参数资料
型号: SP6652LEDEB
厂商: Exar Corporation
文件页数: 10/16页
文件大小: 0K
描述: EVAL BOARD FOR SP6652LED
标准包装: 1
系列: *
APPLICATIONS INFORMATION
over frequency f zero = 80kHz (from the
Bode plot) divided by the loop gain band -
width, given as 20kHz, which is used in the
following equation:
Error Amp Gain
= f zero / (loop gain bandwidth)
= 80kHz / 20kHz
=4
The error amp transconductance is about
1mS, so this sets the R Z resistor to be:
Rz = 4/1mS = 4KΩ
We will use R Z = 4KΩ for the 3.3V output
compensation.
The zero for loop compensation is placed
at the first modulator pole of 4 kHz to pro -
vide a loop response of -20/dB/decade at
the crossover frequency. The compensa -
tion capacitor Cc can be calculated from
the crossover frequency pole1 and the R Z
value:
C C = 1/(2π? R Z ?pole1) = 1/(2π? 4K ?4kHz)
= 10nF
From the Typical Performance Charac -
teristics load step curves, the 2.5V output
and 3.3V output are stable with R Z = 4KΩ
and C C = 10nF. For 1.8V to 0.85V output,
the values R Z = 2KΩ and C C = 10nF are
recommended.
Figure 4. Typical SP6652 circuit layout.
Oct10-07 RevJ
SP6652 1A, High Efficiency, Current Mode PWM Buck Regulator
 0
? 2007 Sipex Corporation
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