MAX1945SEUI Maxim Integrated Products, MAX1945SEUI Datasheet - Page 15

• Manufacturer Part Number: MAX1945SEUI
• Description: DC/DC Switching Regulators 1MHz 1% 6A Step-Down PWM
• Manufacturer: Maxim Integrated Products

Specifications of MAX1945SEUI

• Lead Free Status / Rohs Status: Lead free / RoHS Compliant

Set the error-amplifier compensation zero formed by R
and C
f
The following design example is for the application cir-
cuit shown in Figures 1 and 2:
V
I
C
R
gm
gmc = 18.2s
f
R
fp
10
fz
0.04) = 22.1kHz.
Pick the closed-loop unity-gain crossover frequency (f
at 60kHz. Determine the switching regulator DC gain:
G
then:
(1.8
Table 2 shows the recommended values for R
for different output voltages.
Table 2. Compensation Values for Output Voltages (500kHz)
PLOAD
OUT(MAX)
SWITCH
OUT
ESR
OUT
OUT
DC
ESR
DC
C
-6 ✕
R
EA
0.04))/180kΩ ≈ 340pF, choose C
C
C
✕
= gmc
VOUT (V)
= 1/(2π
= (C
= 0.04Ω
= 1.8V
= V
= 180µF
= (V
C
= 50µs
(0.3 + 0.04) = 2.6kHz.
= 1/(2π
60kHz)/(50
, at maximum load. Calculate C
R
C
= 500kHz
equal to the load-impedance pole frequency,
C
C
OUT
OUT
OUT
= 6A
✕
(1%), choose R
✕
/I
R
✕
OUT(MAX)
✕
✕
C
C
OUT
(R
OUT
f
C
C
C
✕
______________________________________________________________________________________
OUT
)/(gm
OUT
= (C
10
1MHz, 1% Accurate, 6A Internal Switch
= 18.2
✕
-6 ✕
+ R
OUT
(R
R
EA
= 1.8V/6 A = 0.3Ω
ESR
OUT
110kΩ
330pF
0.8
ESR
✕
✕
0.8
C
✕
0.3 = 5.46
V
) = 1/(2 π
= 180kΩ, 1%
✕
R
+ R
))/R
FB
OUT
5.46
✕
ESR
C
G
)/R
C
= (180uF
500kHz Switching
✕
DC
) = 1/(2 π
= 330pF, 10%
C
C
2.6kHz) ≈ 190kΩ
as:
✕
✕
fp
180
LOAD
147kΩ
330pF
✕
C
1.2
(0.3 +
✕
✕
and C
10
180
) =
-6 ✕
c
C
C
✕
)
Following procedure outlines the compensation
process of the MAX1945 for 1MHz operation with all
ceramic output capacitors (Figure 3). The basic regula-
tor loop consists of a power modulator, an output-feed-
back divider, and an error amplifier. The switching
regulator has a DC gain set by gmc
gmc is the transconductance from the output voltage of
the error amplifier to the output inductor current. The
load impedance of the switching modulator consists of
a pole-zero pair set by R
(C
power train of the switching regulator:
Regulator DC Gain:
G
Load-Impedance Pole Frequency:
fp
Load-Impedance Zero Frequency:
fz
where, R
feedback divider has a gain of G
V
fier has a DC gain, G
tion capacitor, C
error amplifier, R
C
dominant pole frequency as:
Determine the compensation zero frequency as:
For best stability and response performance, set the
closed-loop unity-gain frequency much higher than the
load impedance pole frequency. In addition, set the
closed-loop unity-gain crossover frequency less than
one-fifth of the switching frequency. However, the maxi-
Step-Down Regulators
FB
C
DC
ESR
LOAD
OUT
and R
180kΩ
is equal to 0.8V. The transconductance error ampli-
330pF
= ∆V
1.8
= 1/(2
), and its ESR. The following equations define the
= 1/(2
OUT
OUT
C
✕
set a compensation zero. Calculate the
= V
/∆V
✕
π
fp
fz
π
✕
EA
COMP
OUT
OEA
EA
C
✕
C
, and the output resistance of the
= 1/(2π
OUT
C
= 1/(2π
/I
OUT
EA(DC)
(20MΩ), set the dominant pole.
OUT(MAX)
= gmc
287kΩ
220pF
✕
2.5
✕
R
✕
OUT
ESR
(R
✕
, of 70dB. The compensa-
C
OUT
C
✕
C
)
, the output capacitor
C
, and gmc = 18.2. The
R
✕
FB
✕
OUT
R
+R
R
= V
OEA
C
ESR
1MHz Switching
)
✕
FB
)
))
R
/V
365kΩ
220pF
OUT
OUT
3.3
, where
, where
15