EL7566DREZ-T7 Intersil, EL7566DREZ-T7 Datasheet - Page 12

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EL7566DREZ-T7

Manufacturer Part Number
EL7566DREZ-T7
Description
IC REG 6A DC/DC STP-DWN 28HTSSOP
Manufacturer
Intersil
Type
Step-Down (Buck)r
Datasheet

Specifications of EL7566DREZ-T7

Internal Switch(s)
Yes
Synchronous Rectifier
Yes
Number Of Outputs
1
Voltage - Output
0.8 ~ 6 V
Current - Output
6A
Frequency - Switching
370kHz
Voltage - Input
3 ~ 6 V
Operating Temperature
0°C ~ 85°C
Mounting Type
Surface Mount
Package / Case
28-TSSOP Exposed Pad, 28-eTSSOP, 28-HTSSOP
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Power - Output
-

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
EL7566DREZ-T7
Manufacturer:
Intersil
Quantity:
4 050
INDUCTOR
The NMOS positive current limit is set at about 8A. For
optimal operation, the peak-to-peak inductor current ripple
ΔI
inductance value:
The peak current the inductor sees is:
When inductor is chosen, it must be rated to handle the peak
current and the average current of I
OUTPUT CAPACITOR
Output voltage ripple and transient response are the
predominant factors when choosing the output capacitor.
Initially, output capacitance should be sized with an ESR to
satisfy the output ripple ΔV
When a step load change, ΔI
the initial voltage drop can be approximated by ESR*ΔI
The output voltage will continue to drop until the control loop
begins to correct the output voltage error. Increasing the
output capacitance will lessen the impact of load steps on
output voltage. Increasing loop bandwidth will also reduce
output voltage deviation under step load conditions. Some
experimentation with converter bandwidth and output
filtering will be necessary to generate a good transient
response (Reference Figure 15).
As with the input capacitor, it is recommended to use X5R or
X7R type of ceramic capacitors. SPCAP or POSCAP type
Polymer capacitors can also be used for the low ESR and
high capacitance requirements of these converters.
Generally, the AC current rating of the output capacitor is not
a concern because the RMS current is only 1/8 of ΔI
LOOP COMPENSATION
Current-mode control in system forces the inductor current
to be proportional to the error signal. This has the advantage
of eliminating the double pole response of the output filter,
and reducing complexity in the overall loop compensation. A
simple Type 1 compensator is adequate to generate a
stable, high-bandwidth converter. The compensation resister
is decided by:
I
R
L
ΔV
LPK
C
L
=
O
should be less than 1A. The following equation gives the
=
(
------------------------------------------- -
=
V
=
V
----------- -
VFB
IN
IN
I
I
ΔI
O
O
×
L
+
V
×
ΔI
×
ΔI
--------
O
F
-------------------------------------------------------------------------------------------------
2
ESR
L
L
C
)
×
×
×
F
V
2
S
O
×
π
GM
×
(
ESR
PWM
O
12
O
requirement:
×
+
, is applied to the converter,
GM
R
OUT
EA
O
.
)
×
C
OUT
L
.
O
.
EL7566
where:
• GM
• ESR is the ESR of the output capacitor
• C
• GM
• F
• Once R
Design Example
A 5V to 2.5V converter with a 6A load requirement.
The input capacitor or combination of capacitors has to be
able to take about 1/2 of the output current, e.g., 3A.
Panasonic EEFUD0J101XR is rated at 3.3A, 6.3V, meeting
the above criteria.
ΔI
L = 2.7µH. Coilcraft's DO3316P-272HC has the required
current rating.
L = 2.7µH yields about 1A inductor ripple current. If 25mV of
ripple is desired, C
Panasonic's EEFUD0G151XR 150µF has an ESR of 12mΩ
and is rated at 4V.
ESR is not the only factor deciding the output capacitance.
As discussed earlier, output voltage droops less with more
capacitance when converter is in load transient. Multiple
iterations may be needed before final components are
chosen.
50kHz is the intended crossover frequency. With the
conditions R
R
8200pF.
R
C
L
1. Choose the input capacitor
2. Choose the inductor. Set the converter switching
3. Choose the output capacitor
4. Loop compensation
C
OUT
C
L
GM
GM
best performance, set this value to about one-tenth of the
switching frequency.
=
C
= 1A yields 2.3µH. Leave some margin and choose
= 10.5kΩ and C
OUT
frequency at 500kHz:
=
(
------------------------------------------- -
V
is the intended crossover frequency of the loop. For
V
PWM
PWM
EA
EA
1.5
=
IN
IN
is output capacitance
V
------- -
is the transconductance of the error amplifier,
= 120µS
I
×
×
C
O
O
V
ΔI
is the transconductance of the PWM comparator,
= 120S
C
is chosen, C
C
O
OUT
L
)
and C
×
×
F
V
×
S
O
OUT
R
--------------- -
C
R
OUT
C
= 8900pF, round to standard value of
C
are calculated as:
's ESR needs to be less than 25mΩ.
C
is decided by:
May 8, 2006
FN7102.7

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