Datasheet

15
16
Q1
Q
EXT
DRVE
DRVC
R
P
V
IN
V
OUT
Q
EXT
R
P
V
IN
V
OUT
15
16
Q1
DRVE
DRVC
bq24450
www.ti.com SLUS929C –APRIL 2009– REVISED FEBRUARY 2012
Selecting the External Pass Transistor
All the examples so far have used a PNP transistor for the external pass element. But the driver transistor in the
bq24450 can be configured to drive many different types of pass transistors. This section will look at some of the
different configurations that are possible. In all configurations, though, these factors hold:
1. The external pass device must have sufficient voltage rating for the application, and must have the current
and power handling capabilities to charge at the desired rate at the maximum input to output differential in
the application.
2. The device must have enough current gain at the required charging current to keep the drive current below
25mA.
The choice of the pass device and the configuration of the internal driver transistor have an effect on the
following:
1. The minimum and maximum practical charging current.
2. The open-loop gains of the current and voltage loops, and hence the value of the compensation capacitor at
the COMP pin. In battery charging applications, dynamic response is not a requirement, and the values of
C
COMP
given below should give stable operation under all conditions.
3. The IC's power dissipation and thus its self-heating. The IC typically has a thermal resistance of 100°C/W.
An external resistance R
P
can be added to share some of the power dissipation and reduce the IC's
self-heating.
4. The minimum differential voltage ΔV (from the input to the battery) required to operate.
The next section addresses a few topologies, and gives values for the charge current range, the minimum input
to output differential ΔV, power dissipation P
D
in the IC, R
P
and C
COMP
for each of the topologies. (In the
expressions below, h
FE
is the current gain of the external transistor).
Common-Emitter PNP
I
MAX-CHG
range: 25mA to 1000mA
Minimum ΔV: 0.5V
R
P
= (V
IN(MIN)
– 2.0V) ÷ I
MAX-CHG
× h
FE(MIN)
P
D
= (V
IN(MAX)
– 0.7V) ÷ h
FE
× I
MAX-CHG
– (I
MAX-CHG
)
2
÷ (h
FE
)
2
× R
P
C
COMP
= 0.1 μF
PNP in a Quasi-Darlington With Internal Driver
I
MAX-CHG
range: 25mA to 1000mA
Minimum ΔV: 2V
R
P
= (V
IN(MIN)
– V
OUT(MAX)
– 1.2 V) ÷ I
MAX-CHG
× h
FE(MIN)
P
D
= (V
IN(MAX)
– V
OUT
– 0.7V) ÷ h
FE
× I
MAX-CHG
– (I
MAX-CHG
)
2
÷ (h
FE
)
2
× R
P
C
COMP
= 0.01μF to 0.047μF
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