MP1527
2A, 1.3MHz
Step-Up Converter
Monolithic Power Systems
C3 ≈ 1.9x103 C2 VOUT2 / (VIN R32)
Example
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:
Given:
Input Voltage (VIN): 5V
Output Voltage (VOUT): 12V
Maximum Load Current (ILOAD-MAX): 500mA
Output Capacitor (C2): 10µF (ESR=10mꢀ
Maximum)
Inductor Value (L): 4.7µH
Find the frequency of the right-half-plane zero:
f
ZESR = 1 / (2π*C2*RESR
)
f
RHPZ = VIN2 / (2π*L*VOUT*ILOAD-MAX
fRHPZ = (5V)2 /
)
The second compensation capacitor is
required if:
(2π*4.7µH*12V*500mA)=141KHz
4*fC ≥ fZESR
The frequency of the right-half-plane zero is
less than 750khz, so use equation (1) to
determine the compensation resistor R3:
or
2
4*GCS*GEA*VIN*VFB*R3
(2π*C2*RESR
/
(2π*C2*VOUT
)
≥
1
/
R3 ≈ 48*VIN*VOUT*C2 / (L*ILOAD-MAX
R3 ≈ 48*5*12*10µF/(4.7µH*500mA) =12.3Kꢀ
(use 10Kꢀ)
)
)
Simplifying:
(8.4x10-3*VIN*R3*RESR )/ VOUT2 ≥ 1
Find the compensation capacitor C3:
C3 ≈ 1.9x103*C2*VOUT2 / (VIN*R32)
If this is the case, calculate the second
compensation capacitor by the equation:
C3 ≈ 1.9x103*10µF (12V)2 / (5 * 10Kꢀ2) = 5.4nF
R3*C4 = C2*RESR
or
(use the nearest standard value, 5.6nF)
Determine if the second compensation
capacitor is required:
C4 = (C2*RESR) / R3
8.4x10-3 * 5V * 5.6Kꢀ * 10mꢀ / 12V2 = 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 ]
