2a.2 example 2 (for positive offsets), 5 the correcting component definition, 2a2 example 2 (for positive offsets) - 1 0 – Fluke 900 User Manual

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FLUKE 900 SERVICE MANUAL

According to table 1 in Section S.4.2.5 , a 47K resistor should be used (it gives OS 5).
Other blocks will be also affected by this resistor. Therefore, all of them have to be
checked to determine whether they are still within specification. If any block is pushed
out of spec with this resistor (for example if 240 ns has offset 0 it will be 4 after
correction) the OS calculation procedure has to be repeated. OS is acceptable if aU
blocks are within +-4 margin after correction.
If the OS and resistor cannot be defined (if the smallest possible OS in one block

pushes another one out of spec) go to Section S.4.2.6.2.

5.4.2A.2 Example 2 (for positive offsets)

Assume line 5 has the biggest offset in 160 ns block.

pin#

U~H

.160

D-H

ns. .

U-L D-L

.

.

.

.

5

+ 3

+ 4

+ 6 +6

The biggest offset are D-L and U-L. They have to be moved down 2 units to make it
+4. Offsets should be moved down by 5 or 4 in order to create the safety margins as
in the previous example. If we choose 4 (either of the two values may be selected) it
will be:

. .160 ns. .

U-H D-H

U-L

D-L

-1 0

+2

+2

The needed capacitor can be found using table 2 in Section 5.4.2.3. It is 3.3 pF. AU
other blocks should be now checked as described in 5.4.2.4.I.
If the OS and capacitor cannot be defined (if the smaUest possible OS in one block

pushes another one out of spec) go to Section 5.4.2.6.2.

5.4.2.5 The Correcting Component Definition

If a pin has an excessive positive offset, a capacitor has to be added. In case of a too large
negative offset a resistor should be used.

The value of the component to be added can be defined with the foUowing tables:

NOTE: AU values intiie foUowing tables are only approximated; the tolerance is+/- 1.

5 - 1 0

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