参数资料
型号: IRU3034CS
厂商: International Rectifier
文件页数: 8/12页
文件大小: 0K
描述: IC CTRL/REG PWM SWITCH 8-SOIC
标准包装: 95
应用: 控制器,Intel Pentium?,II,P55C
输入电压: 12V
输出数: 1
输出电压: 2 V ~ 3.5 V
工作温度: 0°C ~ 70°C
安装类型: 表面贴装
封装/外壳: 8-SOIC(0.154",3.90mm 宽)
供应商设备封装: 8-SOIC
包装: 管件
产品目录页面: 1384 (CN2011-ZH PDF)
其它名称: *IRU3034CS
IRU3034-CS
IRU3034-CS-ND
IRU3034 & (PbF)
APPLICATION INFORMATION
Introduction
The IRU3034 device is an application specific product
designed to provide an on-board switching supply for the
new generation of microprocessors requiring separate
Core and I/O supplies where the load current demand
from the I/O supply requires this regulator to also be a
switching regulator such as the motherboard applica-
tions with AGP slot or the Pentium II with on-board 5V
to 3.3V converter. The IRU3034 provides an easy and
low cost switching regulator solution for Vcore and 3.3V
supplies with true short circuit protection.
Switching Controller Operation
The operation of the switching controller is as follows:
After the power is applied, the output drive pin (Drv) goes
to 100% duty cycle and the current in the inductor
charges the output capacitor causing the output voltage
to increase. When output reaches a pre-programmed
set point the feedback pin (V FB ) exceeds 1.25V causing
the output drive to switch Low and the V HYST pin to switch
High which jumps the feedback pin higher than 1.25V
resulting in a fixed output ripple which is given by the
following equation:
? Vo = (Rt/Rh) × 11
Where:
Rt = Resistor connected from V OUT to the V FB pin of
IRU3034.
Rh = Resistor connected from V FB pin to V HYST pin.
For example, if Rt=1K and Rh=422K, then the output
ripple is:
? Vo = (1/422) × 11 = 26mV
The advantage of fixed output ripple is that when the
output voltage changes from 2V to 3.5V, the ripple volt-
age remains the same which is important in meeting the
Intel maximum tolerance specification.
Soft-Start
The soft-start capacitor must be selected such that dur-
ing the start-up when the output capacitors are charging
up, the peak inductor current does not reach the current
limit threshold. A minimum of 0.1 μ F capacitor insures
this for most applications. During start-up the soft-start
capacitor is charged up to approximately 6V keeping
the output shutdown before an internal 10 μ A current
source start discharging the soft-start capacitor which
slowly ramps up the inverting input of the PWM com-
parator, V FB . This insures the output to ramp up at the
same rate as the soft-start cap thereby limiting the input
current. For example, with 0.1 μ F and the 10 μ A internal
current source the ramp up rate is:
( ? V/ ? t) = I/Css = 10/0.1 = 100V/s or 0.1V/ms
Assuming that the output capacitance is 6000 μ F, the
peak input current will be:
I IN(pk) = Css × ( ? V/ ? t) = 6000 μ F × (0.1V/ms) = 0.6A
The soft start capacitor also provides a delay in the turn
on of the output which is given by:
T D = Css × K
Where:
K = 30ms/ μ F
For example for Css=0.1 μ F,
T D = 0.1 × 30 = 3ms
Switcher Current Limit Protection
The IRU3034 uses an external current sensing resistor
and compares the voltage drop across it to a programmed
voltage which is set externally via a resistor (R CL ) placed
between the CS- terminal of the IC and V OUT . Once the
voltage across the sense resistor exceeds the thresh-
old, the soft-start capacitor pulls up to 12V, pulling up
the inverting pin of the error comparator higher than non-
inverting which causes the external MOSFET to shut
off. At this point the CS comparator changes its state
and pulls the soft-start capacitor to Vcc which is 12V
and shutting the PWM drive. After the output drive is
turned off, an internal 10 μ A current source slowly dis-
charges the soft-start capacitor to approximately 5.7V,
before the output starts to turn back on causing a long
delay before the MOSFET turns back on. This delay
causes the catch diode to cool off between the current
limit cycles allowing the converter to survive a short cir-
cuit condition. An example is given below as how to
select the current limiting components. Assuming the
desired current limit point is set to be 20A and the cur-
rent sense resistor Rs=5m ? , then the current limit pro-
gramming resistor, R CL is calculated as:
Vcs = I CL × Rs = 20 × 0.005 = 0.1V
R CL = Vcs/I B = (0.1V)/(20 μ A) = 5K ?
Where:
I B = 20 μ A is the internal current source of IRU3034
8
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