Maxim Integrated 71M6541 Demo Board - Manual (Page 39)

71M6541 Demo Board REV 3.0 User’s Manual
39 Rev 4.0
5.
) cos( 2
2
180 0
S
XI XV
E E
A A
φ
+ +
=
6. 
) cos(
1 2 ) (
180 0
S XV
XI
A
E E
A
φ
+ +
=
Use above results along with E
60
and E
300
to calculate
φ
S
.
7. 1
) 60 cos(
) 60 cos(
60
−
−
=
IV
A A IV
E S XI XV
φ
1 ) sin( ) 60 tan( ) cos(
− + =
S XI XV S XI XV
A A A A
φ φ
8. 1
) 60 cos(
) 60 cos(
300
−
−
− −
=
IV
A A IV
E S XI XV
φ
1 ) sin( ) 60 tan( ) cos(
− − =
S XI XV S XI XV
A A A A
φ φ
Subtract 8 from 7
9.
) sin( ) 60 tan( 2
300 60 S XI XV
A A E E
φ = −
use equation 5:
10. ) sin( ) 60 tan(
) cos(
2
180 0
300 60 S
S
E E
E E
φ
φ
+ +
= −
11.
) tan( ) 60 tan( ) 2 (
180 0 300 60 S
E E E E
φ + + = −
12.  





+ +
−
= −
) 2 )( 60 tan(
) (
tan
180 0
300 60 1
E E
E E
S
φ
Now that we know the A
XV
, A
XI
, and
φ
S
errors, we calculate the new calibration voltage gain coefficient from the
previous ones:
XV
NEW
A
V CAL
V CAL
_
_
=
We calculate PHADJ from
φ
S
, the desired phase lag:
[ ]
[ ] 





− − − −
− − − +
=
− −
− −
) 2 cos( ) 2 1 ( 1 ) tan( ) 2 sin( ) 2 1 (
) 2 cos( ) 2 1 ( 2 ) 2 1 ( 1 ) tan(
2
0
9
0
9
0
9 2 9
20
T f T f
T f
PHADJ
S
S
π φ π
π φ
And we calculate the new calibration current gain coefficient, including compensation for a slight gain increase
in the phase calibration circuit.
2 9
0
9
0
9 20 20
) 2 1 ( ) 2 cos( ) 2 1 ( 2 1
)) 2 cos( ) 2 1 ( 2 2 2 ( 2
1
1 _
_
− −
− − −
− + − −
− − +
+
=
T f
T f PHADJ PHADJ
A
I CAL
I CAL
XI
NEW
π
π