ISL6742B Datasheet, PDF (16/20 Pages) Intersil Corporation – Fast current sense to output delay

ISL6742B
The current sense signal, which represents the inductor current
after it has been reflected through the isolation and current
sense transformers, and passed through the current sense
burden resistor, is expressed in Equation 17:
V_CS
=
-^NN----^SP-----^---^RN-----^CC----^ST--



I_O
+
D_-----2-_--L-t_{--S-O---W----  VI N}

NN_-----SP--
–
V_{O}



V
(EQ. 17)
Where V_CS is the voltage across the current sense resistor and I_O
is the output current at current limit.
Since the peak current limit threshold is 1V, the total current
feedback signal plus the external ramp voltage must sum to this
value when the output load is at the current limit threshold.
V_e + V_CS = 1
(EQ. 18)
R₉
R₆
R_CS
C₄
1 _VREF
16
2
ISL6742B ¹⁵
3
14
4 _CT
13
5
12
6
11
7 _CS
10
8
9
CT
Substituting Equations 16 and 17 into Equation 18 and solving
for R_CS yields Equation 19:
R_CS = ^N-----^P----_N-^---^N_S-----^C----^T-  _I--_O------+-----_V-L-------_O-O-------_t--_S-¹---_W-----_---_{1-------+-----D-2}------_--

(EQ. 19)
For simplicity, idealized components have been used for this
discussion, but the effect of magnetizing inductance must be
considered when determining the amount of external ramp to
add. Magnetizing inductance provides a degree of slope
compensation and reduces the amount of external ramp
required. The magnetizing inductance adds primary current in
excess of what is reflected from the inductor current in the
secondary.
I_P = ^V-----^I--^N-----_L^---^D_m-----^t--^S----^W---
A
(EQ. 20)
Where V_IN is the input voltage that corresponds to the duty cycle
D and Lm is the primary magnetizing inductance. The effect of
the magnetizing current at the current sense resistor, R_CS, is
expressed in Equation 21:
V_CS = ^-----^I--^P-_N----^_C---^R-_T---^C----^S--
V
(EQ. 21)
If V_CS is greater than or equal to Ve, then no additional slope
compensation is needed and R_CS becomes Equation 22:
R_CS
=
------------------------------------------------------------^N----^C----^T-------------------------------------------------------------
NN_-----SP--




I_O
+
D_---2--t_{-L-S---O-W---}

 ^{V I N}

NN_-----SP--
–
V_{O}



+
V_{-----I--N----L-}_---D_m-----t_--S----W---
(EQ. 22)
If V_CS is less than V_e, then Equation 19 is still valid for the value
of R_CS, but the amount of slope compensation added by the
external ramp must be reduced by V_CS.
Adding slope compensation is accomplished in the ISL6742B
using an external buffer and the CT signal. A typical application
sums the buffered CT signal with the current sense feedback and
applies the result to the CS pin as shown in Figure 17.
FIGURE 17. ADDING SLOPE COMPENSATION
Assuming the designer has selected values for the RC filter (R₆
and C₄) placed on the CS pin, the value of R₉ required to add the
appropriate external ramp can be found by superposition.
V_e – V_CS = -_R2----^D₆-----+-^---^RR----⁶-₉-
V
(EQ. 23)
Rearranging to solve for R9 yields:
R₉ = -^--²----^D------–---_V-^V---_e-^e----–-+---_--^---_V-^V---_C-^C---_S-^S----^----^---^R----⁶-

(EQ. 24)
The value of R_CS determined in Equation 19 must be rescaled so
that the current sense signal presented at the CS pin is that
predicted by Equation 17. The divider created by R₆ and R₉
makes this necessary.
R_CS = ^R-----⁶--_R---+--₉--^R-----⁹-  R_CS
(EQ. 25)
Example:
V_IN = 280V
V_O = 12V
L_O = 2.0µH
N_P/N_S = 20
Lm = 2mH
I_O = 55A
Oscillator Frequency, f_SW = 400kHz
Duty Cycle, D = 85.7%
N_CT = 50
R6 = 499Ω
Solve for the current sense resistor, R_CS, using Equation 19.
R_CS = 15.1Ω.
Determine the amount of voltage, V_e, that must be added to the
current feedback signal using Equation 16.
V_e = 153mV
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FN8565.1
November 3, 2015