Hi all,
thanks for your replies.
Charles, I checked the datasheet for the MAX1242 and I do not seem to
find the input impedance..
Frank, I add values to calculate an average value, this should work fine.
Gaston, thanks for your view as well, but I think I don't really get it.
Even if the LED is not linear, so what? The current meter should still
show the correct value, right? I mean, I cannot adjust it in any other
way than turning one trim pot. So if it shows the correct value for one
setup, it should also show the correct value for all others.
So the only reasonable thing to do is to calibrate it at a higher
voltage as I understand it, right.
I'll let you guys know.
Many thanks,
Jens
Am 17.03.2011 13:47, schrieb GastonP:
On Mar 16, 6:42 pm, Jens Boos<[email protected]> wrote:
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 :-)
There is a failure in the reasoning here. You have a voltage source we
can assume with low impedance, then a led with a given voltage drop, a
trimpot to adjust the current through the network then two amp meters
in series and then the ground.
Well, LEDs as references are not linear, and less than all, unknown
LEDs.
In all your measurements you have assumed that it has a fixed voltage
between its terminals, like a battery, when it does not, and also a
1:5 current change does not help.
Take that LED out of the circuit to do the calibration, make sure that
the voltage source you use to do the calibration is as high as
possible (which makes the voltage drops in *both* amp meters to be
negligible, and see how having all linear elements in it will do the
trick.
Regards
Gaston
--
You received this message because you are subscribed to the Google Groups
"neonixie-l" group.
To post to this group, send an email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/neonixie-l?hl=en-GB.
