8

FN6991.0

December 21, 2009

where D is the percent of on time during a switching

cycle. Setting Q = 1 and solving for Se yields

Equation 10:

Since Sn and Se are the on time slopes of the current

ramp and the external ramp, respectively, they can be

(ΔI) during the on time and Ve is the voltage that must

be added by the external ramp.

For a flyback converter, Vn can be solved for in terms of

input voltage, current transducer components, and

primary inductance, yielding Equation 12:

the minimum input voltage, and D is the maximum duty

cycle.

The current sense signal at the end of the ON time for

CCM operation is Equation 13:

the output current at current limit. Equation 13 assumes

the voltage drop across the output rectifier is negligible.

Since the peak current limit threshold is 1.00V, 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 as shown in Equation 14.

Substituting Equations 12 and 13 into Equation 14 and

Adding slope compensation is accomplished in the

ISL7884xASRH using an external buffer transistor and

the RTCT signal. A typical application sums the buffered

RTCT signal with the current sense feedback and applies

the result to the CS pin as shown in Figure 6.

Assuming the designer has selected values for the RC

required to add the appropriate external ramp can be

found by superposition.

The factor of 2.05 in Equation 16 arises from the peak

amplitude of the sawtooth waveform on RTCT minus a

base-emitter junction drop. That voltage multiplied by

the maximum duty cycle is the voltage source for the

Equation 17:

rescaled so that the current sense signal presented at the

CS pin is that predicted by Equation 13. The divider

Example:

Ns/Np = 10

Lp = 8.0µH

Duty Cycle, D = 28.6%

Equation 15.

S

e

S

n

1

π

---

0.5

+

⎝⎠

⎛⎞ 1

1D

–

-------------

1

–

⎝⎠

⎛⎞

=

(EQ. 10)

V

e

V

n

1

π

---

0.5

+

⎝⎠

⎛⎞ 1

1D

–

-------------

1

–

⎝⎠

⎛⎞

=

(EQ. 11)

V

e

DT

⋅

⋅⋅

L

p

----------------------------------------------------

1

π

---

0.5

+

⎝⎠

⎛⎞ 1

1D

–

-------------

1

–

⎝⎠

⎛⎞

=

V

(EQ. 12)

V

CS

N

⋅

N

P

------------------------ I

O

1D

–

() V

⋅⋅

sw

2L

s

--------------------------------------------

+

⎝⎠

⎜⎟

⎛⎞

=

V

(EQ. 13)

V

e

V

CS

+

1

=

(EQ. 14)

R

CS

1

Df

⋅⋅

L

p

-------------------------------

1

π

---

0.5

+

1D

–

------------------ 1

–

⎝⎠

⎜⎟

⎜⎟

⎛⎞

⋅

N

s

N

p

-------

I

O

1D

–

() V

⋅⋅

2L

s

--------------------------------------------

+

⎝⎠

⎜⎟

⎛⎞

⋅

+

-----------------------------------------------------------------------------------------------------------------------------------------------------

=

(EQ. 15)

CS

RTCT

R6

C4

R9

FIGURE 6. SLOPE COMPENSATION

V

e

2.05D R

6

⋅

R

6

R

9

+

----------------------------

=

V

(EQ. 16)

R

9

2.05D

V

e

–

() R

6

⋅

V

e

----------------------------------------------

=

Ω

(EQ. 17)

R

′

CS

R

6

R

9

+

R

9

--------------------- R

CS

⋅

=

(EQ. 18)

ISL78840ASRH, ISL78841ASRH, ISL78843ASRH, ISL78845ASRH