CS5165H datasheet(15/20 Pages) CHERRY

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Application Information: continued

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:

RDROOP =

= 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):

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:

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: VIN = +5V, VOUT = +2.8V, ILOAD = 14.2A, L = 1.2µH,

Freq = 200KHz

Ripple current =

= 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 =

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

Inductor Peak Current

Peak Current = Maximum Load 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 without

saturating.

Response Time to Load Increase

(limited by Inductor value unless Maximum On-Time is

exceeded)

Response Time =

Example: VIN = +5V, VOUT = +2.8V, L = 1.2µH, 14.2A

change in Load Current

Response Time =

= 7.7µs

1.2µH × 14.2A

(5V-2.8V)

L × ∆ IOUT

(VIN-VOUT)

Ripple

Current

∆ VOUT

∆ IOUT

[(5V-2.8V)x 2.8V]

[200KHz × 1.2µH × 5V]

[(VIN - VOUT) × VOUT]

(Switching Frequency × L × VIN)

0.0039 × 284 × 1.37

717.86

RDROOP × W × t

14.2A

0.05

ILOAD

0.05

56mV

14.2A

56mV

IOUT

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