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| 型号： | ISL6754AAZA-T |
| 厂商： | Intersil |
| 文件页数： | 13/19页 |
| 文件大小： | 0K |
| 描述： | IC REG CTRLR PWM CM/VM 20-QSOP |
| 标准包装： | 2,500 |
| PWM 型： | 电流/电压模式 |
| 输出数： | 6 |
| 频率 - 最大： | 2MHz |
| 占空比： | 100% |
| 电源电压： | 9 V ~ 16 V |
| 降压： | 无 |
| 升压： | 无 |
| 回扫： | 无 |
| 反相： | 无 |
| 倍增器： | 无 |
| 除法器： | 无 |
| Cuk： | 无 |
| 隔离： | 无 |
| 工作温度： | -40°C ~ 105°C |
| 封装/外壳： | 20-SSOP（0.154"，3.90mm 宽） |
| 包装： | 带卷 (TR) |
ISL6754
V e = ------------------------------------------ ? -------- ? --- + D – 0.5 ?
N P ? π
?
N CT ? L O
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, Fm,
without slope compensation, is:
V n can be solved for in terms of input voltage, current
transducer components, and output inductance yielding:
T SW ? V O ? R CS N S 1
V (EQ. 15)
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 inductor current, when reflected through the isolation
transformer and the current sense transformer to obtain the
current feedback signal at the sense resistor yields:
Fm = --------------------
N S ? R CS ? D ? T SW ? N S ? ?
N P ? N CT ?
V CS = ------------------------ ? I O + --------------------- ? V IN ? -------- – V O ? ?
1
SnTsw
(EQ. 9)
2L O ? N P ? ?
V
(EQ. 16)
where Sn is the slope of the sawtooth signal and Tsw is the
duration of the half-cycle. When an external ramp is added,
the modulator gain becomes:
where V CS is the voltage across the current sense resistor
and I O is the output current at current limit.
Fm = --------------------------------------- = ----------------------------
m c SnTsw
1 1
( Sn + Se ) Tsw
(EQ. 10)
Since the peak current limit threshold is 1.00V, the total
current feedback signal plus the external ramp voltage must
sum to this value.
where Se is slope of the external ramp and
V e + V CS = 1
(EQ. 17)
m c = 1 + -------
Se
Sn
(EQ. 11)
Substituting Equations 15 and 16 into Equation 17 and
N P ? N CT 1
R CS = ------------------------ ? ------------------------------------------------------
I O + -------- T SW ? --- + ---- ?
L ? π 2 ?
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
solving for R CS yields:
N S V O 1 D
O
Ω
(EQ. 18)
damped if the Q-factor is set to 1, and over-damped for
Q > 1, and under-damped for Q < 1. An under-damped
condition can result in current loop instability.
For simplicity, idealized components have been used for this
discussion, but the effect of magnetizing inductance must be
Q = -------------------------------------------------
1
π ( m c ( 1 – D ) – 0.5 )
(EQ. 12)
considered when determining the amount of external ramp
to add. Magnetizing inductance provides a degree of slope
compensation to the current feedback signal and reduces
where D is the percent of on time during a half cycle. Setting
Q = 1 and solving for S e yields:
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.
S e = S n ? ? --- + 0.5 ? ------------- – 1 ?
Δ I P = -------------------------------
1 1
? ? π ? 1 – D ?
(EQ. 13)
V IN ? DT SW
L m
A
(EQ. 19)
V e = V n ? ? --- + 0.5 ? ------------- – 1 ?
? ? π
? 1 – D
?
Δ I P ? R CS
Δ V CS = --------------------------
V
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 .
1 1
(EQ. 14)
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.
13
where V IN is the input voltage that corresponds to the duty
cycle D and L m is the primary magnetizing inductance. The
effect of the magnetizing current at the current sense
resistor, R CS , is:
(EQ. 20)
N CT
FN6754.1
September 29, 2008
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