12 FN7102.7
May 8, 2006
INDUCTOR
The NMOS positive current limit is set at about 8A. For
optimal operation, the peak-to-peak inductor current ripp le
ΔIL should be less than 1A. The following equation gives the
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 IO.
OUTPUT CAPACITOR
Output voltage ripple and transie nt response are the
predominant factors when choo sing the output capacitor.
Initially, output capacitance should be sized with an ESR to
satisfy the output ripple ΔVO requirement:
When a step load change, ΔIO, is applied to the converter,
the initial voltage drop can be approximated by ESR*ΔIO.
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 ΔIL.
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:
where:
•GM
PWM is the transconductance of the PWM comparator ,
GMPWM = 120S
• ESR is the ESR of the output capacitor
•C
OUT is output capacitance
•GM
EA is the transconductance of the error amplifier,
GMEA = 120µS
•F
C is the intended crossover frequency of the loop. For
best performance, set this value to about one-tenth of the
switching frequency.
• Once RC is chosen, CC is decided by:
Design Example
A 5V to 2.5V converter with a 6A load requirement.
1. Choose the input capacitor
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.
2. Choose the inductor. Set the converter switching
frequency at 500kHz:
ΔIL = 1A yields 2.3µH. Leave some margin and choose
L = 2.7µH. Coilcraft's DO3316P-272HC has the required
current rating.
3. Choose the output capacitor
L = 2.7µH yields about 1A inductor ripple current. If 25mV of
ripple is desired, COUT's ESR needs to be less than 25mΩ.
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.
4. Loop compensation
50kHz is the intended crossover freque ncy. With the
conditions RC and CC are calculated as:
RC = 10.5kΩ and CC = 8900pF, round to standard value of
8200pF.
LVIN
(VO)VO
×–
VIN ΔILFS
××
--------------------------------------------
=
ILPK IO
ΔIL
2
--------
+=
ΔVOΔILESR×=
RC
IO
VFB
------------ FC2πESR(ROUT)COUT
×+×××
GMPWM GMEA
×
-------------------------------------------------------------------------------------------------
×=
ROUT
VO
IO
--------
=
CC1.5 COUT
ROUT
RC
----------------
××=
LVIN
(VO)VO
×–
VIN ΔILFS
××
--------------------------------------------
=
EL7566