Application Note AN-1150 - International Rectifier

at which the converter is expected to start-up. Assuming V AC,ON =65VAC and noload
condition at start-up,
VBOP(
HI )
( RBOP
1
+ RBOP2
)
RBOP3
=
( 2. V V −V
)
R
R
BOP3
BOP3
AC,
ON
1.56V
(3MΩ + 3MΩ)
=
( 2.65VAC
−1.56V - 2V)
= 105.9kΩ
BOP(
HI )
Bridge
Next, assuming a target V AC,OFF =55VAC, C BOP has to be selected. First V BOP,AVG
is calculated at V AC,OFF :
V
V
BOP,
AVG
BOP,
AVG
= 2. VAC,
OFF
( RBOP3
)
( π / 2).( RBOP
1
+ RBOP2
+ RBOP3
)
2.55V
AC
.105.9kΩ
=
( π / 2).(3MΩ + 3MΩ + 105.9kΩ
)
V
BOP, AVG
= 0. 86
V
Then, forcing V BOP,MIN (=V BOP,AVG - ∆V BOP /2) = 0.76V, we can calculate the
required ∆V BOP at V AC,OFF . At V AC,OFF =55VAC, this yields
∆V BOP = 2*(0.86-0.76) = 0.2V
In order to calculate C BOP , we just have to force the magnitude of the transferfunction
at f=2*f AC =126Hz to be equal to 0.2V calculated above (maximum f AC is
the design condition that needs to be considered to ensure that the IC is
guaranteed to terminate operation at V AC,OFF . At a lower f AC , when there is higher
ripple, the IC will cease operation at a higher V AC ). Thus:
where:
1
ω
1+
( )
ω O
2
R
= 0.2V
×
R
RBOP3
1
∆VBOP = 2 ⋅V
AC , OFF
⋅ ⋅
= 0. 2V
RTOT
ω 2
1 + ( )
ω
ω
O
=
( R
TOT
BOP3
ω o is then calculated to be:
×
ω = 2π
⋅ (2 ⋅ f
AC
)
R
TOT
BOP1 + RBOP
2
) ⋅
1
2 ⋅V
AC,
OFF
R
BOP3
O
⋅ C
BOP
6.1059MΩ
= 0.2V
× ×
0.1059MΩ
ω = 2 π ⋅ (2 ⋅ f ) = 2π
⋅ (2 ⋅ 63) = 791.68
AC
1
= 0.148
2 ⋅55Vac
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17