail
op-amp for that.

I have SPICE'd the differential amplifier and it seems to work ... but I am
not sure if SPICE takes component heating into account in it's simulations
(I am using Protel CircuitMaker).  I can supply a circuit if anyone feels
the need for it to help me answer my question, which is ...

My main question of concern before I go ahead and prototype this is, should
I be too concerned about the voltage drop changing across the resistor as it
heats up, or is the difference negligable ?  I am not too concerned with
high resolution - even 1/8 or 1/4 amp resolution is fine for my application.

Any help would be appreciated.

Rgs
Ian

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2003\02\17@111823 by Larry Bradley

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A couple of things to suggest:

Use a smaller shunt resistor (.005 ohms - 40 mv at 8 amps) and set up the
opamp for a gain of 100 to give you 4 volts out. This will reduce your
heating - now  you are dissipating .04*8 = 0.32 watts.

Finding such a small value resistor may be a real pain, however. You can
wind your own using copper wire tables to determine what gauge and length
you need. Problem here is that copper has a poor temperature co-efficient.

Use a low-offset rail-to-rail opamp such as the LMC6462 - .25 mv offset, vs
2 mv offset of the 358. They are a couple of bucks each from Digikey, as i
recall.

If you want to proceed as per your original design, the 0.1 5 watt in
parallel should work OK. Looking at the Dale web site for power resistors,
it appears that the temperature coeffiecient is around 600 ppm per deg
C.  If the resistor heats up by 50 degrees, that will be 30,000 ppm or 3%.
I don't know if you will get that much of a temperature rise using a 10
watt resistor dissipating only 3 watts or so.

If you want to use the LM358, just use a simple charge pump such as an
ICL7662 or equivalent with 2 10uf tantalum capacitors and you have instant
-ve voltage for the 358.

You don't really need to worry about the 2 mv offset of the 358 - you can
compensate for that in your program. Same with the resistor tolerance - you
can calibrate the beast.

Have fun!

Larry


At 01:09 PM 2/17/2003 +1100, you wrote:
{Quote hidden}

Larry Bradley
Orleans (Ottawa), Ontario, CANADA

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2003\02\17@112447 by Alan B. Pearce

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>I am planning to op-amp this to get about 0 to 3V out for the
>PIC AD input. This is easily achieved using a cheap op-amp
>such as an LM358, esp. if a negative rail is supplied to get
>the output down to zero.  If I forego the negative rail, I
>cannot quite get down to zero (not that that would be any
>great hassle anyway) - I figured I would need a more expensive
>rail-to-rail op-amp for that.

Check out the Maxim Semiconductor web site for chips they make for doing
"hot side" current measurement. You are still going to need some form of
high wattage low resistance shunt, but as I understand it, these chips will
convert the voltage across the shunt to a ground referenced voltage for you.
See http://www.maxim-ic.com/quick_view2.cfm/qv_pk/3597 for an example. There
may be others suitable as well, I noticed that there is one which has what
looks like a PWM output instead. Also check out the application note link at
the bottom of the page.

The only thing I would suggest about the resistors you are proposing is to
check out the temperature co-efficient of the resistor. Without knowing what
accuracy you want, the resistors you propose may be sufficient, but if you
want better accuracy, then you may have to use more resistors to keep the
temperature change down to a level where the resistance change becomes your
biggest source of inaccuracy. OTOH you may be able to have a table in the
PIC to compensate for such a change if this is needed.

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2003\02\17@112909 by Ned Konz

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On Sunday 16 February 2003 06:09 pm, Ian McLean wrote:
> From my calculations, the power dissipation is I2R = 64*0.05 = 3.2W
> - so 2 x 5W resistors better than one to avoid overheating.
>
> Now, the voltage developed across the resistors is = IR = 8 x 0.05
> = 400mV. I am planning to op-amp this to get about 0 to 3V out for
> the PIC AD input. This is easily achieved using a cheap op-amp such
> as an LM358, esp. if a negative rail is supplied to get the output
> down to zero.  If I forego the negative rail, I cannot quite get
> down to zero (not that that would be any great hassle anyway) - I
> figured I would need a more expensive rail-to-rail op-amp for that.

Why in the world would you want to dissipate so much power in your
sense resistor? You're just going to be cooking all the other
components and wasting input power.

There are lots of relatively cheap op amps that have much better
offset specs than LM358s.

Is this a high volume project where a few cents of op amp price makes
a difference? If it's a one-off, you could even spend $1 or $2 more
and get a real instrumentation amp that also has the gain setting
resistors in it.

For instance, the (dual) LMC6482 is only $2.29 in singles from
DigiKey. It's got rail-to-rail inputs and outputs, will run off 3 to
15V supplies, is available in DIP as well as modern packages, and has
a Vos of 750uV.

With a gain of 100, you'd be looking at 30mV full scale, or about 1/4W
dissipation in the sense resistor. The 3 milliohm current sense
resistor could be a piece of fine wire if you didn't care about the
tempco. Or (better) just spend 42 cents and get a 5 milliohm 1W sense
resistor:
http://www.digikey.com/scripts/us/dksus.dll?Detail?Ref=93462&Row=90481

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2003\02\17@142543 by Ian McLean

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Thank you.  I think I may just go with the easier solution here and use the
Maxim chip MAX4080.  It both ground references the voltage and provides a
full 5V swing output.

Thanks for the link.

{Original Message removed}

2003\02\17@172734 by Olin Lathrop

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> My main question of concern before I go ahead and prototype this is,
should
> I be too concerned about the voltage drop changing across the resistor
as it
> heats up, or is the difference negligable ?  I am not too concerned with
> high resolution - even 1/8 or 1/4 amp resolution is fine for my
application.

So check the data sheet for the resistor (RTFM)!!

250mA out of 8A is about 3%.  The data sheet should be able to tell you if
the resistor is guaranteed to change by less than that over temperature or
not.  Then there is the issue of accuracy too, which you didn't mention.


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