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IRU1015
6
Rev. 1.1
06/29/01
2) With the output capacitance being 1500
μ
F:
Where:
t = 2
μ
s is the regulator response time
To set the output voltage, we need to select R1 and R2:
3) Assuming R1 = 121
, 1%
Select R2 = 215
, 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
DROP
L
Input voltage drop allowed in step 4
V
L
Maximum regulator dropout voltage
I
L
Load current step
Select a single 1500
μ
F 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:
54
1.2
The Sanyo MVGX series is a good choice to achieve
both price and performance goals. The 6MV1500GX,
1500
μ
F, 6.3V has an ESR of less than 36m
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.
V
ESR
V
ESL
V
C
T
LOAD
CURRENT
LOAD CURRENT RISE TIME
1015plt1-1.0
C
IN
=1.2
μ
F
0.15
×
50
ESR = (5 - 3.45 - 1.2 - 0.15)
1.2
ESR =(
VIN
- V
OUT
-
V - V
DROP
)
I
Vc = t
C 1500
×
I 2
×
1.2
ESR
≤
= 45m
R2 =
o
- 1
p
×
121 =
o
p
×
121 = 213
V
REF
V
- 1
1.25