68
Z31980_00_03
Installation
Pressing OK updates the BSMS field value in edlock table. A subsequent lock procedure
should be able to find the lock signal in the usual way.
Fine-tuning the flock AU program.
In the AU program, one can find the line:
static const double scale = 19.79;
It defines the ratio between the ppm scale and the BSMS field values. In principle, any
Fourier should be close to that value. It is recommended, though, to calibrate it for any
instruments as the shim current boards are not completely identical to each other.
To determine this value, one has to acquire two proton spectra. For this the Lock should
be OFF. After acquisition of the first spectrum, one changes the lock field by 1000 units
(+/- doesn't matter). With the new field value another spectrum is acquired. Now, deter-
mine the ppm shift of the two spectra in respect to each other, e.g. in the dual display
mode. This value should be close to 20ppm. Please correct the line of the AU program
shown above to reflect the accurate value (take the absolute value, no negative sign).
Third approach: Acquisition of a proton spectrum
An easier and unambiguous approach is to acquire a proton signal using a large spec-
trum width SW. The easiest sample is “Doped Water” with just one large resonance line
from H
2
O. The solvent is D
2
O. Therefore, the lock frequency in ppm on the 2H scale is
very close to that of the proton spectrum on the 1H scale (= 4.7ppm). First try to lock on
D
2
O in order to transfer the lock parameters for D
2
O to the lock system. The lock will fail
because we are in a situation there the signal is outside of the lock window. Otherwise,
we wouldn’t need to search for the lock. With the “Sweep” set to off, change the field
value such that the 1H resonance line is at 4.7ppm. This is also the right field value for
the lock signal. Now, inspecting the lock window, it is possible to set the phase correctly.
Save both values in the “edlock” window to finish the procedure.
The procedure is illustrated using Cyclosporine as a sample. There is always a reso-
nance line from the protonated lock solvent because chemical exchange is easy
between 1H and 2H. In cyclosporine, it is the largest signal on the left-handed side of the
spectrum resulting from C6D6H.
If the field value is off, it may look like this.
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