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

--
Ned Konz
http://bike-nomad.com
GPG key ID: BEEA7EFE

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