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CS5165

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nominal output voltage before a load turn−on. The total

voltage drop due to a load step is ΔV−40 mV and the

deviation from the nominal output voltage is 40 mV smaller

than it would be if there was no droop resistor. Similarly at

full load the regulator output is pre−positioned at 16 mV

below the nominal voltage before a load turn−off. The total

voltage increase due to a load turn−off is ΔV−16 mV and the

deviation from the nominal output voltage is 16 mV smaller

than it would be if there was no droop resistor. This is

because the output capacitors are pre−charged to value that

is either 40 mV above the nominal output voltage before a

load turn−on or, 16 mV below the nominal output voltage

before a load turn−off (see Figure 15).

Obviously, the larger the voltage drop across the droop

resistor ( the larger the resistance), the worse the DC and

load regulation, but the better the AC transient response.

Design Rules for Using a Droop Resistor

The basic equation for laying an embedded resistor is:

RAR + ò

or R + ò

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 35 μm (1.37

mils) for one ounce copper. ρ = 717.86 μΩ−mil

For a Pentium II load of 14.2 A the resistance needed to

create a 56 mV drop at full load is:

Response Droop + 56 mV

IOUT

+ 56 mV

14.2 A +

3.9 mW

The resistivity of the copper will drift with the

temperature according to the following guidelines:

DR + 12% @ TA +) 50°C

DR + 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):

W +

ILOAD

0.05

where:

W = minimum width (in mils) required for proper power

dissipation, and ILOAD Load Current Amps.

The Pentium®II maximum load current is 14.2 A.

Therefore:

W + 14.2 A

0.05 +

284 mils + 0.7213 cm

Droop Resistor Length Calculation

L +

RDROOP

+ 0.0039

284

1.37

717.86

+ 2113 mil + 5.36 cm

Output Inductor

The inductor should be selected based on its inductance,

current capability, and DC resistance. Increasing the

inductor value will decrease output voltage ripple, but

degrade transient response.

Inductor Ripple Current

Ripple Current +

[(VIN * VOUT)

VOUT]

(Switching Frequency

VIN)

Example: VIN = +5.0 V, VOUT = +2.8 V, ILOAD = 14.2 A,

L = 1.2 μH, Freq = 200 kHz

Ripple Current +

[(5.0 V * 2.8 V)

2.8 V]

[200 kHz

1.2 mH

5.0 V] +

5.1 A

Output Ripple Voltage

VRIPPLE + Inductor Ripple Current

Output Capacitor ESR

Example:

VIN = +5.0 V, VOUT = +2.8 V, ILOAD = 14.2 A, L = 1.2 μH,

Switching Frequency = 200 kHz

Output Ripple Voltage = 5.1 A × Output Capacitor ESR

(from manufacturer’s specs)

ESR of Output Capacitors to limit Output Voltage Spikes

ESR + D

VOUT

DIOUT

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 100 mV for a 14.2 A Load Step,

ESR = 0.1/14.2 = 0.007 Ω

Inductor Peak Current

Peak Current + Maximum Load Current )

Ripple Current

Example: VIN = +5.0 V, VOUT = +2.8 V, ILOAD = 14.2 A,

L = 1.2 μH, Freq = 200 kHz

Peak Current + 14.2 A ) (5.1 2) + 16.75 A

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 +

DIOUT

(VIN * VOUT)

Example: VIN = +5.0 V, VOUT = +2.8 V, L = 1.2 μH, 14.2 A

change in Load Current

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