CS5165GDWR16 View Datasheet(PDF) - Cherry semiconductor
Part Name
Description
MFG CO.
CS5165GDWR16
Cherry semiconductor
'CS5165GDWR16' PDF : 19 Pages View PDF
Application Information: continued
where:
A= W × t = cross-sectional area
ρ= the copper resistivity (µΩ - 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 = 717.86µΩ-mil
For a Pentium®II load of 14.2A the resistance needed to cre-
ate a 56mV drop at full load is:
56mV 56mV
RDROOP = IOUT = 14.2A = 3.9mΩ
The resistivity of the copper will drift with the temperature
according to the following guidelines:
∆R = 12% @ TA = +50˚C
∆R = 34% @TA = +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):
ILOAD
W=
0.05
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)x 2.8V]
Ripple current = [200KHz × 1.2µH × 5V] = 5.1A
Output Ripple Voltage
VRIPPLE = Inductor Ripple Current × Output Capacitor ESR
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
ESR =
∆ VOUT
∆ IOUT
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 = 0.007Ω
where:
W = minimum width (in mils) required for proper power
dissipation, and ILOAD Load Current Amps.
The Pentium®II maximum load current is 14.2A.
Therefore:
14.2A
W = 0.05 = 284 mils = 0.7213cm
Droop Resistor Length Calculation
( ) Inductor Peak Current
Peak Current = Maximum Load Current +
Ripple
Current
2
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 without
saturating.
RDROOP × W × t 0.0039 × 284 × 1.37
L=
ρ
=
717.86
= 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.
Response Time to Load Increase
(limited by Inductor value unless Maximum On-Time is
exceeded)
L × ∆ IOUT
Response Time =
(VIN-VOUT)
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
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