Datasheet

 
SLUS544E − SEPTEMBER 2003 − REVISED MARCH 2009
14
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APPLICATION INFORMATION
Determining the value for the slope compensation resistor:
Design Example:
N
CT(p)
= 1 V
OUT
= 12 Np = 7 R
SENSE
= 5.23
F
S(out)
= 500000
N
CT(s)
= 50 L
OUT
= 3.2 x 10
−6
Ns = 5 V
EA(cl)
= 1.98
F
S(out)
= 500000
Where,
• N
CT(p)
= Number of primary turns on the Current Transformer − [Turns]
• N
CT(s)
= Number of Secondary turns on the current transformer − [Turns]
• V
OUT
= Nominal output voltage of the converter − [V]
• L
OUT
= Inductance value of each output inductor − [H]
• N
P
= Number of primary turns on the main transformer − [Turns]
• N
S
= Number of secondary turns on the main transformer − [Turns]
• R
SENSE
= Value of current sense resistor on secondary of current sense transformer − [Ohms]
• V
EA(cl)
= Maximum Value of the E/A output voltage − [Volts]
• F
S(out)
= Switching frequency of each output − [Hz]
Determine the correct value for the slope resistor, R
SLOPE
, to provide the desired amount of slope
compensation.
N
CT
+
N
CT(p)
N
CT(s)
, Current Transformer Turns Ratio
1. Transform the Secondary Inductor Downslope to the Primary
S
L(prime)
+
V
OUT
L
OUT
N
s
N
p
,S
L(prime)
+ 2.679 Ańms
2. Calculate the Transformed Slope Voltage at Sense Resistor:
VS
L(prime)
+ S
L(prime)
N
CT
R
SENSE
,VS
L(prime)
+ 0.2802 Vńms
3. Calculate the R
SLOPE
value to give a compensating ramp equal to the transformed slope voltage given
above.:
M + 1.0
Desired ratio between the compensating ramp and the output inductor downslope ramp, transformed to the
primary sense resistor
R
SLOPE
+
10
4
ǒ
M VS
L(prime)
)
,R
SLOPE
+ 35.69 kW