IRU1015

6

Rev. 1.1

06/29/01

2) With the output capacitance being 1500µF:

Where:

∆t = 2µs is the regulator response time

To set the output voltage, we need to select R1 and R2:

3) Assuming R1 = 121Ω, 1%

Select R2 = 215Ω, 1%

Selecting both R1 and R2 resistors to be 1% toler-

ance results in the least amount of error introduced

by the resistor dividers leaving a

≈ ±2.5% error bud-

get for the IRU1015 reference which is well within the

initial accuracy of the device.

Finally, the input capacitor is selected as follows:

4) Assuming that the input voltage can drop 150mV be-

fore the main power supply responds, and that the

main power supply response time is

≈ 50ms, then

the minimum input capacitance for a 1.2A load step

is given by:

The ESR should be less than:

Where:

∆V L Maximum regulator dropout voltage

∆I L Load current step

Select a single 1500µF the same type as the output

capacitors exceeds our requirements.Figure 6 shows

the completed schematic for our example.

Figure 5 - Typical regulator response to the

fast load current step

An example of a regulator design to meet the AMD speci-

fication for 486DX4-120MHz is given below.

Assume the specification for the processor as shown in

Table 1:

Type of

Vout

Imax

Max Allowed

Processor

Nominal

Output Tolerance

AMD 486DX4

3.45 V

1.2 A

±150 mV

Table 1 - GTL+ specification for Pentium Pro

The first step is to select the voltage step allowed in the

output due to the output capacitor’s ESR:

1) Assuming the regulator’s initial accuracy plus the re-

sistor divider tolerance is

≈ ±86mV (±2.5% of 3.45V

nominal), then the total step allowed for the ESR and

the ESL, is -64mV.

Assuming that the ESL drop is -10mV, the remaining

ESR step will be -54mV. Therefore the output ca-

pacitor ESR must be:

The Sanyo MVGX series is a good choice to achieve

both price and performance goals. The 6MV1500GX,

1500µF, 6.3V has an ESR of less than 36mΩ typi-

cal. Selecting a single capacitor achieves our design

goal.

The next step is to calculate the drop due to the ca-

pacitance discharge and make sure that this drop in

voltage is less than the selected ESL drop in the

previous step.

T

LOAD

CURRENT

LOAD CURRENT RISE TIME

1015plt1-1.0

= 400µF

1.2 × 50

0.15

ESR =

= 0.167Ω

(5 - 3.45 - 1.2 - 0.15)

1.2

ESR =

∆I

∆Vc =

=

= 1.6mV

∆t × ∆I

2 × 1.2

C

1500

ESR ≤

= 45mΩ

54

1.2

R2 =o

o

o

o

o

- 1p

p

p

p

p

× 121 =o

o

o

o

o

- 1p

p

p

p

p

× 121 = 213Ω

3.45

1.25