Application Information: continued
Inductor Ripple Current
[(VIN - VOUT) × VOUT
where:
A= W × t = cross-sectional area
]
ρ= the copper resistivity (µΩ - mil)
Ripple current =
(Switching Frequency × L × VIN)
L= length (mils)
W = width (mils)
t = thickness (mils)
Example: VIN = +5V, VOUT = +2.8V, ILOAD = 14.2A, L = 1.2µH,
Freq = 200KHz
Ripple current =
[(5V-2.8V)x 2.8V]
[200KHz × 1.2µH × 5V]
For most PCBs the copper thickness, t, is 35µm (1.37 mils)
= 5.1A
for one ounce copper. ρ = 717.86µΩ-mil
For a Pentium®II load of 14.2A the resistance needed to cre-
ate a 56mV drop at full load is:
Output Ripple Voltage
56mV
IOUT
56mV
14.2A
VRIPPLE = Inductor Ripple Current × Output Capacitor ESR
RDROOP
=
=
= 3.9mΩ
Example:
The resistivity of the copper will drift with the temperature
according to the following guidelines:
VIN = +5V, VOUT = +2.8V, ILOAD = 14.2A, L = 1.2µH,
Switching Frequency = 200KHz
Output Ripple Voltage = 5.1A × Output Capacitor ESR
∆R = 12% @ TA = +50˚C
∆R = 34% @TA = +100˚C
(from manufacturer’s specs)
ESR of Output Capacitors to limit Output Voltage Spikes
∆ VOUT
ESR =
Droop Resistor Width Calculations
∆ IOUT
The droop resistor must have the ability to handle the load
current and therefore requires a minimum width which is
calculated as follows (assume one ounce copper thickness):
This applies for current spikes that are faster than regulator
response time. Printed Circuit Board resistance will add to
the ESR of the output capacitors.
ILOAD
W=
0.05
In order to limit spikes to 100mV for a 14.2A Load Step,
ESR = 0.1/14.2 = 0.007Ω
Ripple
where:
Inductor Peak Current
Current
W = minimum width (in mils) required for proper power
Peak Current = Maximum Load Current +
(
)
2
dissipation, and ILOAD Load Current Amps.
The Pentium®II maximum load current is 14.2A.
Therefore:
Example: VIN = +5V, VOUT = +2.8V, ILOAD = 14.2A, L = 1.2µH,
Freq = 200KHz
Peak Current = 14.2A + (5.1/2) = 16.75A
14.2A
0.05
A key consideration is that the inductor must be able to
deliver the Peak Current at the switching frequency without
saturating.
W =
= 284 mils = 0.7213cm
Droop Resistor Length Calculation
Response Time to Load Increase
RDROOP × W × t
0.0039 × 284 × 1.37
(limited by Inductor value unless Maximum On-Time is
L =
=
= 2113 mil = 5.36cm
717.86
ρ
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)
15
CHERRY [ CHERRY SEMICONDUCTOR CORPORATION ]
