Re: Re: Op-Amp ----- any experts around this afternoon?

[Ake Hedman <akhe@xxxxxxxxxxxxx>]

Nick,

Do you have access to a negative voltage? In that case you can use a=20
summing op-amp setup and offset the level. (=20
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampvar5.html
)

If you just have +12V you can use the diod I mentioned again to offset=20
the output with 0.6V (60 Kelvin)  or two for  1.2V and so on. With four=20
of them you should have about 0.33V out at 0 Celsius. Not perfect but=20
pretty close.

Cheers
/Ake

Nick wrote:

>
> > just a  correction. There is no need
> > for the diode as the ouput is in
> > Kelvin. Subtract 2.73 from the reading
> > and you get Celsius. Sorry for that.
> >
> > /Ake
>
> Ake, I must be getting there because I was just about to ask why I
> needed the diode!  You beat me to it with your second post.
>
> Thanks for the link.  That shows three resistors which if I am
> calculating correctly, should be:
>
> Rf =3D 20k
> Rin =3D 2.2k
>
> to get the gain of ten (Equally I guess I could use 200k and 22k,
> and I'm not sure how to choose).  Then:
>
> Rz =3D 0.5k
>
> as it suggests a value "equal to the parallel combination of Rf
and
> Rin" which, if I remember rightly is the sum of the reciprocals.
>
> How am I doing so far?  If I am understanding ok so far, then I
> think I have the gain aspect sorted, giving me 1 volt per 10 degrees.
>
> I am not sure how to achieve the "offset" so that the 0-5v
values
> that Comfort expects are at a sensible place within the device's
> spectrum.  In other words, what I have so far, I think, will still
> give 0v at 0 degrees kelvin and 5v at +50 degrees Kelvin.
>
> What I need is 0v at -10 degrees Centigrade and 5v at +40 degrees
> Centigrade.
>
> Hope that makes sense.  Thanks again,
>
> Nick
>
> PS, once I get this sorted, I'll publish the results as it sounds
> like a very cheap way of getting temperatures into Comfort (etc).
>
>
>
>
>
>
>
>
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Ake Hedman (YAP - Yet Another Programmer)
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