MP1527
2A, 1.3MHz
Step-Up Converter
Monolithic Power Systems
C3
≈
1.9x10
3
C2 V
OUT2
/ (V
IN
R3
2
)
In some cases, if an output capacitor with high
capacitance and high equivalent series
resistance (ESR) is used, then a second
compensation capacitor (from COMP to
SGND) is required to compensate for the zero
introduced by the output capacitor ESR. The
extra capacitor is required if the ESR zero is
less than 4x the crossover frequency. The
ESR zero frequency is:
f
ZESR
= 1 / (2π*C2*R
ESR
)
The second compensation capacitor
required if:
4*f
C
≥
f
ZESR
or
4*G
CS
*G
EA
*V
IN
*V
FB
*R3
(2π*C2*R
ESR
)
/
(2π*C2*V
OUT2
)
≥
1
/
Example
Given:
Input Voltage (V
IN
): 5V
Output Voltage (V
OUT
): 12V
Maximum Load Current (I
LOAD-MAX
): 500mA
Output Capacitor (C2): 10µF (ESR=10mΩ
Maximum)
Inductor Value (L): 4.7µH
Find the frequency of the right-half-plane zero:
f
RHPZ
= V
IN2
/ (2π*L*V
OUT
*I
LOAD-MAX
)
f
RHPZ
= (5V)
2
/
(2π*4.7µH*12V*500mA)=141KHz
The frequency of the right-half-plane zero is
less than 750khz, so use equation (1) to
determine the compensation resistor R3:
R3
≈
48*V
IN
*V
OUT
*C2 / (L*I
LOAD-MAX
)
R3
≈
48*5*12*10µF/(4.7µH*500mA) =12.3KΩ
is
Simplifying:
(8.4x10 *V
IN
*R3*R
ESR
)/
-3
V
OUT2
(use 10KΩ)
≥
1
Find the compensation capacitor C3:
C3
≈
1.9x10
3
*C2*V
OUT2
/ (V
IN
*R3
2
)
C3
≈
1.9x10
3
*10µF (12V)
2
/ (5 * 10KΩ
2
) = 5.4nF
If this is the case, calculate the second
compensation capacitor by the equation:
R3*C4 = C2*R
ESR
or
C4 = (C2*R
ESR
) / R3
(use the nearest standard value, 5.6nF)
Determine if the second
capacitor is required:
compensation
8.4x10
-3
* 5V * 5.6KΩ * 10mΩ / 12V
2
= 0.016
≤
1
Therefore no second compensation capacitor
is required.
MP1527 Rev 1.8_8/31/05
Monolithic Power Systems, Inc.
12
MPS [ MONOLITHIC POWER SYSTEMS ]
