参数资料
型号: LTC1735CGN-1#TRPBF
厂商: Linear Technology
文件页数: 22/28页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM CM 16-SSOP
标准包装: 2,500
PWM 型: 电流模式
输出数: 1
频率 - 最大: 335kHz
占空比: 99.4%
电源电压: 3.5 V ~ 30 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: 0°C ~ 85°C
封装/外壳: 16-SSOP(0.154",3.90mm 宽)
包装: 带卷 (TR)
LTC1735-1
APPLICATIO S I FOR ATIO
V ITH Scale Factor =
=
= 0 . 084 V / A
A V =
? V ITH 1 . 37 V
ThenextstepistocalculatetheI TH pinvoltage,V ITH ,scale
factor. The V ITH scale factor reflects the I TH pin voltage
required for a given load current in continuous inductor
current operation. V ITH controls the peak sense resistor
voltage, which represents the DC output current plus one
half of the peak-to-peak inductor current. The no load to
full load V ITH range is from 0.3V to 2.4V, which controls
the sense resistor voltage from 0V to the ? V SENSE(MAX)
voltage of 75mV. For the circuit shown in Figure 8, the
calculated V ITH scale factor is:
V ITH Range ? Sense Re sistor Value
? V SENSE ( MAX )
( 2 . 4 V – 0 . 3 V ) ? 0 . 003
0 . 075 V
V ITH from 0.40V at light load to 1.77V at full load, a 1.37V
change. During Burst Mode operation, the LTC1735-1
output voltage is controlled by a comparator, not the error
amplifier. Even though the error amplifier is not used in
Burst Mode operation, it is necessary to assume linear
operation for all error amplifier gain calculations.
To create the ± 30mV input offset error, the voltage gain of
the error amplifier must be limited. The desired gain is:
= = 22 . 8
Input Offset Error 2 ( 0 . 03 V )
Connecting a resistor to the output of the transconductance
error amplifier will limit the voltage gain. The value of this
resistor is:
= = 17 . 54 k
Assuming continuous inductor current, V ITH is:
R ITH =
A V
Error Amplifier g m
22 . 8
1 . 3 ms
V ITH = ? ? I OUTDC +
? ? V ITH Scale Factor ?
? ? ? I L ? ?
? ? 2 ? ?
+ V ITH Offset
At full load current:
To center the output voltage variation, V ITH must be
centered so that no I TH pin current flows when the output
voltage is nominal. V ITH(NOM) is the average voltage be-
tween V ITH at maximum output current and minimum
output current:
V ITH ( MAX ) = ? ? 15 A +
?
?
V ITH ( NOM ) =
+ V ITH ( MIN )
= + 0 . 40 V = 1 . 085 V
? ?
? ? ?
= 1 . 77 V
5 A P ? P ?
2
?
? 0 . 084 V / A ? + 0 . 3 V
? ?
V ITH ( MAX ) – V ITH ( MIN )
2
1 . 77 V – 0 . 40 V
2
At minimum load current:
The Thevenin equivalent of the gain limiting resistance
V ITH ( MIN ) = ? ? 0 . 2 A +
2 A P ? P ?
? ? 0 . 084 V / A ? + 0 . 3 V
? ?
? ?
? ? ?
= 0 . 40 V
2 ?
?
value of 17.54k is made up of a resistor R5 that sources
current into the I TH pin and resistor R1 that sinks current
to SGND.
To calculate the resistor values, first determine the ratio
k =
= = 3 . 79
Noticethat?I L ,thepeak-to-peakinductorcurrent,changes
from light load to full load. Increasing the DC inductor
current decreases the permeability of the inductor core
material, which decreases the inductance and increases
? I L . The amount of inductance change is a function of the
inductor design.
If the circuit shown in Figure 8 sustained continuous in-
ductor current operation, the error amplifier would control
22
between them:
V INTVCC – V ITH ( NOM ) 5 . 2 V – 1 . 085 V
V ITH ( NOM ) 1 . 085 V
V INTVCC is equal to V EXTVCC or 5.2V if EXTV CC is not used.
Resistor R5 is:
R 5 = ( k + 1 ) ? R ITH = ( 3 . 79 + 1 ) ? 17 . 54 k = 84 . 0 k
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