MP1527
2A, 1.3MHz
Step-Up Converter
Monolithic Power Systems
There is also a right-half-plane zero (fRHPZ) that
exists in all continuous mode (continuous
mode means that the inductor current does not
drop to zero on each cycle) step-up
converters. The frequency of the right half
plane zero is:
R3 = VIN*RLOAD-MIN*C2 / (10GCS*GEA*VFB*L)
The minimum load resistance (RLOAD-MIN) is
equal to the regulated output voltage (VOUT
divided by the maximum load current ILOAD-MAX
)
.
Substituting that into the above equation:
2
2
f
RHPZ = VIN *RLOAD / (2π*L*VOUT
)
R3 = VIN*VOUT*C2 /(10GCS*GEA *VFB*L*ILOAD-MAX
)
where L is the value of the inductor.
Putting in the known constant values:
To stabilize the regulation control loop, the
crossover frequency (The frequency where the
loop gain drop to 0dB or gain of 1, indicated as
fC) should be at least one decade below the
right-half-plane zero and should be at most
75KHz. fRHPZ is at its lowest frequency at
maximum output load current (RLOAD is at a
minimum)
(1) R3 ≈ 48*VIN*VOUT*C2 / (L*ILOAD-MAX
For fC = 75KHz,
fC = (GCS*GEA*VIN*VFB*R3) / (2π*C2*VOUT
)
2
)
Solving for R3,
R3 = (2π*fC*C2*VOUT2 / (GCS*GEA*VIN*VFB)
The crossover frequency is calculated by the
equation:
Using 75KHz for fC and putting in the other
known constants:
fC = AVDC*fP1*fP2 / fZ1
(2) R3 ≈ 2.2x108*C2*VOUT2 / VIN
or
2
The value of the compensation resistor is
limited to 10Kꢀ to prevent overshoot on the
output at turn-on. So if the value calculated for
R3 from either equation (1) or equation (2) is
greater than 10kꢀ, use 10Kꢀ for R3.
fC = GCS*GEA*VIN*VFB*R3 / (2π*C2*VOUT
)
The known values are:
G
G
V
CS = 4.3S
EA = 400µS
FB = 1.22V
Choose C3 to set the zero frequency fZ1 to
one-fourth of the crossover frequency fC:
Putting in the known constants:
fC = 3.3x10-4 *VIN *R3/ (C2*VOUT
fZ1 = fC / 4
or
2
)
If the frequency of the right-half-pane zero
fRHPZ is less than 750KHz, then the crossover
frequency should be 1/10 of fRHPZ, and
determine the compensation resistor (R3) with
equation (1). If fRHPZ is greater than or equal to
750KHz, set the crossover frequency to 75KHz
with equation (2).
2
1 /(2π*C3*R3) = GCS*GEA*VIN*VFB*R3 / (8π*C2*VOUT
Solving for C3:
)
C3 = 4*C2*VOUT2 / (GCS*GEA*VIN*VFB*R32)
Entering the known values gives:
For fC = fRHPZ / 10, then
MP1527 Rev 1.8_8/31/05
Monolithic Power Systems, Inc.
11
MPS [ MONOLITHIC POWER SYSTEMS ]
