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SP6652 Datasheet(PDF) 7 Page - Sipex Corporation

Part # SP6652
Description  1A, High Efficiency, High Frequency Current Mode PWM Buck Regulator
PDF  11 Pages
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Manufacturer  SIPEX [Sipex Corporation]
Direct Link  http://www.sipex.com
Logo SIPEX - Sipex Corporation

SP6652 Datasheet(HTML) 7 Page - Sipex Corporation

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Date:5/25/04
SP6652 1A, High Efficiency, High Frequency Current Mode PWM Buck Regulator
© Copyright 2004 Sipex Corporation
Voltage Loop and Compensation
in PWM Mode
The voltage loop section of the circuit consists
of the error amplifier and the translator circuits
(see functional diagram). The input of the volt-
age loop is the 0.75V reference voltage minus
the divided down output voltage at the feedback
pin. The output of the error amplifier is trans-
lated from a ground referred signal (the COMP
node) to a power input voltage referred signal.
The output of the voltage loop is fed to the
positive terminal of the Current Loop compara-
tor, and represents the peak inductor current
necessary to close the loop.
The total power supply loop is compensated
with a series RC network connected from the
COMP pin to ground. Compensation is simple
due to current-mode control. The modulator has
two dominant poles: one at a low frequency, and
one above the crossover frequency of the loop,
as seen in the graph below, Linearized Modula-
tor Frequency Response vs. Inductor Value.
The low frequency pole for L1=5
µH is 4kHz,
the second pole is 500kHz, and the gain-band-
width is 20kHz. The total loop crossover fre-
quency is chosen to be 200kHz, which is 1/6th of
the clock frequency. This sets the 2nd modula-
tor pole at 2.5 times the crossover frequency.
Therefore the gain of the error amplifier can be
200kHz/20kHz = 10 at the first modulator pole
of 4kHz. The error amp transconductance is
1mS, so this sets the RZ resistor value in the
compensation network at 10/1mS = 10k
Ω. The
zero frequency is placed at the first pole to
provide at total system response of -20dB/de-
cade (the zero from the error amp cancels the
first modulator pole, leaving the 1 pole rolloff
from the error amp pole). The compensation
capacitor becomes:
Cc = 1/(2*
π*Rz*pole1) = 1/(6.28*10kΩ*4kHz)
= 4nF
DETAIL DESCRIPTION: Contunued
L1VAL
2u
3u
4u
5u
6u
7u
8u
9u
10u
Mod_pole1
Mod_pole2
Gbw_modfb
0
4K
8K
12K
16K
20K
1
0
0.4M
0.8M
1.2M
1.6M
2.0M
2
0
10K
20K
30K
40K
50K
3
>>
1
2
3
Conditions: V
IN=5V, VOUT=3.3V, fCLK=1.2MHz, COUT=10µF, and MCV=132mV/µs. The inductor is varied from
2µH to 10µH
Linearized Modulator Frequency Response vs. Inductor Value.



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