LM48511SQ National Semiconductor, LM48511SQ Datasheet - Page 17

• Manufacturer Part Number: LM48511SQ
• Manufacturer: National Semiconductor

Specifications of LM48511SQ

• Operational Class: Class-D
• Audio Amplifier Output Configuration: 1-Channel Mono
• Output Power (typ): 6.7x1@4OhmW
• Audio Amplifier Function: Speaker
• Single Supply Voltage (typ): Not RequiredV
• Dual Supply Voltage (typ): 5V
• Power Supply Requirement: Triple
• Rail/rail I/o Type: No
• Power Supply Rejection Ratio: 88dB
• Single Supply Voltage (min): Not RequiredV
• Single Supply Voltage (max): Not RequiredV
• Dual Supply Voltage (min): 3/4.8V
• Dual Supply Voltage (max): 5.5/8V
• Operating Temp Range: -40C to 85C
• Operating Temperature Classification: Industrial
• Mounting: Surface Mount
• Pin Count: 24
• Lead Free Status / Rohs Status: Not Compliant

Where R
R
CALCULATING REGULATOR OUTPUT CURRENT
The load current of the boost converter is related to the av-
erage inductor current by the relation:
Where "DC" is the duty cycle of the application.
The switch current can be found by:
Inductor ripple current is dependent on inductance, duty cy-
cle, supply voltage and frequency:
FB3
is given by:
I
SW
CO
I
AMP
= I
is the ESR of the output capacitor. The value of
= I
INDUCTOR(AVE)
INDUCTOR(AVE)
R4 =1 / 2
π
+ 1/2 (I
f
CO
x (1 - DC) (A)
C1
RIPPLE
) (A)
(11)
(12)
(13)
17
where f = switching frequency = 1MHz
combining all terms, we can develop an expression which al-
lows the maximum available load current to be calculated:
The equation shown to calculate maximum load current takes
into account the losses in the inductor or turn-off switching
losses of the FET and diode.
DESIGN PARAMETERS V
The value of the FET "ON" voltage (referred to as V
equations 9 thru 12) is dependent on load current. A good
approximation can be obtained by multiplying the on resis-
tance (R
The maximum peak switch current the device can deliver is
dependent on duty cycle.
I
AMP(max)
DS(ON)
I
RIPPLE
= (1–DC)x[I
of the FET times the average inductor current.
= DC x (V
SW(max)
DD
SW
-V
SW
–DC(V-V
AND I
) / (f x L) (A)
SW
SW
)]/2fL (A)
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SW
(14)
(15)
in