Hi you guys,

thanks for the hints, I was already suspecting the shunt resistor to
be too large - and as it turns out, some bad soldering increased the
value so it was much larger than 5 ohms. Thanks for the hints though,
you were dead on.

So now I get more reasonable values, but it is still not working
perfectly. I tried to calibrate it this way:

A LED with a series resistor driven at 8.00mA (it is a rather bright
LED), this current was checked with my multimeter in series with my
amperemeter. I then adjusted the op amp with a trim pot so that my
amperemeter displayed 8.0mA. I then reduced the voltage of the LED so
my multimeter showed only 1.7mA. But sadly, the amperemeter shows
1.4mA.

So there seems to be some error in linearity. This is something I
cannot fix with my setup, since I only have one trim pot. I do not
know what to make of it, do you have any ideas where this error might
come from? You proved right about the first problem, so maybe you can
help me now, too :-)

Background: I use a 10 Bit ADC (MAX1242) at the internal reference of
2.5V, amplify the voltage drop over my 5 ohms shunt resistor by 15.6,
add 20 values and then divide by 64.

Best regards,
Jens


On 16 Mrz., 06:42, "JohnK" <[email protected]> wrote:
> I have trouble picturing the circuit arrangement; BUT 11V across 5 ohms is
>  >2 Amps which is >20 Watts !! Does the resistor get hot? And where do you
> get >2 Amps?
>
> Your resistor is >>5 ohms?
>
> John K.
>
> ----- Original Message -----
> From: "Jens Boos" <[email protected]>
>
> ....clip.....>
> > This confused me. The shunt resistor is 5 ohms, so there should not be
> > any significant voltage drop across the shunt resistor at 2, 3, 4 mA.
> > But the strange part is: The voltage drop is 11V and something, and
> > there is no current flowing with the switch closed.
>
> > Does someone understand what has happened, or is fiurther information
> > required?
>
> ...clip....

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