AS1324
Data Sheet - Application Information
Basic losses in the design of a system should also be considered. Internal battery resistances and copper trace can
account for additional efficiency degradations in battery operated systems. By making sure that CIN has adequate
charge storage and very low ESR at the given switching frequency, the internal battery and fuse resistance losses can
be minimized. CIN and COUT ESR dissipative losses and inductor core losses generally account for less than 2% total
additional loss.
Thermal Shutdown
Due to its high-efficiency design, the AS1324 will not dissipate much heat in most applications. However, in applica-
tions where the AS1324 is running at high ambient temperature, uses a low supply voltage, and runs with high duty
cycles (such as in dropout) the heat dissipated may exceed the maximum junction temperature of the device.
As soon as the junction temperature reaches approximately 150ºC the AS1324 goes in thermal shutdown. In this mode
the internal PMOS & NMOS switch are turned off. The device will power up again, as soon as the temperature falls
below +145°C again.
Checking Transient Response
The main loop response can be evaluated by examining the load transient response. Switching regulators normally
take several cycles to respond to a step in load current. When a load step occurs, VOUT immediately shifts by an
amount equivalent to:
VDROP = ΔIOUT x ESR
(EQ 13)
Where:
ESR is the effective series resistance of COUT.
ΔIOUT also begins to charge or discharge COUT, which generates a feedback error signal. The regulator loop then acts
to return VOUT to its steady-state value. During this recovery time VOUT can be monitored for overshoot or ringing that
would indicate a stability problem.
Design Example
Figure 28 shows the AS1324 used in a single lithium-ion (3.7V typ) battery-powered mobile phone application. The
load current requirement is 600mA (max) but most of the time the device will require only 2mA (standby mode current).
Figure 28. Design Example
2.2µH
4
3
VOUT
2.2V
VIN
3.7V
VIN
SW
CIN
4.7µF
CER
COUT
10µF
CER
22pF
AS1324
1MΩ
5
1
R2
EN
VFB
R1
375kΩ
GND
2
For the circuit shown in Figure 28, efficiency at low- and high-load currents is an important consideration when select-
ing the value for the external inductor, which is calculated as:
VOUT
--------------
fΔIL
VOUT
--------------
VIN
⎛
⎞
⎠
(EQ 14)
L =
× 1 –
⎝
From (EQ 14), substituting VOUT = 2.2V, VIN = 3.7V, ΔIL = 240mA and f = 1.5MHz gives:
2,2V
2,2V
⎞
⎛
(EQ 15)
----------------------------------------------------
------------
= 2,48μH
L =
× 1 –
⎝
⎠
3,7V
(1,5MHz × 240mA)
Therefore, a standard 2.2µH inductor should be used for this design.
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