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15
Application Information: continued
where:
A= W
×
t = cross-sectional area
ρ
= the copper resistivity (μ
- mil)
L= length (mils)
W = width (mils)
t = thickness (mils)
For most PCBs the copper thickness, t, is 35μm (1.37 mils)
for one ounce copper.
ρ
= 717.86μ
-mil
For a Pentium
II load of 14.2A the resistance needed to cre-
ate a 56mV drop at full load is:
R
DROOP
=
=
= 3.9m
The resistivity of the copper will drift with the temperature
according to the following guidelines:
R = 12% @ T
A
= +50
C
R = 34% @T
A
= +100
C
Droop Resistor Width Calculations
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):
W=
where:
W = minimum width (in mils) required for proper power
dissipation, and I
LOAD
Load Current Amps.
The Pentium
II maximum load current is 14.2A.
Therefore:
W =
= 284 mils = 0.7213cm
Droop Resistor Length Calculation
L =
=
= 2113 mil = 5.36cm
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.
Inductor Ripple Current
Ripple current =
Example: V
IN
= +5V, V
OUT
= +2.8V, I
LOAD
= 14.2A, L = 1.2μH,
Freq = 200KHz
Ripple current =
[200KHz
×
1.2μH
×
5V]
= 5.1A
Output Ripple V oltage
V
RIPPLE
= Inductor Ripple Current
×
Output Capacitor ESR
Example:
V
IN
= +5V, V
OUT
= +2.8V, I
LOAD
= 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
V
OUT
ESR =
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.
In order to limit spikes to 100mV for a 14.2A Load Step,
ESR = 0.1/14.2 = 0.007
Inductor Peak Current
Peak Current = Maximum Load Current +
(
)
Example: V
IN
= +5V, V
OUT
= +2.8V, I
LOAD
= 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 without
saturating.
Response Time to Load Increas
e
(limited by Inductor value unless Maximum On-Time is
exceeded)
Response Time =
Example: V
IN
= +5V, V
OUT
= +2.8V, L = 1.2μH, 14.2A
change in Load Current
Response Time =
= 7.7μs
1.2μH
×
14.2A
(5V-2.8V)
L
×
I
OUT
(V
IN
-V
OUT
)
Ripple
Current
2
I
OUT
[(5V-2.8V)x 2.8V]
[(V
IN
- V
OUT
)
×
V
OUT
]
(
Switching Frequency
×
L
×
V
IN
)
0.0039
×
284
×
1.37
717.86
R
DROOP
×
W
×
t
ρ
14.2A
0.05
I
LOAD
0.05
56mV
14.2A
56mV
I
OUT
C