ISL6742 Datasheet, PDF (14/18 Pages) Intersil Corporation – Advanced Double-Ended PWM Controller

ISL6742
Slope Compensation
Peak current-mode control requires slope compensation to
improve noise immunity, particularly at lighter loads, and to
prevent current loop instability, particularly for duty cycles
greater than 50%. Slope compensation may be accomplished
by summing an external ramp with the current feedback signal
or by subtracting the external ramp from the voltage feedback
error signal. Adding the external ramp to the current feedback
signal is the more popular method.
From the small signal current-mode model [1] it can be
shown that the naturally-sampled modulator gain, F_m,
without slope compensation, is expressed in Equation 10:
F_m
=
------¹--------
S_nS_n
(EQ. 10)
where S_n is the slope of the sawtooth signal and t_SW is the
duration of the half-cycle. When an external ramp is added,
the modulator gain becomes Equation 11:
F_m
=
-----------------¹-------------------
(S_n + S_e)t_SW
=
------------¹--------------
m_cS_nt_SW
(EQ. 11)
where S_e is slope of the external ramp and:
m_c
=
1
+
S_-----e-
S_n
(EQ. 12)
The criteria for determining the correct amount of external
ramp can be determined by appropriately setting the
damping factor of the double-pole located at half the
oscillator frequency. The double-pole will be critically
damped if the Q-factor is set to 1, over-damped for Q > 1,
and under-damped for Q < 1. An under-damped condition
may result in current loop instability.
Q
=
------------------------¹-------------------------
π(m_c(1 – D) – 0.5)
(EQ. 13)
where D is the percent of on-time during a half cycle (half
period duty cycle). Setting Q = 1 and solving for S_e yields
Equation 14:
S_e
=
S _{n ⎝⎛}
⎛
⎝
1_--
π
+
0.5_⎠⎞
------¹-------
1–D
–
1_⎠⎞
(EQ. 14)
Since S_n and S_e are the on-time slopes of the current ramp
and the external ramp, respectively, they can be multiplied
by t_ON to obtain the voltage change that occurs during t_ON.
V_e
=
V _{n ⎝⎛}
⎛
⎝
1_π--
+
0.5_⎠⎞
------¹-------
1–D
–
1_⎠⎞
(EQ. 15)
where V_n is the change in the current feedback signal during
the on time and V_e is the voltage that must be added by the
external ramp.
V_n can be solved for in terms of input voltage, current
transducer components, and output inductance yielding
Equation 16:
V_e
=
-^t-^S----^W---_N----^â‹_C---^V-_T---^O--_â‹---_L-^â‹---^R_O----^C----^S--
⋅
N_-----S--
N_P
⎛
⎝
1_--
π
+
D
–
0.5_⎠⎞
V
(EQ. 16)
where R_CS is the current sense burden resistor, N_CT is the
current transformer turns ratio, L_O is the output inductance,
V_O is the output voltage, and N_S and N_P are the secondary
and primary turns, respectively.
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
=
N_-----S-----⋅_---R_{-----C----S--}
N_P ⋅ N_CT
⎛
⎜
⎝
I_O
+
D_------⋅_---t_--S----W----
2L_O
⎛
⎝⎜ ^{V I}
N
⋅
N_-----S--
N_P
–
⎞⎞
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)
Substituting Equations 16 and 17 into Equation 18 and
solving for R_CS yields Equation 19:
R_CS
=
N_-----P-----⋅_---N_-----C----T-
N_S
⋅
-------------------------¹---------------------------
I_O
+
V_-----O--
L_O
t_S
⎛
W⎝
1_π--
+
D_{-2--⎠⎞}
Ω
(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.
ΔI_P
=
V_{-----I--N-----}⋅_---D_-----t_--S----W---
L_m
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:
ΔV_CS
=
Δ_-----I_--P-----⋅_---R_----C----S--
N_CT
V
(EQ. 21)
14
FN9183.2
October 31, 2008