Application Information: continued
Inductor Ripple Current
[(VIN - VOUT) × VOUT
]
Ripple current =
(Switching Frequency × L × VIN)
Example: VIN = +5V, VOUT = +2.8V, ILOAD = 14.2A, L =
1.2µH, Freq = 200KHz
[(5V-2.8V) × 2.8V]
Ripple current =
= 5.1A
[200KHz × 1.2µH × 5V]
Trace 1 Output voltage ripple
Output Ripple Voltage
Trace 2 Buck regulator #1 inductor switching node
Trace 3 Buck regulator #2 inductor switching node
VRIPPLE = Inductor Ripple Current × Output Capacitor
ESR
Figure 25: 30A load waveforms.
Example:
VIN = +5V, VOUT = +2.8V, ILOAD = 14.2A, L = 1.2µH,
Switching Frequency = 200KHz
Output Ripple Voltage = 5.1A × Output Capacitor ESR
(from manufacturer’s specs)
ESR of Output Capacitors to limit Output Voltage Spikes
∆ VOUT
ESR =
∆ IOUT
This applies for current spikes that are faster than regula-
tor response time. Printed Circuit Board resistance will
add to the ESR of the output capacitors.
In order to limit spikes to 100mV for a 14.2A Load Step,
Trace 1 Output voltage ripple
ESR = 0.1/14.2 = 0.007Ω
Trace 2 Buck regulator #1 inductor switching node
Trace 3 Buck regulator #2 inductor switching node
Ripple
Inductor Peak Current
Peak Current = Maximum Load Current +
Figure 26: 15A load transient waveforms.
Current
(
)
2
Figure 26 shows supply response to a 15A load step with a
30A/µs slew rate. The V2TM control loop immediately
forces the duty cycle to 100%, ramping the current in both
inductors up. A voltage spike of 136mV due to output
capacitor impedance occurs. The inductive component of
the spike due to ESL recovers within several microseconds.
The resistive component due to ESR decreases as inductor
current replaces capacitor current.
Example: VIN = +5V, VOUT = +2.8V, ILOAD = 14.2A, L = 1.2µH,
Freq = 200KHz
Peak Current = 14.2A + (5.1/2) = 16.75A
A key consideration is that the inductor must be able to
deliver the Peak Current at the switching frequency with-
out saturating.
The benefit of adaptive voltage positioning in reducing the
voltage spike can readily be seen. The differences in DC
voltage and duty cycle can also be observed. This particu-
lar transient occurred near the beginning of regulator off-
time, resulting in a longer recovery time and increased
voltage spike.
Response Time to Load Increase
(limited by Inductor value unless Maximum On-Time is
exceeded)
L × ∆ IOUT
Response Time =
(VIN-VOUT
)
Output Inductor
The inductor should be selected based on its inductance,
current capability, and DC resistance. Increasing the induc-
tor value will decrease output voltage ripple, but degrade
transient response.
Example: VIN = +5V, VOUT = +2.8V, L = 1.2µH, 14.2A
change in Load Current
1.2µH × 14.2A
Response Time =
= 7.7µs
(5V-2.8V)
18