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LP6212 датащи(PDF) 5 Page - Lowpower Semiconductor inc

номер детали LP6212
подробное описание детали  Output to Input Disconnect at Shutdown Mode
PDF  6 Pages
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производитель  POWER [Lowpower Semiconductor inc]
домашняя страница  http://www.lowpowersemi.com
Logo POWER - Lowpower Semiconductor inc

LP6212 датащи(HTML) 5 Page - Lowpower Semiconductor inc

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Preliminary Datasheet
LP6212
LP6212 –00 Version1.0
Datasheet
May.-2014
www.Lowpowersemi.com - -
Page 5 of 6
Operation Information
The LP6212 uses a 1.2MHz fixed frequency,
current-mode regulation architecture to regulate the
output voltage. The LP6212 measures the output
voltage through an external resistive voltage divider
and compares that to the internal 0.6V reference to
generate the error voltage to the inductor current to
regulate the output voltage. The use of current-mode
regulation improves transient response and control
loop stability.
When the LP6212 is disabled (EN=Low), both power
switches are off. There is no current path from SW to
OUT. Therefore, the output voltage discharges to
ground. When the LP6212 is enabled (EN=High),a
limited start-current charges the output voltage rising to
SW, then TH part operates in force PWM mode for
regulating the output voltage to the target value. At the
beginning of each cycle, the N-channel MOSFET
switch is turned on, forcing the inductor current to rise,
The current at the source of the switch is internally
measured and converted to a voltage by the current
sense amplifier. That voltage is compared to the error
voltage. When the inductor current rises sufficiently, the
PWM comparator turns off the switch, forcing the
inductor current to the output capacitor through the
internal P-Channel MOSFET rectifier, which forces the
inductor current to decrease. The peak inductor current
is controlled by the error voltage. Thus the output
voltage controls the inductor current to satisfy the lode.
Setting the Output Voltage
Set the output voltage by selecting the resistive voltage
divider ratio. The voltage divider drops the output
voltage to the 0.6V feedback voltage. Use a 100K
resistor for R2 of the voltage divider. Determine the
hi g h -si d e re sis t or R 1 by t he e qua t i on:
Vout=(R1/R2+1) x VFB
Vout=(R1/R2+1) x 0.6V



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